This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Forrestel and Slagh, 1 Part I: Deployment Time Calculations
We developed a simulation that models the time it takes to deploy a unit. In order to do that, we created inverse CDF models for each of the five phases that allowed us to enter a random percentile, thus giving us a simulated time. After 5000 iterations, we were able to calculate the average and standard deviation of the time it takes to deploy a unit over the multiple phases. For example, the Inverse CDF for the deployment phase was equal to (122.85*LN(1 %Tile)). In Mathematica, mu and the standard deviation were calculated. Predeployment was the most volatile of the five stages, with values ranging from 46 to over 1100 hours in time. It seems that because of such a large variance, a commander could greatly reduce his total time by making the predeployment process more efficient. Next, we calculated: 5=1 Di , 5=1 i i
5 x _ Di , and x .
T _ We found i =1 Di = 209.61, the sum of the values from the 5 stages (Mathematica). 5=1 x Di = 208.944, the sum of sample means from the 5 stages after 5000 iterations. i
_ x _ T = 208.944, the sample mean of T, which is equal to 5=1 x over 5000 iterations. i
Di _ Forrestel and Slagh, 2 T = 209.61, because T = 5=1Di , and T is the expected value. The sample mean rapidly i
approaches this expected value as we increase the number of iterations, so this became our mu. Part II: A Powerful Model for Calculating Deployment Time
We created an EDF for the array of time values to deploy a unit over 5 stages for 5000 iterations. We fit Normal, Gamma, and Weibull distributions to the EDF in order to see what distribution produced the smallest SSE. Deployment Time
1.2 From this graph as well as the SSE values for EDF Normal Gamma Weibull 1 0.8 Proportion 0.6 the three distributions, it seems that the Gamma distribution provides the best fit 0.4 0.2 0 0 200 400 600 Tim e (Hours) 800 1000 1200 because the Sum Squared Error is the lowest, thus indicating the least deviation from the EDF values. Next, we fit the Normal and Gamma distributions to the EDF so that they had the same mean and variance as our simulated sample. This was necessary because we already knew the mean and variance to deploy the unit, and we only needed to find which distribution could produce a smaller SSE for the given data. Also, we couldn't use the nonfixed mu and sigma models from above because it is incorrect to simply to assume a new mean and variance when we have already calculated the data through our concept simulation. Forrestel and Slagh, 3
Deployment EDF, Fixed mu and Variance
1.2 We compared the SSE with the given mean and variance models to the previous models where mean and 1 0.8 Proportion Prop. 0.6 Normal Model Gamma Model 0.4 0.2 0 0 200 400 600 Time (Hours) 800 1000 1200 variance were not kept constant. It was evident that the SSE of the nonfixed mean and variance models were lower. However, these models are not as realistic as the fixed mean and variance models, because we have already calculated T as well as the variance, and they are not close to T given by the non fixed models. In this case, the Gamma model still provided the least deviation from the mean. For the Gamma model, it was necessary to calculate alpha and beta. In an exponential distribution, (alpha)(beta) = mu and (alpha)(beta) = variance so, (alpha)(beta) = 209.61 and variance = 15939. So, Alpha = 209.61/beta then, 209.61*beta = 15939 Alpha = 2.757, Beta = 76.04 Part III: Cost Analysis/Pleasing the Division Commander
Our Division Commander wanted to decrease deployment time by at least 18 hours and we were given a budget of $125,000 in order to do so. One way to decrease deployment time was to hire additional civilian contract workers during either phase one, phase two, or both. Forrestel and Slagh, 4 Also, we had the option of flying the troops to the POE instead of bus for an addition $40,000, which would reduce deployment time by 12 hours alone. In order to maximize time saved per dollar spent, we created a chart with all of the possible scenarios and their costs. From here, we translated the scenarios into another table which showed the total time saved. This way, we see which scenarios allow us to meet the 18 hour time reduction within the $125,000 budget. Our necessary assumptions included that if we were to use the plane, the total hours saved from the one or two sets of contractors had to be at least 6, in order to get to 18. Again, if the plane was used, we had to be at or less than $85,000 for the price of civilian contractors. If we did not use the plane, it is assumed that we had to make it to 18 saved hours by spending less than $125,000. We knew that it would be important to get the Division Commander exactly what he/she wanted: the most time saved (at least 18 hours), while staying within the $125,000 budget. But it is also important to realize that a commander will always be happy to have a few options to choose from, and in this situation, there are some alternatives worth considering. Option 1: Most Hours Saved, Within Budget Forrestel and Slagh, 5 This was the Division Commander's original intent, to cut at least 18 hours off of the total time. For $125,000, the unit can save 22.56 hours by employing 7 extra phase 1 workers, 1 extra phase 2 worker, and by purchasing the airplane flights. Option 2: Meets 18 Hour Requirement, Least Costly Under these conditions, the mission saves the required 18 hours while spending the least amount of money possible. For $85,000, the unit can save 18.52 hours by employing 3 phase 1 workers, 1 phase 2 workers, and by using the flight. The extra 4.02 hours saved by option 1 come at a cost of almost $10,000/extra hour saved. Option 3: Fear of Flying If the Division Commander decides that flying the troops to the mission area is too dangerous or not the best course of action, the requisite hours can be met within budget without flying. For $120,000, the unit can save 20.73 hours by employing 14 extra phase 1 workers. Again, it seems that the efficiency of phase 1, or predeployment, is crucial to expediting the operation. Documentation We received no additional assistance on this project. ...
View Full Document
- Spring '06