math 528 operations models. example 12.3 order due dates at wozac

MATH 528

Operations Models

Example 12.3

Order Due Dates at Wozac

3

Background Information

The Wozac Company is a drug manufacturing company.

Wozac has recently accepted an order from its best customer for 8000 ounces of a new miracle drug, and Wozac wants to plan its production schedule to meet the customer’s promised delivery date of December 1, 2000.

There are three sources of uncertainty that make planning difficult.

4

Background Information -- continued First, the drug must be produced in batches, and

there is uncertainty in the time required to produce a batch, which could be anywhere from 5 to 11 days. This uncertainty is described by the discrete distribution of this table.

Distribution of Days to Complete a Batch

Days Probability

5 0.05

6 0.10

7 0.20

8 0.30

9 0.20

10 0.10

11 0.05

5

Background Information -- continued Second, the yield (usable quantity) from any batch is

uncertain. Based on historical data, Wozac believes the yield can be modeled by a triangular shaped distribution with minimum, most likely, and maximum values equal to 600, 1000, 1100.

Third, all batches must go through a rigorous inspection once they are completed. The probability that a typical batch passes this inspection is only 0.8. With probability 0.2, the batch fails inspection, and none of it can be used to help fill the order.

6

Background Information -- continued Wozac wants to use simulation to help decide

how many days prior to the due date it should begin production.

7

Solution

The idea is to simulate successive batches – their days to complete, their yield, and whether they pass inspection – and keep a running total of the usable ounces obtained so far.

We then use IF functions to see whether the order is complete or another batch is required.

We simulate only as many batches as are required to meet the order, and we keep track of the days required to produce all of these batches.

In this way we can “back up” to see when production must begin to meet the due date.

8

Developing the Simulation Model The completed model appears on the next slide. It

can be developed as follows. Inputs. Enter all inputs in the shaded cells. Batch indexes. We do not know ahead of time how

many batches will be required to fill the order. We want to have enough rows in the simulation to cover the worst case that is likely to occur. After some experimentation we found that 25 batches are almost surely enough. Therefore, enter the batch indexes 1-25 in column A of the simulation section. The idea, then, is to fill the entire range B29:F53 with formulas. However, we will use IF functions in these formulas so that if enough has already been produced to fill the order, blanks are inserted into the remaining cells.

9

Developing the Simulation Model

10

Developing the Simulation Model -- continued

Days for batches. Simulate the days required for batches in column B. First, enter the formula =RISKDISCRETE(Days,Probs) in cell B29. Then enter the general formula =IF(OR(F29=“Yes”,F29=“”),””,RISKDISCRETE(Days,Probs)) in cell B30 and copy it down to cell B53. Note how the IF function enters a blank in this cell if either of two conditions is true; the order was just completed in the previous batch or it has been completed for some time. Similar logic will appear in later formulas.

Batch yields. Simulate the batch yield in column C. First, enter the formula =RISKTRIANG(B23,C23,D23) in cell C29. Then enter the general formula =IF(OR(F29=“Yes”,F29=“”),””,RISKTRIANG($B$23,$C$23,$D$23)) in cell C30 and copy it down to C53.

11


Pass inspection? Check whether each batch passes inspection with the formulas =IF(RAND()<PrPass,”Yes”,”No”) and IF(OR(F29=“Yes”, F29=“”),””,IF(RAND()<PrPass,”Yes”,”No”)) in cells D29 and D30 and copy the latter down to cell D53. Note that we could use @Risk’s RISKUNIFORM(0,1) function instead of RAND(), but there is no advantage to doing so.

12


Order filled? We keep track of the cumulative usable production and whether the order has been filled in column E and F. First, enter the formulas =IF(D29=“Yes”,C29,0) and =IF(E29>=AmtReqd,”Yes”,”Not yet”) in cells E29 and F29 for batch 1. Then enter the general formulas =IF(OR(F29=“Yes”,F29=“”),””,IF(D30=“Yes”,C30+E29,E29)) and =IF(OR(F29=“Yes”,F29=“”),””,IF(E30>=AmtReqd,“Yes”,”Not yet”)) in cells E30 and F30, and copy them down to row 53.

13


Note that the entry in column F is “Not enough” if the order is not yet complete. In the row that completes, the order, it changes to “Yes”, and then it is blank in succeeding rows.

Summary measures. Calculate the batch and days required in cell I28 and I29 with the formulas =RISKOUTPUT() + COUNT(B29:B53) and =RISKOUTPUT()+SUM(B29:B53) These are the two cells we will use as output cells for @Risk. Also, calculate the day the order should be started to just meet the due dates in cell I30 with the formula =DueDate-I29 This formula uses date subtraction to find an elapsed time. Of course, it assumes that production occurs every day of the week, which we will assume.

14

Using @RISK

We set the number of iterations to 1000 and the number of simulations to 1.

After running @Risk, we obtain the histograms of the number of batches required and the number of days required on the next two slides.

How should Wozac use this information? The key question are how many batches will be required and when to start production.

We have entered several of @Risk’s statistical functions directly in the spreadsheet to help answer these questions.

15

Using @RISK

16

Using @RISK

17

Using @RISK -- continued

For the first question, we use the formula =RISKMAX(I28) in cell I33. It shows that the worst case from the iterations, in terms of batches required is 20 batches.

We can answer the second question in two ways. First, we can calculate summary measures for days

required and then back up from the due date. We do this in the range I35:J39. The formulas in column I are =INT(RISKMEAN(DaysReqd)), =RISKMIN(DaysReqd), =RISKMAX(DaysReqd), =RISKPERCENTILE(DaysReqd,0.05) and =RISKPERCENTILE(DaysReqd,0.95)We then subtract each of these from the due date to obtain the potential starting dates in column J. Wozac should realize the pros and cons of these starting dates.

18


Alternatively, we can use @Risk’s RISKTARGET function to find the probability of meeting the due date for any starting date, such as those in the range H42:H46. We enter the formula =RISKTARGET(DaysReqd,DueDate-H42) in cell I42 and copy it down. This function returns the fraction of iterations where the value in the first argument is less than or equal to the value in the second argument.

What is our recommendations to Wozac? We suggest going with the 95th percentile – begin

production on August 2. Then there is only a 5% chance of failing to meet the due date.

Example 12.5

Room Construction Project

20


Tom Lingley, an independent contractor, has agreed to build a new room on an existing house.

He plans to begin work on Monday morning June 1. The main question is when will he complete his work,

given that he works only weekdays. The owner of the house is particularly hopeful that

the room will be ready by Saturday, June 27, that is, in 20 or fewer working days.

The work proceeds in stages, labeled A through J, as summarized in the table on the next slide.

21

Background Information -- continued

Three of the activities, E, F, and G, will be done by separate independent subcontractors. The expected durations of the activities are shown in the table.

Activity Time Data

Description Index Predecessors Expected Duration

Prepare foundation A None 4

Put up frame B A 4

Order custom windows C None 11

Erect outside walls D B 3

Do electrical wiring E D 4

Do plumbing F D 3

Put in duct work G D 4

Hang dry wall H E, F, G 3

Install windows I B, C 1

Paint and clean up J H 2

22

Background Information -- continued However, these are only best guesses. Lingley knows that the actual activity times

can vary because of unexpected delays, worker illnesses, and so on.

He would like to use a computer simulation to see How long the project is likely to take, and How likely it is that the project will be

completed by the deadline, and Which activities are likely to be critical.

23

Solution

We first need to choose distributions for the uncertain activity times.

Then, given any randomly generated activity times, we will illustrate a method for calculating the length of the project and identifying the activities on the critical path.

24

The Pert Distribution

As always, there are several reasonable candidate probability distributions we could use for the random activity times.

Here we illustrate a distribution that has become popular in project scheduling, called the Pert distribution. As shown on the next slide, it is “rounded” version of the triangular distribution that is specified by three parameters: a minimum value, and a maximum value.

25

The Pert Distribution -- continued

26

The Pert Distribution -- continued

The distribution in the figure uses the values 7, 10, and 19 for these three values, which implies a mean of 11. We will use this distribution for activity C.

Similarly, for the other activities, we choose parameters for the Pert distribution that lead to the means in the table.

In reality, it would be done the other way around. The contractor would estimate the minimum, most likely, and maximum parameters for the various activities, and the means would follow from these.

27

Developing the Simulation Model The key to the model is representing the project

network in activity-on-arc form, as in the diagram below, and then finding Ej for each j, where Ej is the earliest time we can get to node j.

We stated in Chapter 5 that when the nodes are numbered so that all arcs go from lower-numbered nodes to higher-numbered nodes, we can calculate the Ej’s iteratively, starting with E1=0, with the equation Ej = max(Ei + tij).

Here, the maximum is taken overall arcs leading into node j, and tij is the activity time on such an arc.

28

Developing the Simulation Model -- continued Then En is the time to complete the project, where n

is the index of the finish node. This will make it very easy to calculate the project

length. We also need a method for identifying the critical

activities for any given activity times. By definition, an activity is critical if a small increase

in its activity time causes the project time to increase. Therefore, we will keep track of two sets of activity times, and associated project times.

29

Developing the Simulation Model -- continued The first uses the simulated activity times.

The second adds a small amount, such as 0.001 day, to a “selected” activity’s time. By using the RISKSIMTABLE function with a list as long as the number of activities, we can make each activity the “selected” activity in this method.

30

PROJECTSIM.XLS The spreadsheet model appears on the next

slide. This file contains the model.

31

The Spreadsheet

32

Developing the Simulation Model -- continued The details of the model are followed.

Inputs. Enter the parameters of the Pert activity time distributions in the shaded cells and the implied means next to them. As discussed above, we actually chose the minimum, most likely, and maximum values while in @Risk’s Model window to achieve the means in the activity table. Note that some of these distributions are symmetric about the most likely value, whereas other are skewed.

Activity time. Generate random activity times in column I by entering the formula =RISKPERT(E5, F5, G5) in cell I5 and copying it down.

33


Augmented activity times. We want to successively add a small amount of each activity’s time to determine whether it is on the critical path. To do this, enter the formula =RISKSIMTABLE({1, 2, 3, 4, 5, 6, 7, 8, 9, 10}) in cell B16. Then enter the formula =I5+IF(Index=C5,0.001,0) in cell J5 and copy it down.

Event times. We want to use the equation to calculate the node event times in the range B20:B27. There is no quick way to enter the required formulas. We need to use the project network as a guide for each node. Begin by entering 0 in cell B20. Then enter the appropriate formulas in the other cells. For example, the formulas in cells B22, B23, and B27 are =B21+I6, =MAX(B20+I7,B21+I6) and =RISKOUTPUT( )+MAX(B23+I13,B26+I14)

34


To understand these, note the node 3 has only one arc leading into it, and this arc originates at node 2. No MAX is required for this mode’s equation. In contrast, node 4 has two arcs leading into it, from nodes 1 and 2, so a MAX is required. Similarly, node 8 requires a MAX because it has two arcs leading into it. Also, it is the finish node, so we designate its event time cell as an @Risk output cell – it contains the time to complete the project.

Augmented even times. Copy the formulas in the range B20:B27 to the range C20:C27 to calculate the even times when the selected activity’s time is augmented by 0.001.

35


Project time increase?. To check whether the selected activity’s increased activity time increases the project time, enter the formula =RISKOUTPUT( )+IF(C27>B27,1,0) If this calculates to 1, then the selected activity is critical for these particular activity times. Otherwise, it is not. Note that this cell is also designated as an @Risk output cell.

36

Using @RISK


After running @Risk, we request the histogram of project times shown on the next slide.

Recall from the example in Chapter 5 that when the activity times are not considered random, the project time is 20 days. Now it varies from a low of 16.09 days to a high of 25.20 days, with an average of 20.38 days.

Although the 5th and 95th percentiles appear in the figure, it might be more interesting to Tom Lingley to see the probabilities of various project times being exceeded.

37

Using @RISK

38


For example, we entered 20 in the Left X box next to the histogram. The Left P value implies there is about a 59% chance that the project will not be completed within 20 days.

Similarly, the values in the Right X and Right P boxes imply that the chance of the project lasting longer/ than 23 days is slightly greater than 5%.

This is certainly not good news for Lingley, and he might have to resort to the crashing we discussed earlier.

39


The summary measures for the B29 output cell appears in this table.

40


Each “simulation” in this output represents one selected activity being increased slightly.

The Mean column indicates the fraction of iterations where the project time increases as a result of the selected activity’s time increase.

Hence, it represents the probability that this activity is critical.

For example, the first activity (A) is always critical, the third activity (C) is never critical, and the fifth activity (E) is critical about 44% of the time.More specifically, we see that the critical path always includes activities A, B, D, H, J, and one of the three “parallel” activities E, F, and G.

Example 12.1

Bidding for a Government Project

42


The Miller Construction Company is trying to decide whether to make a bid on a construction project.

Miller believes it will cost the company $10,000 to complete the project, and it will cost $350 to prepare a bid.

Four potential competitors are going to bid against Miller. The lowest bid will win the contract.

Based on past history, Miller believes that each competitor’s bid will be a multiple of its cost to complete the project, where this multiple has a triangular distribution with minimum, most likely, and maximum values 0.9, 1.3, and 2.5.

43

Background Information -- continued These four competitor’s bids are also

assumed to be independent of one another. If Miller decides to prepare a bid, then it has

decided that its bid amount will be a multiple of $500 in the range $10,500 to $15,000.

The company wants to use simulation to determine which strategy to use to maximize its expected profit.

44

Solution

The logic is straightforward. We first simulate the competitor’s bids. Then

for any bid Miller makes, we see whether Miller wins the contract, and if so, what its profit is.

45

BIDDING.XLS

The spreadsheet model appears on the next slide.

This file contains the model.

46

The Spreadsheet

47

Developing the Simulation Model

The model can be developed with the following steps. Inputs. Enter the inputs in the shaded cells. These

include Miller’s costs, Miller’s possible bids, and the parameters of the triangular distribution for the competing bids.

Miller’s bid. We can test all of Miller’s possible bids simultaneously with the RISKSIMTABLE function. Do this in cell B15 with the formula =RISKTABLE(PossibleBids).

Competitor’s bids. Generate random bids for the four competitors in the CompBid range by entering the formula =RISKTRIANG($B$9,$B$10,$B$11)*ProjectCost in cell B15 and copying across. Of course Miller will not see these other bids until it has submitted its own bid.

48


Win contract?. See whether Miller wins the bid by entering the formula RISKOUTPUT( )+IF(MillerBid<MIN(CompBids),1,0)in cell B23. Here, 1 means that Miller wins the bid, and 0 means a competitor wins the bid. Note that we are designating this cell as an output cell for @RISK.

Miller’s profit. If Miller submits a bid, the bid cost is lost for sure. Beyond that, the profit to Miller is the bid amount minus the cost of completing the project if the bid is won. Otherwise, Miller makes nothing. So enter the formula=RISKOUTPUT( ) +IF(B23=1,MillerBid-ProjectCost,0)-BidCost in cell C23. We also designate this as an output cell.

49

Using @RISK


The summary results appear on the next slide. For each simulation – that is, each bid amount –

there are two outputs: 1 or 0 to indicate whether Miller wins the contract and Miller’s profit.

A little thought should convince you that each of these can have only two possible values for any bid amount.

50

Using @RISK

51


For example, if Miller bids $12,000, it will either win or lose the contract, and its profit will be either $1650 or -$350.

This is reflected in the histogram of profit for this bid amount shown on the next slide, where there are only two bars.

The two possible values of the outputs appear in the Minimum and Maximum columns of the table on the previous slide.

52


53


The Mean column, on the other hand, indicates the average of these values over the 1000 iterations.

For example, the mean of 0.545 for the “Win Bid?” output for simulation #4 indicates that Miller wins the contract on 54.5% of the iterations when bidding $12,000.

The mean profit of $740 for this bid amount is simply a weighted average of the two possible profits, $1650 and -$350.

54


Specifically, you can check that it is 0.545(1650)+0.455(-350) = 740. The other means in the output can be interpreted similarly.

What should Miller bid? First, it is clear that Miller should bid. Not

bidding means no profit, Whereas all of the possible bids except for the last one lead to a positive expected profit with at most a $350 loss.

55


If Miller is an EMV maximizer, as we discussed in Chapter 10, then the $12,000 bid should be chosen because it has the highest mean profit.

However, if Miller is risk averse, a smaller bid amount might be attractive.

As the bid amounts increase, the upside potential is greater, but the chance of not winning the bid and losing $350 increases.

math 528 operations models. example 12.3 order due dates at wozac

Documents