Question

Megah Holding Sdn. Bhd is in progress to start their housing project. The list of activity and scheduled of the project is proposed as in Table 1.

1. If the indirect costs for each duration is as follows, compute the direct and total cost for each duration. What is the optimum cost-time schedule?

Indirect cost

27 weeks – RM 350

26 weeks – RM 270

25 weeks – RM 160

24 weeks – RM 100

12 weeks – RM 70

2. An incentive of RM20 per week has been offered by your clients if you can shorten the duration of the project. Would you grab the opportunity? If yes, for how many weeks?

Table 1 Housing Project Schedule

Activity	Predecessor	Normal time	Normal cost	Crash cost (slope)	Maximum crash time
A	-	10	40	80	2
B	-	8	10	30	3
C	-	5	80	40	1
D	A	11	50	50	2
E	C	15	100	100	4
F	B, D, E	6	20	30	1

*Time unit = 1 week

Answer 1

The Network diagram will be as follows:

In the above diagram each activity is followed by its normal time and maximum crash time in parenthesis.

First we will try to find out the available paths:

Paths	Normal Duration
A-D-F	27 weeks (10+11+6)
B-F	14 weeks (8+6)
C-E-F	26 weeks (5+15+6)

Thus the expected project duration is 27 weeks and the critical path is A-D-F. Now when we want to reduce the duration of the project, we should select activities from the critical path itself. This is because even after crashing we should try to maintain the same path as the critical path. It cannot change. Secondly those activities from the critical path should be selected which have a lower activity cost slope. Activity cost slope is calculated by dividing the difference between Crash cost and Normal cost by difference between the normal time and the crash time. In the question we are already given the Crash cost slope

The following table shows the Activity cost slope of each activity alongwith its Scope for crashing:

Activities	Activity Cost Slope	Scope of Crashing
A	80	2
B	30	3
C	40	1
D	50	2
E	100	4
F	30	1

i)Table showing crashing of activities and total cost at each duration:

Crash Decision	Incremental Cost	Revised Normal Duration
Crash F by 1 week	30	26 weeks (27-1)
Crash D by 1 week	50	25 weeks (26-1)
Crash D by 1 week & Crash C by 1 week	50 & 40	24 weeks (25-1)
Crash A by 2 weeks & Crash E by 2 weeks	60 (80*2) & 200 (100*2)	22 weeks (24-2)

Note that to reduce the duration to 24 weeks we had to select two activities, D & C. The reason for this is that if D alone was chosen then A-D-F would have no longer remained the critical path but C-E-F. Thus simultaneously we have to crash activity C too so that the A-D-F remains the critical path.

It is not possible to reduce the duration of the project to 12 weeks since the critical path can be reduced maximum to 22 weeks. This is because if you add the scope of crashing of the activities within the critical path which A-D-F it is 5 weeks. Original project duration is 27 days. Thus 27 minus 5 is 22 weeks which is the minimum possible duration of this project.

Duration	Normal Cost	Crash Cost	Indirect Cost	Total Cost
27 weeks	300 (40+10+80+50+100+20)	Nil	350	650 (300+350)
26 weeks	300 (40+10+80+50+100+20)	30	270	600 (300+30+270)
25 weeks	300 (40+10+80+50+100+20)	80 (30+50)	160	540 (300+80+160)
24 weeks	300 (40+10+80+50+100+20)	170 (30+50+50+40)	100	570 (300+170+100)
12 weeks	Not possible	Not possible	Not possible	Not possible

Thus as can be seen from the above table the optimum cost-time schedule is 25 weeks. At this duration the cost is the minimum.

ii) If the client offers an incentive of 20 per week then we should accept the offer and reduce the duration of the project by 2 weeks thus giving final duration of 25 weeks and the total cost after incentive will be RM500 {540-(20*2)}. Reducing the duration by 2 weeks alone gives the least cost for the project.