Question

The R&D department is planning to bid on a large project for the development of a new communication system for commercial planes. The accompanying table shows the activities, times, and sequences required: ACTIVITY IMMEDIATE PREDECESSOR TIME (WEEKS) А B С D E F G H A A А B C.D D.F D E.G.H 4 6 6 3 3 5 2 7 b. What is the critical path? (You may select more than one answer. Single click the box with the question mark to produce a check mark for a correct answer and double click the box with the question mark to empty the box for a wrong answer.) A-C-F-G-1 A-B-E- A-D-H-1 A-D-F-G-1 c. Suppose you want to shorten the completion time as much as possible, and you have the option of shortening any or all of B, C, D and G each one week. Which would you shorten? (You may select more than one answer. Single click the box with the question mark to produce a check mark for a correct answer and double click the box with the question mark to empty the box for a wrong answer.) DB Ос D D G DE F d-1. What is the new critical path? (You may select more than one answer. Single click the box with the question mark to produce a check mark for a correct answer and double click the box with the question mark to empty the box for a wrong answer.) A-C-F-G-1 A-B-E- A-D-H-1 A-D-F-G- d-2. What is the earliest completion time for the new critical path? Completion time weeks

Answer 1

Draw the table listing ES, LS, EF, LF and Slack time of each Job Earliest Start (ES) is calculated by forward pass method as latest of the predecessors Earliest Finish (EF). Latest Finish (LF) is calculated using backward pass method as earliest of the successors Latest Start (LS). Earliest Finish (EF) = Earliest Start (ES) + Activity's duration Latest Start (LS) = Latest Finish (LS) - Activity's duration Slack = Latest Finish - Earliest Finish Activity ES EF LS LF Slack Successor Critical Activity Yes 3 0 10 3 14 0 7 7 0 3 3 3 7 B,C,D E F F,G,H 9 3 9 Immediate Time Predecessor (weeks) 3 А 4 A 6 A 6 B 3 C,D DF 5 D 2 E,G,H 7 0 Yes Yes 9 А B с D E F G Н. 3 0 7 - 3 9 0 G 10 12 17 11 24 9 17 12 17 17 24 14 9 12 15 17 Yes Yes 12 9 17 1 0 6 - 0 Yes

b.

List alternative paths and their respective time.

Path	Time
A-B-E-I	17
A-C-F-G-I	24
A-D-F-G-I	24
A-D-G-I	21
A-D-H-I	18

Time of each path is equal to the sum of activity times.

The longest path is the critical path. We see that two paths, viz. A-C-F-G-I and A-D-F-G-I have the longest time. So, these are critical paths

ANSWER:   A-C-F-G-I and A-D-F-G-I

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c.

ANSWER:  G, C, D

Activities on critical paths need to be shortened. since each of the activities can be shortened by one week. Activity G is o both the critical paths. Shortening it by one week, will shorten the two critical paths by 1 week. so, project duration will be 23 weeks. Duration of all other paths is lesser than 23. Shortening activity C and D by one week each, will again shorten each of the critical paths by one week each. And the project completion time will be 22 weeks.

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d-1.

New critical paths are same as in (b)

ANSWER:   A-C-F-G-I and A-D-F-G-I

This is already explained in part (c)

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d-2.

Completion time = 22 weeks

This is already explained in part (c)