eindhoven university of technology master performance ... · waterlandziekenhuis purmerend august,...

Eindhoven University of Technology

MASTER

Performance analysis of the clinical chemistry laboratoryWaterland Hospital Purmerend

Op het Veld, M.

Award date:2011

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https://research.tue.nl/en/studentTheses/fd3e5fb1-ae86-456c-8b7f-e532ffab1b23

Eindhoven University of Technology

Department of Mathematics and Computer Science

Performance Analysis of the Clinical ChemistryLaboratory

Waterland Hospital Purmerend

Author:Maike Op het Veld

Supervisors:Dr. J.A.C Resing

TU/e

Ir. M.S. van den Broek,Dr.ir. M. van Vuuren

CQM Eindhoven

I. Estany-Stoelwinder,P. Hofman

WaterlandziekenhuisPurmerend

August, 2011

Abstract

This thesis contains the performance analysis of the clinical chemistry lab-oratory in the Waterland Hospital in Purmerend. The object is to improvethe processes at the laboratory. The system is modeled as a large networkof queues. Due to the size of the network, only the part of the laboratory atwhich all samples arrive and are prepared for testing (SAP) is investigatedin detail. The analysis is mainly done by simulation and approximativemethods, as the processes are too complex for exact analysis.

The simulation reveals that the throughput times are good in the currentsituation and the occupation rate of the employees is relatively low. Thesimulation also shows that the occupation rate cannot be increased whilekeeping the throughput times at an acceptable level in the current situation,because the samples mainly arrive in batches. It is also found that there aremany options for decreasing throughput times.

Two machines at the SAP serve the customers in batches. To investigatethe effect of the machine batch policy on the throughput times, a simplifiedmodel of the machines is analyzed by exact and approximative methods.The analysis shows that machines with a low workload should be startedalready when there are only a few samples present, in order to achieveoptimal throughput times. The analysis also shows that the minimal batchsize for which the waiting time is minimal increases when the workload ofthe system increases.

Acknowledgements

With this thesis I complete the master Industrial and Applied Mathematics,with the specialization Statistics, Probability and Operations Research, atthe University of Technology in Eindhoven. I am greatly indebted to a num-ber of people, from whom I have received support during the nine monthsof this project, and during the rest of my study.

First of all, I would like to thank Jacques Resing, Monique Van denBroek and Marcel van Vuuren. The weekly discussions with Jacques atthe TU/e, and with Marcel and Monique at CQM have guided me throughthis project. I am also grateful for the support and enthusiasm from IlonkaEstany and Peter Hofman from the Waterland hospital in Purmerend. Theother employees of the clinical chemistry laboratory, and Ilona den Bestenin particular, have always been very helpful and eager to introduce me intothe laboratory world as well, for which I am very thankful.

Further, I am very grateful for the enthusiasm, interest, and support ofeveryone at CQM, which has contributed very much to the pleasure I foundin performing this project.

As this project also ends my study period, I would also like to give aspecial thanks to my parents and brother, who have always supported mein everything I did and do. Last but not least, I would also like to thankRuud, and my other friends just for being there.

Maike Op het Veld, August 2011

ii

Contents

1 Introduction 11.1 Waterland Hospital Purmerend . . . . . . . . . . . . . . . . . 11.2 Clinical Chemistry Laboratory . . . . . . . . . . . . . . . . . 11.3 Current developments in health care and lean thinking . . . . 21.4 Problem description . . . . . . . . . . . . . . . . . . . . . . . 31.5 Thesis outline . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Clinical Chemistry Laboratory 62.1 Laboratory structure . . . . . . . . . . . . . . . . . . . . . . . 6

2.1.1 Sample arrival and preparation . . . . . . . . . . . . . 62.1.2 Test execution . . . . . . . . . . . . . . . . . . . . . . 82.1.3 Administration . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Result administration . . . . . . . . . . . . . . . . . . . . . . 92.3 Priorities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.4 Employee behavior . . . . . . . . . . . . . . . . . . . . . . . . 102.5 Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Data analysis 133.1 Data format . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 Arrival patterns . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2.1 General description . . . . . . . . . . . . . . . . . . . . 143.2.2 Arrival types . . . . . . . . . . . . . . . . . . . . . . . 15

3.3 Routing patterns in the laboratory . . . . . . . . . . . . . . . 223.4 Process times . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.5 Throughput times . . . . . . . . . . . . . . . . . . . . . . . . 243.6 Data conclusions . . . . . . . . . . . . . . . . . . . . . . . . . 253.7 Research conclusions . . . . . . . . . . . . . . . . . . . . . . . 26

iv

CONTENTS v

4 Sample arrival and preparation 294.1 Arrival types . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Servers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.3 Priorities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304.4 Tasks at the sample arrival and preparation area . . . . . . . 304.5 Routing patterns . . . . . . . . . . . . . . . . . . . . . . . . . 324.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 Simulation model 355.1 Simulation set up . . . . . . . . . . . . . . . . . . . . . . . . . 355.2 Simulation input . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.2.1 (Inter) Arrival times . . . . . . . . . . . . . . . . . . . 385.2.2 Batch sizes . . . . . . . . . . . . . . . . . . . . . . . . 455.2.3 Cito tests . . . . . . . . . . . . . . . . . . . . . . . . . 505.2.4 Process times . . . . . . . . . . . . . . . . . . . . . . . 505.2.5 Disturbances . . . . . . . . . . . . . . . . . . . . . . . 515.2.6 Task priorities . . . . . . . . . . . . . . . . . . . . . . 51

5.3 Verification and validation . . . . . . . . . . . . . . . . . . . . 525.4 Simulation output . . . . . . . . . . . . . . . . . . . . . . . . 52

5.4.1 Key performance indicators . . . . . . . . . . . . . . . 535.4.2 Waiting times . . . . . . . . . . . . . . . . . . . . . . . 54

5.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

6 Simulation scenarios 586.1 Basic scenarios . . . . . . . . . . . . . . . . . . . . . . . . . . 586.2 Basic scenario results . . . . . . . . . . . . . . . . . . . . . . . 60

6.2.1 Cito shorter . . . . . . . . . . . . . . . . . . . . . . . . 606.2.2 No disturbing factors . . . . . . . . . . . . . . . . . . . 626.2.3 Move outpatient clinic . . . . . . . . . . . . . . . . . . 626.2.4 Blood withdrawal round . . . . . . . . . . . . . . . . . 636.2.5 Move sorting activities . . . . . . . . . . . . . . . . . . 636.2.6 BWO 2x . . . . . . . . . . . . . . . . . . . . . . . . . . 636.2.7 Batch size outpatient clinic train . . . . . . . . . . . . 646.2.8 One employee . . . . . . . . . . . . . . . . . . . . . . . 646.2.9 No cito type . . . . . . . . . . . . . . . . . . . . . . . 696.2.10 Preanalyzer . . . . . . . . . . . . . . . . . . . . . . . . 70

6.3 Scenario combinations . . . . . . . . . . . . . . . . . . . . . . 706.4 Sensitivity analysis . . . . . . . . . . . . . . . . . . . . . . . . 71

6.4.1 Sensitivity to arrival parameters . . . . . . . . . . . . 716.4.2 Sensitivity to task priority setting . . . . . . . . . . . 73

CONTENTS vi

6.5 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 756.5.1 Scenarios . . . . . . . . . . . . . . . . . . . . . . . . . 756.5.2 Sensitivity . . . . . . . . . . . . . . . . . . . . . . . . . 78

7 Analysis 807.1 Machine batch policy . . . . . . . . . . . . . . . . . . . . . . . 807.2 M/Ga,N/1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

7.2.1 Model description . . . . . . . . . . . . . . . . . . . . 827.2.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . 89

7.3 MX/Da,N/1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 937.3.1 Model description . . . . . . . . . . . . . . . . . . . . 94

7.4 M/Da,N/c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 957.4.1 Model description . . . . . . . . . . . . . . . . . . . . 967.4.2 Performance measures . . . . . . . . . . . . . . . . . . 1007.4.3 Results . . . . . . . . . . . . . . . . . . . . . . . . . . 108

7.5 Conclusions and suggestions for further research . . . . . . . 1107.5.1 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . 1107.5.2 Suggestions for further research . . . . . . . . . . . . . 113

8 Conclusions 1158.1 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

8.1.1 Laboratory . . . . . . . . . . . . . . . . . . . . . . . . 1158.1.2 Simulation . . . . . . . . . . . . . . . . . . . . . . . . 1168.1.3 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 117

8.2 Suggestions for further research . . . . . . . . . . . . . . . . . 1178.2.1 Laboratory . . . . . . . . . . . . . . . . . . . . . . . . 1188.2.2 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 118

A Simulation Input and Output 120A.1 (Inter) Arrival distributions . . . . . . . . . . . . . . . . . . . 120A.2 Batch size distributions . . . . . . . . . . . . . . . . . . . . . 121A.3 Cito probabilities . . . . . . . . . . . . . . . . . . . . . . . . . 121A.4 Process times . . . . . . . . . . . . . . . . . . . . . . . . . . . 121A.5 Task priority setting . . . . . . . . . . . . . . . . . . . . . . . 122A.6 Validation by sampling . . . . . . . . . . . . . . . . . . . . . . 124

B Goodness-of-fit tests 125B.1 Continuous distributions . . . . . . . . . . . . . . . . . . . . . 125B.2 Discrete distributions . . . . . . . . . . . . . . . . . . . . . . . 126

CONTENTS vii

C Results from the imbedded Markov chain for M/Ga,N/1 127

Chapter 1

Introduction

This research is performed for the clinical chemistry laboratory of the Water-land Hospital in Purmerend. This chapter gives some background informa-tion about the hospital (Section 1.1) and the clinical chemistry laboratory(Section 2), explains the developments that have led to this research, (Sec-tion 1.3), describes the problem and solution approach, (Section 1.4) andgives an outline of this thesis, (Section 1.5).

1.1 Waterland Hospital Purmerend

The Waterland Hospital is the result of a fusion between two hospitals inthe region Waterland in 1984. The current location in Purmerend is in usesince 1988 and the name of the hospital was Streekziekenhuis Waterlanduntil January 1992. Since 2002, the hospital also has a secondary locationin Volendam called Waterland-Oost.

The Waterland hospital counts 359 beds, which makes it a small tomiddle-sized hospital. Approximately 25.000 inpatients are treated and80.000 outpatients visit the hospital each year. Slightly more than 1.200employees and 160 volunteers work in the hospital.

1.2 Clinical Chemistry Laboratory

The clinical chemistry laboratory is the department in the hospital in whichall body fluids are tested. These tests show for example if a patient iscontaminated with a disease, which blood group a patient has, if medicinesare working or if a patient is pregnant. There are nearly 1000 different teststhat can be performed. Body fluids are tested each day for nearly every

1

CHAPTER 1. INTRODUCTION 2

inpatient and many outpatients come to the hospital for blood withdrawalor to bring body fluid samples that have to be investigated. This meansthat over 3.000 tests are performed per day with more than 800 body fluidsamples. Most of the tests are done by a machine, and several tests are doneby hand. On a week day, eleven or twelve employees work in the laboratory,in the weekend two or three employees are present and during the eveningand night one employee runs the tests that are urgent. There are two types ofrequests; regular requests and urgent requests. The results of the tests thatare marked as urgent should be available within an hour after requesting.The results of regular requests are generally available within one day, exceptfor tests that are done by external institutions.

The laboratory consists of three departments; a sample arrival and prepa-ration department, a department in which all tests are executed and anadministration department. Most analysts can work at any place in thelaboratory, and each day the analysts are assigned to work in one of thedepartments.

1.3 Current developments in health care and leanthinking

Currently, many developments concerning health care are taking place inthe Netherlands. Hospitals are facing a shortage in staff, privatization, andbudget cuts. This has led to more awareness for the efficiency, service andquality of the processes in hospitals. The Waterland Hospital also faces thesame problems and introduced lean thinking in 2009 to reach improvementsin efficiency, quality and service of the processes in the hospital.

Lean manufacturing stems from the car manufacturer Toyota and con-cerns the efficiency of processes. The theory focuses on improving flow andeliminating ’waste’, every process step that does not add value to the endproduct for the consumer is considered to be wasteful. The consumer playsa central role in lean theory, because the most important goal is a satisfiedcustomer. There are five essential steps in lean thinking [11]:

1. Identify which features create value.

2. Identify the sequence of activities called the value stream.

3. Make the activities flow.

4. Let the customer pull products or service through the system.


5. Pursue perfection.

The theory provides several tools to take these steps, such as value streammapping, 5S and kanban systems. For further information about lean think-ing and the tools it provides, the reader is referred to [11].

The goal of introducing lean thinking in the hospital can be formulatedas follows:’Transform the organization from an output controlled organization to aprocess minded and possibly process controlled organization’.Before lean thinking was introduced in the hospital, the main focus forevaluating the efficiency of a process was the result of the process, such ascosts, the number of treated patients and the number of complaints frompatients. By introducing lean thinking, the hospital tries to realize a switchto a focus on the processes itself. This switch cannot be made quickly, it isa time-consuming process that involves all employees of the hospital; it isexpected to take 5 to 10 years.

At the moment, the implementation of lean thinking is still in an initialphase. A lean manager is employed to regulate the activities concerninglean thinking and education projects about lean thinking for employees tobecome a lean coach in his or her department have started. A lean coach isexpected to function as an accelerator for the introduction of lean thinkingin his or her department by introducing changes and motivating and guidingthe employees. The first two rounds of lean education projects are nearlyfinished and each lean coach has to create a lean improvement project in hisor her department. Next to the lean education projects, much informationabout the processes in the hospital is gathered. This information is gatheredby data analysis and actual measurements. For example, at the emergencydepartment, the operating rooms and the radiology department measure-ments concerning process times and throughput times have been performed.There are still many departments for which the processes have to be mappedand improvements concerning quality, service and efficiency can be reached.

1.4 Problem description

The clinical chemistry laboratory is a large and busy department and asnearly every patient that comes to the hospital also needs body fluid tests,it is valuable for nearly all other departments of the hospital. The de-velopments mentioned in the previous section also apply to the laboratory.Therefore the quality, service and efficiency of the processes are a main topic


in the laboratory and the quality officer of the laboratory is educated to be-come a lean coach. One change initiated by the lean coach is to rearrangethe machines, such that employees can easily work together and help eachother when necessary and the work ’flows’ better through the laboratory.This is only the beginning, and there are many other aspects in the lab-oratory that can be improved, therefore this research is performed at thelaboratory.

The next list gives an overview of a few explicit reasons to investigatethe processes in the clinical chemistry laboratory:

1. No good overview of the processes in the laboratory exists.There is no thorough knowledge about for example the actual numberof body fluid samples that are investigated per day, the work load ofthe machines, the throughput time of test results or which tests aredone most frequently.

2. Complaints about throughput times for test results.The laboratory sometimes receives complaints about throughput times,but no statistical information about throughput times is available.

3. The workload varies over the day.At some moments, the work load is very high, while no work is availableat other moments.

4. The laboratory has to shrink by 3 fte.Due to budget cuts, most departments of the hospital have to shrink.

5. New division of workplaces.Machines and workplaces are rearranged, as a result of a previous leanproject.

6. Many urgent request that are not really urgent.Each request can be marked as urgent by the doctor that is requestingit. This should only be done if the result should be available within anhour, however many doctors make a request urgent if they only needit on the same day.

7. New machines can be purchased.For example a new machine that takes over some activities in thearrival and preparation area can be purchased.

This has led to the following research goal and approach:


Research goal:

Investigate and improve the processes in the clinical chemistry laboratory.

Research approach:

1. Process mapping and data analysisCreate an overview of the processes in the laboratory and investigatethroughput times of tests, the number of performed tests, the originof requests, process step times etcetera.

2. Find improvement possibilities in the processesFind a part of the laboratory that appears to have possibilities forimprovement.

3. Investigate improvement possibilitiesInvestigate improvement possibilities with mathematical tools.

1.5 Thesis outline

In Chapter 2, the structure and characteristics of the laboratory are dis-cussed and the laboratory is modeled as a network of queues. Because thenetwork is too large and complex to analyze entirely, a part of the labora-tory is analyzed in detail. To investigate which part of the laboratory is themost interesting to analyze, and to obtain insight in the network concerningarrival streams, routing patterns and processing times, Chapter 3 containsa data analysis of the laboratory.

With the information from the data analysis and from consultation withthe laboratory staff, it is chosen to analyze the part of the laboratory atwhich all samples arrive and are prepared for testing. Chapter 4 explainsthe structure of this part of the laboratory, and models it as a queueingnetwork.

Because the network is too complex for exact analysis, Chapter 5 devel-ops a simulation model. In Chapter 6, test scenarios and the results fromthe scenarios are discussed, and a sensitivity analysis is performed. An exactand approximative analysis concerning the influence of the machine batchpolicy on the throughput times is performed in Chapter 7. The final chaptercontains conclusions and recommendations for further research.

Chapter 2

Clinical ChemistryLaboratory

As mentioned in Section 2, more than 3000 tests with over 800 samples areperformed on an average weekday, all these tests are done in the laboratoryand all samples find their way through the laboratory. This chapter explainsthe structure and the processes in the laboratory. The laboratory structureis clarified in Section 2.1, the procedure for processing the results is explainedin Section 2.2, the different request priorities are discussed in Section 2.3 andthe employee behavior is the subject of Section 2.4.

2.1 Laboratory structure

The clinical chemistry laboratory consists of three main areas:

1. Sample arrival and preparation

2. Test execution

3. Administration.

2.1.1 Sample arrival and preparation

The sample arrival and preparation area (SAP) is the area at which allbody fluids arrive in the laboratory and the samples are prepared for thetests. Two employees receive, register, sort and centrifuge the body fluidsamples such that the samples are ready to go to the test execution area.The samples arrive from several origins:

6

CHAPTER 2. CLINICAL CHEMISTRY LABORATORY 7

• Inpatient care

– Pneumatic mailThere exists a transportation network for body fluid samples fromemergency departments such as the intensive care unit (ICU),cardiac care unit (CCU) and the emergency unit (SEH). Eachmorning at approximately 7:00, the nurses or doctors from theICU and CCU send body fluid samples from their patients to thelaboratory. During the day, any sample from these three depart-ments is sent by pneumatic mail as well. Some other departmentsalso send their samples to the laboratory by pneumatic mail.

– Inpatient blood withdrawalEach morning, all employees of the laboratory, except for the ad-ministrative employee and one or two analysts that have to stayin the laboratory to handle the urgent test requests, go into thehospital to withdraw blood from the patients in all departmentsexcept the ICU, CCU and SEH. During the rest of the day, ana-lysts can also be requested to withdraw blood.

– Brought samplesDuring the day, nurses or doctors from the hospital units physi-cally bring body fluid samples to the laboratory.

• Outpatient clinic WaterlandziekenhuisThe outpatient clinic is located in the hospital and opened from 8:00until 17:00 on week days, only on Tuesdays, the clinic is open until20:00. Patients come to this clinic to withdraw blood or hand in bodyfluid samples. It is located in the hospital, but at a different floorthan the laboratory. The clinic sends the body fluid samples to thelaboratory with a small train that arrives at the sample arrival andpreparation area.

• Outpatient clinic Waterland OostThis clinic is located in Volendam and is only opened from 8:00 until12:30 on each week day. The body fluid samples are collected after12:30 and brought to the laboratory by a carrier.

• External arrivalsThe laboratory from the Waterland hospital cooperates with otherlaboratories by exchanging tests, so each day samples from other lab-oratories arrive in Purmerend. Each Tuesday, an analyst goes to a


psychiatric clinic and takes the body fluid samples of these patients tothe laboratory.

Figure 2.1 gives an overview of the arrivals of body fluid samples in thesample arrival and preparation area.

Sample arrival and

preparation area

Inpatient: Pneumatic mail

Inpatient: Blood withdrawal

Inpatient: Brought samples

Outpatient clinic WLZ

Outpatient clinic W-O

External

Figure 2.1: Arrivals at Sample Arrival and Preparation Area

2.1.2 Test execution

After processing the samples at the arrival and preparation area, the samplesare ready for the test execution phase. The tests can be split into seventypes:

1. Blood gas

2. Coagulation

3. Haematology

4. Transfusion

5. Chemistry

6. Urine

7. External


The blood gas machine is located at the sample arrival and preparationarea, because these tests can only be conducted shortly after blood with-drawal. The other test types all have a separate area in the laboratory atwhich one or more machines are located, except the external tests, as theseare done in other laboratories. The analysts have to operate the machines,check the results and do manual tests. When a machine is finished with atest, an analyst should confirm the result. If the result is too abnormal, theclinical chemist or the team leader has to confirm the result. The samplesare stored in the laboratory for a few days, after which they are eliminated.For an overview of the test execution area, see Figure 2.2.

Sample arrival and

preparation area

Blood gas

Coagulation

Haematology

Transfusion

Chemistry

Urine

External

Confirmation

Samples Results

Test executionSample arrival and preparation

Figure 2.2: Test Execution Area

2.1.3 Administration

The administration activities consist of handling the external tests. Thisconsists of preparing, registering and sending the requests and receiving andprocessing the results. Also other small administrative activities are handledby the administration.

2.2 Result administration

Each inpatient and outpatient is registered in a digital information systemcalled Labosys. All qualified employees of the hospital can enter the digital


information system and request information about patients. Body fluid testrequests and results are also registered in this system. The machines at theclinical chemistry laboratory are linked with the information system. Themachine can recognize the patient that belongs to the body fluid sampleby the code on the sample and puts the result in the information systemimmediately. The laboratory staff only has to confirm the results, afterwhich the doctor that requested the test for the patient can see the confirmedresult in Labosys.

2.3 Priorities

Three different priority types for the test requests exist:

1. Blood gas determination requestThis is the most urgent request type, because this test should be per-formed immediately after blood withdrawal, so it is given priority overall other tests.

2. Cito requestThe result of a cito request should be available within one hour, so itis given priority over all other requests, except for blood gas determi-nation tests.

3. Regular requestThis type contains all other requests. The standard for these requestsis that the result is available within one day. Among the regular typerequests, one can still observe some priority handling, as inpatients aregenerally prioritized over outpatient requests.

2.4 Employee behavior

Each day, the employees of the laboratory are assigned to a work spot. Thiswork spot generally differs each day, so nearly all employees can work at anyplace in the laboratory. At 7:00, one or two employees from the morningshift start their work day and handle all cito requests, set up the machinesand prepare the blood withdrawal round. The rest of the employees starttheir working day with the inpatient round at 7:45, and after this round,everyone starts working at the work spot he or she is assigned to for thatday. This means that the morning shift employees also start working attheir spot. At 16:00 the morning shift employees end their working day,


while the employee from the evening and night shift starts at either 14:00 or17:00. At 17:15, the normal shift employees end their day and the eveningshift employee is the only one left in the laboratory. This employee preparesthe inpatient round for the next day and handles all cito requests during theevening and night. He or she is allowed to leave when the employees for thenext day arrive at 7:00. In the weekend, there are much less requests thanon a week day, so two or three employees handle all requests.

Each day, every employee has two short breaks of 15 minutes and a longbreak of 30 minutes. The night shift has a dinner break, and he or she cango to sleep in between cito requests during the night.

2.5 Model

Based on the information in the previous sections, the processes in the lab-oratory can be modeled as a network of queues. The sample arrival andpreparation area and the test execution area together form a network ofqueues in which the body fluid samples are the customers. The administra-tion area exists separately from this network, because only paper work isdone here. The network formed by the sample arrival and preparation areaand the test execution area is displayed in a global way in Figure 2.3. Eachblock in the figure consists again of more detailed process steps.

Sample arrival and

preparation area

Inpatient: Pneumatic mail

Inpatient: Blood withdrawal

Inpatient: Brought samples


Outpatient clinic W-O

External

Blood gas

Coagulation

Haematology

Transfusion

Chemistry

Urine

External

Confirmation

Samples Results

Sample arrival and preparation area Test execution area

Samples

are stored

Figure 2.3: Sample arrival and test execution network

The network contains some complicating factors that differentiate it froma general queueing network:

• Many different arrival types, mainly in batches


• Three different customer priorities

• Servers have multiple tasksAt many places in the laboratory, an employee has to perform morethan one task. This means that the tasks do not have a server availableall the time, and the server has to choose which task to perform.

The network is also quite large, Figure 2.3 already shows six different ar-rival types and eight different process blocks. Some of the arrival types areagain split in different streams and each of the blocks contains different jobsand routings again, so it is hard to model the complete network in detail.Therefore, this research focuses on a part of the laboratory. To investigatewhich part of the laboratory is the most interesting to investigate, and to ob-tain input for the network concerning arrival streams, routing patterns andprocessing times, the next chapter contains a data analysis of the laboratory.

Chapter 3

Data analysis

This chapter discusses data concerning the laboratory, to find which part ofthe laboratory is most interesting to investigate, and to obtain input for thequeueing network model. First, in Section 3.1, the data format is discussed,after which the arrival patterns are investigated in 3.2, the routing patternsin 3.3, the process times in 3.4 and finally the throughput times in 3.5. Theconclusions concerning the data are discussed in 3.6 and the consequencesfor the research are explained in Section 3.7.

3.1 Data format

The laboratory has a digital information system, called Labosys, in whichall requested tests are registered and the results are saved. Many details areregistered for each test, such as the test type, the urgency type of the test, thetime the request form is read into the system, the time the result is available,the time the result is confirmed and the origin of the request. Unfortunately,much information is not registered, for example the throughput time per partof the laboratory, the actual time of arrival in the laboratory, or the processtimes. In this chapter, a data set containing laboratory data from March2010 until February 2011 is used. Weekdays are structured differently thanthe days in the weekends, so they should be investigated separately. It wouldbecome too extensive to investigate both day structures separately, and thelaboratory has the most interest in improving weekdays, so it is chosen tofocus on weekdays in this research.

There are three dimensions in the data:

1. Lab numbers:Each time tests are requested for a patient, a lab number is assigned

13

CHAPTER 3. DATA ANALYSIS 14

to the patient.

2. Body fluid samples:Per patient, and thus lab number, generally more than one body fluidsample has to be investigated. The number of samples depends onthe tests that are requested. Some tests need a preparation substance,and each machine receives a separate sample, such that tests can bedone simultaneously.

3. Tests:More than one test is performed per body fluid sample in general,every test that can be done on one machine is performed with the samesample. There are nearly 1000 different tests that can be requested.

3.2 Arrival patterns

In this section, a general description of the arrivals of body fluid samples inthe laboratory is given first (Section 3.2.1), after which each arrival streamis discussed in detail (Section 3.2.2)

3.2.1 General description

The arrival time of body fluid samples in the laboratory is not available, butthe registration time in Labosys does give an impression of the arrival timesfor many samples. Figure 3.1 gives an impression of the mean number ofregistrations of body fluid samples per half hour on a weekday for all arrivaltypes in the investigated year, except for the samples from the outpatientclinic in Volendam (BWO), and the samples from the blood withdrawalround. The registration times for these two types differ very much from thearrival times. BWO samples are registered at withdrawal in Volendam inthe morning, while they only arrive in the afternoon in Purmerend. Thesamples from the blood withdrawal round are registered the day and nightbefore the round, while they are only withdrawn and arrive in the laboratoryin the morning. Therefore, these sample types are filtered and gathered inone time slot, 13:30-14:00 for BWO and 8:00-8:30 for the blood withdrawalround. Figure 3.1 shows that the peak moments are caused by BWO andthe blood withdrawal round, and that the morning is the busiest part ofthe day. Figure 3.2(a) shows that the outpatient clinic in Purmerend andthe inpatient care provide the most samples, but the outpatient clinic inVolendam and the external arrivals also provide a significant amount of


050

100

150

Bod

y flu

id s

ampl

es

0:00 06:00 12:00 18:00 24:00Time

Body fluid samples over time on a weekday

Figure 3.1: Mean number of body fluid sample arrivals per half hour on aweekday

samples. It is also interesting to see whether each week day is equally busy,or if a trend exists in the week. Figure 3.3 shows that the mean numberof body fluid samples per day differs between approximately 800 on Fridayto 950 on Tuesday. The mean number of body fluid samples appears todecrease over the week.

The urgent samples arrive through the same arrival streams as the regu-lar samples. On a weekday, on average 116 of the 876 samples that arrive inthe laboratory are cito samples, and 17 samples need blood gas determina-tion. The blood gas samples mainly originate from inpatient care, becausethis is a very urgent test and it is mainly done for the observation of pa-tients. The outpatient clinic in Purmerend occasionally receives a bloodgas sample as well, but the samples from the outpatient clinic in Volendamand from external arrivals never concern a blood gas request. Figure 3.2(b)shows the mean number of cito body fluid samples per origin on a week day.It shows that the most cito samples are provided by inpatient care, becausemost patients that need urgent tests are hospitalized. The outpatient clinicin Purmerend also provides a significant amount of cito samples, but thesesamples mainly concern requests from doctors that need the result on thesame day, instead of within one hour.

3.2.2 Arrival types

Figure 2.1 distinguishes six arrival types, but within some of these arrivaltypes, different streams can be distinguished, these are also described in


010

020

030

040

0B

ody

fluid

sam

ples

Outpatient clinic WLZ Inpatient care External Outpatient clinic BWO

Mean number of body fluid samples per origin on a weekday

(a) All samples

020

4060

80C

ito b

ody

fluid

sam

ples

Outpatient clinic WLZ Inpatient care External Outpatient clinic BWO

Mean number of cito samples per origin on a weekday

(b) Cito samples

Figure 3.2: Mean number of body fluid samples per origin on a weekday

Section 2.1.1.

1. Inpatient careThe body fluid samples can again be split in three categories:

(a) Pneumatic mail

i. Samples from IC and CCU at 7:00.ii. Samples from IC, CCU and SEH during the day.

iii. Samples from other departments during the day.

(b) Blood withdrawal

i. Round at 7:45 at all departments except IC, CCU and SEH.

ii. Samples during the day.


020

040

060

080

01,

000

Bod

y flu

id s

ampl

es

Monday Tuesday Wednesday Thursday Friday

Mean number of body fluid samples per day

Figure 3.3: Average amount of body fluid samples on a week day per day

(c) Brought samples from all departments in the hospital exceptCCU, IC and SEH.

2. Outpatient clinic Waterlandziekenhuis

3. Outpatient clinic Waterland Oost

4. External arrivals

The previous section shows that the largest amount of cito samples stemsfrom inpatient care. Within inpatient care, the cito samples are divided overthe different arrival types. All lab numbers from IC, CCU and SEH containcito samples, because these are the emergency departments. The bloodwithdrawal round generally only contains a small number of cito requests,1 or 2 samples. 1(a)iii, 1(b)ii, and 1c concern cito samples more often,approximately 14% of the time. The number of cito samples per day for theother arrival types is already displayed in Figure 3.2(b).

Each of the arrival types has its own characteristics. A characteristicthat all arrival types share is that the body fluid samples arrive in thelaboratory in batches per lab number. The number of body fluid samplesin a lab number is determined by the tests that are requested for the labnumber. In general, each machine type in the laboratory requires a sep-arate sample, because the machines require different pre-processing steps(for example centrifugation, or a certain temperature), and this makes si-multaneous testing possible. Therefore, the number of body fluid samplesper lab number is generally equal to the number of machine types required


to perform the tests for the lab number. Table 3.1 gives some descriptivestatistics concerning the number of body fluid samples per lab number.

Mean St.dev. Min MaxBody fluid samples per lab number 2.41 1.32 1 11

Table 3.1: Body fluid samples per lab number

0.1

.2.3

Fra

ctio

n

1 2 3 4 5 6 7 8 9 10 11Number of body fluid samples

Body fluid samples per lab number

Figure 3.4: Body fluid samples per lab number

The arrival types can be split in two groups, arrivals that occur once aday around an expected time, and arrivals that occur more often during theday with stochastic interarrival times. The following subsections describemore characteristics of the arrival types per group.

Once a day arrivals

The samples from the outpatient clinic BWO, the blood withdrawal round inthe morning, and the ICU and CCU at 7:00 belong to this type of arrivals.The samples from these origins arrive in batches of lab numbers; Table3.3 gives some statistics concerning the number of lab numbers per arrivalper type, and Figure 3.5 displays the distribution of the batch sizes. Theemployees of the laboratory know when to expect a batch of samples fromthese origins. The actual arrival times are not registered, but it is well-known that the samples from the ICU and CCU arrive a few minutes after7:00, and the samples from BWO arrive around 13:30. The samples from theblood withdrawal round generally arrive in the laboratory between 8:00 and9:30, but no specific arrival time is known. The arrival time of the samplesfrom the blood withdrawal round depends on the number of patients (i.e.


lab numbers) from which blood has to be withdrawn, the department type(blood withdrawal for children takes longer), and the number of employeesthat walk the round. Information concerning these parameters should becollected to find an estimate for the arrival time of the samples from theblood withdrawal round.

Mean St.dev. Min MaxBWO 45.46 11.65 10 79Blood withdrawal round 44.25 14.75 10 74ICU & CCU 7:00 6.87 2.75 1 14

Table 3.2: Lab numbers per arrival - Once a day arrivals

0.0

1.0

2.0

3.0

4.0

5F

ract

ion

0 20 40 60 80Number of lab numbers

Arrival batch size BWO

(a) BWO

0.0

2.0

4.0

6F

ract

ion

0 20 40 60 80Number of lab numbers

Arrival batch size from blood withdrawal round

(b) Withdrawal round

0.0

5.1

.15

.2F

ract

ion

0 5 10 15Number of lab numbers

Arrival batch size from ICU&CCU at 7:00

(c) ICU & CCU 07:00

Figure 3.5: Histogram of lab numbers per arrival - Once a day arrivals

Continuous arrivals

The arrival types 1(a)ii, 1(a)iii, 1(b)ii, 1c, 2, and 4 belong to this group. It isnot possible to distinguish between 1(a)iii, 1(b)ii, and 1c in the data, so theseare considered to be one group for now. The lab numbers from these groupsdo not always arrive individually, Table 3.3 gives some descriptive statisticsfor the batch sizes, except for the outpatient clinic in Purmerend. Theoutpatient clinic has some specific characteristics, so it is treated separatelylater in this section. Figure 3.6 depicts the distribution of the arrival batchsizes.

The mean number of lab number registrations per half hour per weekdayfrom ICU/CCU/SEH is displayed in Figure 3.7(a). The figure shows thatthe number of arrivals is low during the night, relatively high during theday, and it goes down in the evening. The arrivals from external origins aredisplayed in 3.7(b). The information concerning external arrivals is limited,


Mean St.dev. Min MaxICU/CCU/SEH (1(a)ii) 1.28 0.62 1 6External (4) 5.14 6.39 1 24Rest (1(a)iii, 1(b)ii, 1c) 1.37 2.13 1 13

Table 3.3: Lab numbers per arrival - Continuous arrivals

0.2

.4.6

.8F

ract

ion

0 2 4 6 8 10Number of lab numbers

Arrival batch size from ICU&CCU&SEH

(a) ICU/CCU/SEH

0.1

.2.3

Fra

ctio

n

0 10 20 30 40 50Number of lab numbers

Arrival batch size from external

(b) External

0.2

.4.6

.8F

ract

ion

0 5 10 15Number of lab numbers

Arrival batch size from rest

(c) Rest

Figure 3.6: Histogram of lab numbers per arrival - Continuous arrivals

because one cannot distinguish between the different external origins. Thedifferent origins all have specific characteristics, but this is ignored here andthe external arrivals are considered to be one group. The figure shows thatthere might be a group of lab numbers that arrives typically around noon,as there exists a higher peak in Figure 3.7(b). The rest of the lab numbersfrom inpatient care, 1(a)iii, 1(b)ii, 1c, are displayed in Figure 3.7(c). Notethat the number of lab numbers equals 0 between 6 and 9. This is dueto the data filtering for the blood withdrawal round, it is assumed that alllab numbers from the non-urgent departments that are activated between6 and 9 are taken in the blood withdrawal round. The figure shows a cleardecrease of lab number registrations from this type over the day. The peakis in the morning, which may be caused by lab numbers that were forgottenin the blood withdrawal round. The later it gets on a day, the likelier it isto wait for the blood withdrawal round on the next day if the tests are notcito, so the number of lab numbers decreases over the day.


The arrival time at the laboratory differs a lot from the registration times ofthe lab numbers at the outpatient clinic WLZ, so the arrival process at thelaboratory cannot be obtained from this information. Because it is too time-


0.5

11.

5La

b nu

mbe

rs

0:00 06:00 12:00 18:00 24:00Time

Lab numbers from ICU, CCU, and SEH over time on a weekday

(a) ICU/CCU/SEH

02

46

Lab

num

bers

0:00 06:00 12:00 18:00 24:00Time

External lab numbers over time on a weekday

(b) External

0.5

11.

52

2.5

Lab

num

bers

0:00 06:00 12:00 18:00 24:00Time

Rest of lab numbers over time on a weekday

(c) Rest

Figure 3.7: Lab number arrivals per type

consuming to obtain accurate data for the arrival times at the laboratory,


and the process of withdrawal at the outpatient clinic and the transportto the laboratory is quite clear, the processes at the outpatient clinic aremodeled separately in Section 5.2.1. The output of this process is equal tothe arrival process at the laboratory.

The arrivals at the outpatient clinic are input for the model of the out-patient clinic. Figure 3.8 displays the average number of registrations at theoutpatient clinic during a weekday. It is clear from this figure that the out-patient clinic is the busiest in the morning. During lunch time, the arrivalintensity goes down, then after lunch, the intensity increases slightly again,after which it decreases again towards the end of the day.

05

1015

Lab

num

bers

0:00 06:00 12:00 18:00 24:00Time

Lab numbers per half hour from outpatient clinic WLZ

Figure 3.8: Arrivals Outpatient Clinic WLZ

3.3 Routing patterns in the laboratory

All samples for the laboratory arrive at the sample arrival and preparationarea. After the pre-process steps at this area, the samples proceed to the testexecution step. Figure 2.2 shows that there are seven different test types.Each sample is destined for one of the test execution areas, depending on thetests that have to be done with the sample. Figure 3.9(a) displays the meannumber of body fluid samples per test type on a weekday. It is clear that thehaematology (302 samples) and chemistry department (344 samples) processmost of the samples, and the other departments process between 20 and 50samples per weekday. Unfortunately, the data also contains on average 80samples per day from which the test type is unknown, it is not registered.

Part of the samples in Figure 3.9(a) concern cito samples, not all ma-chines can process cito samples, only chemistry, coagulation and tests can becito. Figure 3.9(b) shows the distribution of cito samples over the different


test types. It shows that the haematology and chemistry department alsotest the most cito samples.

010

020

030

040

0B

ody

fluid

sam

ples

Coagulation Transfusion Haematology Chemistry Urine Blood gas External

Number of body fluid samples per test type on a weekday

(a) Regular samples

010

2030

4050

Cito

bod

y flu

id s

ampl

es

Coagulation Haematology Chemistry

Cito samples per test type on a weekday

(b) Cito samples

Figure 3.9: Average number of body fluid samples per test type/machineon a weekday

3.4 Process times

The process times are not registered. For some machines, the process timesare known, these are mainly deterministic. The process times for the humanprocess steps are not known, only estimates by experience can be given.


3.5 Throughput times

The exact throughput times for the laboratory are not available, but an es-timate can be obtained by subtracting the registration time from the resultconfirmation time. As explained earlier, the registration time does not coin-cide with the arrival time in the laboratory, but it does give an estimate formost samples. For the blood withdrawal round and BWO, the registrationtimes are adjusted because they differ too much from the arrival times.

Even though each machine has different process times, the throughputtimes are only considered per test type (see Section 2.1.2). This is done be-cause there are too many different machines. Not only the observed through-put times are interesting, but also the sum of the individual process steptimes, i.e. the theoretical throughput times. It is then interesting to seewhether the theoretical throughput times differ much from the observedthroughput times, and for which test types the largest deviations occur.Each block in Figure 2.3 consists again of several process steps and/or ma-chines; the process times of these steps and machines are not registered.Therefore, an overview of all process steps in the blocks is created and esti-mates for process times are made in consultation with the laboratory staff.Theoretical throughput times are then created by taking the sum of theestimated process times.

Figure 3.10(a) shows the observed throughput times and the theoreti-cal throughput times for regular samples and blood gas samples. The fig-ure shows that the largest deviation between the observed and theoreticalthroughput times occurs for the transfusion tests. After investigating thethroughput times for transfusion in detail, and consulting the laboratorystaff, it can be concluded that this is mainly caused by the fact that manytransfusion tests are done on the day after arrival at the laboratory. If thesamples arrive in the afternoon, it is often decided to store the samples anddo the tests on the next day. The chemistry tests also have a large deviationbetween theoretical and observed throughput times, the reason for this thesame as for the transfusion tests, some tests are only done on the next day.

Cito tests are only done on a few machines, so for cito tests, the through-put times are considered per machine. Also for cito tests, the observedthroughput times are compared with theoretical throughput times. Figure3.10(b) shows the results. The D-dimere and APTT machine perform testsof the coagulation test type, the Sapphire performs tests from the Haematol-ogy type and the Cobas6000 and the Axsym perform Chemistry tests. TheSapphire and Cobas6000 do approximately 90% of all cito tests, and thesemachines have small deviations between theoretical and observed through-


put times. Therefore, it can be concluded that the cito throughput timesappear to be quite good. The throughput times only differ much from thetheoretical times for samples that are tested on the Axsym, this is partiallycaused by the fact that this machine has no urgency entrance. When a citosample arrives, the sample has to wait until all samples in the machine arefinished. The process times for this machine are quite long, approximately30 minutes, so the waiting times can become quite large. The other machinesall have a separate entrance for cito tests.

3.6 Data conclusions

It can be concluded that there are many different arrival types, of which apart arrives at fixed moments in large batches of lab numbers, and anotherpart arrives in smaller batches or singly continuously over the day. Allsamples arrive in a batch of body fluid samples per lab number, on average2.4 body fluid samples belong to one lab number. Figure 3.11 gives anoverview of the mean number of lab number arrivals on a weekday perarrival type. The most samples arrive during the day, between 7:00 and17:15. The haematology and chemistry department test 74% of all samples,the other samples are divided over the other 5 departments. Figure 3.11also shows the mean number of samples per laboratory department on aweekday.

Cito samples mainly stem from inpatient care and the outpatient clinicin the WLZ, and on average 13% of all samples is cito. The cito samplesarrive through the same channels as the normal samples, but the pneumaticmail is the most common way for cito samples to arrive in the laboratory.The haematology and chemistry department also test the most cito samples,90% of the cito samples goes to these departments, the rest is processed atthe coagulation department.

Only estimates of observed and theoretical throughput times can beevaluated. The throughput times appear quite good, especially for citosamples. The regular samples have long throughput times at the transfusionand chemistry department, but this is mainly caused by the policy of testingsamples a day later than the arrival in the laboratory. These departmentshave not experienced any complaints about this.


3.7 Research conclusions

The data results have been discussed with the laboratory staff. Many resultshave given more insight, and the throughput times have not been evaluatedas dramatically bad. It does not appear that one of the test type areascauses the most problems, and it is perceived that the machines in the testexecution area have more than enough capacity to handle the samples. Theidea lives that the sample arrival and preparation area is an area at whichmuch improvement may be possible, due to the following reasons:

• The sample arrival and preparation area consists of very many differ-ent activities, which all have to be performed by two employees. Theseemployees are disturbed by phone calls and people entering the labo-ratory, because the sample arrival and preparation area is located atthe entrance of the laboratory. The large number of different activitiesalso causes that the employees experience this work place as a busyplace, whilst the clinical chemist thinks the activities can be performedby one employee instead of two employees. The work load fluctuationsare also experienced most at the sample arrival and preparation, whichis mainly caused by the batch arrivals. There are many moments atwhich no work is available, but also many moments at which manysamples arrive at once.

• Currently, there are some plans for changing the laboratory that alsoaffect the sample arrival and preparation area. The outpatient clinicmay be moved next to the laboratory and the sample and arrival prepa-ration area may get closer to the test execution area. The effect ofthese plans are not investigated yet, so there is much interest for this.

• The throughput times of all samples benefit from improvements at thesample arrival and preparation area, because all samples arrive at thesample arrival and preparation area.

• A machine that takes over some of the activities at the sample arrivaland preparation area is available, and the laboratory staff is interestedin buying this. Therefore it is interesting for them to see what theeffect of this machine would be.

For all these reasons, it is concluded that the research continues withinvestigating the sample arrival and preparation area. The next chaptersfirst describe the sample arrival and preparation area in detail, and theninvestigate improvement possibilities.


0

100

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300

400

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700

Coagulation Transfusion Haematology Chemistry Blood gas Urine

Thro

ugh

pu

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me

(min

ute

s)

Test type

Observed and theoretical throughput times

Outpatient WLZ - O

Outpatient WLZ - T

Inpatient - O

Inpatient - T

External - O

External - T

Outpatient BWO - O

Outpatient BWO - T

O = ObservedT = Theoretical

(a) Regular and blood gas samples

0

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D-dimere APTT Sapphire Cobas6000 Axsym

Thro

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s (m

inu

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Observed and theoretical throughput times for cito samples

Outpatient WLZ - O

Outpatient WLZ - T

Inpatient - O

Inpatient - T

External - O

External - T

Outpatient BWO - O

Outpatient BWO - T

O = ObservedT = Theoretical

(b) Cito samples

Figure 3.10: Throughput times for regular and cito samples


Sample arrival and

preparation area

361 lab numbers

876 body fluid samples

Blood gas

Coagulation

Haematology

Transfusion

Chemistry

Urine

External

Outpatient clinic WLZ(2)

Outpatient clinic W-O(1)

External(2)

147 lab numbers

45 lab numbers

44 lab numbers

Inpatient

Pneumatic mail:

ICU/CCU 7:00 (1) : 7 lab numbers

ICU/CCU/SEH (2) : 39 lab numbers

Blood withdrawal round(1):

51 lab numbers

Rest of other departments(2):

28 lab numbers

17 samples

34 samples

302 samples

344 samples

32 samples

31 samples

48 samples

Unknown: 68 samples

1: Once a day arrivals

2: Continuous arrivals

Figure 3.11: Average number of lab numbers/samples per arrival/test typeon a weekday

Chapter 4

Sample arrival andpreparation

The sample arrival and preparation area, (SAP), is already globally de-scribed in Section 2.1.1, this chapter describes the processes in detail andmodels it as a queueing network in which the body fluid samples are thecustomers. Section 4.1 describes the arrivals of the network, Section 4.2discusses the servers, Section 4.3 the customer priorities, Section 4.4 the ac-tivities the servers have to perform, and Section 4.5 the routing patterns inthe network. The last section summarizes the characteristics in a queueingnetwork and explains the next research steps.

4.1 Arrival types

Every sample destined for the laboratory arrives at the SAP, so the arrivalsof the network are equal to the description in Section 2.1.1 and 3.2.

4.2 Servers

There are two server types at the SAP, employees and machines. The em-ployees are the main servers, they coordinate everything.

On a weekday, two employees are assigned to work at the SAP. Themorning shift employee handles the urgent requests that arrive in the lab-oratory from 7:00 until one of the two employees that are assigned to workat the SAP that day returns from the blood withdrawal round. The twoemployees that have to work at the SAP handle all requests during theday. Both employees have a morning break of 15 minutes, a lunch break of

29

CHAPTER 4. SAMPLE ARRIVAL AND PREPARATION 30

30 minutes and an afternoon break of 15 minutes. The SAP should neverbe unattended, so the employees take their breaks successively. After 17:15,only urgent samples arrive at the SAP and the night shift employee processesthese samples until the morning shift employee for the next day arrives.

Three machine are located at the SAP, two centrifuges and a blood gasanalyzer. The centrifuges can process 72 samples at a time, and it takes ex-actly 10 minutes for one batch. The blood gas analyzer takes approximately2 minutes for the analysis of one sample. Both machines need an employeeto start service, the blood gas analyzer is available again immediately afterfinishing service, and the centrifuges are available after they are unloaded.

4.3 Priorities

At the SAP, the same priorities as in the rest of the laboratory can berecognized, see also Section 2.3. In order of urgency the priorities can belisted as follows,

• Blood gas determination sample

• Cito sample

• Regular sample.

4.4 Tasks at the sample arrival and preparationarea

There are four main process steps at the sample arrival and preparation areathat can be distinguished,

1. Arrival handling

2. Registration

3. Dividing

4. Processing.

Arrival handling

Any sample that arrives at the laboratory needs some type of handling:


• Pneumatic mailThe tube in which the samples are sent needs to be taken out of themail box, the contents have to be taken out and the tube has to besent back to the originating department.

• Carrier for external arrivalsThe carrier puts down a box with the samples, the samples have to betaken out and if necessary, new supplies are given to the carrier.

• Train from outpatient clinicThe train has to be unloaded and sent back to the outpatient clinic.

• Extra blood withdrawalAn employee from the SAP has to go into the hospital to withdrawblood, and after returning in the laboratory, the supplies have to becleared.

• Brought samplesThe samples that are brought have to be accepted by the employeesof the laboratory.

• Inpatient roundWhen an employee returns from the inpatient round, one puts thesamples on the table at the SAP and clears the supplies.

Registration

The registration forms are read into Labosys and identification stickers areput on the samples.

Dividing

The samples are divided into groups that need the same process steps later.

Processing

This step consists of five different tasks:

• Blood gas determinationThis activity is actually part of the test execution, but because theanalysis has to be done as soon as possible after blood withdrawal, itis located at the SAP and the employees that work at the SAP haveto execute the test.


– Start the machine

– Confirm the results in Labosys

• Centrifuge

– Load the samples into the centrifuge and start it

– Unload the machine

• Blood divisionAll samples for external tests have to be divided over several tubes, tosplit the contents of the sample.

• Bringing cito samplesThe results from the cito samples have to be available within one hour,so the samples are brought to the machines at the test execution areaby the SAP-employees as soon as the samples are ready at the SAP.

• SortingAll non-urgent samples that arrive at the SAP are sorted by number,such that they can be found again easily later if for example anothertest has to be done with the sample.

The process times of the tasks are not registered and there is no differ-ence between process times for cito and regular samples. All the tasks thatare described in this section are performed by one of the two employees. Itoften occurs that there are customers waiting for more than one task, andthe employees have to decide which task to perform first. There are somestandard rules concerning task priorities, such as prioritizing all tasks con-cerning blood gas and cito samples. Though, no complete overview of taskpriorities exists.

4.5 Routing patterns

The route that samples follow at the SAP is determined by the origin ofthe sample and the tests that have to be performed with the sample. Everysample needs some type of arrival handling at the SAP. The registrationstep is already performed for samples from the blood withdrawal round andthe outpatient clinics, so these samples skip the registration step. Bloodgas samples and urine samples are immediately picked out after arrival orregistration, so they do not need to be divided. All samples other than urine


samples from BWO, also skip the dividing step, these samples go straightto the processing step after arrival.

The routing within the last processing step is determined by the origin,by the destination of the sample and the precedence relations of the jobs.Centrifugation should always take place before blood division, sorting orbringing cito samples. The machine type on which the sample has to betested determines whether the sample needs centrifugation; approximately80% of all samples need centrifugation. Only samples that are externallytested need to be divided, and only cito samples are brought to the labora-tory. A complete overview of the routing patterns can be found in Figure4.1.

4.6 Conclusion

The previous sections can be summarized by a queueing network with sevendifferent arrival types, three different customer priorities, two machines,and two processors that have to perform tasks at 16 different stations. It ischosen to model the arrivals as seven different queues with separate stations.This is done because each arrival type has a separate queue in practice aswell, the arrival handling differs per arrival type, and this makes the routingdetermined by the arrival type observable. The arrival distributions andprocess time distributions are not determined yet.

The network is displayed in Figure 4.1. The arrows at the left correspondwith sample arrivals, the circles correspond with the tasks the employeeshave to do, the trapezia display the machines and the two rectangles displaythe employees. The customer priorities cannot be recognized in the figure,but every batch or lab number that contains an urgent sample, or an indi-vidual urgent sample, is given priority in the queue for the correspondingstation.

This queueing network is quite complicated, and obtaining results forthis network by exact analysis is complex. Therefore, a simulation modelis developed in Chapter 5 and scenarios are tested with this simulationmodel in Chapter 6. Because it is also interesting to analyze some of thecharacteristics of the queueing network by exact analysis, Chapter 7 analyzesparts of the queueing network with some simplifications.


Inpatient:

Pneumatic mailA1

A2

A3

A4

A5

A6

A7

B1 D1

Arrival handling Registration Dividing Processing

Inpatient:

Brought samples

Inpatient:

Blood withdrawal

External

Inpatient:

Blood withdrawal

round

Outpatient:

WLZ

Outpatient:

BWO

D5

D6

D7

C1

Sample arrival and preparation model

D2E1

D3 D4

E2

E2

Employee 2Employee 1

Job descriptions: Machines:

A1 up to A7: Receive the corresponding arrived samples D3: Load and start a centrifuge E1: Blood gas analyzer

B1: Register the requests and put identification stickers on the samples D4: Turn off centrifuge and unload it E2: Centrifuge

C1: Divide the samples in groups that need the same processing in the next step D5: Divide blood over several tubes

D1: Start blood gas analyzer D6: Bring cito samples

D2: Confirm the result in Labosys and turn off blood gas analyzer D7: Sort the samples by number

Figure 4.1: Simulation model

Chapter 5

Simulation model

A simulation model for the sample arrival and preparation area is developedin this section. Section 5.1 explains the set up of the model, Section 5.2discusses the input for the model, Section 5.3 contains the verification andvalidation of the model and Section 5.4 discusses the output from the model.

5.1 Simulation set up

As already explained in Section 3.1, this research focuses on weekdays, be-cause the laboratory staff has the most interest in this and it is too extensivetoo investigate both weekdays and the weekend. During the night, only oneemployee is present and only urgent tests are performed, this all appears togo fluently, so only the day is considered in the simulation. The simulationruns from 7:00, which is equal to the time that the morning shift employeesstart their work day, to 17:15, when the normal shift employees end theirday.

The simulation model is based on the queueing network model describedin Chapter 4, and the model structure is displayed in Figure 4.1. Theprocesses can best be described by a discrete event system, as the systemis completely determined by a sequence of events, such as sample arrivalsor a server finishing an operation, and by the changes that take place atthese moments. The structure of a discrete event system is displayed inFigure 5.1. The following four events with the corresponding consequencesare distinguished for the SAP simulation model:

1. ArrivalA batch of lab numbers arrives at the SAP. The arrows at the left in

35

CHAPTER 5. SIMULATION MODEL 36

Initialization

Sort event list by time

Take the first event in

the list

Process event:

Add/delete events from

event list

Change state values

Figure 5.1: Discrete event model

Figure 4.1 display the arrival events.Consequences:

(a) Put a batch of lab numbers in the line for the correspondingreceival task (A1 up to A7).

(b) Create a new arrival event if necessary.

(c) Check whether (one of) the employees are (is) available to do atask.

(d) If an employee is available, choose the task and sample with high-est priority (see Appendix A.5) and create an ”Employee finishesa task” event.

2. Employee finishes a taskAn employee finishes one of the 16 different tasks, displayed by thecircles in Figure 4.1.Consequences:

(a) i. If the task concerns turning on a machine, then create acorresponding ”Machine finishes an operation” event.

ii. If the task concerns a disturbance, then restart the task theemployee was doing before the disturbance including a switchtime, and create an ”Employee finishes a job” event.


iii. If the sample is finished at the SAP and ready for test exe-cution, then save the details of the sample.

iv. If none of the three previous cases hold, put a batch of labnumbers/a batch of body fluid samples/a body fluid samplein the line for the next task

(b) If the job concerns turning off or unloading a machine, set themachine to available again.

(c) If an employee is available, check whether a sample is waitingfor any of the tasks. If a sample is waiting, choose the task andsample with highest priority (see Appendix A.5) and create an”Employee finishes a task” event.

3. Machine finishes an operationOne of the three machines at the SAP has finished an operation; inFigure 4.1 the machines can be recognized by trapezia. The machinesstart as soon as an employee is finished starting it up, and it is readyto use again after an employee is finished with unloading it or turningit off.Consequences:

(a) Put a body fluid sample in the line for task D2 if it is the bloodgas analyzer, or a batch of body fluid samples in the line for taskD4 if it is a centrifuge.

(b) Check whether (one of) the employees are (is) available to do atask.

(c) If an employee is available, then choose the task and sample withhighest priority (see Appendix A.5) and create an ”Employeefinishes a task” event.

4. Disturbing factorBoth employees are disturbed by phone calls, other employees, or otherfactors. This event type can occur at any time, either when an em-ployee is busy or when an employee is idle. The disturbance is linkedwith a specific employee.Consequences:

(a) Interrupt the employee who is disturbed and remember withwhich task the employee was occupied.

(b) Create an ”Employee finishes a task” event, corresponding withfinishing handling the disturbance.


Urgent requests can be recognized immediately in practice, so they aregiven priority from arrival. Blood gas and cito samples are generally part ofa lab number. The batch in which the lab number arrives is given priority atarrival, and the lab number that contains the urgent request is given priorityat registration. After registration, only the samples from the lab numberthat are urgent are treated with priority, the other samples from the labnumber are treated as a regular sample from this stage.

5.2 Simulation input

To construct the simulation model, some input parameters and distributionshave to be determined. In this section, arrival times and interarrival timedistributions are estimated in Section 5.2.1, batch size distributions are es-timated in Section 5.2.2, the input for cito samples is discussed in Section5.2.3, process time distributions are determined in 5.2.4, the disturbancesare modeled in 5.2.5 and the task priority setting is explained in 5.2.6

5.2.1 (Inter) Arrival times

Each of the arrival types (see Section 3.2) have a different (inter)arrival timedistribution. This section explains how these distributions are modeled inthe simulation. First, the arrival times for samples that arrive once a dayare estimated, after this the interarrival time distributions for samples thatarrive more frequently than once a day are estimated. Table A.1 in AppendixA.1 shows an overview of the input distributions and parameters for thearrivals in the simulation.

Once a day arrivals

There is no general way to estimate the arrival times for the samples thatarrive once a day, so they are treated separately.

1. Pneumatic mail from ICU and CCU at 7:00The samples from ICU and CCU arrive separately shortly after 7:00.The registration data concerning these arrivals is collected per depart-ment. An estimate for the arrival times is generated by setting a fixedtime, 7:00, and adding an exponential time with a mean estimatedby the mean of the registration times of the data minus the meanprocess times for arrival handling and registration (the process timedistributions are estimated in Section 5.2.4).


2. Blood withdrawal round samplesThe samples for the blood withdrawal round are registered much ear-lier than the arrival time in the laboratory, as these are registeredduring the day and night before. Therefore, the registration timesgive no information concerning arrival times in the laboratory. Theblood withdrawal round starts at 7:45, the employees go to the hospi-tal departments in groups to withdraw blood from the patients. Whena group of employees is finished with blood withdrawal at one depart-ment, they check whether colleagues at other departments need help,and on most days, everyone eventually returns in one group at thelaboratory. The children’s department is the only department thatis visited separately, so the employee(s) that visit(s) this departmentreturn(s) separately. This means that two arrival times have to be de-termined, the arrival time for the blood withdrawal at the children’sdepartment, and the arrival time of all other departments together.

To estimate the arrival times, the process times of the blood with-drawal round are estimated. The staff has written down the time theyreturned in the laboratory after the blood withdrawal round, whichdepartments they visited, the total number of samples they have with-drawn on those departments, and the number of employees they visitedthe departments with. With this information, the mean blood with-drawal time per patient at the children’s department and the otherdepartments is estimated. For these estimates, it is used that each labnumber, i.e. patient, consists of 2.41 samples on average, see Table 3.1.The mean number of employees that go to the children’s department,and to the other departments is determined from this information aswell.

It is assumed that the time for one blood withdrawal is exponentiallydistributed. The process time for blood withdrawal from several pa-tients after each other is then the sum of exponential variables, so anErlang variable. The variation in the number of employees that dothe round is very small, so the number of employees in the simulationis fixed at the mean that is found by the collected information. Thetwo process times for the blood withdrawal round are now estimatedby an Erlang(n, 1/kµ), where n is the number of patients to withdrawblood from, 1/µ is the mean blood withdrawal time per patient, and kis the number of employees that visit the departments. n is estimatedseparately in Section 5.2.2. The arrival times for the samples in thelaboratory can be generated by adding the estimated process times to


the departure time, which is assumed to be fixed at 7:45.

3. Samples from outpatient clinic BWOThe samples from the outpatient clinic BWO are registered in Labosysat withdrawal in Volendam, while they only arrive at the laboratoryin the afternoon, so the registration times are useless. It is well-knownthat the samples from BWO arrive around 13:45, so the arrival timesare simulated by setting a fixed time, 13:30, and adding an exponentialtime with a mean of 15 minutes.

Continuous arrivals

The arrivals that occur more than once a day are:

1. Pneumatic mail with samples from IC, CCU and SEH during the day,

2. Samples from other departments during the day via pneumatic mail,brought to the laboratory or by extra blood withdrawal,

3. Samples from external sources,

4. Samples from outpatient clinic WLZ.

The available data concerns the registration times instead of arrivaltimes, but the time between the registration times is expected to give agood indication of the interarrival times. To obtain an estimate for theinterarrival time distribution for the first three arrival types, the same ap-proach can be applied. The estimation of the interarrival time distributionfor the samples of the outpatient clinic is influenced by the outpatient clinictrain, which is discussed later in this section.

Much data is available, so one could choose to use the empirical dis-tribution. A disadvantage of the empirical distribution is that it takes upmuch memory in the simulation program. A theoretical distribution that isfitted to the data takes up much less memory and it smoothes out irregular-ities that the empirical distribution may have. Another advantage of fittinga theoretical distribution is that it is easier to generate values outside therange of the data and to perform sensitivity analysis with the simulationmodel. Sensitivity analysis is easier with a fitted distribution, because themean and variation of the distribution can be adjusted easily. A way to fita theoretical continuous distribution is to fit a phase-type distribution onthe first two moments of a variable, this method is developed by Tijms [17]and is applied very often in queueing theory. This fit is especially useful


for analytical computations on the system, because it fits phase-type dis-tributions, and the exponential components of these distributions have thememory-less property.

The method that Tijms [17] developed is based on the mean, E[X], andthe coefficient of variation, cX , given a random variable X. Two classesof distributions are used, Erlang distributions and hyperexponential distri-butions. Denote by Erlang(n, 1/µ) an Erlang distribution with mean n/µ,and by Hn(p1, .., pn; 1/µ1, ..., 1/µn) a hyperexponentially distributed vari-able, where pi denotes the probability that the variable is an exponentialvariable with mean 1/µi. Lemma 1 describes the fitting method developedby Tijms.

Lemma 1 Let X be a random variable with mean E[X] and coefficient ofvariation cX . Then the random variable Y matches the first two momentsof X if Y is chosen as follows:

1. If 1/k ≤ c2X ≤ 1/(k − 1) for certain k = 1, 2, 3, ..., then

Y ={

Erlang(k − 1, 1/µ) w.p. p,Erlang(k, 1/µ) w.p. 1− p,

where

p =1

1 + c2X[kc2X −

√k(1 + c2X)− k2c2X ], µ =

k − pE[X]

.

2. If cX ≥ 1, thenY = H2(p1, p2, 1/µ1, 1/µ2)

wherep1µ1

= p2µ2,

p1 = 12

(1 +

√c2X−1

c2X+1

), p2 = 1− p1,

µ1 = 2p1E[X] , µ2 = 2p2

E[X] .

Goodness-of-fit tests

To check whether this method fits the data well, goodness-of-fit tests areperformed. The data is continuous, so goodness-of-fit tests based on thediscrepancy between the empirical and the fitted distribution are suitable.An explanation of the applied goodness-of-fit tests can be found in AppendixB.1.


Applying the parametric bootstrap method for the goodness-of-fit testbased on the Cramer-von Mises test statistic results in bad p-values, whichimplies that the fitted distribution gives a bad fit to the underlying distri-bution. This suggests to reject the proposed fit, but when other fits areevaluated, such as the fit of a Weibull distribution, these are also rejected.Therefore, another goodness-of-fit measure is applied as well. This is doneby plotting the quantiles of both distributions in a quantile-quantile (q-q)plot. If two distributions are similar, the points in the q-q plot are on theline y=x. Figure 5.2 shows the q-q plots for the different arrival streamsfor the data quantiles against the quantiles from the theoretical distributionobtained by the method of Tijms (the q-q plot for the outpatient clinic WLZis discussed in the next subsection). All plots show that the distributionsappear to coincide pretty good, except for the tails. The deviations in thetail have probably caused the bad test results.

The question is now whether the tails in the data truly represent theunderlying distribution, or that the outliers are partially caused by mea-surement errors. The interarrival times are derived from the time betweentwo registrations. This already causes some errors because this is only an ap-proximation of the interarrival times. It is also very likely that some outliersin the time between two registrations are caused by forgotten registrations,instead of true large interarrival times.

It is hard to find any theoretical fit to this data that is not rejected bya goodness of fit test. Taking this into account, the theoretical fit by themethod of Tijms is accepted for the simulation, because it does appear tofit the data quite well according to the q-q plot, except for the outliers.Note that this does leave room for improvement of the simulation, becausea better fit could possibly be obtained by another method.


As mentioned in Section 3.2.2, the process of blood withdrawal at the out-patient clinic and the transport to the laboratory is modeled separately toobtain the arrival process for the samples of the outpatient clinic WLZ. Thissection first describes the processes and then explains how this is modeledin the simulation.

The patients arrive individually at the outpatient clinic in the WLZ, theyare registered and wait in the queue, after which their blood is withdrawnby one of the three or four employees working at the outpatient clinic. Afterwithdrawal, the samples are collected in a train. The train is only sent tothe laboratory whenever a cito sample arrives, or when the train is almost


0 50 100 150 200

050

100

150

200

250

300

Theoretical quantiles by Lemma 1

Dat

a qu

antil

es

QQ−plot ICU/CCU/SEH

(a) ICU/CCU/SEH

0 50 100 150

050

100

150

200

250


Dat

a qu

antil

es

QQ−plot other departments

(b) Other departments

0 100 200 300

100

200

300

400


Dat

a qu

antil

es

QQ−plot external

(c) External

0 10 20 30

05

1015

2025

3035


Dat

a qu

antil

esQQ−plot outpatient WLZ

(d) Outpatient WLZ 9:00-12:00

Figure 5.2: QQ-plots for fitted interarrival time distributions

fully loaded with regular samples. The cito samples are sent separately ina train, without any other regular samples. There are 3 trains available,which all have capacity for 36 body fluid samples. Because the train is onlysent when it is nearly full, or when a cito sample arrives, the regular samplesfrom the outpatient clinic laboratory arrive in batches of 34 to 36 samplesand the cito samples arrive separately.

The process of withdrawal at the outpatient clinic can be seen as a multi-server queue, a G/G/c system. Right after the multiserver, three transporta-tion machines (i.e. the trains) are waiting and these serve the customers (i.e.the body fluid samples) with a prioritized batch policy. The process time ofthe machines (i.e. travel time) is deterministic. The machine can be mod-


eled as a GX/Da,b/3 system. The output process of this system is equal tothe arrival process of body fluid samples from the outpatient clinic WLZ inthe laboratory. Figure 5.2.1 displays the model.

Patient

arrival

Blood withdrawal by

3 or 4 employeesTransportation to

the laboratory

Blood withdrawal at the outpatient clinic

Figure 5.3: Outpatient clinic WLZ modeled as a queueing system

This system is simplified in the simulation model. Because it is tooextensive to collect process times for blood withdrawal, it is assumed thatthe output process of the blood withdrawal is equal to the arrival process ofpatients, so the blood withdrawal step is not modeled in the simulation. Forsome queueing systems, such as the M/M/c queue, this is true, and for theM/G/c queue the output process approaches a Poisson process as c → ∞[17]. By ignoring the withdrawal step, the model for the outpatient clinicin the simulation only considers the train. The time of a one-way journeyfor the train is exactly 2 minutes and 10 seconds. When the train arrives atthe laboratory, it has to be unloaded and sent back again. After 2 minutesand 10 seconds, the train arrives at the outpatient clinic again, after whichit is available for transport. Figure 5.4 displays how the outpatient clinic ismodeled in the simulation.

The input for this model is the arrival time distribution of customersat the outpatient clinic. Figure 3.8 clearly shows a trend over the day inthe registrations at the outpatient clinic WLZ. Therefore, the day is splitin time slots (8:00-9:00, 9:00-12:00, 12:00-13:00, 13:00-14:00, 14:00-15:00,15:00-16:00, 16:00-17:00) and the interarrival time distribution is estimatedper time slot. The simulation does not switch to a new time slot exactly atthe boundary. Each time a sample arrives, it is checked whether the currenttime exceeds the start time of the next time slot, if so, the next interarrivaltime is generated with the parameters for the new time slot.

Remark that the empirical distribution is discrete, because the regis-


2'10

Sample arrivals in

batches per lab

number

Train leaves:

1: At least 34 regular samples

2: One or more cito samples

2'10

Unload train

in laboratory

Train available for

transport again

Outpatient clinic in simulation model

Figure 5.4: Outpatient clinic WLZ in the simulation

tration times are rounded in minutes. Since the interarrival times for theoutpatient clinic are smaller than the interarrival times for the other sam-ples, the effect of rounding off the registration times to minutes shows in theempirical distribution function; Figure 5.5 displays the distribution functionof the registrations between 9:00 and 12:00. Since the interarrival times arecontinuous in reality, it is better to fit a theoretical distribution than to usethe empirical distribution. The arrival process of patients is modeled by themethod explained in Lemma 1. The quantile-quantile plot for the fit anddata concerning the interarrival times between 9:00 and 12:00 is shown inFigure 5.2. The plot shows that the fitted distribution deviates from theempirical distribution especially for larger values, but appears quite goodfor smaller values.

5.2.2 Batch sizes

Batches play an important role in the arrivals of samples at the SAP. Manylab numbers do not arrive singly, but in batches, for example the samplesfrom the blood withdrawal round, samples from the outpatient clinic BWOand most of the samples from pneumatic mail. As explained in Chapter 3,each lab number again consists of a batch of body fluid samples. In thissection, the batch size distributions for batches of lab numbers and for thebatches of body fluid samples are estimated. Table A.2 in Appendix A.2shows an overview of the input parameters for the simulation.


0 20 40 60 80

0.0

0.2

0.4

0.6

0.8

1.0

Interarrival time (minutes)

F_n

(x)

Empirical CDF for interarrival times outpatient clinic WLZ

Figure 5.5: Empirical CDF of interarrival times for outpatient clinic WLZ9:00-12:00

Lab number batches

There are several ways to obtain an estimate for the batch size distributionsfor the lab number batches. Because much data is available, one could usethe empirical distribution. As already mentioned in the section concerninginterarrival times, storing the empirical distribution takes up much memoryin the simulation. Fitting a theoretical distribution avoids this and makes iteasier to perform a sensitivity analysis. A discrete distribution could be fit-ted on the first two moments by applying the method developed in [3]. Thismethod fits a distribution with mean E[X] and coefficient of variation cXdepending on the value of a := c2X−1/E[X]. Four classes of distributions areused: Poisson, mixed binomial, mixtures of negative-binomial and mixturesof geometric distributions with balanced means. Let GEO(p) denote a ran-dom variable with probability distribution pi = (1− p)pi, i = 0, 1, 2, ..., andlet NB(k, p) denote a random variable which is the sum of k independentGEO(p) variables. A Poisson(λ) is a Poisson distributed random variablewith mean λ, and a BIN(k, p) variable is binomially distributed, where k isthe number of trials and p the succes probability. Adan et al. [3] introducethe following Lemma:

Lemma 2 Let X be a random variable on the non-negative integers withmean E[X] and coefficient of variation cX and let a = c2X − 1/E[X]. Then


the random variable Y matches the first two moments of X if Y is chosenas follows:

1. If −1/k ≤ a ≤ −1/(k + 1) for certain k = 1, 2, 3, ..., then

Y ={BIN(k, p) w.p. q,BIN(k + 1, p) w.p. 1− q,

where

q =1 + a(1 + k) +

√−ak(1 + k)− k

1 + a, p =

E[X]k + 1− q

.

2. If a = 0, then Y = Poisson(λ) with λ = E[X].

3. If 1/(k + 1) ≤ a ≤ 1/k for certain k = 1, 2, 3, ..., then

Y ={NB(k, p) w.p. q,NB(k + 1, p) w.p. 1− q,

where

q =a(1 + k)−

√(1 + k)(1− ak)

1 + a, p =

E[X]k + 1− q + E[X]

.

4. If a ≥ 1 then

Y ={GEO(p1) w.p. q1,GEO(p2) w.p. q2,

where

p1 =E[X](1 + a+

√a2 − 1)

2 + E[X](1 + a+√a2 − 1)

, q1 =1

1 + a+√a2 − 1

,

p2 =E[X](1 + a−

√a2 − 1)

2 + E[X](1 + a−√a2 − 1)

, q2 =1

1 + a−√a2 − 1

.


Mean regularsample

Mean Citosample

Mean bloodgas sample

95th perc.regular

95th perc.cito

Theoretic fit 16.0(±0.4) 12.8(±0.1) 8.5(±0.1) 42.7(±1.4) 25.2(±0.2)Empirical 16.1(±0.4) 12.3(±0.1) 8.4(±0.1) 43.7(±1.3) 23.9(±0.3)

Work load Cito within15/5 minutes

Round before9:30

Number ofswitches

Theoretic fit 55%(±0.3) 43%(±0.5) 100%(±0.1) 265(±2)Empirical 54%(±0.4) 45%(±0.5) 100%(±0.0) 263(±2)

Table 5.1: Sensitivity of theoretical fit vs. empirical


It is also checked by a goodness-of-fit test whether the fitted distributionfits the underlying data well. Because the data is discrete, a chi-square testis used. An explanation of the test can be found in Appendix B.2.

The results for the chi-squared tests for the fitted distributions accord-ing to Lemma 2 are displayed in Table A.2 in the appendix. According tothe results, only the fitted distribution for batch sizes from IC/CCU/SEH,CCU at 7:00, and the samples from other departments during the day arenot rejected at a 95% significance level. The probability-probability plotsin Figure 5.6, in which the cumulative probability distributions of the fittedand the empirical distributions are plotted show that the fit does not appearto be very bad for most types. Though, this is still not a very convincingargument for using the fitted distributions. Therefore, it is also tested howsensitive the simulation program is for the choice of the batch size distribu-tion. 1000 days are simulated with empirical batch size distributions, and1000 days with theoretical batch size distributions. Table 5.1 shows somemean output parameters, of which the first five concern the throughputtimes, and the corresponding 95% confidence intervals. The results showthat the differences are very small. The only confidence intervals that donot overlap concern the throughput times for cito samples.

Because the system does not appear to be very sensitive to the batch sizedistribution, and the advantages of a theoretical fit concerning the memory-usage and sensitivity analysis, it is chosen to use the theoretical distributionfit.

Body fluid sample batches

In general, one lab number consists of more than one body fluid sample. Thenumber of samples for one lab number mainly depends on the requested tests


and the corresponding machines on which the tests have to be performed.Generally, each machine type requires a separate sample such that testson different machine types can be done simultaneously. This means thatthe number of samples is equal to the number of machine types requiredto execute the requested tests. Exceptions may occur if it is decided laterthat a patient also needs another test, for which a blood tube from anothermachine is used instead of withdrawing a new blood tube. Not only the sizeof the batch is determined by the required machine types, also the necessaryprocess steps, such as centrifugation and blood division, is determined bythe machine types the samples are destined for. Therefore, not only the sizeof the batch but also the corresponding machine types have to be generated.

The method that is applied generates the number of samples and thecorresponding machine types in one routine. There are 17 machine typesfor conducting the tests at the test execution area, including an externalmachine type and a blood gas type. The method differentiates between theorigins and between lab numbers that contain cito samples and lab numberswithout cito samples. This differentiation is necessary, because the requestedtests, and thus the requested machines for the tests, depend on the originand the status of the patient. Patients that need cito tests need urgentcare, which means they generally need other tests than patients that are noturgent.

The origin of the request also influences the tests that are requested,patients that are in the hospital need many medicine observation tests, ortransfusion tests, while patients from outpatient clinics need more tests fordiseases for example. Therefore 5 origins are distinguished, which are theoutpatient clinic WLZ, the outpatient clinic BWO, inpatients from IC, CCUor SEH, inpatients from other departments and external requests. For eachof the origins, lab numbers that contain cito tests and lab numbers withoutcito tests are distinguished.

Let pkij denote the fraction of lab numbers in the data from origin j andpriority type k that contained a sample for machine type i. Whenever a labnumber from origin j and priority type k arrives in the simulation, repeatthe following routine until the lab number contains at least one sample:

1. Draw 17 uniform random variables Un.

2. If Un ≤ pknj , the lab number contains a sample for machine type n,else the lab number does not contain a sample for machine type n.

3. If Un > pknj for every n, go back to step 1


This method results in nearly the same distribution of body fluid samplesas in the data. It does not match completely because it can occur thatthe uniform variables are larger than pkij for each machine i, which happenswith probability

∏i(1− pkij), for origin j and priority type k, and this draw

is ignored. It is ignored, because once an arrival occurs, it is sure thatthe batch size is at least 1. It is very hard to derive the exact conditionalprobability that the batch contains a sample for machine type i while thebatch size contains at least one sample. The probability that all Un arelarger than Un > pknj is at most 0.04 for all origins and priority types. Thisdoes mean that the number of lab numbers that contains a sample fromorigin i is slightly overestimated for each i, but this is accepted because theerror is very small.

5.2.3 Cito tests

Any lab number that arrives in the laboratory may contain a cito sample.The cito arrivals are not modeled separately, but they are seen as a part ofthe arrivals. For each of the arrival types, the probability that an arrivinglab number contains a cito sample is estimated by the mean fraction of labnumbers that contains a cito sample on a weekday between 7:00 and 17:15for that arrival type in the data set. Table A.3 in Appendix A.3 shows theprobabilities that a lab number that arrives contains a cito sample per originin the simulation.

5.2.4 Process times

The process times of the tasks and machines in Figure 4.1 are not registered.In order to set up the simulation model, estimations for the process timeshave to be made. This is done by clocking the times at the SAP by hand fortwo days. Two different weekdays are chosen, a Wednesday and a Monday,and all process steps are clocked. Of course, this method is subject to somedisadvantages, such as little data, the influence of the presence of ’clockers’on the employees and different employees may have different process times,but it does give an idea of the process times. It is checked after measurementwhether the two days were comparably busy as other days, by looking atthe number of requests in Labosys in comparison with other days. Bothmeasurement days appear to be comparable with an average day.

After collecting the data concerning process times, the means are deter-mined and the process times are estimated by an exponential variable withthis average, except for turning on the centrifuge and registration. The


process time for turning on the centrifuge is dependent on the number ofsamples that are loaded into the centrifuge. The registration step for exter-nal samples generally takes longer than for other samples because the formsare different from the normal forms from the WLZ hospital, so a differentprocess time is determined for these forms. An overview of the mean processtime per task is given in Table A.4 in Appendix A.4.

5.2.5 Disturbances

There is no information available concerning disturbances, so these are mea-sured by hand in the same way the process times are measured. On the twosame weekdays, the time and length of the disturbances is recorded by hand.With this data, the mean interarrival time of disturbances and the meanlength of disturbances is determined. It is assumed that the interarrivaltime and the length of the disturbance are exponentially distributed.

In the simulation the time between two disturbances for employee 1is equal to 10 minutes on average, and the duration of the disturbances ishyperexponentially distributed, with probability 0.20 it is a long disturbancewhich takes on average 1.5 minutes, and with probability 0.80 it is a shortdisturbance which takes on average 0.5 minutes. Employee 2 is disturbedevery 30 minutes on average, and the disturbance takes 1 minute on average.

5.2.6 Task priorities

The employees have to do 16 different tasks, it occurs often that samplesare waiting at more than one task, so for the simulation, it is important toget an idea of the priority setting between the tasks. The employees havesome guidelines for this, but not everything is prescribed. Some priorities arevery clear, such as processing a blood gas determination request as soon as itarrives, or processing cito requests before regular requests. Other prioritiesare not that obvious, so a task priority list is created in consultation withthe laboratory staff and by observation on the measurement days. In thesimulation, each time an employee finishes a task, or a sample arrives, theemployee picks a tasks by walking through a list and picking the first taskfor which samples are waiting and other conditions such as availability of themachine or concerning the previous task hold true. In reality, the employeesmay not always follow the same routine, but it appears that this list doesgive a good impression of how it goes in general. The task priority list canbe found in Appendix A.5.


5.3 Verification and validation

During and after the construction of a simulation model, the model shouldbe verified and validated. The first step in the model check is verification,it should be verified that the program works as it is supposed, it should bein accordance with the model. This is mainly done by debugging.

Once the verification step is completed, the model should be validated.By validation, one checks whether the simulation model is a sufficiently closeapproximation of the reality, [5]. Validation can be done by comparing theoutput of the model with data. Data concerning throughput times at theSAP is not available, therefore, the model is validated by comparing thethroughput times of the simulation with a few sampled times and by theopinion of the staff. Table A.5 displays the sampled throughput times andthe mean throughput times of 100 simulated days. The mean throughputtimes from the simulation appear to be of the same order as the sampledtimes.

The validation by the staff is done by seven people: four analysts, theteam leader, the clinical chemist and the quality officer of the laboratory.Each of the seven people that are consulted for the validation step havea professional view on the laboratory, and in particular on the SAP. Thelaboratory has four analysts that all coordinate a part of the laboratory, oneof them coordinates the SAP. Each of these four analysts works regularly atthe SAP. The laboratory is lead by the clinical chemist and the team leader,which is an analyst as well. The quality officer regulates the quality of allparts of the laboratory. In a few meetings, these seven people have givenfeedback concerning the simulation model and its output, which eventuallyresulted in a final model in which all seven employees have stated their trustthat the model depicts the reality closely.

5.4 Simulation output

The simulation model produces the following output:

1. Sample details

(a) Arrival time

(b) Origin

(c) Destination machine

(d) Priority type


(e) Waiting times at each station

(f) Finish time

2. Station details, split in an urgent part for cito and blood gas samples,and a regular part for the other samples

(a) Queue length

(b) Waiting times of the samples in the queue

3. Employee

(a) Start time, end time and task type of each task the employeeperforms on a day

5.4.1 Key performance indicators

With all output parameters, many different output measures can be calcu-lated. The goal is to compare scenarios with the simulation model, so ninekey performance indicators are defined in consultation with the laboratorystaff:

1. Mean throughput time for regular samples

2. Mean throughput time for cito samples

3. Mean throughput time for blood gas determination samples

4. 95th percentile of the throughput time for regular samples

5. 95th percentile of the throughput time for cito samples

6. Workload, percentage of working time on a day

7. Percentage of cito body fluid samples that is processed within 15 min-utes if centrifugation is necessary, and within 5 minutes if no centrifu-gation is necessary.

8. Percentage of body fluid samples from the blood withdrawal roundthat is processed before 9:30, because the results of these samplesshould be available before 11:00 for the doctor round.

9. Number of switches between different jobs per employee, because manyswitches increase the work load experience for the employees.


KPI-value - 100 95% CI - 100 KPI-value - 1000 95% CI - 1000Mean regular sample 15.64 (± 0.90) 16.01 (± 0.40)Mean Cito sample 12.72 (± 0.26) 12.77 (± 0.09))Mean blood gas sample 8.50 (± 0.43) 8.51 (± 0.12)95th perc. regular 41.31 (± 3.62) 42.67 (± 1.40)95th perc. cito 25.02 (± 0.84) 25.19 (± 0.22)Work load 55.04% (± 1.05) 55.03% (± 0.26)Cito within 15/5 minutes 42.28% (± 1.62) 42.91% (± 0.48)Round before 9:30 99.96% (± 0.04) 99.92% (± 0.06)Number of switches 264 (± 4) 265 (± 2)

Table 5.2: KPI values & Confidence intervals

With these KPI, the scenarios can be compared easily in a global way. Ifmore detailed information concerning a scenario is requested, more detailedmeasures, such as mean throughput time per origin, or machine, can beobtained as well.

When comparing the scenarios, not only the value of the KPI, but alsothe confidence intervals concerning the KPI give much information. Table5.2 shows the mean of the KPI and the corresponding 95% confidence in-tervals for 100 and 1000 simulated days. The confidence intervals appearrelatively small for 100 runs already, only the confidence interval for themean throughput time of a regular sample is relatively a bit larger. Theconfidence intervals for 1000 runs are of course smaller, but the mean re-sults are very close to the results from 100 days. Because the differences arequite small, and the simulations would take long if each scenario needs 1000runs (approximately 5 minutes per 100 days), the scenarios are tested by100 days. As the confidence intervals are already quite small for 100 runs,the confidence intervals are not displayed in the result tables in the followingchapters. This is also done to keep the result tables clear.

5.4.2 Waiting times

Other than the defined KPI’s, the mean waiting time for each task is also aninteresting aspect of the system to investigate. Figure 5.7 displays the meanwaiting time per task for regular samples and cito samples obtained from asimulation over 100 days. For task descriptions, see Figure 4.1. The figureshows that the registration (B1), blood dividing (D5) and sorting (D7) aretasks that cause the highest waiting times for regular samples. The longwaiting time for registration is mainly caused by the fact that lab numbersthat contain cito samples are given priority at registration, and the next


process steps for these samples are also performed before the regular labnumbers are registered. The sorting and blood dividing step are given lowpriority amongst the tasks (see Appendix A.5), therefore the waiting timesfor these tasks are long as well.

The waiting times for cito samples are all under 1 minute, except forreceiving external lab numbers (A4) that contain cito samples and turningon the centrifuge (D3). The waiting time for turning on the centrifuge iscaused by different reasons for cito and regular samples. The waiting timefor regular samples is mainly caused by waiting for other samples until thereare more than 20 samples waiting, or until a cito sample arrives. The wait-ing time for cito samples is caused by waiting for a centrifuge to becomeavailable, because a cito sample is put in a centrifuge immediately if a cen-trifuge is available. The sensitivity towards the priority setting of the tasksis investigated in Section 6.4.

5.5 Conclusions

The simulation model is developed, input distributions are determined andthe model is validated. The model is now ready to use for testing scenarios,and the scenarios can be compared with the defined KPI. The scenariotesting is done in the next chapter.


0.0 0.2 0.4 0.6 0.8 1.0

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Theoretical CDF by lemma 2

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(b) CCU 7:00

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(c) Outpatient BWO

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(e) Other departments

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(f) External

Figure 5.6: P-P plots for fitted batch size distributions


0 1 2 3 4 5 6 7Mean waiting time (minutes)

Cito

Regular

D7D6D5D4D3D2D1C1B1A7A6A5A4A3A2A1

D7D6D5D4D3D2D1C1B1A7A6A5A4A3A2A1

Figure 5.7: Mean waiting time per task and priority type

Chapter 6

Simulation scenarios

The simulation model is constructed to test scenarios, this chapter discussesthe scenarios and the results. Section 6.1 explains the basic scenarios, Sec-tion 6.2 discusses the results from the basic scenarios, results from the com-bination of several scenarios are discussed in Section 6.3, Section 6.4 teststhe sensitivity of the system to higher arrival rates and the task prioritysetting and in Section 6.5, a summary of the conclusions of this chapter isgiven.

6.1 Basic scenarios

The simulation model is used for investigating fifteen basic scenarios:

1. Current situation

2. Cito shorterThe walking distance to the laboratory for bringing cito requests isshortened. This may be realized by moving the SAP into the labora-tory. The time is shortened from 1.62 minutes to 0.1 minutes.

3. No disturbing factorsThe SAP-staff is disturbed by phone calls, other staff etcetera. Thisscenario tests the effect of eliminating all disturbing factors.

4. Move the outpatient clinicThe outpatient clinic is located at a different floor than the laboratory.This scenario tests the effect of moving the outpatient clinic next tothe laboratory. This means that the samples are not transported by

58

CHAPTER 6. SIMULATION SCENARIOS 59

train anymore, but they arrive one by one at the SAP immediatelyafter withdrawal.

5. New structure for the blood withdrawal roundCurrently, the employees that run the blood withdrawal round all helpeach other after they are finished with their own departments, thismeans that most of the samples arrive in one batch at the SAP after theround. Two scenarios with another structure for the blood withdrawalround are evaluated:

(a) Every department separatelyIn this scenario, each employee just withdraws blood from pa-tients in the department he or she is assigned to and then returnsto the laboratory without helping out in other departments. Thismeans that the samples do not arrive in one large batch anymore,but they arrive in smaller groups.

(b) Every department separately, with SAP employees on the small-est departmentThis scenario is similar to the previous one, only in this one,employees from the SAP withdraw blood from the departmentswith the fewest patients, such that they return to the laboratoryquickly and they can start processing the samples.

6. Move sorting activitiesOne of the activities in the SAP is sorting. Sorting is done such thatthe samples are ordered by number, and if one has to look up a patient,this can be done easily. At the moment, sorting is not only done in theSAP, but also after test execution and before loading the outpatientclinic train. This scenario evaluates the effect of eliminating the sortingactivity at the SAP, this means that the sorting is now done after testexecution.

7. BWO twice a dayAt the moment, all samples from the outpatient clinic in Volendam arebrought by a carrier between 13:30 and 14:00. This scenario evaluatesthe effect of sending a carrier at 11:00 and 13:30, such that the batchsize is smaller.

8. Varying batch size in the outpatient clinic train to 15(a) or 5(b)The normal procedure for sending an outpatient clinic train is to fullyload it and then send it away. These scenarios test whether it mightbe beneficial to send the train earlier, so not to load it completely.


9. One employeeOn a regular day, two employees are assigned to work at the SAP.This scenario only assigns one employee to the SAP and evaluates theeffect.

10. No cito typeThis scenario treats all requests in the same way, except for bloodgas determination requests. Cito requests are not given priority overregular requests.

11. PreanalyzerThere is a machine available that can take over the dividing, sort-ing, blood dividing and centrifuging from the employees. There aredifferent settings for the centrifuge in the preanalyzer:

(a) BatchThe centrifuge starts running as soon as a fixed number of samplesis present.

(b) TimeThe centrifuge starts running after a fixed time.

(c) No centrifugeA smaller preanalyzer without built-in centrifuge is considered,so the centrifuge handling is still done by the SAP staff.

6.2 Basic scenario results

In Table 6.1 the results of 100 simulation days for the basic scenarios de-scribed in the previous section are shown. The results are discussed in detailin the following subsections.

6.2.1 Cito shorter

Shortening the process time for bringing cito samples to the laboratoryactually shows one of the effects of moving the SAP closer to the laboratory.This is an idea that the laboratory staff is seriously thinking about, and thefirst negotiations with the hospital management concerning the costs havetaken place already. In this scenario, it is assumed that everything else staysthe same. The mean process time is reduced from 1.62 to 0.1 minutes.

As expected, the mean throughput time and 95th percentile for citosamples decrease, though they decrease with more than 1.5 minute, so not


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only the process time is shorter, also the waiting time is reduced. Also themean throughput time for a regular and a blood gas sample and the 95thpercentile for a regular sample decrease, due to shorter waiting times. Themean waiting time per process step decreases for regular samples from 2.3minutes to 1.8 minutes and for cito samples it decreases from 0.62 minutesto 0.57 minutes. The amount of work and thus the workload decreases,because the same number of samples arrives while the process time for oneprocess step decreases.

6.2.2 No disturbing factors

This scenario eliminates all disturbing factors, so no interruptions of anykind occur anymore. The laboratory staff experiences that phone calls,questions and other disturbances from their regular SAP activities are veryinterrupting, and expects that taking away the disturbances has a tremen-dous effect on the results. Though, the results of the simulation imply thatthe effect of these disturbances are not that large on the defined KPI’s. Theperformance measures all change positively, but not as much as expected.Apparently according to the simulation, the disturbances mainly cause amental effect.

6.2.3 Move outpatient clinic

The outpatient clinic is currently located at the floor below the laboratory.If the outpatient is moved up next to the laboratory, the arrival stream fromthe outpatient clinic would become more constant. Currently, the regulararrivals from the outpatient clinic occur in batches of 34 to 36 samples,and the cito samples arrive separately. In this scenario, it is assumed thatthe samples from the outpatient clinic are given to the SAP immediatelythrough a hatch. The priority list is adjusted as well in this scenario, thearrival handling for regular samples from the outpatient clinic is given thelowest priority, such that they are handled at the moments no other tasksare waiting.

The mean throughput time for regular samples increases, which is mainlycaused by an increase of mean throughput time for the samples from theoutpatient clinic, this increases from 13.3 minutes, to 15.5 minutes. This iscaused by an increase in the mean waiting time for the arrival handling, itincreases from 1.1 minute, to 3.5 minutes. This may imply that the scenariogives bad results, but the opposite is true, because the results for the normalscenario do not capture the waiting time at the outpatient clinic. In the


scenario with the moved outpatient clinic there is no waiting time, whilein the normal scenario, the mean waiting time for regular samples at theoutpatient clinic is nearly 26 minutes. The scenario also has much influenceon the mean number of switches, this increases from 264 to 398.

6.2.4 Blood withdrawal round

The changes in the blood withdrawal round scenarios give no remarkableresults for the throughput times at the SAP. When the departments aresplit, the mean throughput times increase because the SAP-employees cancome back later, while the samples are already waiting in the laboratory.When the SAP-employees return first, the mean throughput time for regularsamples is smaller than for the current situation, and the other throughputmeasures have similar values as for the normal scenario. Therefore, it can beconcluded that the effect of these changes appear small in the SAP. Outsidethe SAP, the mean time until a sample from the blood withdrawal roundarrives in the laboratory increases from 31.7 minutes to 34.6 minutes. Thisoccurs because the employees do not help out each other anymore, and whena department has much samples on a day, these samples are all withdrawnby a few employees. These scenarios could better be investigated in detailby modeling the blood withdrawal round in a more detailed way, but thatis too extensive for this research.

6.2.5 Move sorting activities

In this scenario, the samples skip the sorting step at the SAP. Of course,this means that other employees have to spend more time on sorting, butcurrently sorting is done at three different places. Sorting is one of the taskswith the lowest priorities, so by deleting this step, the mean throughputtime decreases very much. The mean waiting time for sorting is 4.0 minutesin the normal scenario, this is nearly equal to the decrease in the meanthroughput time. The cito samples do not have to be sorted, and theyare always prioritized over the regular samples, so no effect on cito samplescan be notified. Also the mean number of switches and the mean workloaddecrease significantly.

6.2.6 BWO 2x

This scenario only shows the effect of spreading the samples from outpa-tient clinic BWO over 2 batches, one arrives around 11:00, the other arrives


around 13:30. The idea is that this decreases the workload in the afternoon,but the results show no remarkable effect for any of the KPI’s.

6.2.7 Batch size outpatient clinic train

The structure of withdrawal and transport of body fluid samples at theoutpatient clinic is described in detail in Section 5.2.1. In the normal sit-uation, the regular samples are only sent to the laboratory when the trainis nearly full. The cito samples are sent separately immediately after with-drawal. There are 3 trains with capacity for 36 samples, and the travel timeis approximately 5 minutes, while only 2.4 samples arrive per 3 minutes onaverage. It appears that there is enough capacity to send the trains moreoften. In that way, the samples have shorter waiting times at the outpatientclinic and the samples arrive at the SAP in smaller batches.

Table 6.1 shows the effect for the SAP performance measures for chang-ing the batch size to 15 samples and 5 samples. The table shows that thesescenarios reduce the mean throughput time for regular samples slightly andhave a large influence on the mean number of switches, because the trainshave to be unloaded more often with smaller batch sizes. What these resultsdo not show are the waiting times at the outpatient clinic.

Figure 6.1 shows the effect of the minimal batch sizes varying from 1to 36 samples on the throughput and waiting times, the workload in theSAP and the mean number of switches. The figure shows that the waitingtimes at the outpatient clinic decrease when the minimal batch size getssmaller, but the number of switches increases, just as the mean workload.The throughput times are not affected significantly. This implies that thetotal mean throughput times for regular samples from the outpatient clinictrain can be reduced tremendously by sending the train earlier, though theworkload experience may increase at the SAP, because the train arrives moreoften and the number of task switches increases.

6.2.8 One employee

The scenario in which only one employee carries out all activities at the SAPinstead of two employees is very important for the laboratory management.If the results indicate that one employee suffices, it would mean that anemployee is free to help out at other parts in the laboratory or that thelaboratory could do with one employee less when keeping the requirementthat the laboratory has to shrink by 3fte in mind.

Table 6.1 shows that this scenario causes all throughput time indicators


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Waiting time clinic regular samples Waiting time clinic cito samples

Mean throughput time regular samples Mean throughput time cito samples

Effect of batch size on throughput and waiting times

(a) Throughput and waiting times

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to increase, the mean throughput time increases with 182% for regular sam-ples, with 20% for cito samples and with 27% for the blood gas samples,


the 95th percentiles increase by 298% for regular samples and by 18% forcito samples. The percentage of samples that meets the time requirementsdecreases by 14% for cito samples and by 11% for blood withdrawal roundsamples. The directions of the changes are expected beforehand, becausefewer operators are available to carry out the activities, so samples havelonger waiting times. The mean waiting time for regular samples increasesfrom 2.3 minutes to 9.6 minutes, and the mean waiting time for cito samplesincreases from 0.62 minutes to 1.10 minutes. The time increase for the citosamples is much less than for the regular samples, as the cito samples stillreceive priority over other samples.

The number of samples in the system increases, which may increasethe workload experience and pressure. Figure 6.2 shows the mean numberof samples at the SAP for the normal scenario and the scenario with oneemployee per time slot of one minute over the 100 simulation days.

020

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Average number of samples in system: One employee

(b) One employee

Figure 6.2: Mean number of samples in the system


The number of switches increases as well, from 264 to 376, and theworkload percentage increases from 55% to 81%. It is remarkable that theworkload does dot double; the activities do not cost the same amount oftime as for two employees. Figure 6.3 shows how the time on a work day isdivided over tasks, breaks and time in which no work is available. The timein which no work is available is split in two categories, one considers theabsences that lasted less than 2.5 minutes, and the other category considersthe absence times longer than 2.5 minutes. This division is made becauseif the time that no work is available mainly consists of short time slots inwhich no work is available, the absence times may not be experienced astrue absence of work by the employees.

The figure shows that the largest amount of time in which no work isavailable consists of time slots that are larger than 2.5 minutes. The figurealso shows that some activities take less time when working alone. Bringingcito samples takes less time for one employee, because he or she can bring citosamples in larger batches that arise because the samples have longer waitingtimes. Disturbances are also shorter for one employee, due to the assumptionthat the disturbances partially depend on the employee. For example, if oneemployee less is working at the SAP, other employees will not come lookingfor this employee for questions anymore. The blood withdrawal round takesless time, because the work is still divided over the same number of people,and the employee of the SAP only has to do his part. Remark that breaksare not included in the scenario with one employee, because the SAP cannever be unattended. This means that employees from other laboratoryareas have to take over in the break for the SAP employee.

It can be concluded that the results indicate that one employee would beable to do all activities. There is still 19% of the time left at which no jobsare available and no waiting queues that cannot be served arise. Thoughthe throughput times increase tremendously, especially for regular samples.

The laboratory management wishes to increase the workload percentage,but it does not want to increase throughput times too much. This goal maybe hard to achieve, because many of the samples arrive around the sametime which causes a trade-off between workload percentages and throughputtimes. At moments many samples arrive, many operators are necessary toaccomplish low throughput times, but at the moments in between the busyperiods, less work is available, so only a few operators should be present tohave a high workload percentage. This can be illustrated by an exampleconcerning an M/M/1 queue.

Consider an M/M/1 queue with service rate µ = 1 per minute and aconstant arrival rate λ per minute. It is well-known that the mean sojourn


050

01,

000

Min

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Normal One employee

Rest Bring cito

Start centrifuge Disturbances

Switch time Bloodwithdrawal round

Break Short breaks (<2.5min)

Long breaks (>2.5min)

Figure 6.3: Time use per scenario

time in an M/M/1 queue is equal to 1µ(1−ρ) , [2]. If the goal is to keep the

mean sojourn time smaller than 10 minutes, 11−ρ < 10, the occupation rate

of the employees ρ, must be smaller than 9/10.Now the constant arrival rate is changed into a time-dependent arrival

rate, for example, 75% of the customers arrive in the morning and 25% ofthe customers arrive in the afternoon. The system in the morning can berepresented by an M/M/1 queue with µ = 1 and arrival intensity 3λ/2, andthe system in the afternoon by an M/M/1 queue with µ = 1 and arrivalintensity λ/2. This results in a system with on average the same numberof arrivals over the day as the previous system with a constant arrival rate.The mean sojourn time in the morning is given by E[Sm] = 1/(1 − 3λ/2)and in the afternoon by E[Sa] = 1/(1− λ/2). The total mean sojourn timeover the day is given by E[S] = 3E[Sm]/4+E[Sa]/4. The goal is to keep themean sojourn time less than 10 minutes, this implies that λ = ρ < 0.615.

This example illustrates that it may not be achievable for the laboratoryto have high workload percentages and low throughput times, because of


the high number of arrivals in the morning. This means that the laboratoryshould make the arrival stream more constant over the day, if the goal is tohave high occupancy rates with low throughput times.

6.2.9 No cito type

Ignoring the fact that some samples are more urgent than other samples hassome interesting effects. The results show the throughput performance mea-sures per sample type, even though the type is ignored now. The cito samplesare still brought to the test execution area, because otherwise comparing thenormal scenario and the scenario without cito types would be influenced bythe fact that also a process step is deleted. The mean throughput time forregular samples decreases by 0.9 minutes, while the mean throughput timefor cito samples increases by 2.7 minutes. The number of switches decreaseswith nearly 50 switches, and the workload percentage decreases by 2%. Ap-parently, because the employees can serve the whole queue of customers ata task before switching to a new task, the the workload and the number ofswitches decrease.

The effect on the throughput times can be explained by an exampleconcerning an M/M/1 queue with 2 types of customers with non-preemptivepriorities. Non-preemptive priorities mean that a higher priority customermay not interrupt service of a lower priority customer. Type 1 customershave priority over type 2 customers. The customers arrive according to aPoisson stream with rate λi, i = 1, 2, and the service times are exponentialwith mean 1/µ. Define ρi = λi/µ; ρ1 + ρ2 equals the total workload ofthe system. It is shown in [2] that the mean throughput time for a type 1customer is given by

E[S1] =(1 + ρ2)/µ

1− ρ1,

and for a type 2 customer is given by

E[S2] =(1− ρ1(1− ρ1 − ρ2))/µ(1− ρ1)(1− ρ1 − ρ2)

.

If ρ1 = 0.2 and ρ2 = 0.6, E[S1] = 2 and E[S2] = 6. For an M/M/1 queuewith 1 customer type, the mean sojourn time is calculated by

E[S] =1/µ

1− ρ.

If λ = 0.8 and µ = 1, the workload is equal to ρ = 0.8, which is equal to thesystem with 2 customer priorities. The mean sojourn time in this system


is equal to 5. This shows that when a priority is introduced in the system,the mean throughput time of the highest priority decreases and the lowestpriority increases.

This is also what happens at the SAP. The mean throughput time forregular samples is equal to 15.6 minutes, and 12.7 minutes for cito samplesin the normal scenario. Ignoring the cito priority results in a total meanthroughput time for regular and cito samples of 14.7 minutes.

6.2.10 Preanalyzer

The scenarios concerning the preanalyzer give remarkable results. The ex-pectation is that the machine accelerates the process, but the results inTable 6.1 show that this is not true. The throughput time measures forregular samples decrease slightly, while the throughput times for the citosamples increase. This happens because the machine cannot give priority tothe cito samples and start the centrifuge immediately. The centrifuge onlystarts when enough samples are present, or when a fixed time has passed.

Note that applying the time measure gives better results for cito samples,while the batch measure gives better results for the regular samples, this iscaused by the fact that regular samples arrive in larger batches. The numberof switches and the workload decrease due to the preanalyzer. The optionof a preanalyzer without a centrifuge already gives better results for the citothroughput times, but then positive effects on the number of switches andthe workload become smaller.

The idea of buying a preanalyzer is to reduce the number of tasks atthe SAP, but in the current situation, the 2 employees can handle the workeasily. The situation in which a preanalyzer might become interesting iswhen 1 employee works at the SAP in combination with a preanalyzer.This is investigated in the next section of this chapter.

6.3 Scenario combinations

Several combinations of two or more scenarios are investigated as well, Table6.2 displays the results. The table shows that combining the scenario withone employee with shorter process times for bringing cito samples, movingthe sorting activities and eliminating all disturbances (2, 3, 6, 9) gives resultsthat are comparable or better than the current situation. A combination ofscenarios that is easier and cheap to realize is the scenario with one employeeand moving the sorting activities (9, 6), this results in throughput times that


are a bit worse than in the current situation, but it does clear one employeefor other activities.

If two employees keep working at the SAP, the throughput times canbe improved significantly by combining scenarios such as moving sortingactivities, shortening process times for bringing cito samples and eliminatingdisturbance factors. Though, it would cost money to realize this, or otheremployees get more work, while the workload for the 2 employees at theSAP would decrease to a level around or even below 40%.

The scenarios with the preanalyzer and one employee do not appear at-tractive, the regular throughput times are better than the scenario with justone employee, but cito samples take much longer, while it is most impor-tant that these samples have short throughput times. The preanalyzer maybecome interesting if it can differentiate between cito and regular samples.

Because the scenario with one employee gives high throughput times,a scenario with two employees in the morning and one employee in theafternoon is investigated. The throughput times for regular samples andcito samples decrease compared to the scenario with one employee, but theyare still much above the scenario with two employees.

6.4 Sensitivity analysis

It is not only interesting to investigate the scenarios from the previous sec-tion, but also how the system reacts on higher arrival rates and larger arrivalbatches, and to a different task priority setting.

6.4.1 Sensitivity to arrival parameters

Table 6.3 shows that when all arriving lab number batches are doubled,the throughput times increase, the workload increases, but the number ofswitches decreases, and the samples still appear to be handled in a reasonableamount of time. The number of switches decreases, because the samples arehandled in larger batches at the tasks. Note that the throughput timesincrease very much, while the workload percentage is still under 75%, thisillustrates again that the combination of high workload percentages and lowthroughput times is hard to achieve when the samples arrive in peaks duringthe day.

When interarrival times are half as long as in the normal scenario, itmeans that samples from the arrival types that occur more often on a day,arrive twice as often. The total number of samples in the system does notdouble, because the once a day arrival batches stay equal. The scenario


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353

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370

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shows that the workload percentage and the number of switches are moresensitive to this type of change. The regular throughput times also increase,but by a similar amount as for the double batches, the cito throughputtimes increase by a smaller amount than for the double batch sizes. Thisis probably caused by the fact that even though the cito samples arrivemore often, they still arrive individually, so they can be handled separatelyimmediately.

Combining the larger batches with smaller interarrival times results invery high workloads, high throughput times and queues that cannot beserved anymore.

It is also interesting to investigate the effect of the batch sizes and inter-arrival times when the number of samples that arrive stays equal. Table 6.3shows the results for doubling the batch sizes and also doubling the interar-rival times. The mean throughput times increase compared with the normalsituation, but the workload decreases. When the batch sizes and interarrivaltimes are halved, the throughput times for regular and cito samples decreaseslightly, while the workload increases by 20%. This implies that it is betterfor the throughput times when the arrivals occur more continuously overthe day in smaller batches. However, because some of the process steps aredone in batches, the workload does increase when the batches are smaller.

6.4.2 Sensitivity to task priority setting

For a tandem queue with coupled processors, the optimal policy to obtainlow throughput times is to give priority to the second station in the network,see [14]. The employees at the SAP give priority to the beginning of thenetwork, it is interesting to investigate the effect of putting the priority onthe tasks at the end of the network.

Table 6.4 shows that the results for giving priority to the tasks at theend of the network over the tasks at the beginning of the network are bad.This probably happens because many process steps are performed in batchesinstead of individually, the process times do not depend on the number ofcustomers in the batch, and the process times for the tasks that can be donein batches are much larger than the tasks that are done one-by-one. If onefor example always brings cito samples to the laboratory before dividing orregistrating the samples that are already in the network as well, the samplesare all brought separately as soon as they are waiting to be brought tothe laboratory. The last sample has to wait until all its predecessors areprocessed separately. If the task priority setting would be to first receivethe samples, then register everything, divide everything and then bring the


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samples to the laboratory (if necessary), the samples are brought in batches.The first sample has to wait until the last sample in the batch is ready tobe brought to the laboratory, but then they are all brought at once. Thisis eventually shorter, because the long process time of bringing cito samplesonly occurs once.

The tasks concerning registration, blood dividing and sorting experiencethe highest waiting times in the current situation, see Section 5.4.2. There-fore, it is also investigated what the effect of increasing the priority of thesetasks is. The priority for registration is now set to the highest after handlingthe blood gas and cito samples, and blood dividing and sorting are the nexttwo priorities. Table 6.4 shows that the effect on the throughput times isnegligible, and the confidence intervals of the KPIs all overlap. The waitingtime for the registration decreases from 5.8 to 3.6, while the waiting timesfor sorting and blood dividing stay at the same level, and the waiting timesfor turning on the centrifuge and the arrival handling steps all increase by0.1 to 0.5 minutes. Also when just blood dividing or sorting is given high-est priority after cito and blood gas samples, the mean waiting time forblood dividing or sorting decreases, while other waiting times increase andthroughput times stay equal or increase slightly. Apparently, increasing thepriority of these tasks decreases the waiting times of the concerning tasksbut it also results in longer waiting times at other places. No better taskpriority setting is found in this research.

6.5 Conclusions

The main conclusions resulting from the scenarios and the sensitivity anal-ysis are discussed in this section.

6.5.1 Scenarios

The conclusions concerning the scenarios with one employee are describedseparately, because the results are very important for the laboratory staff.The conclusions concerning the other scenarios are discussed in the secondpart of this section.

One employee

The results from the scenario with only one employee show that it is possibleto let one person run the SAP, as the queues and waiting times do not ex-plode and the occupation rate is equal to 81%, but the throughput times do


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100%

252

Tab

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4:Se

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prio

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sett

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(*In

min

utes

)


increase to a level 3 times as high for the regular samples. The large increasein the throughput times while the occupation rate is still reasonable is causedby the fact that most samples arrive in large batches. This implies that thelaboratory staff should weigh the importance of the low throughput timesto the importance of high occupation rates, or change the arrival process,because a combination of both low throughput times and high occupationrates does not appear to be achievable in the current situation.

By combining the scenario with one employee with other scenarios, thehigh throughput times and the workload decrease again. By combining itwith moving the sorting activities to another place in the laboratory, thethroughput times decrease to a level just above the current situation andthe occupation rate to 70%. This appears to be a good option, because thisclears one employee for other activities, while the throughput times do notincrease very much. Note that by moving the sorting activities, the otheremployees in the laboratory have more work, but as sorting is currently doneat 3 places in the laboratory, it is expected that it has a small effect.

A combination that results in throughput times at a similar level as in thecurrent situation, is combining the scenario with one employee with shorten-ing the walking distance for cito samples and moving the sorting activities.This combination is harder to realize in practice, as it involves moving theSAP closer to the machine testing area. It should also be noted that theoccupation rate of the employee decreases to 59%, which is only slightlyhigher than in the current situation. By also eliminating all interruptions,the throughput times are reduced to a level under the current situation, andthe occupation rate is at the same level.

Other scenarios

Moving the sorting activities appears to be a very attractive and simplescenario, it does not involve much changes, but it has a very large effecton the throughput times and workload. Another scenario that is easy toperform and effective is decreasing the batch sizes of the outpatient clinictrain. By sending the trains earlier, the mean waiting times at the outpatientclinic are decreased tremendously. The best scenario concerning the waitingtimes at the outpatient clinic is to move the outpatient clinic next to thelaboratory. This does not only eliminate the waiting times at the outpatientclinic, but it also enables the employees from the laboratory to help at theoutpatient clinic at busy periods and vice versa when the laboratory is busy.

The realization of moving the outpatient clinic is costly, just as shorten-ing the walking distance by moving the SAP closer to the machine test area.


The effect of shortening the walking distance is very high, both regular andcito throughput times benefit significantly. Another advantage of shorteningthe walking distance, is that the employees of the laboratory can cooperateand help each other better. This also accommodates the situation that oneemployee works at the SAP and is helped out at busy periods by anotheremployee from the test execution area.

The preanalyzer machine is not an interesting option for the laboratoryconcerning the KPI, as it cannot differentiate between the priority types.The results for the scenario concerning the blood withdrawal round are notvery good, but it may be worth to investigate the blood withdrawal roundseparately in detail, such as the routing of the employees and the distributionof patients over the employees.

Treating all samples equally, instead of recognizing priorities gives in-teresting results. The mean throughput time for regular samples decreases,and it increases for cito samples. By treating everything in the same way,the workload decreases slightly, but the number of switches, and thus theworkload experience may decrease significantly.

6.5.2 Sensitivity

The sensitivity analysis is done on the arrival parameters and the task pri-ority setting. The conclusions are discussed separately here.

Arrival parameters

The results show that the system has still room for increasing the arrivalintensities. When the batch sizes are doubled, the throughput time perfor-mance measures increase, and the occupation rate increases to 74%, so thesystem can still handle this situation. If the interarrival times halve, thethroughput times also increase, but not as much as for the double batchsizes. The occupation rate increases more than for the doubled batches, butit is still 84%, which means that the system is not overloaded. The occu-pation rate is more sensitive to the interarrival times and cito throughputtimes are more sensitive to doubling the batch sizes. The situation withdoubled batch sizes combined with halved interarrival times does overloadthe system.

It is also investigated what happens when the total amount of samplesthat arrive over the day is kept constant, but the interarrival times andbatch sizes are varied. It is found that the throughput times increase whenthe batch sizes and interarrival times are larger, while the occupation rate


decreases. When the interarrival times and batches are smaller, the occu-pation rate increases, while the throughput times decrease slightly. Thisimplies that smaller batches are better for the laboratory.

Task priority setting

Currently, the priority is mainly given to the first part of the network, andthis appears to be better than giving priority to the end of the network. Bygiving priority to the end of the network, the tasks and machine operationswith a fixed service time that is independent on the batch size are performedon smaller batches and more often, while other tasks are waiting and thethroughput times increase.

By increasing the priority of the tasks that have high waiting times in thecurrent situation, the mean waiting times for the tasks with higher prioritydecrease, but this decrease is compensated by an increase of waiting timesat other jobs. This eventually results in similar throughput times as in thecurrent situation. No improvement for the throughput times by changingthe task priority setting is found in this research.

Chapter 7

Analysis

This chapter discusses a short analysis of some aspects of the queueingsystem at the sample arrival and preparation area. The system is very largeand complex, so only a small part is considered and simplifying assumptionsare made. The analysis focuses on the batch policy for the centrifuge andoutpatient clinic train at the SAP.

7.1 Machine batch policy

A very important characteristic for the SAP is that batches occur at manyplaces. The customers mainly arrive in batches, and the machines servebatches of customers. The batch policy at the machines may influence theprocesses very much. The laboratory does not give much attention to themachine batch policies, the routines to handle the machines have emergedover time. This section analytically investigates the effect of the machinebatch policy on the waiting times for the outpatient clinic trains and thecentrifuges.

These machines have some specific characteristics:

• Batch service

• Deterministic service time for the centrifuges, and a service time with alarge deterministic part and a small stochastic part for the outpatientclinic trains

• Batch arrivals,

• Multiple processors, there are two centrifuges and three trains

80

CHAPTER 7. ANALYSIS 81

• Customer priorities

The workload for machines at the SAP is very low. The workload canbe calculated by

ρ =λE[G]E[B]

cN, (7.1)

where λ is the arrival intensity of batches of samples, E[G] is the meanservice time of the machine, E[B] is the mean arrival batch size, c is thenumber of servers and N is the maximum batch size of the machine. For theoutpatient clinic train, λ is at most 1/2.5, E[B] = 2.4, E[G] = 5 minutes,c = 3, and N = 36. This results in an approximate workload for theoutpatient clinic train of 0.05. The arrival process for the centrifuge is harderto describe, because samples come from different origins. On average 395samples have to go into the centrifuge between 7:00 and 17:15. If the arrivalsare assumed to occur at a constant rate over the day and arrive individually,the arrival intensity per minute is 395/615 and E[B] = 1. E[G] = 10 anddeterministic, N = 72, and c = 2. The workload for one centrifuge is thenapproximately equal to 0.04.

Currently, the batch policy for the outpatient clinic train is to send thetrain when it is almost fully loaded with regular samples, or when a citosample arrives, which is sent separately. The centrifuge is started whenmore than approximately 20 regular samples are waiting, or when a citosample arrives. The regular and cito samples are processed together in thecentrifuge.

It is difficult to develop a model with all the characteristics of the ma-chines, so first a basic queueing model with general batch service, singlePoisson arrivals, one machine and no priorities is developed in Section 7.2.With this model, the effect of adding bulk arrival is discussed in Section 7.3and the system with multiple servers is discussed in Section 7.4.

7.2 M/Ga,N/1

In this section, the model is first described and literature is discussed. Thereare two well-known techniques for analyzing this system, so for both meth-ods, a description of the approach is given and performance measures aredetermined. The last subsection contains some results.


7.2.1 Model description

In the M/Ga,N/1 system, customers arrive individually according to a Pois-son proces with rate λ, and they wait in the queue until service. The servicetime is arbitrary and the customers are served in batches of at most N cus-tomers. When the server becomes idle, it starts service if at least a (≤ N)customers are waiting in the queue, and remains idle as long as less than acustomers are in the queue. If the service time would be exponential insteadof arbitrary and the batch size would always equal N , the model wouldsimplify to an M/MN/1. This system is equivalent to an Er/M/1 queue[7]. The system with general service times and a < N is harder to analyze.The analysis can be done by applying several techniques, the technique ofadding a supplementary variable is discussed first, after which the imbeddedMarkov chain technique is discussed as well.

Supplementary variable

The M/Ga,N/1 system can be described by a two-dimensional Markov pro-cess {Lq(t), X(t)}, where Lq(t) is equal to the number of customers in thequeue at time t, and X(t) the elapsed service time. Chaudhry and Temple-ton [7] develop expressions for some performance measures of this system.A summary of their results is given by Chaudhry et al. in [6]. Let the su-perscript d denote that a quantity is considered at a post-departure epoch,which means that a batch of customers leaves the system. The service timesare i.i.d random variables with distribution function FG(t), with mean 1/µ.Let ρ = λ/Nµ. The probability generating function (p.g.f.) for the proba-bilities of the number of customers in the system after a departure is givenby

πd(s) =G(λ− λs)

∑N−1j=0 πdj (sN − sj)

sN − G(λ− λs), (7.2)

where πdj = P (Ld = j) and G(s) is the Laplace Stieltjes transform of theservice time distribution FG(x).

Let πn,1 and πn,0 denote the probabilities that there are n customerswaiting in the queue and the server is busy (1) or idle (0). Chaudhry andTempleton [7] show that

πn,0 =∑n

i=0 πdi

Nρ+∑a−1

i=0 (a− i)πdin = 0, 1, . . . , a− 1


and

π0,1 =1−

∑a−1r=0 πr,0Nρ

(N∑i=0

πdi

)(1− α0),

πn,1 =1−

∑a−1r=0 πr,0Nρ

πdn+N − αnN∑i=0

πdi −n∑j=1

πdN+jαn−j

+πn−1,1, n ≥ 1,

where αn denotes the probability of n arrivals during a service period, whichis given by αn =

∫∞t=0

(λt)n

n! e−λtfG(t)dt.The p.g.f. of the number of customers in the queue is given by

πq(s) =a−1∑i=0

πi,0si +

µ(

1−∑a−1

i=0 πi,0

)(1− G(λ− λs)

)πd(s)

λ(1− s)(G(λ− λs). (7.3)

Performance measures based on the supplementary variable tech-nique

The mean number of customers in the system at departure moments, E[Ld]can be derived from the pgf in (7.2), by evaluating the derivative of πd(s)in the point s = 1. An expression for the mean number of customers in thequeue at random moments, found by Chaudhry and Templeton [7], looks asfollows,

E[Lq] =a−1∑i=0

iπi,0 +

(1−

a−1∑i=0

πi,0

)(G(2)(0)λ2/2Nρ−Nρ+ E[Ld]

).

With the mean number of customers in the queue, the mean waiting timecan be calculated by applying Little’s law,

E[W ] = E[Lq]/λ. (7.4)

To find E[Ld] and E[Lq], the N unknown probabilities in the p.g.f. atthe moment a batch of customers leaves the system have to be determined.To find these probabilities, the roots of the denominator have to be locatedfirst. It can be shown by Rouche’s theorem that the denominator has N − 1zeros inside and one on the unit circle, | s |= 1. Because π(s) is well-defined,the numerator must equal zero at these points as well [7].

Determining the roots can be quite complicated, but [10] introduces afixed point iteration method to find the roots under special conditions. The


denominator of (7.10) equals sN − G(λ(1− s)). If G(λ(1− s)) is assumed tohave no zeros in | s |≤ 1, the N roots of sN = G(λ(1− s)) in | s |≤ 1 satisfy

s = wG1/N (λ(1− s)), (7.5)

where wN = 1. For each feasible wk, i.e. wNk = 1, k = 0, 1, . . . , N − 1, itcan be shown that (7.5) has one unique root in | s |≤ 1. Janssen and VanLeeuwaarden [10] suggest to solve the equations by successive substitutionsas follows,

s(n+1)k = wkG

1/N (λ(1− s(n)k )), k = 0, 1, . . . , N − 1 (7.6)

Once the roots sk, k = 0, . . . , N−1 are determined, the limiting probabilitiesπdi for i = 0, . . . , N − 1 can be obtained by solving the system of N linearlyindependent equations. Solving this system of equations fails numericallyfor large N , [6], therefore other techniques are proposed as well, see forexample the polynomial expansion technique of Powell [13].

For most service time distributions, (7.2) and (7.3) result in quite com-plex expressions. For some service time distributions, the p.g.f.’s and theexpressions for E[Ld] and E[Lq] simplify [6], but in general it is hard toderive expressions for E[Ld] and E[Lq]. Because it is not that easy for manyservice time distributions, the step of finding the roots and the unknownprobabilities in (7.2) also involves some complex calculations, and solvingthe system of equations fails numerically for large N , a simple and practi-cal method based on the imbedded Markov chain and explicit equilibriumprobabilities is developed in the next subsection.

Imbedded Markov Chain

The distribution of the number of customers in the queue at the momenta batch of customers leaves, πd, is important in Chaudry and Templeton’sapproach. This distribution is also used often in the analysis of an M/G/1queue, see for example [2]. It is the imbedded Markov chain of the pro-cess. Bailey already presented the imbedded Markov chain technique forthe M/Ga,N/1 queue in 1954 in [4]. The derivation of the results in thissection is mainly based on results concerning the M/G/1 queue in [2] and[15].

Let Ldk denote the number of customers left behind by the kth batch ofcustomers that leaves the system. The number of customers left behind bythe (k+ 1)th batch of customers is equal to Ak+1, the number of customersthat arrived during service, plus the number of customers left behind by the


kth batch of customers minus N , if more than N customers are left behindby the kth batch. This results in

Ldk+1 = Ak+1 + max {Ldk −N, 0}. (7.7)

The sequence {Ldk}∞k=0 is the imbedded Markov chain. Denote the transitionprobabilities by pij = P (Ldk+1 = j|Ldk = i) and let αn denote the probabilitythat n customers arrive during a service time. Given the duration of a servicetime, t, the number of arrivals during the service is Poisson distributed, withparameter λt. Therefore,

αn =∫ ∞t=0

(λt)n

n!e−λtfG(t)dt. (7.8)

The transition probabilities are given by,

pij = αj if i ≤ Npij = αj−i+N if i > N, j ≥ i−Npij = 0 if j ≤ i−N.

Note that the transition probabilities do not depend on the minimal batchsize a. The minimal batch size influences the time that the machine waitsbefore starting service, but not the number of customers left behind by adeparting batch of customers. The transition matrix for N = 5 takes theform

P =

α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 · · ·α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 · · ·α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 · · ·α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 · · ·α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 · · ·α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 α11 · · ·0 α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 α10 · · ·0 0 α0 α1 α2 α3 α4 α5 α6 α7 α8 α9 · · ·0 0 0 α0 α1 α2 α3 α4 α5 α6 α7 α8 · · ·0 0 0 0 α0 α1 α2 α3 α4 α5 α6 α7 · · ·0 0 0 0 0 α0 α1 α2 α3 α4 α5 α6 · · ·0 0 0 0 0 0 α0 α1 α2 α3 α4 α5 · · ·...

......

......

.... . . . . . . . . . . . . . . . . . . . .

(7.9)


To find the limiting distribution πd for the number of customers at de-parture moments, the system of equations

πdj =∑i

πdi Pij , j ≥ 0

with the normalization equation∑

j πdj = 1, has to be solved. After some

derivations that can be found in Appendix C, the following probability gen-erating function (p.g.f.) for the number of customers in the queue just aftera batch of customers leaves the system is found

πd(s) =G(λ(1− s))

∑N−1i=0 πdi (sN − si)

sN − G(λ(1− s)), (7.10)

where G is the Laplace Stieltjes transform of the service time distributionThe p.g.f. (7.2) found by Chaudhry and Templeton [7] coincides with the

p.g.f. (7.10). The method developed by Chaudhry and Templeton also givesexpressions for the number of customers in the queue at random moments(7.3). It is hard to derive these expressions with the technique consideringthe imbedded Markov chain.

As already mentioned in the previous subsection, the p.g.f. are very com-plex for most service time distributions, and deriving performance measuresalso involves complex calculations. To avoid this, one could also cut off thetransition probability matrix at state y >> N in (7.9), and solve the systemπd∗P = πd∗ under

∑i π

d∗ = 1, to obtain an estimate of the equilibriumdistribution. For y sufficiently large, and workloads that are not too high,the results are very accurate when comparing them with simulation results,because the probability of going to a state with many customers is low.

Performance measures based on explicit equilibrium probabilities

It would be ideal if the equilibrium distribution of the number of customersin the queue at the moment a batch of customers leaves the system wouldbe equal to the equilibrium distribution at a random moment, just as forthe M/G/1 queue [2]. Unfortunately, this does not hold for the M/Ga,N/1queue. The limiting distribution at the moment a customer arrives is equalto the limiting distribution of the number of customers in the system, be-cause the PASTA property holds. For an M/G/1 queue, the limiting dis-tribution of the number of customers at a departure moment is equal tothe limiting distribution of the number of customers at an arrival moment,because of the updown-crossing argument [2]. Because customers leave in


batches, this argument does not hold in the M/Ga,N/1 queue. The numberof transitions per unit time from state n to n + 1 customers is not equalto the number of transitions per unit time from n + 1 to n, so the limitingdistributions of the number of customers at departure and arrival momentsis not equal. Therefore, the departure distribution is not equal to the dis-tribution of the number of customers in the system at any time and theperformance measures are calculated in a different way.

Note that a cycle can be recognized in this system. The machine becomesidle, and then waits if necessary until at least a customers are present, thenthe machine starts service, after which the machine becomes idle again.Define t1 to be the time at which the machine has finished serving a batchof customers, and t2 to be the time at which the machine starts service.The cycle is displayed in Figure 7.1. Let Tij denote the time between tiand tj , and Lij the number of waiting customers between ti and tj , wherei, j ∈ [1, 2]. The mean number of customers in the queue is then given by

E[Lq] =E[L12]E[T12] + E[L21]E[T21]

E[T12] + E[T21], (7.11)

and by Little’s law, the mean waiting time is given by

E[W ] = E[Lq]/λ. (7.12)

t1 t2 t1,t2 t1 t2 t1 t2 t1

Cycle 1 Cycle 2 Cycle 3 Cycle 4

T12 T21 T21 T12 T21 T12 T21

Figure 7.1: Time cycle

The distribution of the number of customers at t1 is given by πd, theequilibrium distribution of the Markov chain Ldk. The equilibrium proba-bilities πdi can be interpreted as the relative number of times a batch ofcustomers leaves behind i customers. With these probabilities, the equilib-rium distribution of the number of customers in the queue right after themachine starts service, denoted by πs, can be computed. If i customers arein the queue when a batch of customers leaves the system, i.e. at t1, then


max {i−N, 0} customers are left behind in the queue when the machinestarts service, at t2. So

πs0 =N∑i=0

πdi (7.13)

πsi = πdi+N if i > 0 (7.14)

The next step is to calculate E[L12], E[L21], E[T12] and E[T21]. E[T21] is theeasiest case, this is just the expected service time E[G] because T21 ∼ G.E[L21] consists of the mean number of customers that are in the queuewhen service starts, plus the mean number of customers that arrive duringservice divided by 2. The mean number of customers in the queue whenservice starts can be calculated with the equilibrium distribution πs, and isequal to

∑∞i=1 π

si i. Given that the service time is equal to t, the number

of customers that arrive during t is Poisson distributed. The mean numberof customers that arrive during service divided by 2 can be obtained fromλE[G2]2E[G] , because E[G2]

2E[G] is equal to the mean residual processing time [2]. Thus,it is found that

E[T21] = E[G] (7.15)

E[L21] =∞∑i=1

πsi i+λE[G2]2E[G]

. (7.16)

The duration of T12 depends on the batch policy of the machine, the arrivaltimes of customers and the number of customers that are waiting in thequeue when the machine becomes idle. If i ≥ a customers are waiting in thequeue at t1, then T12 is 0. If i < a customers are waiting, the machine waitsuntil a− i customers arrive and then starts service. The arrivals are Poissondistributed, so the expected duration of T12 if i customers are waiting whenthe machine becomes available is equal to max{(a− i)/λ, 0}. E[T12] cannow be calculated by using πd,

E[T12] =a−1∑i=0

πdi(a− i)λ

. (7.17)

E[L12] also depends on the number of customers that are already in thequeue at t1, and on a. Let Cj denote the probability that at some point,j customers are waiting during T12. This can only occur when j or lesscustomers are waiting at t1, and j < a. Therefore, Cj is proportional to


∑ji=0 π

di . It follows that

E[L12] =a−1∑j=0

jCj .

The sum of the Cj ’s should equal 1, because it is a probability distribution,therefore

Cj =∑j

i=0 πdi∑a−1

l=0

∑li=0 π

di

.

The results of these formulas are compared with simulation results, be-cause the formulas are applied to the estimated equilibrium probabilitiesobtained cutting off the transition matrix. The results have shown that thedeviations are very small. For low workloads, for y ≥ 5N the investigatedparameter settings give results that deviate at most 5% from the simulationresults. For larger workloads, y should increase to 10N or more, and thehigher the workload gets, the larger the deviations become.

7.2.2 Results

In this section, only deterministic service times are investigated, because themachines at the SAP have (nearly) deterministic service times. The formulascan also be applied to other service time distributions. It is interesting toinvestigate how the mean waiting time depends on the minimal service batchsize a, the maximal batch size N , the arrival intensity λ and the service timet. The arrival rate is per minute and the service time is also in minutes.The workload of the system is equal to

ρ =λt

N(7.18)

Figure 7.2 shows how the mean waiting time and the mean service batchsize (E[BS]) depends on the minimal batch size, and the influence of theparameters λ, N , and t. The starting point at the time axis for the minimalbatch size equal to 0 is at least equal to t/2. This is true because if themachine always starts service as soon as it finishes, and a customer canalways enter service when the machine becomes available, the mean waitingtime is just equal to t/2. For higher workloads, which may be caused by smallN , large t, or large λ, the starting point increases, because the possibilityof not being able to enter service when the machine becomes available thefirst time increases.

Figure 7.2(a) shows that for small λ, it is rewarding to start service if atleast 1 customer is present, as the graph for λ = 0.4 shows a minimum at


a = 1. For larger λ, the first part of the graph appears to be horizontal, butthe graph actually also reaches a minimum in this part. The graph appearshorizontal, because the strategy for small a results in very similar strategies,in which the machine starts service immediately after finishing service mostof the times because more than a customers are waiting when the machinebecomes available. For λ = 0.8, the minimum is reached in a = 2, for λ = 1in a = 3, for λ = 1.2 in a = 4 and for λ = 1.6 in a = 7. This implies that thelocation of the minimum is larger when λ is larger. The increasing part ofthe graph is steeper when λ is small, this is due to the fact that for smallerλ, the mean interarrival time of customers is larger.

Figure 7.2(c) shows that when N = 36, for λ = 1 and t = 5, and thusρ = 5/36, all waiting customers can enter service the first time the machineis available most of the times and the starting point at the time axis is at 2.5minutes. For smaller N , the probability that not all customers fit increases,and the starting point at the time axis increases accordingly. The figureshows that the point at which the graph starts to increase steeply is notinfluenced very much by N , it is approximately at the same point for allinvestigated values of N . The part in which the minimum is attained isinfluenced by N , the minimum for N = 36, 10, 8, and 7 is attained at a = 3,and for N = 6 at a = 4. So the smaller N becomes, the larger the minimalbatch size at which the minimum waiting time is attained becomes.

The influence of t appears to be a combination of the effects of N andλ (Figure 7.2(e)), as the starting point of the graph shifts upwards when tbecomes larger, and the location of the minimum shifts. The starting pointat the time axis is obviously influenced by t, as the starting point is at leastequal to t/2, and when t increases and all other parameters stay equal, ρincreases and the probability of not being able to enter service the first timea machine becomes available increases, so the starting point increases evenmore. The location of the minimum is larger when t is larger, for t = 2 theminimum is attained in a = 1, for t = 4 in a = 2, for t = 5 in a = 3, fort = 6 in a = 4, for t = 7 in a = 4 and for t = 8 in a = 4.

It can be concluded that the optimal minimal batch size for minimalwaiting times is influenced by all three parameters N , t and λ. The larger λis, the larger the optimal minimal batch size becomes. For t, the same rela-tion with the minimum holds; the larger t becomes, the larger the optimalminimal batch size becomes. Other than this, a larger t also shifts the meanwaiting times upwards. The influence of the maximal batch size appears tobe smaller than the influence of the other parameters, only when ρ becomeslarge, the location of the minimum starts shifting to a larger value of a. Itcan also be noted that a < N gives better mean waiting times than a = N


0 1 2 3 4 5 6 7 8 9 102

3

4

5

6

7

8

9

10

11

12Mean waiting times vs. Batch size, t=5, N=10

Minimal batch size

Mea

n w

aitin

g tim

e

lambda=0.4lambda=0.8lambda=1lambda=1.2lambda=1.6

(a) Influence of λ

0 1 2 3 4 5 6 7 8 9 102

3

4

5

6

7

8

9

10

11Mean batch size vs. Minimal batch size, t=5, N=10

Minimal batch size

Mea

n ba

tch

size

lambda=0.4lambda=0.8lambda=1lambda=1.2lambda=1.6

(b) Influence of λ

0 1 2 3 4 5 6 7 8 9 102

2.5

3

3.5

4

4.5

5Mean waiting times vs. Batch size, t=5, lambda=1

Batch size

Mea

n w

aitin

g tim

e

N=36N=10N=8N=7N=6

(c) Influence of N

0 1 2 3 4 5 6 7 8 9 105

6

7

8

9

10

11Mean batch size vs. Minimal batch size, t=5, lambda=1

Minimal batch size

Mea

n ba

tch

size

N=36N=10N=8N=7N=6

(d) Influence of N

0 1 2 3 4 5 6 7 8 9 100

1

2

3

4

5

6Mean waiting times vs. Batch size, lambda=1, N=10

Minimal batch size

Mea

n w

aitin

g tim

e

t=2t=4t=5t=6t=8

(e) Influence of t

0 1 2 3 4 5 6 7 8 9 102

3

4

5

6

7

8

9

10

11Mean batch size vs. Minimal batch size, lambda=1, N=10

Minimal batch size

Mea

n ba

tch

size

t=2t=4t=5t=6t=8

(f) Influence of t

Figure 7.2: Mean waiting times and batch sizes vs. minimal batch sizes

for all investigated cases.

SAP machines

A machine with the same size, arrival intensity and service time as themachines at the SAP can give insightful information concerning the machinesat the SAP, even though the machines at the SAP have batch arrivals,multiple machines and customer priorities. The effect of the batch sizes canalready give some intuition concerning the situation at the SAP. Therefore,a system with N = 36, λ = 1/3 and t = 5 is tested, because it has a similar


workload and size as the outpatient trains. Figure 7.3 shows how the meanwaiting time depends on the minimal batch size. The centrifuges can beimitated by a machine with N = 72, λ = 395/1230, and t = 10; Figure 7.3shows the results. As expected, because ρ is small for the machines at thelaboratory, these results show that it would be better for the mean waitingtimes to start the machine when only 2 samples are waiting at the centrifugeand 1 sample at the outpatient clinic train.

Of course, this only shows the effect on the mean waiting time for thesamples, while it takes more time for the staff to handle the machines whenthe batch sizes are smaller. Therefore, a measure for the amount of work persample for the staff is taken into account. This can be done by consideringthe mean operation time per sample for the employee. Turning on themachines is assumed to consist of a fixed time Tf , plus a variable time persample Tv. The mean process time per sample, E[Ps] is then equal to

E[Ps] =E[Tf ] + E[Tv]E[BS]

E[BS]=

E[Tf ]E[BS]

+ E[Tv],

where BS is the mean batch size. E[BS] can be calculated by

E[BS] =a∑i=1

πdi a+N−1∑i=a+1

πdi i+∞∑i=N

πdiN.

If it is assumed that one minute of extra mean operation time is valued as500 times worse than one minute of extra waiting time, the compensatedmean waiting time (E[Wc]) can be found by

E[Wc] = E[W ] + 500E[Ps].

The assumption that the extra mean operation time per sample is valuedaround 500 times worse than extra waiting time is quite realistic, becausethe time of the employee is much more important. If the mean process timeper sample increases by one minute, this means that the employee spendsn minutes in total more on processing the samples for the machine, wheren is equal to the number of samples that are put in the machine. Thiscan increase very fast, because many samples have to go in the machinesper day. Figure 7.3(b) shows the results for this correction. The optimalminimal batch size is now 6 for both the centrifuge and the outpatient clinictrain.

Even though the model is not complete, the results do give some intuitionthat it may be better for the SAP to start the machines earlier than they


0 1 2 3 4 5 6 7 8 9 102

4

6

8

10

12

14

16Mean waiting times vs. Batch size SAP machines

Minimal batch size

Mea

n w

aitin

g tim

e

CentrifugeOutpatient clinic train

(a) Mean waiting time

0 1 2 3 4 5 6 7 8 9 1014

16

18

20

22

24

26

28

30

32

34Compensated mean waiting times vs. Batch size SAP machines

Minimal batch size

Mea

n w

aitin

g tim

e pl

us w

ork

per

sam

ple

CentrifugeOutpatient clinic train

(b) Compensated mean waiting time

Figure 7.3: Mean waiting times for machines at the SAP

do now. The results indicate that the mean waiting times can be reducedtremendously by decreasing the minimal batch size. This is also what thescenarios concerning the batch size for the outpatient clinic train showed inSection 6.2.

7.3 MX/Da,N/1

Chaudry and Templeton [7] write that the study of queues with bulk arrivaland bulk service is not easy. There is much literature available concerningMX/G/1 systems, for example the MX/M/1, and the MX/Gn/1, wherethe subscript n indicates that the service time depends on the number ofcustomers in the system, are discussed in [7]. Though, literature on thecombination of bulk service and bulk arrival queues is sparse, and involvesin general very complex calculations. Recently, Claeys et al. [8] presentedresults concerning a discrete time bulk arrival and bulk service queue with


threshold-based strategies. They develop expressions for queue lengths andcustomer delays.

Because the exact and extensive analysis of this system is beyond thescope of this thesis, a short discussion and approximation of the systemMX/Da,N/1 based on the model for the M/Ga,N/1 system is given here.The system with deterministic service times is considered, because the ser-vice time of the centrifuges and the outpatient clinic train are deterministic,or nearly deterministic.


The model description concerning the imbedded Markov chain from Sec-tion 7.2.1 can be extended to an approximation model for an MX/Da,N/1queueing system. The customers still arrive according to a Poisson process,however they do not occur individually anymore, but in batches of size X,where X is a random variable with distribution function F. Before analyz-ing this system, the definition and some properties of a compound Poissonvariable are introduced.

Let X1, X2, . . . be a sequence of independent and identically distributedrandom variables having distribution function F , Xi ≥ 1, and suppose thatthis sequence is independent of M , a Poisson variable with mean λ. Therandom variable A =

∑Mi=1Xi, is a compound random variable with Poisson

parameter λ and component distribution F . Let

βj = P (Xi = j), j ≥ 1

andPj = P (A = j), j ≥ 0.

Corollary 2.5.4 in [15] gives a recursive expression for the values of Pn,

P0 = e−λ

Pn = λn

∑nj=1 jβjPn−j , n ≥ 1.

If the service time is denoted by t, the number of arrivals during a servicetime is Poisson distributed with arrival intensity λt. The number of samplesper arrival is equal to the random variable X, with distribution functionF . The number of customers that arrive during a service time is then givenby the compound Poisson variable A =

∑Mi=1Xi, where M is Poisson with

mean λt.


In the model without batch arrivals, if there are less than a customerswaiting in the queue, the machine waits until exactly a customers are presentand starts service on all of them. Because of the batch arrivals in the systemof this section, it occurs that more customers arrive at once, and more thana customers are in the queue, instead of exactly a, when the machine startsservice. As long as the number of customers is smaller than the maximumservice batch N , all customers can enter service. Though if more thanN customers are waiting, the customers that cannot enter service have towait. Therefore, (7.7) does not hold for this situation and the transitionprobabilities are harder to obtain.

For the machines at the SAP, the effect of ignoring that more than Ncustomers can be in the queue when the machine starts service is small,because the capacity of the machines is very high and the arrival intensityis low. An approximation for the transition probability matrix is then givenby replacing the probabilities αj in (7.9) with compound Poisson arrivalprobabilities. By cutting off the matrix and solving the system πd∗P = πd∗,under

∑i π

d∗ = 1, an approximation for the limiting equilibrium probabili-ties can be obtained. Performance measures can be calculated in a similarway as in the previous section, where all arrival probabilities are replacedby compound Poisson arrival probabilities.

7.4 M/Da,N/c

Much literature about multiserver queueing models with bulk service can befound. Results concerning the M/Ma,N/c model are derived by Neuts andNadarjan [12], and Sim and Templeton [16]. They obtain expressions forperformance measures, such as waiting time distributions, and service batchsizes. The queueing system with Coxian-2 or Erlang-r service time distribu-tions is studied by Adan and Resing [1], they derive a closed form expres-sion for the waiting time distribution. Because the analysis of multiserverqueueing models with general bulk service with threshold-based strategies incontinuous and discrete time (and general arrival distributions) is very com-plicated, some approximation models have been developed. Wu et. al. [19]constructed approximations for the performance of G/Gk/1 and G/Gk/csystems. These approximations are based on the queueing times in theM/Mk/1 and M/Mk/c models. This approximation might be extended toan approximation for a G/Ga,N/c model. This does appear to be compli-cated, because a main part of their method is based on the wait-to-batchtime. This is equal to (k − 1)/λ for fixed batch sizes of size k, where λ is


equal to the arrival intensity, but for an (a,N) strategy the wait-to-batchtime is not so easy to find.

The service times for the machines at the SAP are (nearly) determin-istic. It is chosen to develop an approximation model specifically for theM/Da,N/c queue based on the same method that is used for M/Ga,N/1queue.


The M/D/c system is extensively studied in literature, and Crommelin [9]was one of the first authors to present results concerning this queueing sys-tem. The main approach for the analysis of this queueing system is tolook at the system at discrete moments, t, t + D, t + 2D, . . ., and exploitthe knowledge that all customers in service have left the system at timet+D, t+ 2D, t+ 3D, . . .. This approach is not followed in this section, butan approximation model based on the imbedded Markov chain and explicitequilibrium probabilities is developed.

The M/Da,N/c system is investigated with the imbedded Markov chainin a similar way as the M/Ga,N/1 queue. The service time is deterministic,and denoted by t. The focus is again on the moment a batch of customersleaves the system, when the machine becomes available for service again.Instead of looking at all moments a batch of customers leaves the system,only the process concerning one machine is considered. Because all machinesare identical, the other machines follow the same process. The state spaceis described by {Ldk,Wk}, where

• Ldk equals the number of customers left behind by the kth batch ofcustomers served by the machine under consideration, and

• Wk equals the number of busy machines at the moment the kth batchof customers leaves the system, excluding the machine that servedbatch k.

Because the service time is deterministic, the machines finish service in thesame order they started service; this is a very important characteristic inthe approach. The analysis is done for a = N first, and is then expanded toa < N .

Let w denote the machine for which the process {Ldk,Wk} is followed.When the (k + 1)th batch of customers leaves the system, the number ofbusy machines Wk+1 is determined by Ak+1, the number of customers thatarrived during the service of w, and the number of customers that were left


behind by w when it started service. The number of customers left behindby w is equal to 0 if Ldk ≤ N , and to Ldk − N otherwise. The number ofbusy machines is limited by the maximal number of machines that can beoccupied when w finishes, i.e. to c− 1. Therefore,

Wk+1 = min {bmin {Ldk −N, 0}+Ak+1

Nc, c− 1}.

The number of customers left behind by the k+ 1th batch is equal to Ak+1,plus the number of customers left behind by the kth batch minus N , ifLdk > N , minus Wk+1 times N . Thus it follows that

Ldk+1 = Ak+1 + min {Ldk −N, 0} −Wk+1N.

The sequence {Ldk,Wk}∞k=0 is the imbedded Markov chain concerning onemachine. Denote the transition probabilities by pijml = P (Ldk+1 = j,Wk+1 =l|Ldk = i,Wk = m). The number of arrivals during a service time t is againPoisson(λt) distributed, so the probability distribution for the number ofarrivals is equal to

αn =(λt)n

n!e−λt. (7.19)

The transition probabilities are given by,

pijml = αj+lN if i ≤ N, j < a,m < c− 1, l ≤ c− 1pijml = αj+lN if i ≤ N, j ≥ 0,m = c− 1, l ≤ c− 1pijml = αj+lN−i+N if i > N, j ≥ i− (l + 1)N,m = c− 1, l = b(i−N)/Ncpijml = αj+lN−i+N if i > N, j ≥ 0,m = c− 1, b(i−N)/Nc < l ≤ c− 1pijml = 0 else.


The transition matrix P looks as follows for a = 3 and c = 3,

P =

(0, 0) (1, 0) (2, 0) (0, 1) (1, 1) (2, 1) (0, 2) (1, 2)(0, 0) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(1, 0) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(2, 0) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(0, 1) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(1, 1) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(2, 1) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(0, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(1, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(2, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(3, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(4, 2) 0 α0 α1 α2 α3 α4 α5 α6 · · ·(5, 2) 0 0 α0 α1 α2 α3 α4 α5 · · ·(6, 2) 0 0 0 α0 α1 α2 α3 α4 · · ·

......

.... . . . . . . . . . . . . . . . . .

(7.20)

For the case that a < N , the transition probabilities cannot be givenas easily anymore. This is due to the fact that it is unknown how manycustomers are served by the machines that are not followed. If Ldk < cN ,the transition probabilities to {Ldk+1,Wk+1} for Wk+1 ≥ 1, depend on thenumber of customers served by the Wk+1 machines. Two extreme simpli-fications can be made. Method 1 assumes that each machine only takesup to a customers of Ak+1, such that in total a customers are served bythe machine if less than a customers are left over from Ldk. If more thana customers are left over from Ldk, the machine takes all customers that fitin the machine. Method 2 assumes that all customers of Ak+1 that fit areserved by the Wk+1 machines. The first simplification underestimates thenumber of customers that are served by the Wk+1 machines, and the secondsimplification overestimates the number of customers that are served by theWk+1 machines.

The transition matrix P looks as follows for the first simplification fora = 2, N = 4 and c = 3,


P =

(0, 0) (1, 0) (0, 1) (1, 1) (0, 2) (1, 2) (2, 2) (3, 2)(0, 0) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(1, 0) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(0, 1) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(1, 1) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(0, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(1, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(2, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(3, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(4, 2) α0 α1 α2 α3 α4 α5 α6 α7 · · ·(5, 2) 0 α0 α1 α2 α3 α4 α5 α6 · · ·(6, 2) 0 0 α0 α1 α2 α3 α4 α5 · · ·(7, 2) 0 0 α0 α1 α2 α3 α4 α5 · · ·(8, 2) 0 0 α0 α1 α2 α3 α4 α5 · · ·(9, 2) 0 0 0 α0 α1 α2 α3 α4 · · ·(10, 2) 0 0 0 0 α0 α1 α2 α3 · · ·(11, 2) 0 0 0 0 α0 α1 α2 α3 · · ·(12, 2) 0 0 0 0 α0 α1 α2 α3 · · ·(13, 2) 0 0 0 0 0 α0 α1 α2 · · ·

......

......

.... . . . . . . . . . . .

(7.21)

For the second simplification, the transition matrix P looks as followsfor a = 2, N = 4 and c = 3,


P =

(0, 0) (1, 0) (0, 1) (1, 1) (0, 2) (1, 2) (2, 2) (3, 2)(0, 0) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(1, 0) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(0, 1) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(1, 1) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(0, 2) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(1, 2) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(2, 2) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(3, 2) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(4, 2) α0 α1 α2 + α3 + α4 α5 α6 + α7 + α8 α9 α10 α11 · · ·(5, 2) 0 α0 α1 + α2 + α3 α4 α5 + α6 + α7 α8 α9 α10 · · ·(6, 2) 0 0 α0 + α1 + α2 α3 α4 + α5 + α6 α7 α8 α9 · · ·(7, 2) 0 0 α0 + α1 α2 α3 + α4 + α5 α6 α7 α8 · · ·(8, 2) 0 0 α0 α1 α2 + α3 + α4 α5 α6 α7 · · ·(9, 2) 0 0 0 α0 α1 + α2 + α3 α4 α5 α6 · · ·(10, 2) 0 0 0 0 α0 + α1 + α2 α3 α4 α5 · · ·(11, 2) 0 0 0 0 α0 + α1 α2 α3 α4 · · ·(12, 2) 0 0 0 0 α0 α1 α2 α3 · · ·(13, 2) 0 0 0 0 0 α0 α1 α2 · · ·

......

......

.... . . . . . . . . . . .

(7.22)

Instead of deriving p.g.f. for the number of customers in the queueand the number of busy machines, the transition matrix is cut off at statey >> N , and the system πd∗P = πd∗ under

∑i π

d∗ = 1 is solved. This givesan approximation for the limiting distribution of the number of customersand the number of busy machines at the moment a batch of customersleaves the system, πd. This distribution only concerns one machine, but themachines are identical, so the distribution is equal for all machines.

7.4.2 Performance measures

For the system with only one machine, the mean waiting times are found byconsidering the cycle formed by two successive moments a machine becomesavailable for service again (see Figure 7.1). This cycle cannot be recognizedin this system anymore, because an extra dimension is added, the number ofbusy machines. Figure 7.4 shows the process of the number of busy machinesover time for a system with 3 machines.

A cycle can be recognized in this process as well, if the system is stable,


1

2

3

0e s

s e

es

s

s e,s

e

e

s e

e s

Time

# busy

machiness = start service

e = end service

Figure 7.4: Number of busy machines over time

the process always returns to the situation in which a machine ends serviceand 0 machines are busy. Denote the points at which a machine starts orends service by {W,S}, where W denotes the number of other busy machineswhen a machine starts or ends service, and S denotes whether the machinestarts s, or ends e, service. The cycle can be split in time slots by definingthe start of a new time slot by a machine that starts or ends service. Letthe number of busy machines denote a level in the cycle. The possible timeslots per level u can be listed as follows,

Start End{u, e} → {u, s} 0 ≤ u < c{u, e} → {u− 1, e} 0 < u < c{u− 1, s} → {u, s} 0 < u < c{u− 1, s} → {u− 1, e} 0 < u ≤ c

To find the mean number of customers in the queue, and thus in a cycle,the mean number of customers in each possible time slot (E[L[{i,f},{j,g}]]),the expected duration of each time slot (E[T[{i,f},{j,g}]]), and the expectednumber occurrences of each time slot in a cycle, (P[{i,f},{j,g}]), are necessary,where i, j ∈ [0, 1, . . . , c−1] and f, g ∈ [e, s]. A formula for the mean numberof customers in the queue can be obtained from

E[Lq] =

∑i,j,f,g P[{i,f},{j,g}]E[T[{i,f},{j,g}]]E[L[{i,f},{j,g}]]∑

i,j,f,g P[{i,f},{j,g}]E[T[{i,f},{j,g}]]. (7.23)

By Little’s law, it follows that

E[W ] = E[Lq]/λ.


Limiting probability distributions for the number of customers atthe beginning and end of the time slots

The limiting distribution of the number of customers and the number of busymachines at the moment a batch of customers leaves the system, πd(l,m),gives information about the number of customers that are waiting at thepoints marked by an e in Figure 7.4, denoted by {i, e}. The limiting proba-bility for the situation that j customers are waiting at the moment a machineends service, given that i machines are busy, is given by

P e(j|i) =πd(j, i)∑∞j=0 π

d(j, i). (7.24)

The limiting probability distribution for the number of customers at themoment a machine starts service can also be derived from πd(l,m). LetM s denote the number of busy machines, and Ls the number of waitingcustomers when the machine starts service. The number of customers leftbehind at the moment the machine starts service can be found as follows,

• if M s < c− 1, the machine always leaves 0 customers behind, becausethere can only be less than a customers in the system when not allmachines are busy,

• if M s = c− 1:

– if Ls ≤ N , the machine leaves 0 customers behind when it startsservice,

– otherwise Ls > N , and Ls −N customers are left behind.

This leads to the following limiting probabilities for the number of cus-tomers in the queue j, when a machine starts service, given that i machinesare busy:

P s(j|i) = 1 if i < c− 1 (7.25)

P s(j|i) =∑N

j=0 πd(j,c−1)∑∞

j=0 πd(j,c−1)

if i = c− 1, j = 0 (7.26)

P s(j|i) = πd(j+N,c−1)∑∞j=0 π

d(j,c−1)if j > 0. (7.27)

Expected duration and mean number of customers per time slot

For each time slot, the expected duration and the mean number of customersis approximated, such that these can be used for (7.23).


T[{i,s},{i,e}] and L[{i,s},{i,e}]

The expected duration of the time slots that end by a machine that endsservice, {i, e}, depends on the elapsed service time of the machine that is inservice longest. Keeping track of the elapsed service time would make thestate space continuous, which complicates the system and the calculations.Therefore an approximation for these time slot durations is given.

First, the time slots starting with a machine that starts service {i, s},and ending by a machine that ends service, {i, e}, are considered. For[{0, s}, {0, e}], the expected duration is just equal to the service time, t.For i > 0 (i < c), a simple approximation for the mean duration of thetime slots could be obtained from t/(i + 1). This approximation assumesthat the i machines that started after the machine that finishes service now,split the service time of this machine in equal parts. An approximation forE[T[{i,s},{i,e}]] is given by

E[T[{i,s},{i,e}]] = t/(i+ 1), i < c.

The number of customers during these time slots consists of the meannumber of customers that are already waiting at {i, s}, plus the mean num-ber of arrivals during the time slot divided by 2. The mean number ofcustomers waiting at {i, s} can be obtained with the conditional limitingprobability distribution P s(j|i). The number of arrivals during a given timet is Poisson distributed with parameter λt. For i < c−1 it is known that lessthan a customers are waiting at the end of the time slot, but this is ignoredhere and the mean number of customers that arrive can be estimated byλE[T 2

[{i,s},{i,e}]]

2E[T[{i,s},{i,e}]]. However, the distribution of T[{i,s},{i,e}] is not determined,

only the expectation, so the following approximation for the mean numberof customers is used, λE[T[{i,s},{i,e}]]

2 . It is clear that this underestimates thetrue number of arrivals, because the time T[{i,s},{i,e}] is not deterministic. Asimple approximation for E[L[{i,s},{i,e}]] is then given by

E[L[{i,s},{i,e}]] =a−1∑j=0

P s(j|i)j +λE[T[{i,s},{i,e}]]

2. (7.28)

T[{i,e},{i−1,e}] and L[{i,e},{i−1,e}]

The time slots that start with a machine ending service {i, e}, and end witha machine ending service {i − 1, e}, can be treated in a similar way as the


time slots of the form [{i, s}, {i, e}]. Again, the expected duration may beapproximated by

E[T[{i,e},{i−1,e}]] = t/(i+ 1), 0 < i < c.

The number of customers may also be approximated in the same way, butnow the conditional limiting probability distribution P e(j|i) is used, becausethe time slot starts with {i, e}. An approximation for the mean number ofcustomers is then given by

E[L[{i,e},{i−1,e}]] =a−1∑j=0

P e(j|i)j +λE[T[{i,e},{i−1,e}]]

2.

T[{i,e},{i,s}] and L[{i,e},{i,s}]

For all time slots of the form [{i, e}, {i, s}], the average duration can beobtained by using a similar technique as for T12 for one machine in Section7.2.1. The difference lies in the fact that [{i, e}, {i, s}] for i > 0 only occurswhen at least a customers are present before a machine that is in serviceends service. Therefore, the average duration is not equal to the mean timeit takes for a − j customers to arrive when j customers are waiting, butto the time it takes until a − j customers arrive conditioned on the timebeing smaller than T[{i,e},{i−1,e}]. Let Yk be an Erlang(k, 1/λ) variable, withmean k/λ. If j customers are waiting when the machine ends service, andi machines are busy, the expected duration of the time slot T[{i,e},{i,s}] isgiven by

E[T[{i,e},{i,s}]] =a−1∑j=0

P e(j|i)E[Ya−j |Ya−j < T[{i,e},{i−1,e}]], i < c. (7.29)

The distribution of T[{i,e},{i−1,e}] is not determined, but an approximation ofE[[Ya−j |Ya−j < T[{i,e},{i−1,e}]] can be given by E[[Ya−j |Ya−j < E[T[{i,e},{i−1,e}]]].For i = 0, no other machine can end service, so the expected time is justequal to a−j

λ . Therefore, an approximation of (7.29) is given by

E[T[{i,e},{i,s}]] =a−1∑j=0

P e(j|i)E[[Ya−j |Ya−j < E[T[{i,e},{i−1,e}]]], 0 < i < c,

E[T[{0,e},{0,s}]] =a−1∑j=0

P e(j|i)a− jλ

.


The expected number of customers in this time slot type is also de-termined in the same way as E[L12] for M/Ga,b/1 (Section 7.2.1). Therecan only be j customers in the queue during [{i, e}, {i, s}] when j or lesscustomers are waiting at {i, e}, and j < a. Therefore,

E[L[{i,e},{i,s}]] =a−1∑j=0

jCij ,

where

Cij =∑j

k=0 Pe(k|i)∑a−1

l=0

∑lk=0 P

e(k|i).

T[{i,s},{i+1,s}] and L[{i,s},{i+1,s}]

The mean duration of a time slot of the form [{i, s}, {i + 1, s}] is equalto the mean time it takes for a customers to arrive, conditioned on beingsmaller than the time until a machine that is in service ends service. This istrue because these time slots can only occur when less than c− 1 machinesare busy at the beginning of the time slot, and P s(0|i) = 1 for i < c − 1,i.e. there are always 0 customers present at the beginning of this time slot.If the time slot ends by the start of a machine, a customers have arrived.Therefore,

E[T[{i,s},{i+1,s}]] = E[[Ya|Ya < T[{i,s},{i,e}]], i < c− 1, (7.30)

where Ya is again an Erlang(a, 1/λ) variable. Also for T[{i,s},{i,e}], the dis-tribution is not determined, so an approximation of (7.30) is given by

E[T[{i,s},{i+1,s}]] = E[[Ya|Ya < E[T[{i,s},{i,e}]]], i < c− 1. (7.31)

Because 0 customers are present at the beginning of the time slot and acustomers at the end, the mean number of customers is determined by

E[L[{i,s},{i+1,s}]] =a− 1

2, i < c− 1.

Number of occurrences of a time slot in a cycle, P [{i, f}, {j, g}]

Approximations for the expected duration and the mean number of cus-tomers for each time slot are determined, so the next step is to find thenumber of the occurrences of a time slot in a cycle. This is done by firstinvestigating the transitions and equilibrium distribution for the number ofbusy machines, and then splitting this up in the different time slots.


If one only considers the transitions from i, to i+1 or i−1 busy machinesin Figure 7.4, a chain can be recognized. This chain is not Markovian,because the transitions depend on Ls, the number of customers left behindby the machine that is in service longest, and not just on the previous state.Even though the chain is not Markovian, it is approximated by a Markovchain and the transition probabilities of this Markov chain are estimatednext. The transition diagram of the chain is displayed in Figure 7.5.

0 1 2 c

P01 P12 P23 Pc-1c

Pcc-1P32P21P10

Figure 7.5: Transition diagram for the number of busy machines

The transitions from level i in the chain are determined by Ls, the num-ber of customers left behind by the machine that is in service longest, andAs, the number of arrivals during the service of the machine that is in servicelongest. If the chain is at level i(< c),

• if As ≥ ia− Ls, a transition to state i+ 1 is made, and

• if As < ia− Ls, a transition to state i− 1 is made.

From level c, a transition to level c− 1 is made with probability 1.Ls is not kept track of, but probabilities for this number can be obtained

from P s(j|i) or πd. Because it is unknown at which level the machine that isin service longest started, the total probability distribution for the number ofcustomers and the level is used, instead of the conditional distribution. Theprobability of leaving behind 0 customers is equal to

∑c−2i=0

∑a−1k=0 π

d(k, i) +∑Nk=0 π

d(k, c−1), and the probability of leaving y customers behind to πd(y+N, c−1). The transition probabilities for the chain can be conditioned on Ls,and by assuming that the levels of the chain and Ls are independent (whichis not true in general), the transition probabilities can be approximated.


The transition probabilities can then be written as

p01 = 1,pi(i+1) =

(∑c−2l=0

∑a−1k=0 π

d(k, l) +∑N

k=0 πd(k, c− 1)

)P (As ≥ ia)

+∑∞

y=1 πd(y +N, c− 1)P (As ≥ ia− y) if 0 < i < c,

pi(i−1) =(∑c−2

l=0

∑a−1k=0 π

d(k, l) +∑N

k=0 πd(k, c− 1)

)P (As < ia)

+∑∞

y=1 πd(y +N, c− 1)P (As < ia− y) if 0 < i < c− 1,

pc(c−1) = 1.

These transition probabilities only depend on the previous state, so a Markovchain is formed. The equilibrium distribution of this Markov chain, ω, givesinformation about the time the chain spends in the different levels for thenumber of busy machines.

The next step is to translate the time that the chain spends in a certainlevel to the number of occurrences of the possible time slots at that level ina cycle. At each level u, other than 0 or c, a time slot can have four possibleforms, [{u− 1, s}, {u− 1, e}], [{u− 1, s}, {u, s}], [{u, e}, {u, s}], [{u, e}, {u−1, e}]. Note that the forms are determined by the level from which the stepto level u is made and the next step after level u, Figure 7.6 depicts this. Atime slot at level 0 can only be of the form [{0, e}, {0, s}], and at level c ofthe form [{c− 1, s}, {c− 1, e}]

Level u

(u,e)

(u-1,s) (u-1,e)

(u,s)

Figure 7.6: Possible time slots at level u

The number of occurrences of the time slots in a cycle are proportionalto the equilibrium probabilities of the approximative Markov chain in thefollowing way:

P[{0,e},{0,s}]/ω0 =P[{u,e},{u,s}]/ωu+1p(u+1)upu(u+1) =


P[{u,e},{u−1,e}]/ωu+1p(u+1)upu(u−1) =P[{u−1,s},{u,s}]/ωu−1p(u−1)upu(u+1) =P[{u−1,s},{u−1,e}]/ωu−1p(u−1)upu(u−1) =

P[{c−1,s},{c−1,e}]/ωc

A cycle can only contain 1 time slot of the form [{0, e}, {0, s}], so P[{0,e},{0,s}] =1. With this information and the equality of the ratios given above, the ex-pected number of occurrences of each time slot in a cycle can be determined.This means that approximations for all ingredients necessary to calculateE[Lq] and E[W ] by equation (7.23) are found.

7.4.3 Results

To investigate the quality of the approximations, the mean number of cus-tomers according to the formulas of the model in the previous section arecompared with the mean of the outcomes of 50 simulations in which 20.000arrivals have occured. Table 7.1 shows the results for a = N . The deviationsfor the investigated values lie between 0.04% and 34.12%. This appears tobe quite a good result when comparing these deviations with the approxi-mation method of Wu et al. [19] for the M/DN/1 queue. The deviationsin their method appear to be of a higher order. Their approximation forc servers cannot be compared with the approximation for a = N in thisresearch, as the approximation model for multiple servers in [19] estimatesthe queueing times for a = 1.

The detailed results show that the number of occurrences of the time slotsis estimated very closely, which indicates that the process of the number ofbusy machines over time is approximated well by the constructed Markovchain. It also implies that the dependence between the level of the processand the number of customers that are left behind by the machine that is inservice longest is small.

The approximations for the number of customers per time slot also ap-pear to be quite good for the time slots that end by the start of a machine.The approximations for the number of customers in the time slots that endby a machine that finishes service have more room for improvement. Thelargest improvement in the approximations may be obtained by improvingthe estimations for the duration of the time slots, as these deviate most fromthe simulation results.

For a < N , the first simplification (Method 1) leads to an overestimationof the true number of customers in the queue, as it assumes that at mosta customers of Ak+1 are served by the machines. The second simplification


c=

2N

=20

N=

10

t=

4t=

10

t=

4t=

10

λ=

2.5

λ=

7.5

λ=

1λ

=3

λ=

1.2

5λ

=3.7

5λ

=0.5

λ=

1.5

E[L

]-

Form

ula

(7.2

3)

9.5

09.6

49.5

08.8

34.4

94.9

14.4

94.7

0E

[L]-

Sim

ula

tion

9.5

0(±

0.1

3)

9.6

2(±

0.0

9)

9.5

0(±

0.1

1)

9.6

0(±

0.1

0)

4.4

5(±

0.1

2)

4.8

1(±

0.0

7)

4.4

5(±

0.0

8)

4.7

5(±

0.0

7)

Dev

iati

on

0.0

5%

0.2

3%

0.0

4%

8.0

1%

0.7

9%

1.9

4%

0.7

8%

1.0

2%

c=

3N

=20

N=

30

t=

5t=

15

t=

5t=

10

λ=

5λ

=10

λ=

1λ

=2.5

λ=

6λ

=16

λ=

2λ

=8

E[L

]-

Form

ula

(7.2

3)

9.3

810.2

511.2

58.6

716.7

716.6

716.6

815.6

9E

[L]-

Sim

ula

tion

9.4

5(±

0.2

8)

9.8

1(±

0.1

5)

9.5

0(±

0.1

9)

9.5

0(±

0.2

5)

14.4

8(±

0.1

9)

15.3

3(±

0.3

1)

14.4

8(±

0.3

3)

15.3

3(±

0.2

9)

Dev

iati

on

0.8

2%

4.4

6%

18.4

6%

8.7

2%

15.7

5%

8.7

2%

15.1

9%

2.3

3%

c=

4N

=15

N=

5t=

5t=

10

t=

3t=

10

λ=

2λ

=10

λ=

2λ

=5

λ=

2λ

=6

λ=

0.5

λ=

1.5

E[L

]-

Form

ula

(7.2

3)

7.9

07.7

86.7

47.3

72.2

64.0

52.2

72.3

3E

[L]-

Sim

ula

tion

7.0

0(±

0.1

4)

7.3

7(±

0.0

9)

7.0

0(±

0.0

8)

7.3

7(±

0.1

3)

3.4

3(±

0.0

9)

3.6

5(±

0.0

8)

1.9

9(±

0.0

4)

2.2

8(±

0.0

9)

Dev

iati

on

12.8

6%

5.6

3%

3.7

8%

0.0

7%

34.1

2%

11.0

5%

14.6

3%

1.8

3%

Tab

le7.

1:A

ccur

acy

of(7

.23)

fora

=N


(Method 2) leads to an underestimation of the true number of customers inthe queue, as it assumes that all customers of Ak+1 that fit in a machine areserved by the machines. Especially for small values of a and large arrivalintensities, the error is expected to be large. This is also what is shown forthe case N = 20, c = 3, λ = 4.5, and t = 10 in Figure 7.7(a), where themean number of customers is compared for the simulation, Method 1 andMethod 2. For a small arrival intensity, the error occurs for larger values ofa, which is shown in Figure 7.7(b). Method 1 does appear to give a goodapproximation for small values of a and small arrival intensities.

Options for approximations of the transition matrix that lie betweenMethod 1 and Method 2 are investigated, but no approximation is foundthat works well for heavily and lightly loaded systems.

Method 1 does appear to work well for small workloads, and the machinesat the SAP have small workloads. Therefore, models with the same numberof machines, arrival intensities, service times and maximal batch sizes as themachines at the SAP are investigated with this method. For both models,the mean waiting time is minimized when the machines are started as soonas 1 sample is present.

7.5 Conclusions and suggestions for further research

7.5.1 Conclusions

The goal of the analysis is to investigate the effect of the minimal batch sizeson the mean waiting times for the machines at the SAP that serve samples inbatches. It is complex to develop an analytical model that suits the machinesat the SAP completely, so no complete model for the machines is analyzed.Though, the analysis of some characteristics of the machines does give someintuition for the machines at the SAP. The M/Ga,N/1 model is studiedin detail, and the relation between the model parameters and the optimalminimal batch size is investigated. Approximations for the M/Da,N/c modelare developed, and the MX/Da,N/1 is only discussed briefly.

M/Ga,N/1

The M/Ga,N/1 is already studied by many researchers and is analyzed withdifferent techniques, which has led to expressions for the p.g.f. of the num-ber of customers at departure moments and the number of customers at arandom moment. With these p.g.f. some performance measures of the sys-tem can be derived. The involved calculations and derivations are complex,


0 2 4 6 8 10 12 14 16 18 205

10

15

20

25

30

35Mean number of customers, N=20, t=10, lambda=4.5, c=3

Minimal batch size

Mea

n nu

mbe

r of

cus

tom

ers

Method 2Method 1Simulation

(a) N = 20, c = 3, λ = 4.5, t = 10

0 5 10 15 201

2

3

4

5

6

7

8

9

10Mean number of customers, N=20, t=10, lambda=1, c=3

Minimal batch size

Mea

n nu

mbe

r of

cus

tom

ers

Simulation outputMethod 2Method 1

(b) N = 20, c = 3, λ = 1, t = 10

Figure 7.7: Mean number of waiting customers

the expressions contain roots that may be hard to locate, and the system ofequations that has to be solved can become numerically unstable. There-fore, simple and practical expressions for the performance measures basedon explicit equilibrium probabilities are derived.

To find the equilibrium probabilities, the transition matrix is cut-off at


a state y >> N , which leads to quite accurate results when y is sufficientlylarge and ρ not too close to 1.

For deterministic service times, it is found that the minimal batch sizefor which the minimal waiting time is achieved is influenced by the arrivalintensity λ, the maximal batch size N and the service time t. For λ and t,the optimal minimal batch size gets larger when λ or t get larger. For N ,the opposite holds, as the optimal minimal batch size gets smaller when Ngets smaller.

It can be concluded that for all investigated cases, the optimal minimalbatch size is smaller than the maximal batch size. This gives rise to theidea that this may also hold for the machines at the SAP, even though thesemachines have other characteristics that are not taken into account in thismodel. By investigating two models with similar workloads and maximalbatch sizes as the machines at the SAP, it follows that the minimal batchsize for which the mean waiting time is minimal is equal to 2 for the modelsimilar to the centrifuge and 1 for the model similar to the outpatient clinictrain.

The SAP does not only consider the mean waiting times as a perfor-mance measure, but also the mean process time per sample that involvesan employee. The mean process time per sample increases when batch sizesbecome smaller. Therefore, the waiting time is compensated for the meanprocess time per sample. The results show that the optimal minimal batchsize becomes larger, for both the centrifuge and the outpatient clinic trainit becomes 6.

It can be concluded that the results imply that the laboratory may ben-efit from experimenting with smaller batch sizes in the machines.

MX/Da,N/1

The study of this type of queueing system is complex, and due to timeconsiderations and more interest in the effect of multiple servers, this modelis only discussed shortly and only an approximation model that builds onthe M/Ga,N/1 model is proposed.

If deterministic service times are considered, the number of arrivals dur-ing a service time is equal to a compound Poisson variable. By ignoring thefact that it may occur that more than N customers are in the queue whena machine starts service, the same transition matrix as for the M/Ga,N/1queue holds. The entries of the matrix are equal to compound Poisson prob-abilities. With the transition matrix, the same performance measures as forthe M/Ga,N/1 could be derived, in which all terms concerning arrivals are


adjusted for the compound Poisson arrivals.

M/Da,N/c

Much literature can be found on multiserver queueing models with bulk ser-vice, most studies that contain exact results consider exponential or phase-type service time distributions. Many studies that consider general arrivaldistributions or general service times develop approximations, see for exam-ple [19]. This research also develops an approximation model, the modelconsiders the M/Da,N/c system, because the service times for the machinesat the SAP are (nearly) deterministic.

The M/Da,N/c is investigated by analyzing the imbedded Markov chainof one machine in a similar way as for the M/Ga,N/1 queue. For a = N , anexact transition matrix can be determined. For a < N , only approximationsare found, because the number of customers served by the other machinesis unknown. By cutting off the transition matrix, an approximation of theequilibrium probabilities is found.

Also an approximation for the mean number of customers in the queueis derived. This is an approximation, because it depends on the numberof customers left behind by the machine that is in service longest. Thisnumber is not kept track of, because this would complicate the state spacevery much. The approximations are quite accurate for a = N , as the resultsdo not deviate very much from the simulation results. For a < N , the twoproposed approximations for the transition matrix appear to cause that theapproximations deviate much for large ρ and small a.

7.5.2 Suggestions for further research

The analysis of this chapter can be improved and extended in several ways,some ideas are listed here.

• The arrival probabilities in the MX/Da,N/1 could be improved byincorporating the possibility that more than N customers are waitingin the queue when a machine starts service. The model could also beextended to a model with general service times.

• The approximations concerning the time slots in the analysis of theM/Da,N/c queue can be improved by determining the time distribu-tions of the time slots.


• The transition matrices for a < N in the M/Da,N/c model can beimproved by determining a distribution for the number of customersserved by the other machines.

• The approximation model of Wu. et al. may possibly be extendedto an approximation for G/Ga,N/c by using the queueing time of anM/Ma,N/c system.

• The SAP machines also have customer priorities, this could be incor-porated in the models.

Chapter 8

Conclusions

In this chapter, the conclusions resulting from this research are summarizedand suggestions for further research are given.

8.1 Conclusions

This research is performed at the laboratory for several reasons. A fewreasons are that the laboratory has to shrink, workplaces are rearranged, nogood overview of the processes in the laboratory exist, there is high workloadvariation, and there are complaints about throughput times.

8.1.1 Laboratory

The processes in the laboratory can be modeled as a network of queues.The sample arrival and preparation area and the test execution area to-gether form a network of queues in which the body fluid samples are thecustomers. The queueing network is very large and complex, as there areseven different arrival types, of which many arrive in batches, more than 100different process steps, two server types, and three customer priority types.Therefore, it is hard to model the complete network in detail.

The data analysis and the discussions with the laboratory staff indicatethat the sample arrival and preparation area (SAP) is an area at which muchimprovements are possible, so the SAP is investigated. All samples arriveat this part of the laboratory, the area consists of many different activi-ties, there are discussions concerning the occupation rate of the employeesin this part of the laboratory, and there are many ideas for improvement.The analysis is mainly done by simulation, due to the complexity of the

115

CHAPTER 8. CONCLUSIONS 116

processes. The machines at the SAP that serve the customers in batchesare also investigated by exact and approximative methods.

8.1.2 Simulation

One of the main conclusions that can be drawn from the simulation is thatthe combination of high occupation rates and low throughput times is notachievable in the current situation. Due to the fact that many arrivals occurin batches, the number of employees should be high to accommodate lowthroughput times. For the moments between the batch arrivals, the numberof employees should be low to achieve high occupation rates.

In the current situation, the occupation rate is only 55%, so the labora-tory staff would like to see this at a higher level. By putting one employee atthe SAP, the occupation rate increases to 81%, but the throughput times forregular samples increase to approximately 3 times the current level. Thisillustrates that the current situation does not allow a combination of lowthroughput times and high occupation rates. By combining this scenariowith moving the sorting activities, the throughput times decrease to a leveljust above the current situation. The occupation rate also decreases, but itis still 70%. When also the walking distance for bringing the cito samplesis shortened, the throughput times would decrease to a level similar to thecurrent situation and the occupation rate would decrease to 59%. If also allinterruptions are eliminated, the throughput times decrease to a level underthe current situation and the occupation rate decreases to 55% again.

The scenarios show that there are many options for improvement of thethroughput times at the SAP. A very simple but effective idea is to movethe sorting activities. Another simple and effective scenario is to reducethe number of samples in the outpatient clinic train, as this reduces thewaiting times at the outpatient clinic very much. For the waiting times atthe outpatient clinic, the best idea is to move the outpatient clinic next tothe laboratory. Another scenario that is very effective, is to shorten thewalking distance to the laboratory, which decreases throughput times for allsample types.

The sensitivity analysis of the simulation confirms that it would be betterfor the throughput times that the arrival batches are smaller, because inthe situation that the batches and the interarrival times are halved, thethroughput times are better. The occupation rate does increase, becausesome tasks have to be performed more often.

The sensitivity analysis also shows that the system could handle thesituation in which the batch sizes would double, and also the situation in


which the interarrival times would halve. The combination of doubling batchsizes and halving interarrival times results in an overloaded system.

The task priority setting in the current situation appears to be good, asno alternative setting is found for which the throughput times decrease.

8.1.3 Analysis

The goal of the analysis is to investigate the effect of the minimal batch sizeson the mean waiting times for the machines at the SAP that serve samples inbatches. It is complex to develop an analytical model that suits the machinesat the SAP completely, so no complete model for the machines is analyzed.Though, the analysis of some characteristics of the machines does give someintuition for the machines at the SAP. The M/Ga,N/1 model is studiedin detail, and the relation between the model parameters and the optimalminimal batch size is investigated. Approximations for the M/Da,N/c modelare developed, and the MX/Da,N/1 is only discussed briefly.

The results for the M/Ga,N/1 system show that for all investigated cases,the optimal minimal batch size is smaller than N . This indicates that thismight also hold for the machines at the SAP. When models with the sameload, maximal batch size, and service time as the SAP machines are con-sidered, it follows that a very low a would be optimal in those cases; a = 2for the centrifuge and a = 1 for the outpatient clinic train. Not only wait-ing times are important for the SAP, but also the process time per sample.Therefore the mean waiting time is compensated for the process time. Dueto the compensation, the optimal minimal batch size increases, but it is stillonly 6 for both the centrifuge and the outpatient clinic train. This indicatesthat it may be rewarding for the laboratory to experiment with smallerminimal batch sizes.

TheMX/Da,N/1 andM/Da,N/c are investigated by analyzing the imbed-ded Markov chain in a similar way as for the M/Ga,N/1 queue. Only theapproach is described for the MX/Da,N/1 queue. Approximations for theperformance measures and results are derived for the M/Da,N/c queue.Some approximations give quite accurate results, but there is still room forimprovement.

8.2 Suggestions for further research

There are many options for further research, both concerning the laboratoryand the analysis.


8.2.1 Laboratory

In this research, the SAP is investigated. There are also other parts of thelaboratory that could be investigated, a few are listed here.

• Machine batch policy for the testing machinesSome of the machines are always fully loaded, it could be investigatedwhat the effects of using a smaller minimal batch size are.

• New workplace arrangementsAll of the employees are assigned to a certain workplace. Some work-places are busier than others, so it may be interesting to investigatewhether other workplace arrangements would be more effective.

• Blood withdrawal roundThe blood withdrawal round is currently not scheduled or structured,and it takes quite a long time before the employees return to thelaboratory and start testing. In this research, only the arrival structurein the laboratory is investigated. It could be investigated whether theblood withdrawal round can go faster by scheduling and structuringit.

8.2.2 Analysis

The SAP is investigated in detail by a simulation, and exact analysis is doneon the batch policy of the machines. There are also many other aspects ofthe processes at the SAP that can be investigated in an analytical way.Some of the characteristics that are interesting to investigate are:

• Arrival typesThe arrivals at the SAP have some special characteristics, which differ-entiate the system at the SAP from arrivals in a regular open queueingnetwork. One characteristic is that some of the arrivals only occur oncea day, which make the problem at that moment more of a schedulingtype then a queueing type. Another characteristic is that the arrivalsmainly consist of batches of customers.

• Customer prioritiesThere are three different customer priority types; blood gas determina-tion requests, cito requests and regular requests. It may be interestingto investigate the effect of the priorities on waiting times and through-put times of the samples in the SAP.


• Processors with multiple tasks and interruptionsThe two employees have to carry out 16 different tasks, while theyare also interrupted by disturbances. This means that the employeeshave to switch between the tasks, and prioritize in the tasks and thecustomers. The effect of the task priority setting may be interestingto investigate in an analytical way.

Appendix A

Simulation Input and Output

A.1 (Inter) Arrival distributions

Table A.1 shows the input parameters and distributions concerning arrivalsfor the simulation.

Arrival type Simulation input

Once a day

IC 7:00 7:00+Exp(1.5)CCU 7:00 7:00+Exp(3.1)Blood withdrawal round - children 7:45+Erlang(n, 9)Blood withdrawal round - rest 7:45+Erlang(n, 6)Outpatient BWO 13:30+Exp(15)

Continuous

ICU/CCU/SEH E[X] = 32.34, cX = 0.90Rest inpatient E[X] = 29.03, cX = 0.91External E[X] = 69.24, cX = 0.66Outpatient 8-9 E[X] = 3.28, cX = 1.22Outpatient 9-12 E[X] = 2.73, cX = 1.38Outpatient 12-13 E[X] = 4.10, cX = 1.69Outpatient 13-14 E[X] = 5.57, cX = 1.66Outpatient 14-15 E[X] = 3.85, cX = 1.22Outpatient 15-16 E[X] = 4.36, cX = 1.23Outpatient 16-17 E[X] = 5.54, cX = 1.15

Table A.1: Arrival input

120

APPENDIX A. SIMULATION INPUT AND OUTPUT 121

A.2 Batch size distributions

Table A.2 shows the input parameters and distributions concerning batchsizes for the simulation.

The blood withdrawal round is divided in 5 groups, which correspondwith departments in the hospital. This is derived from the collected infor-mation concerning the blood withdrawal round.

The arrival batch size of the continuous arrivals is always at least equalto 1. Therefore the data for the batch sizes of the continuous arrival typesis corrected by minus 1, and the distribution is fitted on the corrected data.

Arrival type Simulation input Chi-squarep-value

Once a day IC 7:00 E[X] = 4.52, cX = 0.47 0.03CCU 7:00 E[X] = 2.35, cX = 0.65 0.42Blood withdrawal round - children E[X] = 11.67, cX = 0.48 0.04Blood withdrawal round - rest group 1 E[X] = 24.66, cX = 0.36 0.00Blood withdrawal round - rest group 2 E[X] = 5.10, cX = 0.59 0.01Blood withdrawal round - rest group 3 E[X] = 8.33, cX = 0.54 0.01Blood withdrawal round - rest group 4 E[X] = 2.13, cX = 0.77 0.03Outpatient BWO E[X] = 45.46, cX = 0.26 0.00

Continuous ICU/CCU/SEH E[X] = 0.28, cX = 2.22 0.07Rest inpatient E[X] = 0.37, cX = 3.07 0.16External E[X] = 4.14, cX = 1.55 0.02

Table A.2: Batch size input

A.3 Cito probabilities

Table A.3 shows the probabilities that an arriving lab number contains acito sample per origin in the simulation.

A.4 Process times

Table A.4 shows the mean process times per process step. All process stepsare assumed to be exponentially distributed, except for turning on the cen-trifuge (D3). The process time for turning on the centrifuge consists of afixed part and a part that depends on the number of samples that are loadedinto the centrifuge. The part that depends on the number of samples n, is


Arrival type Cito probabilityIC/CCU/SEH 1Blood withdrawal round 0.03Outpatient clinic WLZ 0.11Rest inpatient 0.14External 0.06Outpatient clinic BWO 0.02

Table A.3: Probability that arriving lab number contains cito sample perorigin

estimated by an Erlang(n,0.04) variable. The registration step (B1) has twoaverages, because registering the forms for external samples take longer.

Task Mean Task MeanA1 0.75 C1 0.06A2 0.10 D1 1.10A3 0.11 D2 1.03A4 0.80 D3 fixed 0.1 + Erlang(n, 0.04)A5 2.64 D4 0.15A6 0.19 D5 0.44A7 0.20 D6 1.62B1 1.00; 1.75∗ D7 0.10

Table A.4: Process time input

A.5 Task priority setting

The employees have to choose between tasks whenever more than one taskis waiting to be performed. Each time an employee finishes a task, or asample arrives occurs in the simulation, the employee walks through thepriority list and picks the first task for which samples are waiting and theother conditions are satisfied. The priority list looks as follows (for taskdescriptions, see Figure 4.1):

1. A1

2. A5

3. A2


4. D1, if the blood gas analyzer is idle

5. D2

6. A6, if it concerns a cito sample

7. B1, if it concerns a cito sample

8. C1, if it concerns a cito sample

9. C1, if the employee was doing C1 previously and regular samples arewaiting for C1

10. D3, if a centrifuge is idle and cito samples are waiting

11. D4, if it contains a cito sample

12. D6

13. D5, if it concerns a cito request

14. A3

15. D4

16. D7, if the previous job is either D3 or D4

17. B1, if the previous job is B1

18. C1, if the previous job is C1

19. A3, if the previous job is A3

20. D7, if the previous job is D7

21. A6

22. A7

23. D3, if 20 or more samples are waiting for the centrifuge and a centrifugeis idle

24. B1

25. C1

26. A4


Origin Process Simulation:mean reg-ular

Samples: regu-lar

Simulation:mean cito

Samples: cito

Pneumaticmail -IC/CCU/SEH

Centrifuge 20.8 21.3; 16.1; 27.4 18.9 19.2; 19.0; 16.3No Centrifuge 8.6 5.1; 16.1; 7.3 7.3 5.9; 12.0Urine 3.1 4.0Blood gas 9.2 11.5; 9.5External 20.7

Inpatientround

Centrifuge 23.9 29.4; 19.1; 17.5 13.1; 21.2No Centrifuge 11.2 18.1; 9.3; 7.0 4.3;Urine 2.7Blood gas 11.3External 27.1 58.1

Inpatient -Otherdepartments

Centrifuge 22.9 21.1; 23.4; 17.5 19.8; 16.1No Centrifuge 9.4 6.6; 5.1; 5.4 4.0; 4.8Urine 5.7 3.6Blood gas 9.0 16.1External 25.0

External

Centrifuge 37.0 41.2 20.1No Centrifuge 24.1 28.7 6.8Urine 14.1Blood gas 12.5External 39.6

Outpatient -WLZ

Centrifuge 19.9 25.2; 26.1; 17.8;15.6; 19.3

15.4 22.2; 14.1; 18.5;18.1; 13.2

No Centrifuge 6.3 2.5; 12.4; 3.9;4.2

3.1 2.2; 10.9; 3.8

Urine 1.4Blood gas 6.9 9.5External 22.1 67.1; 21.0

Outpatient -BWO

Centrifuge 8.1 5.1 3.0No Centrifuge 8.0 5.1 3.9Urine 7.8Blood gasExternal 7.9

Table A.5: Simulation times vs. Sample times

27. D3, if both centrifuges are idle

28. D7

29. D5

A.6 Validation by sampling

Table A.5 shows the samples throughput times vs. the throughput timesobtained from 100 simulated days.

Appendix B


The testing problem for a goodness-of-fit test can be described by

H0 : F = Fθ

againstH1 : F 6= Fθ,

where θ ∈ Θ.

B.1 Continuous distributions

A test based on the empirical distribution is very suitable for continuousdistributions. Let Fn denote the empirical distribution function

Fn(x) =1n

n∑i=1

1{Xi ≤ x}.

The basic idea behind this test is that under H0, for n sufficiently large, Fn,is close to Fθ and any discrepancy measure between Fn and Fθ is suitable asa test statistic [18]. Two very well-known test statistics are the KolmogorovSmirnov (KS) test statistic,

Dn = supt| Fn(t)− Fθ(t) |,

and the Cramer-von Mises test statistic,

Cn =∫

(Fn(t)− Fθ(t))2dFθ(t).

125

APPENDIX B. GOODNESS-OF-FIT TESTS 126

If the null hypothesis holds true, the test statistic is distribution free [18],which means that the distribution of the test statistic does not depend onFθ. To see whether the test statistic is significant, asymptotic results for thelimiting distribution are used.

The parameters θ in the fitted distribution Fθ may be fitted by the data.By plugging in this estimate θ in the test statistic, the distribution of the teststatistic and the significance level change. It is hard to obtain asymptoticresults for the limiting distribution now, an easier method is to estimate thep-value by simulation. This is known as parametric bootstrap, and it worksas follows [18]:

1. Estimate θ by θn from {X1, X2, . . . , Xn} and construct the CDF Fθn.

2. Evaluate the test statistic, for example Dn or Cn.

3. Generate B ∈ N bootstrap samples of size n from Fθn. Denote these

B samples by {X∗1,j , X∗2,j , . . . , X∗n,j}.

4. Compute for allB samplesD∗n,j(X∗1,j , X

∗2,j , . . . , X

∗n,j) or C∗n,j(X

∗1,j , X

∗2,j , . . . , X

∗n,j),

in which θ∗j is the estimate of θ from bootstrap sample j and F ∗n,j theempirical CDF of sample j.

5. An approximation for the p-value is given by

p =

∑Bj=1 1{D∗n,j ≥ Dn}

B.

B.2 Discrete distributions

Goodness-of-fit tests that are very suitable for discrete distributions are chi-square tests. For this test, the observed data is divided over n cells, andfor each cell the observed frequency of observations, Oi, and the expectedfrequency according to the theoretical distribution, Ei, is computed. Thevalue of the test statistic is then equal to

χ2 =n∑i=1

(Oi − Ei)2

Ei.

The p-value can be obtained by comparing the value of the test statisticwith the probability of a χ2-distribution with n− p− 1 degrees of freedomattaining this value, where p is equal to the number of estimated parametersfor the theoretical distribution.

Appendix C

Results from the imbeddedMarkov chain for M/Ga,N/1

To find the limiting distribution πd for the number of customers at departuremoments in Section 7.2, the system of equations

πdj =∑i

πdi Pij j ≥ 0

has to be solved. These equations take the form

πdj =N−1∑i=0

πdi αj +j+N∑i=N

πdi αj−i+N . (C.1)

To solve the system of equations, two generating functions are introduced,

πd(s) =∞∑j=0

πdj sj , A(s) =

∞∑j=0

αjsj .

Multiplying C.1 by sj on both sides, and summing over j yields

πd(s) =N−1∑i=0

πdiA(s) +∞∑j=0

j+N∑i=N

πdi αj−i+Nsj

=N−1∑i=0

πdiA(s) + s−N∞∑j=0

j+N∑i=N

πdi αj−i+Nsj−i+N

=N−1∑i=0

πdiA(s) + s−N∞∑i=N

πdi si∞∑

j=i−Nαj−i+Ns

j−i+N

127

=N−1∑i=0

πdiA(s) + (πd(s)−N−1∑i=0

πdi si)A(s)/sN

=A(s)(sN

∑N−1i=0 πdi + πd(s)−

∑N−1i=0 πdi si)

sN. (C.2)

By rewriting, it is found that,

πd(s)(1−A(s)s−N ) = A(s)(N−1∑i=0

πdi −N−1∑i=0

πdi sis−N )

= A(s)N−1∑i=0

πdi (1− s−N+i)

=A(s)

∑N−1i=0 πdi (sN − si)sN

, (C.3)

from which it follows that

πd(s) =A(s)

∑N−1i=0 πdi (sN − si)sN −A(s)

. (C.4)

There exists a relation between A(s) and G(s),

A(s) =∞∑j=0

αjsj

=∞∑j=0

∫ ∞t=0

(λt)j

j!e−λtfG(t)dtsj

=∫ ∞t=0

∞∑j=0

(λts)j

j!e−λtfG(t)dt

=∫ ∞t=0

e−λt(1−s)fG(t)dt

= G(λ(1− s)), (C.5)

where G is the Laplace Stieltjes transform of the service time distribution.This leads to the following expression for πd(s),

πd(s) =G(λ(1− s))

∑N−1i=0 πdi (sN − si)

sN − G(λ(1− s)). (C.6)

128

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