Customer feedback Critical Path

QUESTION: PERT/CPM TECHNIQUES
Company A is installing a Web-based customer-feedback system to meet customer needs for quick response when rolling out new products. A new product line is rolling out in 34 weeks and the customer-feedback system must be installed and running in time for the new product launch.
Table 1.1 in the attached worksheet "PERT/CPM Tables" shows three estimates of the time it will take Company A to complete each of the activities and project tasks in the customer-feedback system project (optimistic, probable, pessimistic).
Company B is installing a similar feedback system to accompany a new product line and originally had the same time line as Company A. Company B just announced that its new product line is ahead of schedule and will be launched in 30 weeks instead of the initially projected 34 weeks. Because the Web-based customer feedback system must be installed and running in time for the revised date of the product launch, something must be done to shorten implementation time required for the customer-feedback system project. Refer to Table 1.2 in the attached worksheet "PERT/CPM Tables" that lists the expected time it will take to complete the activities if shortened as much as possible (crash time), the cost to complete the activity using normal resources (normal cost), and the cost of completing the activity on an accelerated basis (crash cost).

A. Prepare a PERT chart (network diagram) to identify the critical path for project completion.

A network diagram is a graphical representation of the events in a project. The critical path, highlighted in RED ARROWS, is the path that has the highest duration.

1. Determine each of the following:
a. Expected duration of the entire project:
The expected duration, of the project, is approximately 343.5 (354) days. This is given by the end box in the network diagram. (So sorry about this slight error. If you notice from the above diagram that the end has 33.5 instead of the 34.5 I put.

b. Slack for project task A:
Slack is the number of days for which an activity can be delayed without forcing the whole project to be delayed past its due date.
The slack, for task A, will be 6.5. (The latest start, of an activity, is always the latest finish of the succeeding activity. My mistake was that I put the wrong LF for activity E. I do apologize. Please see attached excel file.
ACTIVITY Earliest Start Earliest finish Latest start Latest finish Slack
A 0 3 6.5 9.5 6.5 = Latest finish - earliest finish
B 0 7 0.5 7.5 0.5
C 3 7.5 9.5 14 6.5
D 7 18.5 7.5 19 0.5
E 7.5 12.5 14 19.00 6.5
F 7 17 7 17 0
G 17 23.5 17 23.5 0
H 18.5 29.5 19.00 30 0.5
I 23.5 30 23.5 30 0
J 30 33.5 30 33.5 0

c. Slack for project task H:
The slack, for task H, will be 0.5. (See attached Excel file. I mixed up the LS and ES of the activity. The earliest start (ES) is always supposed to be the highest of the EF of the one or more preceding activities. The latest start is the difference between the latest finish and the duration of the event.)

2. Determine the probability of completing this project in time for the product launch in 34 weeks:

Task Preceding Activity Optimistic Time to Complete
(weeks) Probable Time to Complete
(weeks) Pessimistic Time to Complete
(weeks) Expected Time to Complete (weeks) Variance
(weeks) CRITICAL PATH
START
A START 2 3 4 3.00 = (1/6)*(2+(4*3)+4) 0.33
B START 5 6 13 7.00 1.33 1.33 7.00
C A 3 4 8 4.50 0.83
D B 10 11 15 11.50 0.83
E C 4 5 6 5.00 0.33
F B 8 10 12 10.00 0.67 0.67 10.00
G F 4 6 11 6.50 1.17 1.17 6.50
H D,E 8 10 18 11.00 1.67
I G 3 6 12 6.50 1.50 1.50 6.50
J H,I 2 3 7 3.50 0.83 0.83 3.50
2.35 33.50

The standard deviation, of the critical path will be determined through:

The probability that the project will be completed within 34 days will be given by:

From the normal distribution table, a score of 0.2 will give a probability of 0.5793 (the difference may be attributable to the 0.01 of the normal distribution table. Although the difference is about 0.04 it may be the required tenths difference).

B. Assume that Company B has the same expected completion times for project activities as Company A. Use your results from part A1 and the data in table 1.2 to determine the following:

Identify the following by using your results from part B:
1. Activities to be crashed in order to complete the project within 30 weeks:
The activities that can be crashed at the lowest cost are Tasks D and F. (only two tasks need to be crashed. The comment, from your reviewer may be because first task F has the lowest crash cost per week and task F is along the second path. If the critical path was crashed then the path containing D would then become the longest. It would have to be crashed also so that it is lower than the current critical path).

2. Number of weeks each of these activities should be crashed to meet the deadline with the lowest possible increase in cost:
The weeks that tasks D and F can be crashed to 8 and 6.5 respectively.

(Instructor says that weeks given are much too high. Please carefully review the PERT chart.)

In a project, the critical path is the pathway which can be reduced while reducing the duration of the project. Also in reducing the critical path duration care should also be taken that other paths do not surpass the initial critical path as a result of the crash. From above, you will recall that the critical path is BFGIJ. These activities are the only ones that can be advisably crashed without increasing the total project cost unduly.
ACTIVITY NORMAL DURATION Crash NORMAL COST CRASH COST CRASH WEEKS TO CRASH AFTER CRASH CRASH COST NORMAL CRASH COST PER WEEK
(weeks)
A 3 2 8,400 11,200 5,600 =(11,200/2)
B 7 5 28,000 40,000 7 28,000 8,000
C 4.5 3 18,000 24,000 8,000
D 11.5 7 36,800 44,800 Yes 3 8.5 19,200 6,400
E 5 3 14,000 16,800 5,600
F 10 6 20,000 24,000 Yes 3.5 6.5 14,000 20,000 4,000
G 6.5 5 18,200 28,000 5,600
H 11 8 44,000 64,000 8,000
I 6.5 4.5 26,000 36,000 6.5 26,000 8,000
J 3.5 3 21,000 36,000 3.5 21,000 12,000
34 30 33,200

C. Determine the least increase in cost overall for crashing the project from 34 to 30 weeks:
ACTIVITY NORMAL DURATION WITH CRASH NORMAL COST CRASH COST TOTAL COST
A 3 3 8,400 8,400
B 7 7 28,000 28,000
C 4.5 4.5 18,000 18,000
D 11.5 7.5 36,800 19,200 56,000
E 5 5 14,000 14,000
F 10 7 20,000 14,000 34,000
G 6.5 6 18,200 18,200
H 11 11 44,000 44,000
I 6.5 6.5 26,000 26,000
J 3.5 3.5 21,000 21,000
34 =sum of critical path BFGIJ 30.0 =sum of critical path BFGIJ 234,400 267,600
The least increase will be 33,200 (267,600 - 234,000). (Review the process that should be used to determine the amount).

(Instructor says this section will need to be revised to match the changes in section C)