Question

Activity A	Duration 6	Predecessor -	Crash weeks -	Crashing cost per week -
B	4	A	2	$480
C	5	A	1	$380
D	6	A	3	$450
E	5	B	2	$430
F	4	B,C	3	$350
G	5	C,D	2	$410
H	6	D	2	$570
I	5	E,F	2	$540
J	6	E,F,G	2	$530
K	3	G,H	1	$550
L	4	I,J	1	$560
M	5	I,J,K	-	$620
N	6	H,K,J	1	$650
O	4	M,N	-	- The time for each activity tocomplete, the possible weeks by which that time can be reduced and the cost per week to decrease the activity time. The cost to complete the project in normal time is $10,000 (crashing) 1. Draw the project network for this project?what are the slacks in the steps? 2. what is the minimum duration of the project? 3. what are the critical paths of this project? 4. How much will be the most economical way to complete the project in 27 weeks?

Answer 1

1. Project Network and ES, EF, LS and LF are shown in the snapshot below:

Numbers notation on each node/activity:

Top Left - ES (Early Start)

Top Right: EF (Early Finish)

Bottom Left: LS (Late Start)

Bottom Right: LF (Late Finish)

Center: Activity Names

Bottom Middle: Activity times (in weeks)

Slack = (LF - EF) or (LS - ES)

All the above values can be tabulated and calculations can be done for Slack as shown below:

Activity	Duration	Predecessor	Crash weeks	Crashing cost per week	ES	EF	LS	LF	Slack
Start	0	-	-	$0	0	0	0	0	0
A	6	-	-	$0	0	6	0	6	0	A
B	4	A	2	$480	6	10	8	12	2
C	5	A	1	$380	6	11	7	12	1
D	6	A	3	$450	6	12	6	12	0	D
E	5	B	2	$430	10	15	12	17	2
F	4	B,C	3	$350	11	15	13	17	2
G	5	C,D	2	$420	12	17	12	17	0	G
H	6	D	2	$570	12	18	14	20	2
I	5	E,F	2	$540	15	20	19	24	4
J	6	E,F,G	2	$530	17	23	17	23	0	J
K	3	G,H	1	$550	18	21	20	23	2
L	4	I,J	1	$560	23	27	25	29	2
M	5	I,J,K	-	$620	23	28	24	29	1
N	6	H,K,J	1	$650	23	29	23	29	0	N
O	4	M,N	-	$0	29	33	29	33	0	O
Finish	0	O	-	$0	33	33	33	33	0

2. Minimum duration of the project = 33 weeks

3. Critical Paths:

All those activities which have ZERO SLACK lie on the CRITICAL PATH. [That's the rule and definition of critical path, since activities on critical path HAVE TO BE started and finished on their scheduled times]

So, Critical Path activities are: A, D, G, J, N, O (as also highlighted in the snapshot below:

So, Critical Path is: A - D - G - J - N - O

Critical path time = 6 + 6 + 5 + 6 + 6 + 4 =

There can be only 1 critical path with those activities.

4. Crashing the non-critical activities will not have much impact. So, we should crash the critical activites and ensure that the non-critical activities don't become critical activities. Also, we crash starting with the least expensive activity to the most expensive.

Out of the critical activities, least expensive is G with $420 as crashing cost per week. So, we crash G by maximum weeks, i.e., by 2 weeks.

Now, Project Duration become = 33 - 2 = 31 weeks

Next candidate for crashing is D with $450 as crashing cost per week, So, we crash D by maximum weeks, i.e., by 3 weeks.

Now, Project Duration become = 31 - 3 = 28 weeks

Next candidate for crashing is J with $530 as crashing cost per week, So, we crash J by 1 week only since we need to complete the project in target 27 weeks which gets achieved by crashing J just by 1 week.

Now, project duration = 28 - 1 = 27 weeks.

So, New cost = Normal Cost + Crashing Cost

=> New Cost = $10,000 + $420*2 + $450*3 + $530*1 = $12,720