Question

For the network shown D(4) B(12) E(6) G(4) C(6) F(5) a-1. Determine the critical path A-C-E-G A-C-F-G A-B-D-G A-B-E-G a-2. Determine the early completion time in weeks for the project. Early completion time For the data shown, reduce the project completion time by three weeks. Assume a linear cost per week shortened weeks NORMAL NORMAL CRASH CRASH ACTIVITY TIME COST TIME COST $ 7,300 2$13,400 18,200 6,550 6,800 5,400 9,000 8,400 7 12 6 12,400 10 5,400 4,500 3,200 6,700 7,600 2

Answer 1

The critical path is the longest path in the project. There are three paths in the project and their length are

ABDG = 7+12+4+4 = 27

ACEG = 7+6+6+4 = 23

ACFG = 7+6+5+4 = 22

a-1

The critical path is A-B-D-G

a-2

The earliest completion time for the project is the duration of critical path.

Early completion time is 27 weeks.

For crashing the project we need to minimize crash cost and also adhere to the duration of the crash. We need to crash it by 3 weeks. Since the critical path exceed the next longest path by more than 3 weeks, it will not change the critical path. This means we need to crash either of A, B, D, and G with 3 weeks in total.

First we need to determine the cost per week for crash for each of them. This is done by subtracting the normal cost from crash cost and dividing it with the difference between normal and crash time. Cost per week of crash is

A = (13400-7300)/(7-2) = 1220

B = (18200-12400)/(12-10) =2900

D = (6800-4500)/(4-2) = 1150

G = (8400-7600)/(4-3) = 800

b-1

We see that the cheapest crash is G. Hence we will start with G. But we can only crash one week with G. So we will choose the next cheapest option for crashing. That is D. and we can crash D by 2 weeks. Thus the answer is

G-D-D

b-2

The total cost to crash is the additional cost that will be borne. That is 800 + 1150 + 1150 = 3100

Resulting cost = $3100