1. Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost
A | - | 10 | 20000 | 7 | 30000 |
B | - | 8 | 15000 | 6 | 20000 |
C | B | 5 | 8000 | 4 | 14000 |
D | B | 6 | 11000 | 4 | 15000 |
E | B | 8 | 9000 | 5 | 15000 |
F | E | 5 | 5000 | 4 | 8000 |
G | A,D,C | 12 | 3000 | 8 | 4000 |
Indirect cost = 2800Solution:The given problem is
Activity | Immediate Predecessors | Normal Time | Normal Cost | Crash Time | Crash Cost |
A | - | 10 | 20000 | 7 | 30000 |
B | - | 8 | 15000 | 6 | 20000 |
C | B | 5 | 8000 | 4 | 14000 |
D | B | 6 | 11000 | 4 | 15000 |
E | B | 8 | 9000 | 5 | 15000 |
F | E | 5 | 5000 | 4 | 8000 |
G | A,D,C | 12 | 3000 | 8 | 4000 |
Edge and it's preceded and succeeded node
Edge | Node1 `->` Node2 |
B | 1`->`2 |
A | 1`->`4 |
C | 2`->`3 |
D | 2`->`4 |
E | 2`->`5 |
G | 3`->`6 |
d | 4`->`3 |
F | 5`->`6 |
The network diagram for the project, along with activity time, is
| | | | | | | |
| | | | | F(5) `F : 5->6` | G(12) `G : 3->6` |
| | |
| | | C(5) `C : 2->3` | d(0) `d : 4->3` |
| | | | |
| | | | | | | |
Forward Pass Method
Forward Pass Method
`E_1=0`
`E_2=E_1 + t_(1,2)` [`t_(1,2) = B = 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) = F = 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) = G = 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 `B,D,d,G`
The total project time is 26
The network diagram for the project, along with E-values and L-values, is
| | | | | | | |
| | | | | F(5) `F : 5->6` | G(12) `G : 3->6` |
| | |
| | | C(5) `C : 2->3` | d(0) `d : 4->3` |
| | | | |
| | | | | | | |
Critical activity | Crash cost per week (Rs) |
B (1 - 2) | `(20000-15000)/(8-6)=2500` |
A (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` |
G (3 - 6) | `(4000-3000)/(12-8)=250` |
d (4 - 3) | - |
F (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) |
B (0 - 1) | `(20000-15000)/(8-6)=2500` |
D (1 - 3) | `(15000-11000)/(6-4)=2000` |
d (3 - 2) | - |
G (2 - 5) | `(4000-3000)/(12-8)=250` |
The critical activity G 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) = B = 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) = F = 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) = G = 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 `B,D,d,G`
The total project time is 22
The network diagram for the project, along with E-values and L-values, is
| | | | | | | |
| | | | | F(5) `F : 5->6` | G(8) `G : 3->6` |
| | |
| | | C(5) `C : 2->3` | d(0) `d : 4->3` |
| | | | |
| | | | | | | |
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) |
B (0 - 1) | `(20000-15000)/(8-6)=2500` |
D (1 - 3) | `(15000-11000)/(6-4)=2000` |
d (3 - 2) | - |
G (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) = B = 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) = F = 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) = G = 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 : `B,C,G`
(2) `1-2-4-3-6` and critical activities : `B,D,d,G`
(3) `1-2-5-6` and critical activities : `B,E,F`
The total project time is 21
The network diagram for the project, along with E-values and L-values, is
| | | | | | | |
| | | | | F(5) `F : 5->6` | G(8) `G : 3->6` |
| | |
| | | C(5) `C : 2->3` | d(0) `d : 4->3` |
| | | | |
| | | | | | | |
Total cost = Direct normal cost + Indirect cost for 21 weeks
`=71000+4xx250+1xx2000+21 xx 2800=132800`
The activity B 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) = B = 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) = F = 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) = G = 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 : `B,C,G`
(2) `1-2-4-3-6` and critical activities : `B,D,d,G`
(3) `1-2-5-6` and critical activities : `B,E,F`
The total project time is 19
The network diagram for the project, along with E-values and L-values, is
| | | | | | | |
| | | | | F(5) `F : 5->6` | G(8) `G : 3->6` |
| | |
| | | C(5) `C : 2->3` | d(0) `d : 4->3` |
| | | | |
| | | | | | | |
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` `(B,C,G)`, activity `D` in the path `1-2-4-3-6` `(B,D,d,G)`, activity `E` in the path `1-2-5-6` `(B,E,F)`, 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) = B = 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) = F = 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) = G = 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 : `B,C,G`
(2) `1-2-4-3-6` and critical activities : `B,D,d,G`
(3) `1-2-5-6` and critical activities : `B,E,F`
(4) `1-4-3-6` and critical activities : `A,d,G`
The total project time is 18
The network diagram for the project, along with E-values and L-values, is
| | | | | | | |
| | | | | F(5) `F : 5->6` | G(8) `G : 3->6` |
| | |
| | | C(4) `C : 2->3` | d(0) `d : 4->3` |
| | | | |
| | | | | | | |
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 | G=4 | 71000 | `4xx250=1000` | 72000 | `22xx2800=61600` | 133600 |
21 | D=1 | 71000 | `4xx250+1xx2000=3000` | 74000 | `21xx2800=58800` | 132800 |
19 | B=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