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Home > Operation Research calculators > Project crashing to solve Time-Cost Trade-Off with varying Indirect cost example
15. Project crashing : Activity i-j, Name of Activity, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost example ( Enter your problem )
( Enter your problem )
  1. Example-1
 		Other related methods
  1. Network diagram : Activity, Predecessors
  2. Network diagram : Activity i-j
  3. Network diagram : Activity i-j, Name of Activity
  4. Critical path, Total float, Free float, Independent float : Activity, Predecessors, Duration
  5. Critical path, Total float, Free float, Independent float : Activity i-j, Duration
  6. Critical path, Total float, Free float, Independent float : Activity i-j, Name of Activity, Duration
  7. Project scheduling : Activity, Predecessors, to, tm, tp
  8. Project scheduling : Activity i-j, to, tm, tp
  9. Project scheduling : Activity i-j, Name of Activity, to, tm, tp
  10. Project crashing : Activity, Predecessors, Normal Time & Cost, Crash Time & Cost and Indirect Cost
  11. Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and Indirect Cost
  12. Project crashing : Activity i-j, Name of Activity, Normal Time & Cost, Crash Time & Cost and Indirect Cost
  13. Project crashing : Activity, Predecessors, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost
  14. Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost
  15. Project crashing : Activity i-j, Name of Activity, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost
14. Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost
(Previous method)

1. Example-1


1. Project crashing to solve Time-Cost Trade-Off with varying Indirect cost
1-2A460390
1-4B61504250
1-3C238160
2-4D51503250
3-4E21002100
2-5F71155175
4-5G41002240

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
ActivityNameNormal
Time		Normal
Cost		Crash
Time		Crash
Cost
1-2A460390
1-4B61504250
1-3C238160
2-4D51503250
3-4E21002100
2-5F71155175
4-5G41002240


Edge and it's preceded and succeeded node
EdgeNode1 `->` Node2
A1`->`2
C1`->`3
B1`->`4
D2`->`4
F2`->`5
E3`->`4
G4`->`5



The network diagram for the project, along with activity time, is
2
 A(4) `A : 1->2`
 D(5) `D : 2->4`
 F(7) `F : 2->5`
1
 C(2) `C : 1->3`
3
 D(5) `D : 2->4`
5
 B(6) `B : 1->4`
 D(5) `D : 2->4`
 E(2) `E : 3->4`
 G(4) `G : 4->5`
4



Forward Pass Method

Forward Pass Method
`E_1=0`

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

`E_3=E_1 + t_(1,3)` [`t_(1,3) = C = 2`]`=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)` [`t_(4,5) = G = 4`]`=13 - 4``=9`

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

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

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

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

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

`=4`

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

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

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

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

`=0`



(b) The critical path in the network diagram has been shown. This has been done by double 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 `A,D,G`

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`
 2 `E_(2)=4`
`L_(2)=4`
 A(4) `A : 1->2`
 D(5) `D : 2->4`
 F(7) `F : 2->5`
 1 `E_(1)=0`
`L_(1)=0`
 C(2) `C : 1->3`
 3 `E_(3)=2`
`L_(3)=7`
 D(5) `D : 2->4`
`E_(3)=2`
`L_(3)=7`
 5 `E_(5)=13`
`L_(5)=13`
`E_(1)=0`
`L_(1)=0`
 B(6) `B : 1->4`
 D(5) `D : 2->4`
 E(2) `E : 3->4`
 G(4) `G : 4->5`
`E_(5)=13`
`L_(5)=13`
`E_(4)=9`
`L_(4)=9`
 4 `E_(4)=9`
`L_(4)=9`



Critical activityCrash cost per week (Rs)
A (1 - 2)`(90-60)/(4-3)=30`
C (1 - 3)`(60-38)/(2-1)=22`
B (1 - 4)`(250-150)/(6-4)=50`
D (2 - 4)`(250-150)/(5-3)=50`
F (2 - 5)`(175-115)/(7-5)=30`
E (3 - 4)-
G (4 - 5)`(240-100)/(4-2)=70`



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


To begin crash analysis, the crash cost slope values for critical activities is
Critical activityCrash cost per week (Rs)
A (0 - 1)`(90-60)/(4-3)=30`
D (1 - 3)`(250-150)/(5-3)=50`
G (3 - 4)`(240-100)/(4-2)=70`



The critical activity A 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)` [`t_(1,2) = A = 3`]`=0 + 3``=3`

`E_3=E_1 + t_(1,3)` [`t_(1,3) = C = 2`]`=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)` [`t_(4,5) = G = 4`]`=12 - 4``=8`

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

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

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

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

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

`=3`

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

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

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

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

`=0`



(b) The critical path in the network diagram has been shown. This has been done by double 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 `A,D,G`

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`
 2 `E_(2)=3`
`L_(2)=3`
 A(3) `A : 1->2`
 D(5) `D : 2->4`
 F(7) `F : 2->5`
 1 `E_(1)=0`
`L_(1)=0`
 C(2) `C : 1->3`
 3 `E_(3)=2`
`L_(3)=6`
 D(5) `D : 2->4`
`E_(3)=2`
`L_(3)=6`
 5 `E_(5)=12`
`L_(5)=12`
`E_(1)=0`
`L_(1)=0`
 B(6) `B : 1->4`
 D(5) `D : 2->4`
 E(2) `E : 3->4`
 G(4) `G : 4->5`
`E_(5)=12`
`L_(5)=12`
`E_(4)=8`
`L_(4)=8`
 4 `E_(4)=8`
`L_(4)=8`



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


To begin crash analysis, the crash cost slope values for critical activities is
Critical activityCrash cost per week (Rs)
A (0 - 1)`xx` (Crashed)
D (1 - 3)`(250-150)/(5-3)=50`
G (3 - 4)`(240-100)/(4-2)=70`



The critical activity D 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)` [`t_(1,2) = A = 3`]`=0 + 3``=3`

`E_3=E_1 + t_(1,3)` [`t_(1,3) = C = 2`]`=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)` [`t_(4,5) = G = 4`]`=10 - 4``=6`

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

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

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

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

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

`=3`

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

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

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

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

`=0`



(b) The critical path in the network diagram has been shown. This has been done by double 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 : `A,D,G`

(2) `1-2-5` and critical activities : `A,F`

(3) `1-4-5` and critical activities : `B,G`

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`
 2 `E_(2)=3`
`L_(2)=3`
 A(3) `A : 1->2`
 D(3) `D : 2->4`
 F(7) `F : 2->5`
 1 `E_(1)=0`
`L_(1)=0`
 C(2) `C : 1->3`
 3 `E_(3)=2`
`L_(3)=4`
 D(3) `D : 2->4`
`E_(3)=2`
`L_(3)=4`
 5 `E_(5)=10`
`L_(5)=10`
`E_(1)=0`
`L_(1)=0`
 B(6) `B : 1->4`
 D(3) `D : 2->4`
 E(2) `E : 3->4`
 G(4) `G : 4->5`
`E_(5)=10`
`L_(5)=10`
`E_(4)=6`
`L_(4)=6`
 4 `E_(4)=6`
`L_(4)=6`



Total cost = Direct normal cost + Indirect cost for 10 weeks
`=713+1xx30+2xx50+100=943`


The crashing activity `G` in the path `1-2-4-5` `(A,D,G)`, activity `F` in the path `1-2-5` `(A,F)`, 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)` [`t_(1,2) = A = 3`]`=0 + 3``=3`

`E_3=E_1 + t_(1,3)` [`t_(1,3) = C = 2`]`=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)` [`t_(4,5) = G = 2`]`=8 - 2``=6`

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

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

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

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

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

`=3`

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

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

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

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

`=0`



(b) The critical path in the network diagram has been shown. This has been done by double 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 : `A,D,G`

(2) `1-2-5` and critical activities : `A,F`

(3) `1-4-5` and critical activities : `B,G`

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`
 2 `E_(2)=3`
`L_(2)=3`
 A(3) `A : 1->2`
 D(3) `D : 2->4`
 F(5) `F : 2->5`
 1 `E_(1)=0`
`L_(1)=0`
 C(2) `C : 1->3`
 3 `E_(3)=2`
`L_(3)=4`
 D(3) `D : 2->4`
`E_(3)=2`
`L_(3)=4`
 5 `E_(5)=8`
`L_(5)=8`
`E_(1)=0`
`L_(1)=0`
 B(6) `B : 1->4`
 D(3) `D : 2->4`
 E(2) `E : 3->4`
 G(2) `G : 4->5`
`E_(5)=8`
`L_(5)=8`
`E_(4)=6`
`L_(4)=6`
 4 `E_(4)=6`
`L_(4)=6`



Total cost = Direct normal cost + Indirect cost for 8 weeks
`=713+1xx30+2xx50+2xx70+2xx30+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 durationCrashing activity and timeDirect Normal CostDirect Crashing CostTotal (Normal + Crashing)Indirect CostTotal Cost
13713``713`400`1113
12A=1713`1xx30=30`743`250`993
10D=2713`1xx30+2xx50=130`843`100`943
8G;F=2713`1xx30+2xx50+2xx70+2xx30=330`1043`50`1093


This material is intended as a summary. Use your textbook for detail explanation.
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14. Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost
(Previous method)


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