Homework Sourse5 The following is a table showing details of a project Job
Solution
The precedence diagram is given below :
A
B
C
D
E
F
G
The possible paths and their cumulative normal duration as follows :
A-G = 10 + 12 = 22
B-C-G = 8 + 5 + 12 = 25
B-D-G = 8 + 6 + 12 = 26
B-E-F-G = 8 + 8 +5 + 12 = 33
The maximum duration upto which AG can be crashed = 7 + 8 = 15 days
The maximum duration upto which path B-C-G can be crashed is = 6 +4 + 8 = 18 days
The maximum duration upto which B-D-G can be crashed =6 + 4 + 8 = 18 days
The maximum duration upto which B-E-F-G can be crashed = 6 +5 +3 + 8 = 22 days
Since highest of the above durations is 22 days, optimum duration by which project can be completed is 22 days
Following actions therefore needs to be taken to achieve optimum project duration of 22 days :
Crash B by 2 days from 8 days to 6 days
Crash C by 1 day from 5 days to 4 days
Crash B by 2 days from 8 days to 6 days
Crash D by 2 days from 6 days to 4 days
Crash B by 2 days from 8 days to 6 days
Crash E by 3 days from 8 days to 5 days
Crash F by 2 days from 5 days to 3 days
Costs associated with above action for each activity as follows :
Activity
Cost ( K000)
Type of cost ( Normal / Crash)
A
10
Normal
B
20
Crash
C
14
Crash
D
15
Crash
E
15
Crash
F
8
Crash
G
12
Normal
Sum of above costs ( K000) = 10 + 20 + 4 + 15 + 15 + 8 + 12 = 94 i.e. K94000
Total indirect cost ( @K400 / day for 22 days ) = K8800`
Thus cumulative cost = Sum of above cost + Total Indirect cost = K94000 + K8800 = K102800
OTPIMUM DURATION = 22 DAYS
ASSOCIATED MINIMUM COST = K102800
| A | B | |
| C | D | E |
| F | ||
| G |
