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Home > Operation Research calculators > Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic) example
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9. Project scheduling : Activity i-j, Name of Activity, to, tm, tp example
( Enter your problem )
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- Example-1
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Other related methods
- Network diagram : Activity, Predecessors
- Network diagram : Activity i-j
- Network diagram : Activity i-j, Name of Activity
- Critical path, Total float, Free float, Independent float : Activity, Predecessors, Duration
- Critical path, Total float, Free float, Independent float : Activity i-j, Duration
- Critical path, Total float, Free float, Independent float : Activity i-j, Name of Activity, Duration
- Project scheduling : Activity, Predecessors, to, tm, tp
- Project scheduling : Activity i-j, to, tm, tp
- Project scheduling : Activity i-j, Name of Activity, to, tm, tp
- Project crashing : Activity, Predecessors, Normal Time & Cost, Crash Time & Cost and Indirect Cost
- Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and Indirect Cost
- Project crashing : Activity i-j, Name of Activity, Normal Time & Cost, Crash Time & Cost and Indirect Cost
- Project crashing : Activity, Predecessors, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost
- Project crashing : Activity i-j, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost
- Project crashing : Activity i-j, Name of Activity, Normal Time & Cost, Crash Time & Cost and varying Indirect Cost
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1. Example-1
1. Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic)
| 1-2 | A | 1 | 1 | 7 | | 1-3 | B | 1 | 4 | 7 | | 1-4 | C | 2 | 2 | 8 | | 2-5 | D | 1 | 1 | 1 | | 3-5 | E | 2 | 5 | 14 | | 4-6 | F | 2 | 5 | 8 | | 5-6 | G | 3 | 6 | 15 |
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` | | 1-2 | 1 | 1 | 7 | 2 | 1 | | 1-3 | 1 | 4 | 7 | 4 | 1 | | 1-4 | 2 | 2 | 8 | 3 | 1 | | 2-5 | 1 | 1 | 1 | 1 | 0 | | 3-5 | 2 | 5 | 14 | 6 | 4 | | 4-6 | 2 | 5 | 8 | 5 | 1 | | 5-6 | 3 | 6 | 15 | 7 | 4 |
The earliest and latest expected time for each activity is calculated by considering the expected time `t_e`
The given problem is
| Activity | Activity | Duration | | 1-2 | A | 2 | | 1-3 | B | 4 | | 1-4 | C | 3 | | 2-5 | D | 1 | | 3-5 | E | 6 | | 4-6 | F | 5 | | 5-6 | G | 7 |
Edge and it's preceded and succeeded node
| Edge | Node1 `->` Node2 | | A | 1`->`2 | | B | 1`->`3 | | C | 1`->`4 | | D | 2`->`5 | | E | 3`->`5 | | F | 4`->`6 | | G | 5`->`6 |
The network diagram for the project, along with activity time, is
| | | | | | | | | | D(1) `D : 2->5` | | E(6) `E : 3->5` |
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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
| | | | | | | | | | D(1) `D : 2->5` | | E(6) `E : 3->5` |
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This material is intended as a summary. Use your textbook for detail explanation.
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