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

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

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

( Enter your problem )

Example-1

Other related methods

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	Normal Time	Normal Cost	Crash Time	Crash 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

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`

A(8) `A : 1->2`

G(5) `G : 5->6`

F(12) `F : 3->6`

D(6) `D : 2->4`

C(5) `C : 2->3`

d(0) `d : 4->3`

B(10) `B : 1->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)=8`
`L_(2)=8`

E(8) `E : 2->5`

5 `E_(5)=16`
`L_(5)=21`

A(8) `A : 1->2`

2 `E_(2)=8`
`L_(2)=8`

`E_(5)=16`
`L_(5)=21`

G(5) `G : 5->6`

F(12) `F : 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`

B(10) `B : 1->4`

4 `E_(4)=14`
`L_(4)=14`

`E_(4)=14`
`L_(4)=14`

`E_(3)=14`
`L_(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

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)=8`
`L_(2)=8`

E(8) `E : 2->5`

5 `E_(5)=16`
`L_(5)=17`

A(8) `A : 1->2`

2 `E_(2)=8`
`L_(2)=8`

`E_(5)=16`
`L_(5)=17`

G(5) `G : 5->6`

F(8) `F : 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`

B(10) `B : 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 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)=8`
`L_(2)=8`

E(8) `E : 2->5`

5 `E_(5)=16`
`L_(5)=16`

A(8) `A : 1->2`

2 `E_(2)=8`
`L_(2)=8`

`E_(5)=16`
`L_(5)=16`

G(5) `G : 5->6`

F(8) `F : 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`

B(10) `B : 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 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)=6`
`L_(2)=6`

E(8) `E : 2->5`

5 `E_(5)=14`
`L_(5)=14`

A(6) `A : 1->2`

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

`E_(5)=14`
`L_(5)=14`

G(5) `G : 5->6`

F(8) `F : 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`

B(10) `B : 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` `(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)=6`
`L_(2)=6`

E(7) `E : 2->5`

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

A(6) `A : 1->2`

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

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

G(5) `G : 5->6`

F(8) `F : 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`

B(10) `B : 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 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)

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