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Example 12.4 Operations Models. 12.112.1 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | 12.10 | 12.11 | 12.12 | 12.13 | 12.14 | 12.15 | 12.16 | 12.1712.212.312.512.612.712.812.912.10.

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Presentation on theme: "Example 12.4 Operations Models. 12.112.1 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | 12.10 | 12.11 | 12.12 | 12.13 | 12.14 | 12.15 | 12.16 | 12.1712.212.312.512.612.712.812.912.10."— Presentation transcript:

1 Example 12.4 Operations Models

2 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Background Information n The FailSafe Company operates a machine that consists of 3 identical module in series. n This means that the machine works as long as all the modules work. n Each of modules depends on a specific component. If the component fails, the module fails – and the machine fails. n To extend the time until machine failure, redundancy is built in at the component level.

3 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Background Information -- continued n Specifically, 25 identical components are placed in parallel in each module. n This means that a module works as long as at least one component in that module is working. n 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. n FailSafe wants to use simulation to find the distribution of the time until machine failure.

4 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Solution n Let C ij be the time until failure for component j in module i. Also let M i be the time until failure for module i, and let T be the time until machine failure. Then from the system configuration we have M i = max(C i1, C i2,…,C i15 ) and T = min (M 1,M 2,M 3 ). n 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. n Therefore, all we need to do is generate the random component times, the C ij s, and use Excels MAX and MIN functions to find the time until machine failure, T.

5 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Choosing a Probability Distribution n 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? n Should it be symmetric and bell shaped, or should it be skewed in one direction or the other? n We will test four possible distributions: normal, lognormal, gamma and Weibull.

6 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Choosing a Probability Distribution -- continued n 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. n They all have the attractive property that they generate positive values only. n To see the possibilities, we Model windows 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.

7 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Choosing a Probability Distribution -- continued n For the normal and lognormal, the parameters are the mean and standard deviation directly. n 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. n If we want them to have a mean 100 and standard deviation 20, we have to manipulate alpha and beta appropriately.

8 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Choosing a Probability Distribution -- continued n The lognormal, gamma, and Weibull distributions with mean 100 and standard deviation 20 appear on the next three slides. n They all look similar to each other and to the normal distribution, and only the Weibull indicates some obvious skewness. n If we were fitting distributions to historical failure time data, these might all provide very good fits. n 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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12 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | REDUNDANCY.XLS n The simulation model appears on the next slide. n This file contains the model.

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14 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model n 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 Model window that yield means of 100 and standard deviations of 20.

15 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued –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.

16 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued –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)))),) 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 arent blank is entered. Otherwise, the function is called with the parameters in the Par1 and Par2 cells. –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.

17 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Developing the Simulation Model -- continued –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 output cell.

18 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | n We set the number of iterations to 1000 and the number of simulations to 4. n After we form the histograms of time until machine failure, one for each distribution, on the following three slides. n Does the input distribution affect the distribution of machine lifetime? At first glance, the answer appears to be No. n The four distributions have similar shapes and their means and standard deviations are similar.

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23 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | continued n However, means and standard deviations might not be as relevant in a real situation as worst-case results. n Therefore, it might be better to compare maximums or 95 th percentiles. n Admittedly, even these do not vary greatly but there are differences.

24 | 12.2 | 12.3 | 12.5 | 12.6 | 12.7 |12.8 | 12.9 | | | | | | | | Conclusions n 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. n Therefore, it pays to spend some time in real simulation applications fitting distributions to any relevant historical data that exist. n Fortunately, as we saw in the previous fitting capabilities make this fairly easy.


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