Assignment 4
Monte Carlo Simulation Exercise
Equipment Installation at Globus Enterprises
The equipment installation group at Globus Enterprises is about to make a cost estimate to determine how much it will cost to install a back-up generator at a government laboratory facility. Over the years, this group has carried out more than 100 such installations and has developed a database reflecting past experience. Data on the distribution of cost for design work, building effort, and testing effort is provided in Table 1.
| Cheapest
($/%) |
Usual
($/%) |
Expensive
($/%) |
|
| Design | 9,000/30 | 10,000/40 | 12,000/30 |
| Build | 60,000/20 | 70,000/60 | 80,000/20 |
| Test | 18,000/20 | 20,000/50 | 24,000/30 |
| Table 1. Historical Data on Cost Distributions
The data in the table picture the cost of an effort and the percentage of times this cost is achieved. For example, 30% of the time, “Design” cost $9,000; 40% of the time it cost $10,000; 30% of the time it cost $12,000. |
00 16 45 84 18
83 28 82 36 91
95 14 80 68 34
54 55 13 20 70
57 68 61 37 30
09 81 24 55 21
Table 2. Two-digit Random Numbers
Questions
1. Conduct a Monte Carlo simulation to create a distribution portraying total estimated project costs. Employ ten iterations in your computation. Display the distribution graphically.
2. On the average, how much does it cost to carry out this project?
3. What is the standard deviation of the distribution that you generated (use the formula: SD = √Σ(Xi – X-bar)2/N, where SD = standard deviation, √ = square root symbol, Σ = the summation sign, Xi = the ith value of X, X-bar = the mean of the X values, and N = the number of values being considered)? What information does the standard deviation offer us that helps us develop a better understanding of risk in this case? (For more help on computing standard deviation, see below.)
4. Roughly what is the probability that the project will cost more than $105,000?
| Computing standard deviation for following numbers: 8, 4, 10, 7, 6 | ||||||||
| X | X-bar | x – X-bar | Squared | |||||
| 8.00 | 7.00 | 1.00 | 1.00 | |||||
| 4.00 | 7.00 | -3.00 | 9.00 | |||||
| 10.00 | 7.00 | 3.00 | 9.00 | |||||
| 7.00 | 7.00 | 0.00 | 0.00 | |||||
| 6.00 | 7.00 | -1.00 | 1.00 | |||||
| Total = | 35.00 | 20.00 | ||||||
| Average = X-Bar = | 7.00 | 4.00 | (Sum Squared)/N = Variance | |||||
| 2.00 | Sqrt(Variance) = Standard Deviation | |||||||
| Computing standard deviation for following numbers: 6, 7, 5.5, 8, 8.5 | ||||||||
| X | X-bar | x – X-bar | Squared | |||||
| 6.00 | 7.00 | -1.00 | 1.00 | |||||
| 7.00 | 7.00 | 0.00 | 0.00 | |||||
| 5.50 | 7.00 | -1.50 | 2.25 | |||||
| 8.00 | 7.00 | 1.00 | 1.00 | |||||
| 8.50 | 7.00 | 1.50 | 2.25 | |||||
| Total = | 35.00 | 6.50 | ||||||
| Average = X-Bar = | 7.00 | 1.30 | (Sum Squared)/N = Variance | |||||
| 1.14 | Sqrt(Variance) = Standard Deviation | |||||||
| Note that the spread of numbers in the first case above is greater than the second case, so that | ||||||||
| standard deviation in the first case (SD = 2.00) is greater than in the second (SD = 1.14) |
