Consider the manufacturing process at Wonder Shed Inc., a manufacturer of storag
ID: 3176551 • Letter: C
Question
Consider the manufacturing process at Wonder Shed Inc., a manufacturer of storage sheds. The manufacturing process involves the procurement of sheets of steel that will be used to form both the roof and the base of each shed. A list of activities fabricate a roof, fabricate a base, and assemble a shed is shown in Table 4.1 below:
Punch the base
We use a beta distribution to represent an activity duration in minutes. We assume that the activity durations are independent. We denote the optimistic, the most likely and the pessimistic times by a, m and b. The mean time and the variance of each activity duration are calculated as follows.
Activity
a
m
b
Mean
Variance
1
15
20
30
20.83333
6.25
2
25
35
40
34.16667
6.25
3
20
25
30
25
2.777778
4
5
10
20
10.83333
6.25
5
10
20
25
19.16667
6.25
6
25
30
40
30.83333
6.25
7
10
15
30
16.66667
11.11111
8
35
40
60
42.5
17.36111
a.
The process contains two paths:
Path 1 : Start 1 3 5 7 8 End
Path 2 : Start 1 2 4 6 7 8 End
The longer path in the mean time is Path 2. By Central Limit Theorem, we assume that the duration of Path 2 is normally distributed. What is the probability that the duration of Path 2 does not exceed 170 minutes?
P(X1<170)=
b. Assume that the durations of Path 1 and Path 2 are independent (even though this assumption does not seem to be so reasonable.) What is the probability that both of Path 1 and Path 2 are done within 150 minutes.
P(X1<150 and X2<150)=
Activity 1 Separate the roof and base materials 2Punch the base
3 Punch the roof 4 Form the base 5 form the roof 6 Subassemble the base 7 Assemble 8 InspectExplanation / Answer
Using given data the duration of path 2 is 155.83 i.e. mean = 155.83
Variance of path 2 = 53.47
std. dev. = 7.31
P(X1 < 170) = P(z < (170-155.83)/(53.47)) = P(z < 0.265) = 0.9737
P(X1<150 and X2<150) = P(X2 < 150) as P2 is the critical path. If P2 is completed in 150 minutes P1 will be completed in same time.
P(X2 < 150) = P(z < (150-155.83)/(53.47)) = P(z < -0.1090) = 0.2126
Mean Variance Std. dev. Path 1 1-3-5-7-8 124.1667 43.75 6.614378 Path 2 1-2-4-6-7-8 155.8333 53.47222 7.31247