MATH 528 Operations Models. Example 12.1 Bidding for a Government Project.

MATH 528

Operations Models

Example 12.1

Bidding for a Government Project

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Background Information

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.

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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.

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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.

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BIDDING.XLS

The spreadsheet model appears on the next slide.

This file contains the model.

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The Spreadsheet

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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.

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Developing the Simulation Model -- continued

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.

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Using @RISK

We set the number of iterations to 1000 and the number of simulations to 10.

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.

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Using @RISK

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Using @RISK -- continued

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.

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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.

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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.

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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.

Example 12.2

Tampering with a Stable Process

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Suppose that you are in the business of drilling a tiny hole in the exact center of a square piece of wood.

In the past, the holes you have drilled were, on average, in the center of the wood, and the x- and y-coordinates each had a standard deviation of 0.1 inch.

Also, the drilling process has been stable – that is, the holes average being in the center of the square, and the deviations from the center of the square follow a normal distribution with mean 0 and standard deviation 0.1 inch.

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Background Information -- continued

This mean, for example, that the x-coordinate is within 0.1 inch of the center for 68% of the holes, the x – coordinate is within 0.2 inch of the center for 95% of the holes, and the x-coordinate is with 0.3 inch of the center for 99.7% of the holes.

This describes the inherent variability in the drilling process.

Without changing the hole-drilling process, you must live with this amount of variation.

Now suppose that you drill a hole and its x- and y- coordinates are x=0.1 and y=0.

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Background Information -- continued A natural reaction is to reduce the x-setting of the drill

by 0.1 to correct for the fact that the x-coordinate was too high.

Then if the next hole has coordinates x = -0.2 and y = 0.1, you might try to increase the x-coordinate by 0.2 and decrease the y-coordinate by 0.1.

Deming’s funnel experiment shows that this method of continually readjusting a stable process – he calls it “tampering” – will actually increase the variability of the coordinates of the position where the hole is drilled. In other words, tampering will generally make a process worse!

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Background Information -- continued To illustrate the effects of tampering, Deming

placed a funnel above a target on the floor and dropped small balls through the funnel in an attempt to hit the target.

As he demonstrated, many balls did not hit the target. His goal, therefore, was to make the balls fall as close to the target as possible.

Deming proposed four rules for adjusting the positioning of the funnel.

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Deming’s Rules

1. Never move the funnel.2. After each ball is dropped, move the funnel –

relative to its previous position – to compensate for any error. To illustrate, suppose the target has coordinates (0,0) and the funnel begins directly over the target. If the ball lands at (0.5,.1) on the first drop, we compensate by repositioning the funnel at (0-0.5,0-0.1) = (-0.5,-1). If the second drop has coordinates (1,-2), we then reposition the funnel at (-0.5 – 1, -0.1 – (-2)) = (-1.5, 1.9).

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Deming’s Rules -- continued

3. Move the funnel – relative to its original position at (0,0) – to compensate for any error. For example, if the ball lands at (0.5,0.1) on the first drop, we compensate by repositioning the funnel at (0-0.5, 0-0.1) = (-0.5, -1). If the second drop has coordinates (1, -2), we then reposition the funnel at (0, -1, 0 – (-2)) = (-1,2).

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Deming’s Rules -- continued

4. Always reposition the funnel directly over the last drop. Therefore, if the first ball lands at (0.5,1), we position the funnel, (0.5, 1). If the second drop has coordinates (1, 2), we position the funnel at (1, 2). This rule might be followed, for example, by an automobile manufacturer’s painting department. With each new batch of paint, they attempt to match the color of the previous batch – regardless of whether the previous color was “correct”.

Do you believe any of the latter three rules will outperform rule 1, the “leave it alone” rule? If so, read on – you might be surprised.

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Solution

To see how these rules work, we assume that the x-coordinate on each drop is normally distributed with mean equal to the x-coordinate of the funnel position and standard deviation of 1.

A similar statement holds for the y-coordinate. Also, we assume that the x- and y- coordinates are selected independently of one another. These assumptions describe the inherent variability in the process of dropping the balls.

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Solution -- continued

To see how the rules work, let F0, X0, F1 be respectively, the x-coordinates of the funnel position on the previous drop, the outcome of the previous drop, and the repositioned funnel position for the next drop.

Then rule 1 never repositions, so that F1 = F0. Rule 2 repositions relative to the previous funnel position, so that F1 = F0 – X0. Rule 3 repositions relative to the original position (at 0), so that F1 = 0 – X0 = -X0. Finally rule 4 repositions at the previous drop, so that F1 = X0. Similar questions hold for the y-coordinate.

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Solution -- continued

For the simulation model, we simulate 50 consecutive drops of the ball from each of the four rules.

Our single output measure is the (straight-line) distance of the final drop from the target.

A rule is presumably a good one if the mean distance is small and the standard deviation of this distance is also small.

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FUNNEL.XLS Given the repositioning equations for the

rules, the simulation model is straightforward. In fact, we use a RISKSIMTABLE function to

test all four rules simultaneously. The spreadsheet model appears on the next

slide. This file contains the model.

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The Spreadsheet

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Developing the Simulation Model The model can be developed with the following steps.

Rule. Enter the formula =RISKSIMTABLE({1,2,3,4}) in cell B3 to indicate that we want to simulate all four rules. Note that if individual values are listed in RISKSIMTABLE, they must be enclosed in curly brackets. No curly brackets should be used if the list is referenced by a range.

Position funnel. Enter 0 in cells B7 and C7 to indicate that the original funnel position is above the target at (0,0). Then implement the positioning equations by entering the formula =IF(Rule=1,B7,IF(Rule=2,B7-D7,IF(Rule=3,-D7,D7))) in cell B8 and copying it to the range B8:C56. Note how this formula references the location of the previous drop. The IF function captures the logic for all four rules.

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Simulate drops Simulate the positions of the drops by entering the formula =RISKNORMAL(B7,1) in cell D7 and copying it to the range D7:E56. This says that the ball’s drop position is normally distributed with mean equal to the funnel’s position and standard deviation 1.

Distance. Calculate the final distance from the target in cell C58 with the formula =RISKOUTPUT( ) +SQRT(SUMSQ(D56:E56)). Here we have used SUMSQ function to get the sum of squares for the distance formula. We have also indicated that this is an output cell for @Risk.

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Using @RISK


Selected summary measures for the final distance from the target for all four rules appears in the table shown here.

We also show histograms of this distance for rules 1, 2, 3 on the next three slides.

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Using @RISK

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Using @RISK

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Using @RISK

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These results prove Deming’s point about tampering. Rule 2 might not appear to be too much worse than

rule 1, but its mean distance and standard deviation of distances are both about 40% higher than rule 1.

Rules 3 and 4 are disastrous. Their mean distances are more than seven times higher than for rule 1, and their standard deviations are also much higher.

The moral of the story, as Deming preached, is that you should not tamper with a stable process. If the process is not behaving as desired, then fundamental changes to the process are required, not a lot of tinkering.

Example 12.3

Order Due Dates at Wozac

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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.

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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

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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.

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

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

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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.

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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.

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Developing the Simulation Model

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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.

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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.

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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.

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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.

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Using @RISK


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.

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Using @RISK

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Using @RISK

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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.

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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 12.4

Component Redundancy at Failsafe

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The FailSafe Company operates a machine that consists of 3 identical module in series.

This means that the machine works as long as all the modules work.

Each of modules depends on a specific component. If the component fails, the module fails – and the machine fails.

To extend the time until machine failure, redundancy is built in at the component level.

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Specifically, 15 identical components are placed in parallel in each module.

This means that a module works as long as at least one component in that module is working.

Each component lasts a random time before failure, where these times are probabilistically independent and each has mean 100 hours and standard deviation 20 hours.

FailSafe wants to use simulation to find the distribution of the time until machine failure.

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Solution Let Cij be the time until failure for component j in

module i. Also let Mi be the time until failure for module i, and let T be the time until machine failure. Then from the system configuration we haveMi = max(Ci1, Ci2,…,Ci15) and T = min (M1,M2,M3).

In other words, a module lasts as long as the best of its components, and the machine lasts as long as the worst of its modules.

Therefore, all we need to do is generate the random component times, the Cij’s, and use Excel’s MAX and MIN functions to find the time until machine failure, T.

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Choosing a Probability Distribution We are told that times until component failure

have mean 100 and standard deviation 20, but which distribution of component failure times should we use?

Should it be symmetric and bell shaped, or should it be skewed in one direction or the other?

We will test four possible distributions: normal, lognormal, gamma and Weibull.

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Choosing a Probability Distribution -- continued The normal is the symmetric bell-shaped distribution.

The other three are less well known, but they are frequently used to model times until failure.

They all have the attractive property that they generate positive values only.

To see the possibilities, we open @Risk’s Model window and select the Insert/Distribution Window menu item. This allows us to choose from many distributions, including the four we are recommending here, and to adjust their parameters.

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Choosing a Probability Distribution -- continued For the normal and lognormal, the

parameters are the mean and standard deviation directly.

For the gamma and Weibull, the parameters are called alpha and beta. These parameters control the exact location and shape of the distributions, but they are not the mean and standard deviation.

If we want them to have a mean 100 and standard deviation 20, we have to manipulate alpha and beta appropriately.

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Choosing a Probability Distribution -- continued The lognormal, gamma, and Weibull distributions with

mean 100 and standard deviation 20 appear on the next three slides.

They all look similar to each other and to the normal distribution, and only the Weibull indicates some “obvious” skewness.

If we were fitting distributions to historical failure time data, these might all provide very good fits.

The question, then is whether it matters which we use in the simulation. Will they all give approximately the same results? We will see shortly.

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Choosing a Probability Distribution -- continued

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REDUNDANCY.XLS The simulation model appears on the next


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The Spreadsheet

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Developing the Simulation Model It can be developed as follows.

Machine configuration. We stated that there are 3 modules in series, each consisting of 15 components. Enter these values in cells B4 and B5. We actually develop a slightly more general model, where the machine can consist of up to 10 modules in series, each of which can have up to 20 modules in parallel.

Parameters of probability distributions. Enter the parameters of the four candidate probability distributions in the shaded range. These are the parameters from the @Risk Model window that yield means of 100 and standard deviations of 20.

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Distribution to use. We can actually test all four distributions simultaneously by using a RISKSIMTABLE function. DO this by entering the formula =RISKSIMTABLE({1, 2, 3, 4}) in cell B16. Then get the name and parameters of this distribution in cells B17 to B19 with the formulas =HLOOKUP(Dist,Ltable,2), =HLOOKUP(Dist,Ltable,3) and =HLOOKUP(Dist,Ltable,4) The latter of these cells, range-named Par 1 and Par 2, will supply the parameters for the random values in the next step.

Component times. In simulation section we enter 1-10 along the top and 1-20 along the sides for the maximum number of modules and components, respectively, that our model can handle. We want to generate enough component lifetimes in this table for the number of modules and components in cells B4 and B5.

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Do this with the following IF Formula, entered in cell B20 and copied to the range B20:K39: IF(AND($A24<=Ncomps,B$23<=Nmods), IF(Dist=1,RISKNORMAL(Par1,Par2), IF(Dist=2,RISKLOGNORM(Par1,Par2),IF(Dist=3,RISKGAMMA(Par1,Par2),RISKWEIBULL(Par1,Par2)))),””) Although this looks intimidating, it is really straightforward. The AND condition checks whether the component and module indices are less than or equal to the values in the Ncomps and Nmods cells. If they aren’t blank is entered. Otherwise, the appropriate @Risk function is called with the parameters in the Par1 and Par2 cells.

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Module times. Calculate the times until module failures in row 45 by entering the formula =IF(B23<=Nmods,MAX(B24:B43),””) in cell B45 and copying it across to column K. Note that the MAX (or MIN) of a range that includes blanks ignores the blanks.

Machine time. Calculate the time until machine failure in cell B47 with the formula =RISKOUTPUT()+MIN(ModLives) This is the only cell we designate as an @Risk output cell.

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Using @RISK


After running @Risk, we form the histograms of time until machine failure, one for each distribution, on the following three slides.

Does the input distribution affect the distribution of machine lifetime? At first glance, the answer appears to be “No”.

The four distributions have similar shapes and their means and standard deviations are similar.

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Using @RISK

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Using @RISK

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Using @RISK

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Using @RISK

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However, means and standard deviations might not be as relevant in a real situation as worst-case results.

Therefore, it might be better to compare maximums or 95th percentiles.

Admittedly, even these do not vary greatly but there are differences.

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Conclusions

We conclude from this example that the input distribution(s) can make a difference in the results, particularly when best-case or worst-case results are of primary interest.

Therefore, it pays to spend some time in real simulation applications fitting distributions to any relevant historical data that exist.

Fortunately, as we saw in the previous chapter, @Risk’s fitting capabilities make this fairly easy.

Example 12.5

Room Construction Project

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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.

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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

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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.

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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.

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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.

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The Pert Distribution -- continued

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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.

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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.

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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.

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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.

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PROJECTSIM.XLS The spreadsheet model appears on the next


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The Spreadsheet

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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.

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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)

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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.

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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.

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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.

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Using @RISK

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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.

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

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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.

MATH 528 Operations Models. Example 12.1 Bidding for a Government Project.

Documents

bid miller

millers bid

competitors bids

spreadsheet slide

lowest bid

government project slide

simulation model

millers possible bids