Home > Operation Research calculators > Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost example

4. Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost example ( Enter your problem )
  1. Example-1
  2. Example-2
Other related methods
  1. Network diagram
  2. Critical path, Total float, Free float, Independent float
  3. Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic)
  4. Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost
  5. Project crashing to solve Time-Cost Trade-Off with varying Indirect cost

1. Example-1



1. Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost
A-1020000730000
B-815000620000
CB58000414000
DB611000415000
EB89000515000
FE5500048000
GA,D,C12300084000

Indirect cost = 2800


Solution:
The given problem is
ActivityImmediate
Predecessors
Normal
Time
Normal
Cost
Crash
Time
Crash
Cost
A-1020000730000
B-815000620000
CB58000414000
DB611000415000
EB89000515000
FE5500048000
GA,D,C12300084000


Edge and it's preceded and succeeded node
EdgeNode1 `->` Node2
B1`->`2
A1`->`4
C2`->`3
D2`->`4
E2`->`5
G3`->`6
d4`->`3
F5`->`6



The network diagram for the project, along with activity time, is
 E(8) `E : 2->5`
5
 B(8) `B : 1->2`
2
 F(5) `F : 5->6`
 G(12) `G : 3->6`
6
1
 D(6) `D : 2->4`
 C(5) `C : 2->3`
 d(0) `d : 4->3`
3
 A(10) `A : 1->4`
4



Forward Pass Method

Forward Pass Method
`E_1=0`

`E_2=E_1 + t_(1,2)` [`t_(1,2) = B = 8`]`=0 + 8``=8`

`E_3=Max{E_i + t_(i,3)} [i=2, 4]`

`=Max{E_2 + t_(2,3); E_4 + t_(4,3)}`

`=Max{8 + 5; 14 + 0}`

`=Max{13; 14}`

`=14`

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

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

`=Max{0 + 10; 8 + 6}`

`=Max{10; 14}`

`=14`

`E_5=E_2 + t_(2,5)` [`t_(2,5) = E = 8`]`=8 + 8``=16`

`E_6=Max{E_i + t_(i,6)} [i=3, 5]`

`=Max{E_3 + t_(3,6); E_5 + t_(5,6)}`

`=Max{14 + 12; 16 + 5}`

`=Max{26; 21}`

`=26`



Backward Pass Method

Backward Pass Method
`L_6=E_6=26`

`L_5=L_6 - t_(5,6)` [`t_(5,6) = F = 5`]`=26 - 5``=21`

`L_4=L_3 - t_(4,3)` [`t_(4,3) = d = 0`]`=14 - 0``=14`

`L_3=L_6 - t_(3,6)` [`t_(3,6) = G = 12`]`=26 - 12``=14`

`L_2=text{Min}{L_j - t_(2,j)} [j=5, 4, 3]`

`=text{Min}{L_5 - t_(2,5); L_4 - t_(2,4); L_3 - t_(2,3)}`

`=text{Min}{21 - 8; 14 - 6; 14 - 5}`

`=text{Min}{13; 8; 9}`

`=8`

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

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

`=text{Min}{14 - 10; 8 - 8}`

`=text{Min}{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 : `1-2-4-3-6` and critical activities are `B,D,d,G`

The total project time is 26
The network diagram for the project, along with E-values and L-values, is
`E_(2)=8`
`L_(2)=8`
 E(8) `E : 2->5`
 5 `E_(5)=16`
`L_(5)=21`
 B(8) `B : 1->2`
 2 `E_(2)=8`
`L_(2)=8`
`E_(5)=16`
`L_(5)=21`
 F(5) `F : 5->6`
 G(12) `G : 3->6`
 6 `E_(6)=26`
`L_(6)=26`
 1 `E_(1)=0`
`L_(1)=0`
 D(6) `D : 2->4`
 C(5) `C : 2->3`
 d(0) `d : 4->3`
 3 `E_(3)=14`
`L_(3)=14`
`E_(6)=26`
`L_(6)=26`
`E_(1)=0`
`L_(1)=0`
 A(10) `A : 1->4`
 4 `E_(4)=14`
`L_(4)=14`
`E_(4)=14`
`L_(4)=14`
`E_(3)=14`
`L_(3)=14`



Critical activityCrash cost per week (Rs)
B (1 - 2)`(20000-15000)/(8-6)=2500`
A (1 - 4)`(30000-20000)/(10-7)=3333.33`
C (2 - 3)`(14000-8000)/(5-4)=6000`
D (2 - 4)`(15000-11000)/(6-4)=2000`
E (2 - 5)`(15000-9000)/(8-5)=2000`
G (3 - 6)`(4000-3000)/(12-8)=250`
d (4 - 3)-
F (5 - 6)`(8000-5000)/(5-4)=3000`



Total cost = Direct normal cost + Indirect cost for 26 weeks
`=71000+26 xx 2800=143800`



To begin crash analysis, the crash cost slope values for critical activities is
Critical activityCrash cost per week (Rs)
B (0 - 1)`(20000-15000)/(8-6)=2500`
D (1 - 3)`(15000-11000)/(6-4)=2000`
d (3 - 2)-
G (2 - 5)`(4000-3000)/(12-8)=250`



The critical activity G with cost slope of Rs 250 per week, is the least expensive and can be crashed by 4 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) = B = 8`]`=0 + 8``=8`

`E_3=Max{E_i + t_(i,3)} [i=2, 4]`

`=Max{E_2 + t_(2,3); E_4 + t_(4,3)}`

`=Max{8 + 5; 14 + 0}`

`=Max{13; 14}`

`=14`

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

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

`=Max{0 + 10; 8 + 6}`

`=Max{10; 14}`

`=14`

`E_5=E_2 + t_(2,5)` [`t_(2,5) = E = 8`]`=8 + 8``=16`

`E_6=Max{E_i + t_(i,6)} [i=3, 5]`

`=Max{E_3 + t_(3,6); E_5 + t_(5,6)}`

`=Max{14 + 8; 16 + 5}`

`=Max{22; 21}`

`=22`



Backward Pass Method

Backward Pass Method
`L_6=E_6=22`

`L_5=L_6 - t_(5,6)` [`t_(5,6) = F = 5`]`=22 - 5``=17`

`L_4=L_3 - t_(4,3)` [`t_(4,3) = d = 0`]`=14 - 0``=14`

`L_3=L_6 - t_(3,6)` [`t_(3,6) = G = 8`]`=22 - 8``=14`

`L_2=text{Min}{L_j - t_(2,j)} [j=5, 4, 3]`

`=text{Min}{L_5 - t_(2,5); L_4 - t_(2,4); L_3 - t_(2,3)}`

`=text{Min}{17 - 8; 14 - 6; 14 - 5}`

`=text{Min}{9; 8; 9}`

`=8`

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

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

`=text{Min}{14 - 10; 8 - 8}`

`=text{Min}{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 : `1-2-4-3-6` and critical activities are `B,D,d,G`

The total project time is 22
The network diagram for the project, along with E-values and L-values, is
`E_(2)=8`
`L_(2)=8`
 E(8) `E : 2->5`
 5 `E_(5)=16`
`L_(5)=17`
 B(8) `B : 1->2`
 2 `E_(2)=8`
`L_(2)=8`
`E_(5)=16`
`L_(5)=17`
 F(5) `F : 5->6`
 G(8) `G : 3->6`
 6 `E_(6)=22`
`L_(6)=22`
 1 `E_(1)=0`
`L_(1)=0`
 D(6) `D : 2->4`
 C(5) `C : 2->3`
 d(0) `d : 4->3`
 3 `E_(3)=14`
`L_(3)=14`
`E_(6)=22`
`L_(6)=22`
`E_(1)=0`
`L_(1)=0`
 A(10) `A : 1->4`
 4 `E_(4)=14`
`L_(4)=14`
`E_(4)=14`
`L_(4)=14`
`E_(3)=14`
`L_(3)=14`



Total cost = Direct normal cost + Indirect cost for 22 weeks
`=71000+4xx250+22 xx 2800=133600`



To begin crash analysis, the crash cost slope values for critical activities is
Critical activityCrash cost per week (Rs)
B (0 - 1)`(20000-15000)/(8-6)=2500`
D (1 - 3)`(15000-11000)/(6-4)=2000`
d (3 - 2)-
G (2 - 5)`xx` (Crashed)



The critical activity D with cost slope of Rs 2000 per week, is the least expensive and can be crashed by 2 weeks

But the time should only be reduced by 1 week, otherwise another path becomes critical.
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) = B = 8`]`=0 + 8``=8`

`E_3=Max{E_i + t_(i,3)} [i=2, 4]`

`=Max{E_2 + t_(2,3); E_4 + t_(4,3)}`

`=Max{8 + 5; 13 + 0}`

`=Max{13; 13}`

`=13`

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

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

`=Max{0 + 10; 8 + 5}`

`=Max{10; 13}`

`=13`

`E_5=E_2 + t_(2,5)` [`t_(2,5) = E = 8`]`=8 + 8``=16`

`E_6=Max{E_i + t_(i,6)} [i=3, 5]`

`=Max{E_3 + t_(3,6); E_5 + t_(5,6)}`

`=Max{13 + 8; 16 + 5}`

`=Max{21; 21}`

`=21`



Backward Pass Method

Backward Pass Method
`L_6=E_6=21`

`L_5=L_6 - t_(5,6)` [`t_(5,6) = F = 5`]`=21 - 5``=16`

`L_4=L_3 - t_(4,3)` [`t_(4,3) = d = 0`]`=13 - 0``=13`

`L_3=L_6 - t_(3,6)` [`t_(3,6) = G = 8`]`=21 - 8``=13`

`L_2=text{Min}{L_j - t_(2,j)} [j=5, 4, 3]`

`=text{Min}{L_5 - t_(2,5); L_4 - t_(2,4); L_3 - t_(2,3)}`

`=text{Min}{16 - 8; 13 - 5; 13 - 5}`

`=text{Min}{8; 8; 8}`

`=8`

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

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

`=text{Min}{13 - 10; 8 - 8}`

`=text{Min}{3; 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) `1-2-3-6` and critical activities : `B,C,G`

(2) `1-2-4-3-6` and critical activities : `B,D,d,G`

(3) `1-2-5-6` and critical activities : `B,E,F`

The total project time is 21
The network diagram for the project, along with E-values and L-values, is
`E_(2)=8`
`L_(2)=8`
 E(8) `E : 2->5`
 5 `E_(5)=16`
`L_(5)=16`
 B(8) `B : 1->2`
 2 `E_(2)=8`
`L_(2)=8`
`E_(5)=16`
`L_(5)=16`
 F(5) `F : 5->6`
 G(8) `G : 3->6`
 6 `E_(6)=21`
`L_(6)=21`
 1 `E_(1)=0`
`L_(1)=0`
 D(5) `D : 2->4`
 C(5) `C : 2->3`
 d(0) `d : 4->3`
 3 `E_(3)=13`
`L_(3)=13`
`E_(6)=21`
`L_(6)=21`
`E_(1)=0`
`L_(1)=0`
 A(10) `A : 1->4`
 4 `E_(4)=13`
`L_(4)=13`
`E_(4)=13`
`L_(4)=13`
`E_(3)=13`
`L_(3)=13`



Total cost = Direct normal cost + Indirect cost for 21 weeks
`=71000+4xx250+1xx2000+21 xx 2800=132800`



The activity B is comman in all critical paths and is 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) = B = 6`]`=0 + 6``=6`

`E_3=Max{E_i + t_(i,3)} [i=2, 4]`

`=Max{E_2 + t_(2,3); E_4 + t_(4,3)}`

`=Max{6 + 5; 11 + 0}`

`=Max{11; 11}`

`=11`

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

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

`=Max{0 + 10; 6 + 5}`

`=Max{10; 11}`

`=11`

`E_5=E_2 + t_(2,5)` [`t_(2,5) = E = 8`]`=6 + 8``=14`

`E_6=Max{E_i + t_(i,6)} [i=3, 5]`

`=Max{E_3 + t_(3,6); E_5 + t_(5,6)}`

`=Max{11 + 8; 14 + 5}`

`=Max{19; 19}`

`=19`



Backward Pass Method

Backward Pass Method
`L_6=E_6=19`

`L_5=L_6 - t_(5,6)` [`t_(5,6) = F = 5`]`=19 - 5``=14`

`L_4=L_3 - t_(4,3)` [`t_(4,3) = d = 0`]`=11 - 0``=11`

`L_3=L_6 - t_(3,6)` [`t_(3,6) = G = 8`]`=19 - 8``=11`

`L_2=text{Min}{L_j - t_(2,j)} [j=5, 4, 3]`

`=text{Min}{L_5 - t_(2,5); L_4 - t_(2,4); L_3 - t_(2,3)}`

`=text{Min}{14 - 8; 11 - 5; 11 - 5}`

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

`=6`

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

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

`=text{Min}{11 - 10; 6 - 6}`

`=text{Min}{1; 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) `1-2-3-6` and critical activities : `B,C,G`

(2) `1-2-4-3-6` and critical activities : `B,D,d,G`

(3) `1-2-5-6` and critical activities : `B,E,F`

The total project time is 19
The network diagram for the project, along with E-values and L-values, is
`E_(2)=6`
`L_(2)=6`
 E(8) `E : 2->5`
 5 `E_(5)=14`
`L_(5)=14`
 B(6) `B : 1->2`
 2 `E_(2)=6`
`L_(2)=6`
`E_(5)=14`
`L_(5)=14`
 F(5) `F : 5->6`
 G(8) `G : 3->6`
 6 `E_(6)=19`
`L_(6)=19`
 1 `E_(1)=0`
`L_(1)=0`
 D(5) `D : 2->4`
 C(5) `C : 2->3`
 d(0) `d : 4->3`
 3 `E_(3)=11`
`L_(3)=11`
`E_(6)=19`
`L_(6)=19`
`E_(1)=0`
`L_(1)=0`
 A(10) `A : 1->4`
 4 `E_(4)=11`
`L_(4)=11`
`E_(4)=11`
`L_(4)=11`
`E_(3)=11`
`L_(3)=11`



Total cost = Direct normal cost + Indirect cost for 19 weeks
`=71000+4xx250+1xx2000+2xx2500+19 xx 2800=132200`



The crashing activity `C` in the path `1-2-3-6` `(B,C,G)`, activity `D` in the path `1-2-4-3-6` `(B,D,d,G)`, activity `E` in the path `1-2-5-6` `(B,E,F)`, each by 1 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) = B = 6`]`=0 + 6``=6`

`E_3=Max{E_i + t_(i,3)} [i=2, 4]`

`=Max{E_2 + t_(2,3); E_4 + t_(4,3)}`

`=Max{6 + 4; 10 + 0}`

`=Max{10; 10}`

`=10`

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

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

`=Max{0 + 10; 6 + 4}`

`=Max{10; 10}`

`=10`

`E_5=E_2 + t_(2,5)` [`t_(2,5) = E = 7`]`=6 + 7``=13`

`E_6=Max{E_i + t_(i,6)} [i=3, 5]`

`=Max{E_3 + t_(3,6); E_5 + t_(5,6)}`

`=Max{10 + 8; 13 + 5}`

`=Max{18; 18}`

`=18`



Backward Pass Method

Backward Pass Method
`L_6=E_6=18`

`L_5=L_6 - t_(5,6)` [`t_(5,6) = F = 5`]`=18 - 5``=13`

`L_4=L_3 - t_(4,3)` [`t_(4,3) = d = 0`]`=10 - 0``=10`

`L_3=L_6 - t_(3,6)` [`t_(3,6) = G = 8`]`=18 - 8``=10`

`L_2=text{Min}{L_j - t_(2,j)} [j=5, 4, 3]`

`=text{Min}{L_5 - t_(2,5); L_4 - t_(2,4); L_3 - t_(2,3)}`

`=text{Min}{13 - 7; 10 - 4; 10 - 4}`

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

`=6`

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

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

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

`=text{Min}{0; 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) `1-2-3-6` and critical activities : `B,C,G`

(2) `1-2-4-3-6` and critical activities : `B,D,d,G`

(3) `1-2-5-6` and critical activities : `B,E,F`

(4) `1-4-3-6` and critical activities : `A,d,G`

The total project time is 18
The network diagram for the project, along with E-values and L-values, is
`E_(2)=6`
`L_(2)=6`
 E(7) `E : 2->5`
 5 `E_(5)=13`
`L_(5)=13`
 B(6) `B : 1->2`
 2 `E_(2)=6`
`L_(2)=6`
`E_(5)=13`
`L_(5)=13`
 F(5) `F : 5->6`
 G(8) `G : 3->6`
 6 `E_(6)=18`
`L_(6)=18`
 1 `E_(1)=0`
`L_(1)=0`
 D(4) `D : 2->4`
 C(4) `C : 2->3`
 d(0) `d : 4->3`
 3 `E_(3)=10`
`L_(3)=10`
`E_(6)=18`
`L_(6)=18`
`E_(1)=0`
`L_(1)=0`
 A(10) `A : 1->4`
 4 `E_(4)=10`
`L_(4)=10`
`E_(4)=10`
`L_(4)=10`
`E_(3)=10`
`L_(3)=10`



Total cost = Direct normal cost + Indirect cost for 18 weeks
`=71000+4xx250+1xx2000+2xx2500+1xx6000+1xx2000+1xx2000+18 xx 2800=139400`

Since total project cost for 18 weeks is more than the cost for 19 weeks. So further crashing is not desirable.
Hence, project optimum time is 19 weeks and cost is 132200.

Crashing schedule of project
Project durationCrashing activity and timeDirect Normal CostDirect Crashing CostTotal (Normal + Crashing)Indirect CostTotal Cost
2671000``71000`26xx2800=72800`143800
22G=471000`4xx250=1000`72000`22xx2800=61600`133600
21D=171000`4xx250+1xx2000=3000`74000`21xx2800=58800`132800
19B=271000`4xx250+1xx2000+2xx2500=8000`79000`19xx2800=53200`132200
18C;D;E=171000`4xx250+1xx2000+2xx2500+1xx6000+1xx2000+1xx2000=18000`89000`18xx2800=50400`139400



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