Question

In: Operations Management

The estimated times and immediate predecessors for the activities in a project at George Kyparis's retinal scanning company are given in the following table.


The estimated times and immediate predecessors for the activities in a project at George Kyparis's retinal scanning company are given in the following table. Assume that the activity times are independent. 

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a) What is the expected completion time of the critical path? What is the expected completion time of the other path in the network? 

b) What is the variance of the critical path? What is the variance of the other path in the network? What is the project duration (in weeks)? 

c) Explain why the probability that the critical path will be finished in 22 weeks or less is not necessarily the probability that the project will be finished in 22 weeks or less.

Solutions

Expert Solution

ACTIVITY

EXPECTED TIME

VARIANCE

A

(9 + (4 * 10) + 11) / 6 = 10

((11 - 9) / 6)^2 = 0.111

B

(4 + (4 * 10) + 16) / 6 = 10

((16 - 4) / 6)^2 = 4

C

(9 + (4 * 10) + 11) / 6 = 10

((11 - 9) / 6)^2 = 0.111

D

(5 + (4 * 8) + 11) / 6 = 8

((11 - 5) / 6)^2 = 1

EXPECTED TIME = (OPTIMISTIC TIME + (4 * MOST LIKELY TIME) + PESSIMISTIC TIME) / 6

VARIANCE = ((PESSIMISTIC TIME - OPTIMISTIC TIME) / 6)^2


CPM ANALYSIS FOR EXPECTED PROJECT DURATION


ACTIVITY

DURATION

ES

EF

LS

LF

SLACK

A

10

0

10

0

10

0

B

10

0

10

2

12

2

C

10

10

20

10

20

0

D

8

10

18

12

20

2


FORWARD PASS: ES = MAXIMUM EF OF ALL PREDECESSOR ACTIVITIES; 0 IF NO PREDECESSORS ARE PRESENT. EF = ES + DURATION OF THE ACTIVITY.

BACKWARD PASS: LF = MINIMUM LS OF ALL SUCCESSOR ACTIVITIES; COMPLETION TIME OF THE PROJECT IF NO SUCCESSORS ARE PRESENT. LS = LF - DURATION OF THE ACTIVITY.

SLACK = LF - EF, OR, LS - ES

CRITICAL PATH = PATH WITH THE LONGEST COMBINED DURATION VALUE AND 0 SLACK.

1. CRITICAL PATH = AC

DURATION OF PROJECT = 20

NON CRITICAL PATH = BD = 10 + 8 = 18

2. VARIANCE OF CRITICAL PATH = 0.111 + 0.111 = 0.222

VARIANCE OF NON CRITICAL PATH = 4 + 1 = 5



3. Because the variance of the noncritical path is 5, we can say that the higher variability of the noncritical path can cause the noncritical path to take more time than the critical path and therefore, we might have BD acting as the critical path.


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