Concept explainers
Ross Hopkins, president of Hopkins Hospitality, has developed the tasks, durations, and predecessor relationships in the following table for building new motels. Draw the AON network and answer the questions that follow.
a) What is the expected (estimated) time for activity C?
b) What is the variance for activity C?
c) Based on the calculation of estimated times, what is the critical path?
d) What is the estimated time of the critical path?
e) What is the activity variance along the critical path?
f) What is the probability of completion of the project before week 36?
a)
To determine: The expected time for activity C.
Introduction: In the activity on nodes (AON) project network diagram, the nodes denote activities and the arrows show only the precedence and succession sequence.
Answer to Problem 17P
The expected time for activity C is 12 weeks.
Explanation of Solution
Given information:
Activity | Immediate predecessor | Time estimates | ||
Optimistic | Most likely | Pessimistic | ||
A | 4 | 8 | 10 | |
B | A | 2 | 8 | 24 |
C | A | 8 | 12 | 16 |
D | A | 4 | 6 | 10 |
E | B | 1 | 2 | 3 |
F | E,C | 6 | 8 | 20 |
G | E,C | 2 | 3 | 4 |
H | F | 2 | 2 | 2 |
I | F | 6 | 6 | 6 |
J | D,G,H | 4 | 6 | 12 |
K | I,J | 2 | 2 | 3 |
Calculation of expected time for activity C:
AON project network diagram shown below, given the precedence relationships of various activities,
Figure 1
With the given predecessor AON diagram is constructed. To Calculate the expected time for activity C given the data that the optimistic time estimate “a” of activity C of 8 weeks, the most likely time estimate “m” of 12 weeks and the pessimistic time estimate “b” of 16 weeks using the formula.
Where ‘a’ is the optimistic time estimate, ‘m’ is the most likely time estimate and ‘b’ is the pessimistic time estimate
Substitute into the equation (1), the values of
Therefore the expected time for activity C is calcualted by adding 8, 48 and16 and dividing the sum with 6 which gives
Hence, the expected time for activity C is 12 weeks.
b)
To calculate: The variance for activity C.
Answer to Problem 17P
The variance for activity C is 1.78 weeks.
Explanation of Solution
Given information:
Activity | Immediate predecessor | Time estimates | ||
Optimistic | Most likely | Pessimistic | ||
A | 4 | 8 | 10 | |
B | A | 2 | 8 | 24 |
C | A | 8 | 12 | 16 |
D | A | 4 | 6 | 10 |
E | B | 1 | 2 | 3 |
F | E,C | 6 | 8 | 20 |
G | E,C | 2 | 3 | 4 |
H | F | 2 | 2 | 2 |
I | F | 6 | 6 | 6 |
J | D,G,H | 4 | 6 | 12 |
K | I,J | 2 | 2 | 3 |
Calculation of variance for activity C:
Calculate the variance for activity C using the formula
Substitute in equation (2) the values of
The variance of activity C is calculated by squaring the value obtained by dividing the difference of 16 and 8 with 6 which gives the resultant as 1.28 weeks.
The variance in activity C is 1.78 weeks.
c)
To determine: The critical path.
Answer to Problem 17P
The critical path AàCàFàHàJàK.
Explanation of Solution
Given information:
Activity | Immediate predecessor | Time estimates | ||
Optimistic | Most likely | Pessimistic | ||
A | 4 | 8 | 10 | |
B | A | 2 | 8 | 24 |
C | A | 8 | 12 | 16 |
D | A | 4 | 6 | 10 |
E | B | 1 | 2 | 3 |
F | E,C | 6 | 8 | 20 |
G | E,C | 2 | 3 | 4 |
H | F | 2 | 2 | 2 |
I | F | 6 | 6 | 6 |
J | D,G,H | 4 | 6 | 12 |
K | I,J | 2 | 2 | 3 |
Calculation of critical path:
Using equation (1) and (2)calculate the expected times and variances for all the activitiesand obtain the values shown in table.
Figure 1, shows the AON diagram which gives the critical path. The critical path is AàCàFàHàJàK.
Hence, the critical path AàCàFàHàJàK.
d)
To Compute: The estimated time of the critical path.
Answer to Problem 17P
The critical path AàCàFàHàJàK which is 40.18 weeks.
Explanation of Solution
Given information:
Activity | Immediate predecessor | Time estimates | ||
Optimistic | Most likely | Pessimistic | ||
A | 4 | 8 | 10 | |
B | A | 2 | 8 | 24 |
C | A | 8 | 12 | 16 |
D | A | 4 | 6 | 10 |
E | B | 1 | 2 | 3 |
F | E,C | 6 | 8 | 20 |
G | E,C | 2 | 3 | 4 |
H | F | 2 | 2 | 2 |
I | F | 6 | 6 | 6 |
J | D,G,H | 4 | 6 | 12 |
K | I,J | 2 | 2 | 3 |
Calculation of estimated time of the critical path:
Figure 1, shows the AON diagram which gives the critical path. The critical path is AàCàFàHàJàK.
The various paths and the expected completion times are listed in the table
Various paths | Completion times |
AàBàEàFàIàK | 37.18 weeks |
AàBàEàGàJàK | 31.18 weeks |
AàCàFàHàJàK | 40.18 weeks |
AàCàGàJàK | 31.51 weeks |
AàDàJàK | 22.84 weeks |
Hence, the estimated time of the critical path AàCàFàHàJàK is 40.18 weeks.
e)
To determine: The variance of the critical path.
Answer to Problem 17P
The variance of the critical path is 10.03 weeks.
Explanation of Solution
Given information:
Activity | Immediate predecessor | Time estimates | ||
Optimistic | Most likely | Pessimistic | ||
A | 4 | 8 | 10 | |
B | A | 2 | 8 | 24 |
C | A | 8 | 12 | 16 |
D | A | 4 | 6 | 10 |
E | B | 1 | 2 | 3 |
F | E,C | 6 | 8 | 20 |
G | E,C | 2 | 3 | 4 |
H | F | 2 | 2 | 2 |
I | F | 6 | 6 | 6 |
J | D,G,H | 4 | 6 | 12 |
K | I,J | 2 | 2 | 3 |
Calculation of variance of the critical path:
The activity variance σ2along the critical path AàCàFàHàJàK is the sum of the variances of individual activities A, C, F, H, J and K.
Therefore,
The activity variance along the critical path is the sum of 1, 1.78, 5.44, 0, 1.78, 0.03 is 10.03 weeks.
Hence, the variance of the critical path is 10.03 weeks
f)
To compute: The probability of completion of the project before week 36 weeks.
Answer to Problem 17P
The probability of completion of the project before week 36 weeks is 9.34%.
Explanation of Solution
Given information:
Activity | Immediate predecessor | Time estimates | ||
Optimistic | Most likely | Pessimistic | ||
A | 4 | 8 | 10 | |
B | A | 2 | 8 | 24 |
C | A | 8 | 12 | 16 |
D | A | 4 | 6 | 10 |
E | B | 1 | 2 | 3 |
F | E,C | 6 | 8 | 20 |
G | E,C | 2 | 3 | 4 |
H | F | 2 | 2 | 2 |
I | F | 6 | 6 | 6 |
J | D,G,H | 4 | 6 | 12 |
K | I,J | 2 | 2 | 3 |
Calculation of probability of completion of the project before week 36 weeks:
Use the normal distribution tables given that the estimated mean time is 40.18 weeks and the standard deviation is square root of 10.03 weeks.
First compute the standard deviation σ by computing the square root of 10.03 weeks.
Calculate the z value as shown below
For a z value of -1.32, the probability is 9.34%.
Therefore, there is a probability of 9.34% that the project may be completed on or before 36 weeks.
Hence, the probability of completion of the project before week 36 weeks is 9.34%.
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Chapter 3 Solutions
Pearson Etext For Principles Of Operations Management: Sustainability And Supply Chain Management -- Access Card (11th Edition)
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