IE 417: Operations Research II

Travis GuzmanAmanda Smith

David Trigg

Presented to: Dr. Sima Parisay

California State Polytechnic University, Pomona

Due: February 7th, 2011

Table of ContentsProblem Statement......................................................................................................................................................3

Summary of the Problem.............................................................................................................................................4

Kendell-Lee Notation....................................................................................................................4

Queuing System Notations:.........................................................................................................4

Formulas........................................................................................................................................5

Rate Diagram................................................................................................................................5

Figure 1: Rate Diagram for M/M/s/GD/∞/∞..........................................................................................5

Part A:..........................................................................................................................................................................6

Figure 2: Queuing System Diagram A......................................................................................................6

Part B:..........................................................................................................................................................................8

Figure 3: Queuing System Diagram B......................................................................................................8

Part B: Performance Measures..................................................................................................................................10

Table 1: Performance Measures Summary............................................................................................10

Part C......................................................................................................................................................................... 11

Table 2: Part A and B Comparison.........................................................................................................11

Part D.........................................................................................................................................................................13

WINQSB- Performance Measures & Cost Analysis.....................................................................................................14

Table 3: Cost Table................................................................................................................................14

Figure 4: WinQSB Input.........................................................................................................................15

Figure 5: WinQSB Output- Performance Measure.................................................................................15

Figure 6: WinQSB Output- System Probability Summary.......................................................................16

WINQSB- Sensitivity Analysis.....................................................................................................................................18

Sensitivity Analysis #1: Arrival Rate.......................................................................................................18

Figure 7: WinQSB Input Parameters for SA #1.......................................................................................18

Figure 8: WinQSB Output Table- SA #1..................................................................................................19

Figure 9: WinQSB Output Utilization Graph- SA #1................................................................................20

Sensitivity Analysis #2: Service Rate......................................................................................................21

Figure 10: WinQSB Input Parameters for SA #2.....................................................................................21

Figure 11: WinQSB Output Table- SA #2................................................................................................22

Figure 12: WinQSB Output Utilization Graph- SA #2..............................................................................23

Figure 13: WinQSB Output Utilization Graph- SA #1 & #2 Comparison.................................................23

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Acknowledgments.....................................................................................................................................................24

Problem Statement

The problem statement can be found on page 1094, problem #3 from Operations Research by Wayne L Winston, 4th edition.

In this problem, all interarrival and service times are exponential.

a. At present, the finance department and the marketing department each have one typist. Each typist can type 25 letters per day. Finance requires that an average of 20 letters per day be typed, and marketing requires that an average of 15 letters per day be typed. For each department, determine the average length of time elapsed between a request for a letter and completion of a letter.

b. Suppose that the two typists were grouped into a typing pool; that is each typist would be available to type letters for either department. For this arrangement, calculate the average length of time between a request for a letter and completion of the letter.

c. Comment on the results of part (a) and (b).

d. Under the pooled arrangement, what is the probability that more than .200 day will elapse between a request for a letter and completion of the letter?

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Summary of the Problem

Before beginning the problem, we must have an understanding of the type of queuing system presented in the problem, as well as associated formulas that will be employed with this type of queuing system.

First, we begin with the type of queuing system given. The Kendell-Lee Notation for Queuing Systems can be described by six characteristics:

Kendell-Lee Notation1/2/3/4/5/6

Arrival process/service process/parallel servers/ queue discipline/max customers/population size

For our problem the Kendell-Lee Notation looks like the following:

M/M/s/GD/∞/∞

Where

M= Interarrival times are independent, identically distributed (iid)

s = number of parallel servers

GD = General queue discipline

∞ = infinite numbers of customers; infinite population size

Queuing System Notations: λ = Arrival rate approaching the system

λe= Arrival rate (effective) entering the system

µ = Maximum (possible) service rate

µe= Practical (effective) service rate

L = Number of customers present in the system

Lq = Number of customers waiting in the line

Ls = Number of customers in service

W = Time a customer spends in the system

Wq= Time a customer spends in the line

Ws= Time a customer spends in service

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ρ = Traffic Intensity

π j = P(j) = Probability that j units are in the system

π0 = P(0) = Probability that there are no units (idle) in the system

Formulas (for M/M/S)

Traffic Intensity: Probability that a Server is Idle:

ρ = λsµ < 1

Number Waiting in Line:

Lq = P ( j>s) ρ(1−ρ)

Number in Service: Probability that Arriving Unit has to Wait:

Ls = λµ

Number in System:L = Lq + Ls

Time Spent in Line: Probability that J Units are in Service:

Wq = Lqλ

Time Spent in System:

Ws = 1µ

Time in System: Probability that Time in System is > t

W = Lλ=P( j>s)sµ−λ

+ 1µ

P (W>t )=e−μt{1+P ( j ≥ s) 1−e−μt (s−1−sρ)

s−1−sρ }***Note: Cost Formulas are located in cost analysis section for clarity****

Rate Diagram

Figure 1: Rate Diagram for M/M/s/GD/∞/∞

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π0=1

∑i=0

s−1 (sρ )i

i !+

(sρ )s

s !(1−ρ)

P( j≥s )=( sρ )s π0s!(1−ρ)

π j=( sρ) j π0j !

Part A:First we extract the information the problem gives us:

s = 2 typist

Service Rate Arrival Rates

µ = 25 letters/day λFinance = 20 letters/day

λMarketing = 15 letters/day

The problem asks, “For each department, determine the average length of time elapsed between a request for a letter and completion of a letter.” In essence this means find W, the total time each letter is in the system for each department. The diagram below will help in analyzing the problem:

Figure 2: Queuing System Diagram A

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Finance Dept.

μ = 25 lettersday

MarketingDept.

μ = 25 lettersday

λFinance=20 lettersday

λMarketing=15 lettersday

μFinance=25 lettersday

μMarketing=25 lettersday

For this part, number of servers is equal to one when we are analyzing traffic intensity. There is only one server in each department, and we are analyzing the departments separately. Keep in mind that when we are analyzing departments separately we are looking at a M/M/1/GD/∞/∞. With the given information, we only need one formula, W, from the M/M/1 system. We proceed to make the following calculations:

FINANCE MARKETING

Traffic Intensity: Traffic Intensity:

ρ = λµ ρ =

λµ

ρ = 20 letters

day

(25 lettersday

) = .80 ρ =

15 lettersday

(1 server )(25 lettersday

) = .60

ρ = 0.80 ρ = 0.60

0< ρ<1 0< 0.80<1so a steady state exists 0< ρ<1 0< 0.60<1so a steady state exists

Time in System: Time in System:

W = 1µ−λ

=¿ 1

25−20=¿

15day W=

1µ−λ

=¿ 1

25−15=¿

110day

We also make the following extra calculations for later comparison purposes:

Number Waiting in Line: Number Waiting in Line:

Lq = p2

(1−ρ) = .802

(1−0.80) = 3.2 letters . Lq = ρ2

(1−ρ) = .602

(1−0.60)= 0.90

letters

Number in System: Number in System:

L = p❑

(1− ρ) =.80❑

(1−0.80)= 4 letters L = p❑

(1− ρ) =.60❑

(1−0.60)= 1.5 letters

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As you can see, it takes the finance department 15 day between a request for a letter

and completion of a letter. It only takes the marketing department 110 day between a

request for a letter and completion of a letter. This is expected since the marketing department receives 5 less letters a day as compared to the finance department and both departments have the same process time.

Part B:

Essentially, Part B asks the same question as Part A, but in this scenario, the typists can type letters for either department.

First we extract the information the problem gives us:s = 2 typistService Rate Arrival Rates

µ = 25 letters/day λFinance = 20 letters/day

λMarketing = 15 letters/day

The diagram below will help in analyzing the problem:

Figure 3: Queuing System Diagram B

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Now, since the typists are working

as a team, the arrival rates are

additive. The new arrival rate is then λFinance +λMarketing =

35 letters/day

Now, the service rate doubles since

two typists are working together;

together their service rate is twice as fast or 2µ =2(25

letters/day) = 50 letters/day

Finance Dept.

μ = 25 lettersday

Then, we proceed to the calculations. Now, we don’t have to perform calculations for separate departments since we are viewing the departments as one team.

FINANCE & MARKETING Traffic Intensity:

ρ = λsµ < 1

ρ = 35 letters

day

(2 server )(25 lettersday

) < 1

ρ = 0.7 < 1 so a steady state does exist

Number Waiting in Line:

Lq = P ( j>s) ρ(1−ρ) =

P ( j>2 ) .70(1−0.70)

= ( .57 )(.70)

(.30) = 1.33 letters

*P(j>2) value gotten from Table 6 on page 1088 ofOperations Research by Wayne L Winston, 4th edition when ρ=.7 and s = 2.

Number in System:

L = Lq + Ls = 1.33 lett .+35 letters

day

25 lettersday

= 2.73 letters

Time in System:

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MarketingDept.

μ = 25 lettersday

λM&F=35 lettersday μM&F=50

lettersday

W = Lλ=¿

2.73lett .

35 lettersday

=¿ .078 days

Thus, when the typists were grouped the new average length between a request for a letter and completion of the letter is .078 days.

Part B: Performance MeasuresPerformance measures were added for Part B of this problem because in this section of the problem the typists reflected two true parallel servers with one arrival rate and one service rate. All performance measures learned in lecture regarding queuing theory thus far were calculated and placed in the summary table below. Thus you can ask any question of the name given and find its corresponding value from this table.

Table 1: Performance Measures Summary

Performance Measures SummaryNotation Name Given Calculation Value

1λ=λE Arrival Rate = Effective AR n/a

35lettersday

2μ Service Rate n/a

50lettersday

3μE Effective Service Rate n/a

35lettersday

4Lq  Number in Queue see Part B 1.33 letters

5Ls  Number in Service

λμ=3525 1.4 letters

6 L  Number in System see Part B 2.73 letters7

Wq  Time in QueueLqλ =

1.3335 .038 days

8Ws  Time in Service

1μ =

125 .04 days

8 W  Time in System Wq + Ws = .038+.04 .078 days

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10

π0

Probability 0 Units in Service  0.1764

11π1

Probability 1 Unit in Service

0.2470

12π2

Probability 2 Units in Service

 0.1729

13 P0 Probability Teller Idle P0 = π0 0.176414

Pw

Probability Arriving Unit Must Wait

.5762

15 U Teller Utilization 1 - π0 = 1-0.1764  0.823616

P(J>1)Probability More than 1 Unit 

1-P(J<1)=1-(π0+π1) = 1-(.1764+.2470)  0.5766

17P(J>2)

Probability More than 2 Units

1-P(J<2)=1-(π0+π1+π2)= 1-(.1764+.2470+.1729) 0.4037 

Part CThe results of parts (a) and (b) can be summarized in the following table:

Table 2: Part A and B ComparisonFinance Department Marketing Department

Part A λ= 20 ltrs/day λ= 15 ltrs/dayW = 1/5 day = .2 days W= 1/10 day = .1 days

Part B λ= 35 ltrs/dayW = .078 days

When we analyzed part (a) by itself, we see that the marketing department has a shorter time in system value (W) than the finance department. This is expected because the marketing department receives less letters a day than the finance department, while both have the same processing capabilities. However, we can note that the effective service rate for each department is equal to the arrival rate of letters to each department. Therefore, within the same time span, the marketing department must type 15 letters while the finance department must type 20.

When the typists for each department work together, we have the lowest time in system value (W) of 0.078 days. Letters arrive at a rate of 35 letters per day while they can be processed at 50 letters per day. Because a letter can go to either typist, the W value is lower since a letter is not waiting on only one typist. This queuing system is the more opportune system.

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π j=( sρ) j π0j !

=(2 x . 7 )1 . 1764

1 !π j=

( sρ) j π0j !

=(2 x . 7 )2. 1764

2 !

P( j≥s )=( sρ )s π0s !(1−ρ)

=(2 x . 7)2 . 17642 !(1−.7 )

The additional performance measures calculated can be compared by placing the diagrams above and below each other. Pay attention to the white boxes that show traffic intensity, number in line, number in system, and time in system.

VS

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Finance Dept.

μ = 25 lettersday

ρ = .80

Lq = 3.2 Letters

L = 4 Letters

W = 1/5 day

ρ = .60

Lq = .9 Letters

L = 1.5 Letters

W = 1/10 day

WAVG = 0.15 day

μFinance=25 lettersday

Finance Dept.

μ = 25 lettersday

λFinance=20 lettersday

μMarketing=25 lettersdayλMarketing=15

lettersday

MarketingDept.

μ = 25 lettersday

Part D

The problem states, “Under the pooled arrangement, what is the probability that more than .200 day will elapse between a request for a letter and completion of the letter?”

In other words, what is the probability that the total time in the system, W, is more than .200 day.

First off, we must locate the necessary formula to calculate the distribution of waiting time. To determine the probability, we need to know the distribution of a letter’s waiting time. On page 1090 the following formula is given:

P (W>t )=e− μt{1+P ( j ≥ s ) 1−e−μt (s−1−sρ)

s−1−sρ }The t value given in the problem statement is .200 day. From Part C, we know all other values in the equation. Now, we just have to plug and chug numbers:

P (W>.200 )=e−25( .2){1+( .57 ) 1−e−25 (.2)(2−1−2( .7 ))

2−1−(2)(.7) }P (W>.200 )=.07

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MarketingDept.

μ = 25 lettersday

λM&F=35 lettersday μM&F=50

lettersday

ρ = .70

Lq = 1.33 Letters

L = 2.73 Letters

W = .078 day

WINQSB- Performance Measures & Cost AnalysisCost values were chosen for this problem and their relative calculations were made in the table below. The following costs given are:

Cs (Cost of Busy Server) at $72/day. This value was chosen based on an 8 hour work day and a typist working at $9/hour.

Ci (Cost of Idle Server) at $72/day. This value was chosen based on an 8 hour work day and an idle typist costing the company $9/hour.

Customer Waiting- It costs the company $1/day for a letter to be waiting in the queue to be typed.

Loss in Good Will- It costs the company $.08/letter of the goodwill of the customer bringing the letters to be typed.

Cost of Space- Since letters are so small, the cost of space is almost negligible at $.02/letter. However, to reflect understanding of material, it was calculated in the cost table.

Loss of Customer- N/A with ∞ Capacity

Table 3: Cost Table

Cost TableCost Calculation In WinQSB? Value per Day

Busy Server/Typist  Csρs = ($72)(.70)(2) Yes $100.80

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Notice these values are the same

Idle Server/Typist  Ci(1-ρ)s = ($72)(1-.70)2 Yes $43.20

Customer/Letter Waiting  ($/unit of time)λeWq =

($1/day)(35 ltrsday

)

(.0384dy) Yes $1.34

Customer/Letter being Served  WinQSB Yes $0.70

Loss in Goodwill ($/part)λePw=

($.08/ltr)(35 ltrsday

)

(.5762) No  $1.61

Cost of Space = Almost Neglible ($/part)Lq= ($.02/letter)(1.33 letters) No $0.03

Loss of Customer -----

TOTAL $147.68

The Queuing System problem was input into WinQSB. The values calculated in the Cost Table (Table 3) and Performance Measure Summary (Table 1) match that of WinQSB.

Figure 4: WinQSB Input

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N/A with ∞ Capacity

Figure 5: WinQSB Output- Performance Measure

Additionally, WinQSB contains a probability and cumulative probability feature that can be seen Figure 6 below:

Figure 6: WinQSB Output- System Probability Summary

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The application/use of the system probability summary in Figure 6 can be utilized in situations like:

Finding the probability that there are more than a specific number of letters in the system. For example, what is the probability that there are more than 5 letters in the system? 7 letters? 10 letters? 20 letters?

P(j>5) = 1-P(j<5) = 1 – 0.8616 = .1384 = 13.84%P(j>7) = 1-P(j<7) = 1 – 0.9322 = .0678 = 6.78%P(j>10) = 1-P(j<10) = 1 – 0.9767 = .0233 = 2.33%

Finding the probability that there are a specific number of letters in the system. For example, what is the probability that there are 0 letters in the system? 11 letters? 15 letters? 20 letters?

P(j=0) = .1765 = 17.65%P(j=11)= .0070 = .7%P(j=15)= .0017 = .17%

Finding the probability that there are less than a specific number of letters in the system. For example, what is the probability that there are less than 3 letters in the system? 5 leters? 10 letters?

P(j<3) = 1-P(j>2) = 1-.5965 = .4035 = 40.65%P(j<5) = 1-P(j>4) = 1-.8023 = .1977 = 19.77%P(j<10)= 1-P(j>10) = 1-.9767 = .0233 = 2.33%

Finding the probability that there are less than or equal to a specific number of letters in the system. For example, what is the probability that there are less than or equal to 5 letters in the system? 10 letters? 20 letters?

P(j<5) = 0.8616 = 86.16%(j<10) = 0.9767 = 97.67%P(j<20) = 0.9993 = 99.93%

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WINQSB- Sensitivity Analysis

Sensitivity Analysis #1: Arrival Rate

We chose to do our first sensitivity analysis on Arrival Rate. Currently, the arrival rate is 35 letters/day. We created a range above and below this number. From past research, we know that a minimum of 25 letters/day arrive for the typists to type and at maximum 50 letters/day. We decided to stop at 50 because when arrival rate equals the service rate (which is currently 50), the system will blow up due to variation.

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Figure 7: WinQSB Input Parameters for SA #1

Just as expected in the screen below, as arrival rate increases:

λ↑, U↑, L↑, Lq↑, W↑, Wq↑, Pw↑

and all costs go up. This was expected because naturally, having more letters in the system causes such a relationship (common sense).

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Figure 8: WinQSB Output Table- SA #1

WinQSB also has the capabilities of showing all the performance measures given in the table above in a graphical form. We chose to graph the system utilization graph. As was already discussed, as arrival rate increase so does the system utilization. This is indicated by the positive slope of the graph in Figure 9.

Figure 9: WinQSB Output Utilization Graph- SA #1

Sensitivity Analysis #2: Service Rate We chose to do our second sensitivity analysis on Service Rate. Currently, the service rate is 25 letters/day. We created a range above and below this number. From past research, we know that a minimum of 10 letters/day can be typed by the typists and a maximum 35 letters/day.

Figure 10: WinQSB Input Parameters for SA #2

From the output, at service rates ranging from 10 letters/day to 17 letters/day the system is unstable. At a service rate of 18/letters per day, the system can begin to handle the 35 letters/day arrival rate.

In the screen below, as service rate increases:

μ↑, U↓, L↓, Lq↓, W↓, Wq↓, Pw↓

and all costs decrease. This was expected because naturally, processing letters at a faster rate causes such a relationship (common sense).

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Figure 11: WinQSB Output Table- SA #2

Again, we chose to graph the system utilization graph. As was already discussed, as service rate increases system utilization decreases. This is indicated by the negative slope of the graph.

Figure 12: WinQSB Output Utilization Graph- SA #2

When we put the two graphs from Figures 9 and 12 side by side in Figure 13, we can see that there is an inverse relationship of system utilization when we either increase arrival rate (left graph) system utilization goes up or increase service rate (right graph) system utilization goes down.

Figure 13: WinQSB Output Utilization Graph- SA #1 & #2 Comparison

Acknowledgments

Bibliography:

Chang, Yih-Long, and Kiran Desai. WinQSB Version 2.0. New York: Wiley, 2003. Print

Winston, Wayne L. Operations Research Application and Algorithms. 4th Edition. New York;

Duxbury, 2003. Print

Software Used:

Microsoft Word 2010 (Windows)

Microsoft Excel 2010 (Windows)

Win QSB, Version 2.0 (Windows)

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