Question

3. The client of a project has requested the project team to crash 8 hours of time. The table below provides the necessary information about the possible project crashing. (a) What is the crash cost for 8 hours of time-savings? (b) Assume the team calls the client and asks for a project extension, reducing the amount of time they need to crash. The project team has a $20 crash budget. Is the budget sufficient to crash 4 hours of time? Activity A B с D E F G H Normal Crash Duration Normal Duration (hours) Cost (S) (hours) 2 10 2 3 15 2 5 25 4 3 20 1 6 30 4 5 1 7 35 6 10 50 7 Crash Immediate Cost (S) Predecessors 0 None 23 A 30 B 24 с 45 с 0 E 50 F 80 D G 1 C EA

Answer 1

Activity	Normal Duration	Normal cost	Crash duration	Crash cost	Slope (Crash cost- Normal cost)/(Normal time- crash time)
A	2	10	2	0	-
B	3	15	2	23	8.00
C	5	25	4	30	5.00
D	3	20	1	24	2.00
E	6	30	4	45	7.50
F	1	5	1	0	-
G	7	35	6	50	15.00
H	10	50	7	80	10.00

H А B с D E F

Path	path length
ABCDH	23
ABCEFGH	34

ABCEFGH IS THE longest path and the critical chain

(a)The critical activities have to be crashed in order of increasing slope

To save 8 days, we have to crash C, E,B, H,G by 1, 2,1,3,1 days respectively

Accordingly, the crash cost is

Slope X Days crashed = 1* 5 + 2* 7.5 + 1*8 + 3*10 + 1*15 = 73

Hence $73 is the crashing cost

(b) No, $20 is not enough to crash by 4 days.

Minimum amount required to crash by 4 days is by crashing B,C,E and the amount is 1*8+ 1*5 + 2*7.5 = $28