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	Immediate Predecessors	Normal Time	Normal Cost	Crash Time	Crash 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

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)=4` `L_(2)=4`	2 `E_(2)=4` `L_(2)=4`
	A(4) `A : 1->2`	D(5) `D : 2->4`	F(7) `F : 2->5`
1 `E_(1)=0` `L_(1)=0`	C(2) `C : 1->3`	3 `E_(3)=2` `L_(3)=7`  D(5) `D : 2->4`	`E_(3)=2` `L_(3)=7`	5 `E_(5)=13` `L_(5)=13`
`E_(1)=0` `L_(1)=0`	B(6) `B : 1->4`	D(5) `D : 2->4`  E(2) `E : 3->4`	G(4) `G : 4->5`	`E_(5)=13` `L_(5)=13`
	`E_(4)=9` `L_(4)=9`	4 `E_(4)=9` `L_(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)=3` `L_(2)=3`	2 `E_(2)=3` `L_(2)=3`
	A(3) `A : 1->2`	D(5) `D : 2->4`	F(7) `F : 2->5`
1 `E_(1)=0` `L_(1)=0`	C(2) `C : 1->3`	3 `E_(3)=2` `L_(3)=6`  D(5) `D : 2->4`	`E_(3)=2` `L_(3)=6`	5 `E_(5)=12` `L_(5)=12`
`E_(1)=0` `L_(1)=0`	B(6) `B : 1->4`	D(5) `D : 2->4`  E(2) `E : 3->4`	G(4) `G : 4->5`	`E_(5)=12` `L_(5)=12`
	`E_(4)=8` `L_(4)=8`	4 `E_(4)=8` `L_(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)=3` `L_(2)=3`	2 `E_(2)=3` `L_(2)=3`
	A(3) `A : 1->2`	D(3) `D : 2->4`	F(7) `F : 2->5`
1 `E_(1)=0` `L_(1)=0`	C(2) `C : 1->3`	3 `E_(3)=2` `L_(3)=4`  D(3) `D : 2->4`	`E_(3)=2` `L_(3)=4`	5 `E_(5)=10` `L_(5)=10`
`E_(1)=0` `L_(1)=0`	B(6) `B : 1->4`	D(3) `D : 2->4`  E(2) `E : 3->4`	G(4) `G : 4->5`	`E_(5)=10` `L_(5)=10`
	`E_(4)=6` `L_(4)=6`	4 `E_(4)=6` `L_(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)=3` `L_(2)=3`	2 `E_(2)=3` `L_(2)=3`
	A(3) `A : 1->2`	D(3) `D : 2->4`	F(5) `F : 2->5`
1 `E_(1)=0` `L_(1)=0`	C(2) `C : 1->3`	3 `E_(3)=2` `L_(3)=4`  D(3) `D : 2->4`	`E_(3)=2` `L_(3)=4`	5 `E_(5)=8` `L_(5)=8`
`E_(1)=0` `L_(1)=0`	B(6) `B : 1->4`	D(3) `D : 2->4`  E(2) `E : 3->4`	G(2) `G : 4->5`	`E_(5)=8` `L_(5)=8`
	`E_(4)=6` `L_(4)=6`	4 `E_(4)=6` `L_(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

---

---