Question

Question 7 The Family Entertainment Group is about to begin the production of its most Ebo Whyte, ensive) movie of the year. The movie's producer, Mr. project. He has identified the eight major activities (labe and H) rcquired to produce the movie. Activities weeks respectively, cannot start until activity A completed. Activities E and F tak until activity B, which takes 3 weeks, is completeo completed, activity G, which takes9 weeks, can begin an weeks, can begin when activities D and F are completed. (a) Draw a fully labelled network together with the activity schedule for this problem. (9 marks) (b) What is the critical path and when will the movie be ready. (4 marks) has decided to use PERT to use PERT/CPM to help plan and control this key lled A, B, C, D, E, F, G, C and D, which take 4 and o which takes 5 weeks, is e 5 and 7 weeks respectively and cannot start When activitics C and E are d activity H, which takes Mr. Whytcis not happy with the completion time and wishes the project to be completed earlier at an extra cost. You have been given the following activity information. ! Crash Time-一ㄒ Normal Cost (GH- (weeks) ACTIVITY Crash Cost (GH¢ millions) millions) 24 36 13 29 30 50 26 36 4 35 57 30 53 35 51 Mr. Whyte have to pay to have the movie ready in 16 (e) How much extra will wecks (12 marks)

Please the question consists of (a) , (b) and (c). I need the crashing part which is the (c) than everything. Is very urgent

Answer 1

As asked, solving for part c

Activity	IP	Normal time	Crash time	Normal cost	Crash cost	Crash slope	Max crash	Crashed	Final time	Start	Finish
A	-	5	3	24	36	6	2	1	4	0	4
B	-	3	2	13	25	12	1	0	3	0	3
C	A	4	2	21	29	4	2	0	4	4	8
D	A	6	3	30	50	6.666666667	3	0	6	4	10
E	B	5	4	26	36	10	1	0	5	3	8
F	B	7	4	35	57	7.333333333	3	0	7	3	10
G	C,E	9	5	30	53	5.75	4	1	8	8	16
H	D,F	8	6	35	51	8	2	2	6	10	16
Total time	16
Normal cost	214
Crash cost	27.75
Total cost	241.75

We begin by determining the critical path. That is A-D-H with a duration of 19 weeks. Next we need to crash one of the activities on this critical path that will be the cheapest option. This is why we determined the crash slope. It is the cost to crash each activity for a week. For example to reduce activity A from 5 weeks to 3 weeks it will cost us additional 36-24 = 12 mil. This means that the crash slope is 6 mil per week.

Now determine the cheapest option along critical path. That is activity A. Once we crash activity A by 1 week, we will also notice that ACG reduces to 17 days. However our critical path remains unchanged.

Next, we crash another activity on critical path. Since the dependency of the path is now bottlenecked by F, we cannot crash activity A anymore to reduce time. Now we need to crash activity H. This will create two critical paths. ADH and BEG. The total duration is 17 weeks

Now we need to crash two activities in order to reach 16 week time. Here we will crash H by another week and G by one week. This will create two critical paths with a total duration of 16 weeks.

The normal cost would have been 214 million. However, the additional cost is the crash cost of 27.75 million.