Critical path, Total float, Free float, Independent float
| A | - | 2 |
| B | - | 4 |
| C | - | 3 |
| D | A | 1 |
| E | B | 6 |
| F | C | 5 |
| G | D,E | 7 |
| H | F,G | 2 |
Solution:| Activity | Immediate Predecessors | Duration |
| A | - | 2 |
| B | - | 4 |
| C | - | 3 |
| D | A | 1 |
| E | B | 6 |
| F | C | 5 |
| G | D,E | 7 |
| H | F,G | 2 |
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 |
| H | 6`->`7 |
The network diagram for the project, along with activity time, is
(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
`E_1=0`
`E_2=E_1 + t_(1,2)``=0 + 2``=2`
`E_3=E_1 + t_(1,3)``=0 + 4``=4`
`E_4=E_1 + t_(1,4)``=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`
`E_7=E_6 + t_(6,7)``=17 + 2``=19`
Backward Pass Method
`L_7=E_7=19`
`L_6=L_7 - t_(6,7)``=19 - 2``=17`
`L_5=L_6 - t_(5,6)``=17 - 7``=10`
`L_4=L_6 - t_(4,6)``=17 - 5``=12`
`L_3=L_5 - t_(3,5)``=10 - 6``=4`
`L_2=L_5 - t_(2,5)``=10 - 1``=9`
`L_1=text{Min}{L_j - t_(1,j)} [j=2, 3, 4]`
`=text{Min}{L_2 - t_(1,2); L_3 - t_(1,3); L_4 - t_(1,4)}`
`=text{Min}{9 - 2; 4 - 4; 12 - 3}`
`=text{Min}{7; 0; 9}`
`=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 : `1-3-5-6-7` and critical activities are `B,E,G,H`
The total project time is 19
The network diagram for the project, along with E-values and L-values, is
(Note: Same diagram as above. Move node position by dragging node.
If you are not able to select node then just click here for
)
For each activity, the total float, free float and independent float calculations are shown in Table
Activity `(i,j)` `(1)` | Duration `t_(ij)` `(2)` | Earliest Start `E_i` `(3)` | `E_j` `(4)` | `L_i` `(5)` | Latest Finish `L_j` `(6)` | Earliest Finish `=E_i+t_(ij)` `(7)=(3)+(2)` | Latest Start `=L_j-t_(ij)` `(8)=(6)-(2)` | Total Float `=L_j-t_(ij)-E_i` `(9)=(8)-(3)` | Free Float `=E_j-E_i-t_(ij)` `(10)=(4)-(3)-(2)` | Independent Float `=E_j-L_i-t_(ij)` `(11)=(4)-(5)-(2)` | Critical Activity? `(12)` |
| 1-2 | 2 | 0 | 2 | 0 | 9 | `0+2=2` | `9-2=7` | `7-0=7` | `2-0-2=0` | `2-0-2=0` | No |
| 1-3 | 4 | 0 | 4 | 0 | 4 | `0+4=4` | `4-4=0` | `0-0=0` | `4-0-4=0` | `4-0-4=0` | Yes |
| 1-4 | 3 | 0 | 3 | 0 | 12 | `0+3=3` | `12-3=9` | `9-0=9` | `3-0-3=0` | `3-0-3=0` | No |
| 2-5 | 1 | 2 | 10 | 9 | 10 | `2+1=3` | `10-1=9` | `9-2=7` | `10-2-1=7` | `10-9-1=0` | No |
| 3-5 | 6 | 4 | 10 | 4 | 10 | `4+6=10` | `10-6=4` | `4-4=0` | `10-4-6=0` | `10-4-6=0` | Yes |
| 4-6 | 5 | 3 | 17 | 12 | 17 | `3+5=8` | `17-5=12` | `12-3=9` | `17-3-5=9` | `17-12-5=0` | No |
| 5-6 | 7 | 10 | 17 | 10 | 17 | `10+7=17` | `17-7=10` | `10-10=0` | `17-10-7=0` | `17-10-7=0` | Yes |
| 6-7 | 2 | 17 | 19 | 17 | 19 | `17+2=19` | `19-2=17` | `17-17=0` | `19-17-2=0` | `19-17-2=0` | Yes |
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then