Home > Operation Research calculators > PERT and CPM example

12. Project crashing : Activity i-j, Name of Activity, Normal Time & Cost, Crash Time & Cost and Indirect Cost example ( Enter your problem )

 11. Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and Indirect Cost (Previous method) 13. Project crashing : Activity, Predecessors, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost (Next method)

### 1. Example-1

1. Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost
 1-4 B 10 20000 7 30000 1-2 A 8 15000 6 20000 2-3 C 5 8000 4 14000 2-4 D 6 11000 4 15000 2-5 E 8 9000 5 15000 5-6 G 5 5000 4 8000 4-3 d 0 0 0 0 3-6 F 12 3000 8 4000

Indirect cost = 2800

Solution:
The given problem is
 Activity Name NormalTime NormalCost CrashTime CrashCost 1-4 B 10 20000 7 30000 1-2 A 8 15000 6 20000 2-3 C 5 8000 4 14000 2-4 D 6 11000 4 15000 2-5 E 8 9000 5 15000 5-6 G 5 5000 4 8000 4-3 d 0 0 0 0 3-6 F 12 3000 8 4000

Edge and it's preceded and succeeded node
 Edge Node1 -> Node2 A 1->2 B 1->4 C 2->3 D 2->4 E 2->5 F 3->6 d 4->3 G 5->6

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

Forward Pass Method

Forward Pass Method
E_1=0

E_2=E_1 + t_(1,2) [t_(1,2) = A = 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) = G = 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) = F = 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 A,D,d,F

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

 Critical activity Crash cost per week (Rs) A (1 - 2) (20000-15000)/(8-6)=2500 B (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 F (3 - 6) (4000-3000)/(12-8)=250 d (4 - 3) - G (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 activity Crash cost per week (Rs) A (0 - 1) (20000-15000)/(8-6)=2500 D (1 - 3) (15000-11000)/(6-4)=2000 d (3 - 2) - F (2 - 5) (4000-3000)/(12-8)=250

The critical activity F 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) = A = 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) = G = 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) = F = 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 A,D,d,F

The total project time is 22
The network diagram for the project, along with E-values and L-values, is
 E_(2)=8L_(2)=8
 E(8) E : 2->5
 5 E_(5)=16L_(5)=17
 A(8) A : 1->2
 2 E_(2)=8L_(2)=8
 E_(5)=16L_(5)=17
 G(5) G : 5->6 F(8) F : 3->6
 6 E_(6)=22L_(6)=22
 1 E_(1)=0L_(1)=0
 D(6) D : 2->4
 C(5) C : 2->3 d(0) d : 4->3
 3 E_(3)=14L_(3)=14
 E_(6)=22L_(6)=22
 E_(1)=0L_(1)=0
 B(10) B : 1->4
 4 E_(4)=14L_(4)=14
 E_(4)=14L_(4)=14
 E_(3)=14L_(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 activity Crash cost per week (Rs) A (0 - 1) (20000-15000)/(8-6)=2500 D (1 - 3) (15000-11000)/(6-4)=2000 d (3 - 2) - F (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) = A = 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) = G = 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) = F = 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 : A,C,F

(2) 1-2-4-3-6 and critical activities : A,D,d,F

(3) 1-2-5-6 and critical activities : A,E,G

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

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

The activity A 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) = A = 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) = G = 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) = F = 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 : A,C,F

(2) 1-2-4-3-6 and critical activities : A,D,d,F

(3) 1-2-5-6 and critical activities : A,E,G

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

(2) 1-2-4-3-6 and critical activities : A,D,d,F

(3) 1-2-5-6 and critical activities : A,E,G

(4) 1-4-3-6 and critical activities : B,d,F

The total project time is 18
The network diagram for the project, along with E-values and L-values, is
 E_(2)=6L_(2)=6
 E(7) E : 2->5
 5 E_(5)=13L_(5)=13
 A(6) A : 1->2
 2 E_(2)=6L_(2)=6
 E_(5)=13L_(5)=13
 G(5) G : 5->6 F(8) F : 3->6
 6 E_(6)=18L_(6)=18
 1 E_(1)=0L_(1)=0
 D(4) D : 2->4
 C(4) C : 2->3 d(0) d : 4->3
 3 E_(3)=10L_(3)=10
 E_(6)=18L_(6)=18
 E_(1)=0L_(1)=0
 B(10) B : 1->4
 4 E_(4)=10L_(4)=10
 E_(4)=10L_(4)=10
 E_(3)=10L_(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 duration Crashing activity and time Direct Normal Cost Direct Crashing Cost Total (Normal + Crashing) Indirect Cost Total Cost 26 71000  71000 26xx2800=72800 143800 22 F=4 71000 4xx250=1000 72000 22xx2800=61600 133600 21 D=1 71000 4xx250+1xx2000=3000 74000 21xx2800=58800 132800 19 A=2 71000 4xx250+1xx2000+2xx2500=8000 79000 19xx2800=53200 132200 18 C;D;E=1 71000 4xx250+1xx2000+2xx2500+1xx6000+1xx2000+1xx2000=18000 89000 18xx2800=50400 139400

This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here

 11. Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and Indirect Cost (Previous method) 13. Project crashing : Activity, Predecessors, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost (Next method)

Dear user,