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

1. Example-1

1. Project crashing to solve Time-Cost Trade-Off with varying Indirect cost
 A - 4 60 3 90 B - 6 150 4 250 C - 2 38 1 60 D A 5 150 3 250 E C 2 100 2 100 F A 7 115 5 175 G D,B,E 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 ImmediatePredecessors NormalTime NormalCost CrashTime CrashCost A - 4 60 3 90 B - 6 150 4 250 C - 2 38 1 60 D A 5 150 3 250 E C 2 100 2 100 F A 7 115 5 175 G D,B,E 4 100 2 240

Edge and it's preceded and succeeded node
 Edge Node1 -> Node2 A 1->2 C 1->3 B 1->4 D 2->4 F 2->5 E 3->4 G 4->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)=4L_(2)=4
 2 E_(2)=4L_(2)=4
 A(4) A : 1->2
 D(5) D : 2->4
 F(7) F : 2->5
 1 E_(1)=0L_(1)=0
 C(2) C : 1->3
 3 E_(3)=2L_(3)=7 D(5) D : 2->4
 E_(3)=2L_(3)=7
 5 E_(5)=13L_(5)=13
 E_(1)=0L_(1)=0
 B(6) B : 1->4
 D(5) D : 2->4 E(2) E : 3->4
 G(4) G : 4->5
 E_(5)=13L_(5)=13
 E_(4)=9L_(4)=9
 4 E_(4)=9L_(4)=9

 Critical activity Crash 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 activity Crash 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)=3L_(2)=3
 2 E_(2)=3L_(2)=3
 A(3) A : 1->2
 D(5) D : 2->4
 F(7) F : 2->5
 1 E_(1)=0L_(1)=0
 C(2) C : 1->3
 3 E_(3)=2L_(3)=6 D(5) D : 2->4
 E_(3)=2L_(3)=6
 5 E_(5)=12L_(5)=12
 E_(1)=0L_(1)=0
 B(6) B : 1->4
 D(5) D : 2->4 E(2) E : 3->4
 G(4) G : 4->5
 E_(5)=12L_(5)=12
 E_(4)=8L_(4)=8
 4 E_(4)=8L_(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 activity Crash 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)=3L_(2)=3
 2 E_(2)=3L_(2)=3
 A(3) A : 1->2
 D(3) D : 2->4
 F(7) F : 2->5
 1 E_(1)=0L_(1)=0
 C(2) C : 1->3
 3 E_(3)=2L_(3)=4 D(3) D : 2->4
 E_(3)=2L_(3)=4
 5 E_(5)=10L_(5)=10
 E_(1)=0L_(1)=0
 B(6) B : 1->4
 D(3) D : 2->4 E(2) E : 3->4
 G(4) G : 4->5
 E_(5)=10L_(5)=10
 E_(4)=6L_(4)=6
 4 E_(4)=6L_(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)=3L_(2)=3
 2 E_(2)=3L_(2)=3
 A(3) A : 1->2
 D(3) D : 2->4
 F(5) F : 2->5
 1 E_(1)=0L_(1)=0
 C(2) C : 1->3
 3 E_(3)=2L_(3)=4 D(3) D : 2->4
 E_(3)=2L_(3)=4
 5 E_(5)=8L_(5)=8
 E_(1)=0L_(1)=0
 B(6) B : 1->4
 D(3) D : 2->4 E(2) E : 3->4
 G(2) G : 4->5
 E_(5)=8L_(5)=8
 E_(4)=6L_(4)=6
 4 E_(4)=6L_(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 duration Crashing activity and time Direct Normal Cost Direct Crashing Cost Total (Normal + Crashing) Indirect Cost Total Cost 13 713  713 400 1113 12 A=1 713 1xx30=30 743 250 993 10 D=2 713 1xx30+2xx50=130 843 100 943 8 G;F=2 713 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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