Question

For the given data, crash the project completely.

ACTIVITY	PREDECESSOR	NORMAL TIME(DAYS)	MAX CRASH TIME	NORMAL COST$	CRASH SLOPE
A	NONE	3	0	60	0
B	NONE	7	1	30	50
C	A	5	3	50	13.33
D	A	6	1	30	20
E	C,B	4	2	40	30

1.What is the normal project duration? = BLANK-1

2. Whatis the normal project cost? = BLANK-2

3. What is the project cost when fully crashed? = BLANK-3

4. Which activities were crashed to reach the final crashed duration? List in order
= BLANK-4

5. What is the project duration when completely crashed? = BLANK-5

Answer 1

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Following is the network diagram

The paths and path-lengths are as follows

A-D =3+6 = 9

A-C-E = 3+5+4 =12

B-E = 7+4 = 11

Hence ACE is the critical path

1. Normal project duration = length of critical path = 12

2. Normal project cost = 60+30+ 50+30+40 = 210

3. Crashing cost = Crash Slope * Crash days = (50*1) + (13.33*3) + (20*1)+ (30*2) = 50+ 40+ 20+ 60 = 170

Therefore, cost when fully crashed = Normal cost + Crashing cost = 210+170 = 380

4.

Activities from critical chain are crashed in order of crashing slope. Out of A-C-E , C has the lowest slope

Activity C is crashed by 3 days. Hence path lengths are

A-D =3+6 = 9

A-C-E = 3+2+4 =9

B-E = 7+4 = 11

Now, B-E is critical path.

E is crashed by two days

Hence path lengths are

A-D =3+6 = 9

A-C-E = 3+2+2 =7

B-E = 7+2 = 9

D and B are crashed by 1 day each in order

Hence path lengths are

A-D =3+5 = 8

A-C-E = 3+2+2 =7

B-E = 6+2 = 8

Hence the order of crashing is  C,E,D,B

5. Project duration after crashing as seen from question 4 is 8 days