Risk & quality management
Wright
MGT553_Assignment4.rtf
1
MGT 5 53. Risk and Quality Mangement
Assignment 4
Monte Carlo Simulation Exercise
E quipment 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
- Conduct a Monte Carlo simulation to create a distribution portraying total estimated project costs. Employ ten iterations in your computation. Display the distribution graphically.
- On the average, how much does it cost to carry out this project?
- 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.)
- 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)
