Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost

1-4	10	20000	7	30000
1-2	8	15000	6	20000
2-3	5	8000	4	14000
2-4	6	11000	4	15000
2-5	8	9000	5	15000
5-6	5	5000	4	8000
4-3	0	0	0	0
3-6	12	3000	8	4000

Indirect cost = 2800

Solution:
The given problem is

Activity	Name	Normal Time	Normal Cost	Crash Time	Crash Cost
1-4	A0	10	20000	7	30000
1-2	A1	8	15000	6	20000
2-3	A2	5	8000	4	14000
2-4	A3	6	11000	4	15000
2-5	A4	8	9000	5	15000
5-6	A5	5	5000	4	8000
4-3	d	0	0	0	0
3-6	A7	12	3000	8	4000

Edge and it's preceded and succeeded node

Edge	Node1 `->` Node2
A1	1`->`2
A0	1`->`4
A2	2`->`3
A3	2`->`4
A4	2`->`5
A7	3`->`6
d	4`->`3
A5	5`->`6

The network diagram for the project, along with activity time, is

			A4(8)	5
	A1(8)	2			A5(5) A7(12)	6
1		A3(6)	A2(5) d(0)	3
	A0(10)	4

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

Forward Pass Method

Forward Pass Method

`E_1=0`

`E_2=E_1 + t_(1,2)``=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)``=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)``=26 - 5``=21`

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

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

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

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

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

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

`=8`

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

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

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

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

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red 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 `A1,A3,d,A7`

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`	A4(8)	5
	A1(8)	2		`E_(5)=16` `L_(5)=21`	A5(5) A7(12)	6
1		A3(6)	A2(5) d(0)	3		`E_(6)=26` `L_(6)=26`
`E_(1)=0` `L_(1)=0`	A0(10)	4	`E_(4)=14` `L_(4)=14`	`E_(3)=14` `L_(3)=14`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

The crash cost slope for various activities is

Critical activity	Crash cost per week (Rs)
A1 (1 - 2)	`(20000-15000)/(8-6)=2500`
A0 (1 - 4)	`(30000-20000)/(10-7)=3333.33`
A2 (2 - 3)	`(14000-8000)/(5-4)=6000`
A3 (2 - 4)	`(15000-11000)/(6-4)=2000`
A4 (2 - 5)	`(15000-9000)/(8-5)=2000`
A7 (3 - 6)	`(4000-3000)/(12-8)=250`
d (4 - 3)	-
A5 (5 - 6)	`(8000-5000)/(5-4)=3000`

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

---

`1^(st)` Crashing :

The critical activity A7 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)``=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)``=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)``=22 - 5``=17`

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

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

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

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

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

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

`=8`

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

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

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

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

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red 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 `A1,A3,d,A7`

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`	A4(8)	5
	A1(8)	2		`E_(5)=16` `L_(5)=17`	A5(5) A7(8)	6
1		A3(6)	A2(5) d(0)	3		`E_(6)=22` `L_(6)=22`
`E_(1)=0` `L_(1)=0`	A0(10)	4	`E_(4)=14` `L_(4)=14`	`E_(3)=14` `L_(3)=14`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

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

---

`2^(nd)` Crashing :

The critical activity A3 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 this reduction will modify the critical path.

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)``=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)``=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)``=21 - 5``=16`

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

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

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

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

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

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

`=8`

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

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

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

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

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red 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 : `A1,A2,A7`

(2) `1-2-4-3-6` and critical activities : `A1,A3,d,A7`

(3) `1-2-5-6` and critical activities : `A1,A4,A5`

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`	A4(8)	5
	A1(8)	2		`E_(5)=16` `L_(5)=16`	A5(5) A7(8)	6
1		A3(5)	A2(5) d(0)	3		`E_(6)=21` `L_(6)=21`
`E_(1)=0` `L_(1)=0`	A0(10)	4	`E_(4)=13` `L_(4)=13`	`E_(3)=13` `L_(3)=13`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

Total cost = Direct normal cost + Indirect cost for 21 weeks
`=71000+1000+1xx2000+21 xx 2800=132800`

---

`3^(rd)` Crashing :

The critical activity A1 with cost slope of Rs 2500 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)``=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)``=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)``=19 - 5``=14`

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

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

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

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

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

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

`=6`

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

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

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

`=text{Min}{0; 1}`

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red 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 : `A1,A2,A7`

(2) `1-2-4-3-6` and critical activities : `A1,A3,d,A7`

(3) `1-2-5-6` and critical activities : `A1,A4,A5`

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`	A4(8)	5
	A1(6)	2		`E_(5)=14` `L_(5)=14`	A5(5) A7(8)	6
1		A3(5)	A2(5) d(0)	3		`E_(6)=19` `L_(6)=19`
`E_(1)=0` `L_(1)=0`	A0(10)	4	`E_(4)=11` `L_(4)=11`	`E_(3)=11` `L_(3)=11`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

Total cost = Direct normal cost + Indirect cost for 19 weeks
`=71000+3000+2xx2500+19 xx 2800=132200`

---

`4^(th)` Crashing :

The crashing activity `A2` in the path `1-2-3-6` `(A1,A2,A7)`, activity `A3` in the path `1-2-4-3-6` `(A1,A3,d,A7)`, activity `A4` in the path `1-2-5-6` `(A1,A4,A5)`, 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)``=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)``=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)``=18 - 5``=13`

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

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

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

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

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

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

`=6`

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

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

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

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

`=0`

(b) The critical path in the network diagram has been shown. This has been done by red 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 : `A1,A2,A7`

(2) `1-2-4-3-6` and critical activities : `A1,A3,d,A7`

(3) `1-2-5-6` and critical activities : `A1,A4,A5`

(4) `1-4-3-6` and critical activities : `A0,d,A7`

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`	A4(7)	5
	A1(6)	2		`E_(5)=13` `L_(5)=13`	A5(5) A7(8)	6
1		A3(4)	A2(4) d(0)	3		`E_(6)=18` `L_(6)=18`
`E_(1)=0` `L_(1)=0`	A0(10)	4	`E_(4)=10` `L_(4)=10`	`E_(3)=10` `L_(3)=10`

(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for )

Total cost = Direct normal cost + Indirect cost for 18 weeks
`=71000+8000+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	A7=4	71000	`4xx250=1000`	72000	`22xx2800=61600`	133600
21	A3=1	71000	`1000+1xx2000=3000`	74000	`21xx2800=58800`	132800
19	A1=2	71000	`3000+2xx2500=8000`	79000	`19xx2800=53200`	132200
18	A2;A3;A4=1	71000	`8000+1xx6000+1xx2000+1xx2000=18000`	89000	`18xx2800=50400`	139400

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