Activity |
Immediate Predecessor(s) |
Normal Time (wks) |
Crash Time (wks) |
Normal Cost ($) |
Crash Cost ($) |
A |
— |
4 |
1 |
5,000 |
8,000 |
B |
— |
5 |
3 |
8,000 |
10,000 |
C |
A |
1 |
1 |
4,000 |
4,000 |
D |
B |
6 |
3 |
6,000 |
12,000 |
E |
B, C |
7 |
6 |
4,000 |
7,000 |
F |
D |
7 |
6 |
4,000 |
7,000 |
a.
Activity | Predecessor | Normal time | Crash time | Normal cost | Crash cost | Cost/ week |
A | - | 4 | 1 | 5000 | 8000 | 1000 |
B | - | 5 | 3 | 8000 | 10000 | 1000 |
C | A | 1 | 1 | 4000 | 4000 | NA |
D | B | 6 | 3 | 6000 | 12000 | 2000 |
E | B,C | 7 | 6 | 4000 | 7000 | 3000 |
F | D | 7 | 6 | 4000 | 7000 | 3000 |
Cost/week of crashing = (Crash cost-normal cost)/(normal time-crash time)
The network diagram is drawn as shown:
The paths with times are given below:
Paths | Time |
A-C-E | 12 |
B-D-F | 18 |
B-E | 12 |
hence path B-D-F taking 18 weeks is the critical path
Cost for normal project time = normal total cost of 31000
plus 1600 per week indirect costs + 1200 per week penalty post week 12
Here project is taking 18 weeks; Hence 6 weeks penalty
So total costs = 31000 + 1600X18 + 6X1200 = 67000
For the minimum cost schedule let us work at reducing the time
For the critical path B-D-F, B has the lowest cost of 1000 | ||||
hence crashing B by 2 units results in project time going to 16 | ||||
Additional cost of 2000 is incurred and there is a saving of 1600X2 = 3200 in indirect cost | ||||
and penalty saving of 1200X2 = 2400 |
Hence net cost saving = 3200+2400-2000 = 3600
Next let us reduce the next activity i.e. D by 3 week costing 6000 | ||
Hence project duration becomes 13 weeks |
And there is a saving of 3X1600 + 3X1200 - 6000 = 2400
Now if we reduce activity F by one week it will cost us 3000 whereas benefit will only be 1600+1200 = 2800. Hence we cannot reduce cost further
So minimum cost = 67000 - 3600 - 2400 = 61000
b. Difference = 67000 - 61000 = 6000
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