Home > Operation Research calculators > Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic) example

3. Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic) example ( Enter your problem )
  1. Example-1
  2. Example-2
Other related methods
  1. Network diagram
  2. Critical path, Total float, Free float, Independent float
  3. Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic)
  4. Project crashing to solve Time-Cost Trade-Off with fixed Indirect cost
  5. Project crashing to solve Time-Cost Trade-Off with varying Indirect cost

1. Example-1



1. Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic)
1-2A117
1-3B147
1-4C228
2-5D111
3-5E2514
4-6F258
5-6G3615


Solution:
Expected time of each activity,
Activity`t_o``t_m``t_p``t_e=(t_o + 4*t_m + t_p)/6``sigma^2=((t_p - t_o)/6)^2`
1-211721
1-314741
1-422831
2-511110
3-5251464
4-625851
5-6361574



The earliest and latest expected time for each activity is calculated by considering the expected time `t_e`

The given problem is
ActivityActivityDuration
1-2A2
1-3B4
1-4C3
2-5D1
3-5E6
4-6F5
5-6G7


Edge and it's preceded and succeeded node
EdgeNode1 `->` Node2
A1`->`2
B1`->`3
C1`->`4
D2`->`5
E3`->`5
F4`->`6
G5`->`6



The network diagram for the project, along with activity time, is
2
 A(2) `A : 1->2`
 D(1) `D : 2->5`
 E(6) `E : 3->5`
5
1
 B(4) `B : 1->3`
3
 G(7) `G : 5->6`
 C(3) `C : 1->4`
 G(7) `G : 5->6`
4
 F(5) `F : 4->6`
6



Forward Pass Method
`E_1=0`

`E_2=E_1 + t_(1,2)` [`t_(1,2) = A = 2`]`=0 + 2``=2`

`E_3=E_1 + t_(1,3)` [`t_(1,3) = B = 4`]`=0 + 4``=4`

`E_4=E_1 + t_(1,4)` [`t_(1,4) = C = 3`]`=0 + 3``=3`

`E_5=Max{E_i + t_(i,5)} [i=2, 3]`

`=Max{E_2 + t_(2,5); E_3 + t_(3,5)}`

`=Max{2 + 1; 4 + 6}`

`=Max{3; 10}`

`=10`

`E_6=Max{E_i + t_(i,6)} [i=4, 5]`

`=Max{E_4 + t_(4,6); E_5 + t_(5,6)}`

`=Max{3 + 5; 10 + 7}`

`=Max{8; 17}`

`=17`


Backward Pass Method
`L_6=E_6=17`

`L_5=L_6 - t_(5,6)` [`t_(5,6) = G = 7`]`=17 - 7``=10`

`L_4=L_6 - t_(4,6)` [`t_(4,6) = F = 5`]`=17 - 5``=12`

`L_3=L_5 - t_(3,5)` [`t_(3,5) = E = 6`]`=10 - 6``=4`

`L_2=L_5 - t_(2,5)` [`t_(2,5) = D = 1`]`=10 - 1``=9`

`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}{12 - 3; 4 - 4; 9 - 2}`

`=text{Min}{9; 0; 7}`

`=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 : `1-3-5-6` and critical activities are `B,E,G`

The total project time is 17
The network diagram for the project, along with E-values and L-values, is
 2 `E_(2)=2`
`L_(2)=9`
`E_(5)=10`
`L_(5)=10`
 A(2) `A : 1->2`
`E_(2)=2`
`L_(2)=9`
 D(1) `D : 2->5`
 E(6) `E : 3->5`
 5 `E_(5)=10`
`L_(5)=10`
 1 `E_(1)=0`
`L_(1)=0`
 B(4) `B : 1->3`
 3 `E_(3)=4`
`L_(3)=4`
 G(7) `G : 5->6`
`E_(1)=0`
`L_(1)=0`
 C(3) `C : 1->4`
`E_(3)=4`
`L_(3)=4`
 G(7) `G : 5->6`
`E_(4)=3`
`L_(4)=12`
 4 `E_(4)=3`
`L_(4)=12`
 F(5) `F : 4->6`
 6 `E_(6)=17`
`L_(6)=17`
`E_(6)=17`
`L_(6)=17`



This material is intended as a summary. Use your textbook for detail explanation.
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