Determine the minimum cost and optimum duration for the project.

Activity	Normal Duration (days)	Crash Duration (days)	Normal Cost (Rs.)	Crash Cost (Rs.)
1-2	7	4	10000	15100
1-3	9	6	7000	9400
2-3	5	2	7000	10000
2-4	6	4	12000	16000
3-4	6	4	9000	12000

843views

written 7.3 years ago by

teamques10 ★ 69k

Activity Name	Activity	Normal Duration (days)	Crash Duration (days)	Normal Cost (Rs.)	Crash Cost (Rs.)	Cost Slope
A	10-20	2	2	1000	1000	0
B	10-30	7	3	500	900	100
C	20-30	6	3	300	420	40
D	20-40	5	4	200	250	50
E	30-40	0	0	0	0	0
F	30-50	9	4	600	900	60
G	40-60	11	6	600	1000	80
H	50-60	6	3	700	910	70

$\text{Cost Slope}=\frac{\text{Crash Cost-Normal Cost}}{\text{Normal Time-Crash Time}}$

Stage 1 : - All Normal

enter image description here

Critical path = 10-20-30-50-60 i.e. A-C-F-H

Project duration = 23 days.

Stage 2 : - All Crash

enter image description here

Critical path :-

1) 10-20-30-50-60 (A-C-F-H)

2) 10-20-40-60 (A-D-G)

Project duration = 12 days

Stage 3 :- Stage 1 + Crash C by 1 day

enter image description here

Critical paths :- 10-20-30-50-60 (A-C-F-H)

10-30-50-60 (B-F-H)

Stage 4 :- Stage 3 + Crash F by 4 days

enter image description here

All critical

Duration = 18 days

Stage 5 : - Stage 4 + Crash F by 1 day & Crash G by 1 day

enter image description here

All critical

Project duration = 17 days

Stage	Description	Duration	Direct Cost	Indirect Cost	Addition Crash Cost	Total
1	All Normal	23	3900	$80 \times 23 = 1840$	-	5740
2	All Crash	12	5830	$80 \times 12 = 960$	-	6340
3	Stage 1 + Crash C by 1 day	22	3900	$80 \times 22 = 1760$	$1 \times 40=40$	5700
4	Stage 3 + Crash F by 4 days	18	3900	$80 \times 18 = 1440$	$40+4 \times 60=280$	5620
5	Stage 4 + Crash,F by 1 day & Crash G by 1 day	17	3900	$80 \times 17 = 1360$	$280+1\times 60+1 \times 80=420$	5680

Optimum cost = Rs. 5620

Optimum duration = 18 days

ADD COMMENT EDIT