Question

The project manager is often faced with the need to reduce the time taken to complete a project. However, the cost usually increased due to the additional resources required. The table below presents normal and crash times and costs for each activity in a project. Immediate predecessor Normal time Normal cost Crash time Crash cost A - 4 50 2 90 B A 6 80 4 160 C A 10 60 8 120 D A 11 50 7 150 E B 8 100 6 160 F C,D 5 40 4 70 G E,F 6 70 4 150 Given the information in the above network and the table, answer the following questions: a) The project manager would like to reduce the duration of the project to 23 days. Identify the activities that you would crash and the resulting total cost for the project. Show workings. (3.5 marks) b) The table below shows the indirect project cost for different durations. What would be the optimal number of days for the project duration? Explain your answer and show workings. Days Indirect Costs 26 500 25 460 24 420 23 400 22 380 21 360 20 320 19 300 18 260 c) Use this data to graphically present the total costs of the project

Answer 1

Task data is captured below

Task Name	Predecessor	Normal Time	Normal Cost	Crash Time	Crash Cost	Cost Slope (CC-NC) / (NT-CT)	Cost Slope Ranking
A	-	4	50	2	90	20	1
B	A	6	80	4	160	40	4
C	A	10	60	8	120	30	3
D	A	11	50	7	150	25	2
E	B	8	100	6	160	30	3
F	C, D	5	40	4	70	30	3
G	E, F	6	70	4	150	40	4
Total			450

Network Diagram

There are three paths in the project

ABEG - 24 days

ACFG - 25 days

ADFG - 26 days

Critical path or project duration is 26 days, due to ADFG.

Total normal cost of project is 450.

a) In order to reduce the project time to 23 days. The reduction is required in all the three paths, as all three of them are more than 23 days.

Minimum cost slope for crashing is avialable in case of A having cost slope ranking 1 i.e. of cost 20 for every day.

Another benefit of chosing A is that it is part of all the three paths, so reduction in the duration of this task due to crashing will have bearing on all the three paths. The critical path won't change as we go on crashing the task A to reducte the duration of critical path.

As can be seen by crashing A by 3 days to 1 day by providing crash cost of 3 * 20 = 60 units of cost.

All the three path duration reduce by 3 days i.e.

ABEG - 24 days to 21 days

ACFG - 25 days to 22 days

ADFG - 26 days to 23 days

Now the project duration is as desired (i.e. of 23 days) so we donot need to put any additional effot to crash any task anymore.

So now the total cost of project to complete it in 23 days is Normal cost + crashing cost of task A by 3 days
= 450 + 60 = 510

b) Total cost of project would add up the indirect cost.

Since indirect cost is dependent on number of days the project requires, reducing the time to complete the project would reduce the indirect cost, however in order to reduce the time we need to invest additional cost to crash some task.
If the cost to crash the task that lowers the duration of project by 1 day is less than the savings in indirect cost that it is worth doing.

Optimal Number of Days

Days->	26	25	24	23	22	21	20	19	18
Indirect Cost	500	460	420	400	380	360	320	300	260
Normal Cost	450
Crashing Cost		20	20	20	20	25	40	40	40
Crashing Task		A	A	A	A	D	G	G	G
Normal+Crash Cost		470	490	510	530	555	595	635	675
Total Cost (Normal + crash + indirect)	950	930	910	910	910	915	915	935	935

To reduce 20,19, 18 we chose G even though it has the cost of 40, as trying to reduce D would require us to reduce C also. Adding crashing cost for both of this it would take (25 + 30) = 55 units of cost which is more than crashing cost of G (40).

For every saving of 1 day we save on indirect cost but spend on crashing cost. Since task A has the top ranking for crashing we crash it till we get the overall benefit.

As can be seen for all 22, 23 and 24 days we spend total cost of 910. So 22 days is the optimal duration as it is less duration with same cost.

However there is point to consider here. 22 days is possible if we crash the Task A by 4 days, means by putting extra effort we reduce the time of task A to 0 that is theretical and is not practical.

Only way it can happen is by doing something like requirement rescoping and alternate arrangement by which A is no more required and also does not create dependency for other task, which is not practical, and depends on the task.

Graph and data is provided on theretical basis but realistic optimal duration would be 23 days.

C)