Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost Example-1

Project crashing to solve Time-Cost Trade-Off with varying Indirect cost
1-2 4 60 3 90
1-4 6 150 4 250
1-3 2 38 1 60
2-4 5 150 3 250
3-4 2 100 2 100
2-5 7 115 5 175
4-5 4 100 2 240

Varying Indirect cost
Month = 15,14,13,12,11,10,9,8,7,6
Cost = 600,500,400,250,175,100,75,50,35,25

Solution:
The given problem is

Activity	Name	Normal Time	Normal Cost	Crash Time	Crash Cost
1-2	A0	4	60	3	90
1-4	A1	6	150	4	250
1-3	A2	2	38	1	60
2-4	A3	5	150	3	250
3-4	A4	2	100	2	100
2-5	A5	7	115	5	175
4-5	A6	4	100	2	240

Edge and it's preceded and succeeded node

Edge	Node1 `->` Node2
A0	1`->`2
A2	1`->`3
A1	1`->`4
A3	2`->`4
A5	2`->`5
A4	3`->`4
A6	4`->`5

The network diagram for the project, along with activity time, is

A0(4)

A3(5)

A5(7)

A2(2)

A3(5)

A1(6)

A3(5)

A4(2)

A6(4)

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

Forward Pass Method

Forward Pass Method

`E_1=0`

`E_2=E_1 + t_(1,2)``=0 + 4``=4`

`E_3=E_1 + t_(1,3)``=0 + 2``=2`

`E_4=Max{E_i + t_(i,4)} [i=1, 2, 3]`

`=Max{E_1 + t_(1,4); E_2 + t_(2,4); E_3 + t_(3,4)}`

`=Max{0 + 6; 4 + 5; 2 + 2}`

`=Max{6; 9; 4}`

`=9`

`E_5=Max{E_i + t_(i,5)} [i=2, 4]`

`=Max{E_2 + t_(2,5); E_4 + t_(4,5)}`

`=Max{4 + 7; 9 + 4}`

`=Max{11; 13}`

`=13`

Backward Pass Method

Backward Pass Method

`L_5=E_5=13`

`L_4=L_5 - t_(4,5)``=13 - 4``=9`

`L_3=L_4 - t_(3,4)``=9 - 2``=7`

`L_2=text{Min}{L_j - t_(2,j)} [j=4, 5]`

`=text{Min}{L_4 - t_(2,4); L_5 - t_(2,5)}`

`=text{Min}{9 - 5; 13 - 7}`

`=text{Min}{4; 6}`

`=4`

`L_1=text{Min}{L_j - t_(1,j)} [j=2, 3, 4]`

`=text{Min}{L_2 - t_(1,2); L_3 - t_(1,3); L_4 - t_(1,4)}`

`=text{Min}{4 - 4; 7 - 2; 9 - 6}`

`=text{Min}{0; 5; 3}`

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red lines by joining all those events where E-values and L-values are equal.
The critical path of the project is : `1-2-4-5` and critical activities are `A0,A3,A6`

The total project time is 13
The network diagram for the project, along with E-values and L-values, is

`E_(2)=4`
`L_(2)=4`

A0(4)

A3(5)

A5(7)

A2(2)

A3(5)

`E_(3)=2`
`L_(3)=7`

`E_(1)=0`
`L_(1)=0`

A1(6)

A3(5)

A4(2)

A6(4)

`E_(5)=13`
`L_(5)=13`

`E_(4)=9`
`L_(4)=9`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

The crash cost slope for various activities is

Critical activity	Crash cost per week (Rs)
A0 (1 - 2)	`(90-60)/(4-3)=30`
A2 (1 - 3)	`(60-38)/(2-1)=22`
A1 (1 - 4)	`(250-150)/(6-4)=50`
A3 (2 - 4)	`(250-150)/(5-3)=50`
A5 (2 - 5)	`(175-115)/(7-5)=30`
A4 (3 - 4)	-
A6 (4 - 5)	`(240-100)/(4-2)=70`

Total cost = Direct normal cost + Indirect cost for 13 weeks
`=713+400=1113`

`1^(st)` Crashing :

The critical activity A0 with cost slope of Rs 30 per week, is the least expensive and can be crashed by 1 weeks

E-values and L-values for next crashed network

Forward Pass Method

Forward Pass Method

`E_1=0`

`E_2=E_1 + t_(1,2)``=0 + 3``=3`

`E_3=E_1 + t_(1,3)``=0 + 2``=2`

`E_4=Max{E_i + t_(i,4)} [i=1, 2, 3]`

`=Max{E_1 + t_(1,4); E_2 + t_(2,4); E_3 + t_(3,4)}`

`=Max{0 + 6; 3 + 5; 2 + 2}`

`=Max{6; 8; 4}`

`=8`

`E_5=Max{E_i + t_(i,5)} [i=2, 4]`

`=Max{E_2 + t_(2,5); E_4 + t_(4,5)}`

`=Max{3 + 7; 8 + 4}`

`=Max{10; 12}`

`=12`

Backward Pass Method

Backward Pass Method

`L_5=E_5=12`

`L_4=L_5 - t_(4,5)``=12 - 4``=8`

`L_3=L_4 - t_(3,4)``=8 - 2``=6`

`L_2=text{Min}{L_j - t_(2,j)} [j=4, 5]`

`=text{Min}{L_4 - t_(2,4); L_5 - t_(2,5)}`

`=text{Min}{8 - 5; 12 - 7}`

`=text{Min}{3; 5}`

`=3`

`L_1=text{Min}{L_j - t_(1,j)} [j=2, 3, 4]`

`=text{Min}{L_2 - t_(1,2); L_3 - t_(1,3); L_4 - t_(1,4)}`

`=text{Min}{3 - 3; 6 - 2; 8 - 6}`

`=text{Min}{0; 4; 2}`

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red lines by joining all those events where E-values and L-values are equal.
The critical path of the project is : `1-2-4-5` and critical activities are `A0,A3,A6`

The total project time is 12
The network diagram for the project, along with E-values and L-values, is

`E_(2)=3`
`L_(2)=3`

A0(3)

A3(5)

A5(7)

A2(2)

A3(5)

`E_(3)=2`
`L_(3)=6`

`E_(1)=0`
`L_(1)=0`

A1(6)

A3(5)

A4(2)

A6(4)

`E_(5)=12`
`L_(5)=12`

`E_(4)=8`
`L_(4)=8`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

Total cost = Direct normal cost + Indirect cost for 12 weeks
`=713+1xx30+250=993`

`2^(nd)` Crashing :

The critical activity A3 with cost slope of Rs 50 per week, is the least expensive and can be crashed by 2 weeks

E-values and L-values for next crashed network

Forward Pass Method

Forward Pass Method

`E_1=0`

`E_2=E_1 + t_(1,2)``=0 + 3``=3`

`E_3=E_1 + t_(1,3)``=0 + 2``=2`

`E_4=Max{E_i + t_(i,4)} [i=1, 2, 3]`

`=Max{E_1 + t_(1,4); E_2 + t_(2,4); E_3 + t_(3,4)}`

`=Max{0 + 6; 3 + 3; 2 + 2}`

`=Max{6; 6; 4}`

`=6`

`E_5=Max{E_i + t_(i,5)} [i=2, 4]`

`=Max{E_2 + t_(2,5); E_4 + t_(4,5)}`

`=Max{3 + 7; 6 + 4}`

`=Max{10; 10}`

`=10`

Backward Pass Method

Backward Pass Method

`L_5=E_5=10`

`L_4=L_5 - t_(4,5)``=10 - 4``=6`

`L_3=L_4 - t_(3,4)``=6 - 2``=4`

`L_2=text{Min}{L_j - t_(2,j)} [j=4, 5]`

`=text{Min}{L_4 - t_(2,4); L_5 - t_(2,5)}`

`=text{Min}{6 - 3; 10 - 7}`

`=text{Min}{3; 3}`

`=3`

`L_1=text{Min}{L_j - t_(1,j)} [j=2, 3, 4]`

`=text{Min}{L_2 - t_(1,2); L_3 - t_(1,3); L_4 - t_(1,4)}`

`=text{Min}{3 - 3; 4 - 2; 6 - 6}`

`=text{Min}{0; 2; 0}`

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red lines by joining all those events where E-values and L-values are equal.
The critical path of the project are :
(1) `1-2-4-5` and critical activities : `A0,A3,A6`

(2) `1-2-5` and critical activities : `A0,A5`

(3) `1-4-5` and critical activities : `A1,A6`

The total project time is 10
The network diagram for the project, along with E-values and L-values, is

`E_(2)=3`
`L_(2)=3`

A0(3)

A3(3)

A5(7)

A2(2)

A3(3)

`E_(3)=2`
`L_(3)=4`

`E_(1)=0`
`L_(1)=0`

A1(6)

A3(3)

A4(2)

A6(4)

`E_(5)=10`
`L_(5)=10`

`E_(4)=6`
`L_(4)=6`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

Total cost = Direct normal cost + Indirect cost for 10 weeks
`=713+30+2xx50+100=943`

`3^(rd)` Crashing :

The crashing activity `A6` in the path `1-2-4-5` `(A0,A3,A6)`, activity `A5` in the path `1-2-5` `(A0,A5)`, each by 2 week

E-values and L-values for next crashed network

Forward Pass Method

Forward Pass Method

`E_1=0`

`E_2=E_1 + t_(1,2)``=0 + 3``=3`

`E_3=E_1 + t_(1,3)``=0 + 2``=2`

`E_4=Max{E_i + t_(i,4)} [i=1, 2, 3]`

`=Max{E_1 + t_(1,4); E_2 + t_(2,4); E_3 + t_(3,4)}`

`=Max{0 + 6; 3 + 3; 2 + 2}`

`=Max{6; 6; 4}`

`=6`

`E_5=Max{E_i + t_(i,5)} [i=2, 4]`

`=Max{E_2 + t_(2,5); E_4 + t_(4,5)}`

`=Max{3 + 5; 6 + 2}`

`=Max{8; 8}`

`=8`

Backward Pass Method

Backward Pass Method

`L_5=E_5=8`

`L_4=L_5 - t_(4,5)``=8 - 2``=6`

`L_3=L_4 - t_(3,4)``=6 - 2``=4`

`L_2=text{Min}{L_j - t_(2,j)} [j=4, 5]`

`=text{Min}{L_4 - t_(2,4); L_5 - t_(2,5)}`

`=text{Min}{6 - 3; 8 - 5}`

`=text{Min}{3; 3}`

`=3`

`L_1=text{Min}{L_j - t_(1,j)} [j=2, 3, 4]`

`=text{Min}{L_2 - t_(1,2); L_3 - t_(1,3); L_4 - t_(1,4)}`

`=text{Min}{3 - 3; 4 - 2; 6 - 6}`

`=text{Min}{0; 2; 0}`

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red lines by joining all those events where E-values and L-values are equal.
The critical path of the project are :
(1) `1-2-4-5` and critical activities : `A0,A3,A6`

(2) `1-2-5` and critical activities : `A0,A5`

(3) `1-4-5` and critical activities : `A1,A6`

The total project time is 8
The network diagram for the project, along with E-values and L-values, is

`E_(2)=3`
`L_(2)=3`

A0(3)

A3(3)

A5(5)

A2(2)

A3(3)

`E_(3)=2`
`L_(3)=4`

`E_(1)=0`
`L_(1)=0`

A1(6)

A3(3)

A4(2)

A6(2)

`E_(5)=8`
`L_(5)=8`

`E_(4)=6`
`L_(4)=6`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

Total cost = Direct normal cost + Indirect cost for 8 weeks
`=713+130+2xx30+2xx70+50=1093`

Since total project cost for 8 weeks is more than the cost for 10 weeks. So further crashing is not desirable.
Hence, project optimum time is 10 weeks and cost is 943.

Crashing schedule of project

Project duration	Crashing activity and time	Direct Normal Cost	Direct Crashing Cost	Total (Normal + Crashing)	Indirect Cost	Total Cost
13		713	``	713	`400`	1113
12	A0=1	713	`1xx30=30`	743	`250`	993
10	A3=2	713	`30+2xx50=130`	843	`100`	943
8	A5;A6=2	713	`130+2xx30+2xx70=330`	1043	`50`	1093

This material is intended as a summary. Use your textbook for detail explanation.
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1. Example-1