Critical path, Total float, Free float, Independent float
| 1-2 | A | 4 |
| 2-3 | B | 6 |
| 2-4 | C | 2 |
| 3-4 | dummy | 0 |
| 3-6 | D | 2 |
| 4-5 | E | 7 |
| 5-6 | F | 4 |
| 6-7 | G | 8 |
| 7-8 | H | 3 |
Solution:The given problem is
| Activity | Activity | Duration |
| 1-2 | A | 4 |
| 2-3 | B | 6 |
| 2-4 | C | 2 |
| 3-4 | d | 0 |
| 3-6 | D | 2 |
| 4-5 | E | 7 |
| 5-6 | F | 4 |
| 6-7 | G | 8 |
| 7-8 | H | 3 |
Edge and it's preceded and succeeded node
| Edge | Node1 `->` Node2 |
| A | 1`->`2 |
| B | 2`->`3 |
| C | 2`->`4 |
| d | 3`->`4 |
| D | 3`->`6 |
| E | 4`->`5 |
| F | 5`->`6 |
| G | 6`->`7 |
| H | 7`->`8 |
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 + 4``=4`
`E_3=E_2 + t_(2,3)``=4 + 6``=10`
`E_4=Max{E_i + t_(i,4)} [i=2, 3]`
`=Max{E_2 + t_(2,4); E_3 + t_(3,4)}`
`=Max{4 + 2; 10 + 0}`
`=Max{6; 10}`
`=10`
`E_5=E_4 + t_(4,5)``=10 + 7``=17`
`E_6=Max{E_i + t_(i,6)} [i=3, 5]`
`=Max{E_3 + t_(3,6); E_5 + t_(5,6)}`
`=Max{10 + 2; 17 + 4}`
`=Max{12; 21}`
`=21`
`E_7=E_6 + t_(6,7)``=21 + 8``=29`
`E_8=E_7 + t_(7,8)``=29 + 3``=32`
Backward Pass Method
`L_8=E_8=32`
`L_7=L_8 - t_(7,8)``=32 - 3``=29`
`L_6=L_7 - t_(6,7)``=29 - 8``=21`
`L_5=L_6 - t_(5,6)``=21 - 4``=17`
`L_4=L_5 - t_(4,5)``=17 - 7``=10`
`L_3=text{Min}{L_j - t_(3,j)} [j=4, 6]`
`=text{Min}{L_4 - t_(3,4); L_6 - t_(3,6)}`
`=text{Min}{10 - 0; 21 - 2}`
`=text{Min}{10; 19}`
`=10`
`L_2=text{Min}{L_j - t_(2,j)} [j=3, 4]`
`=text{Min}{L_3 - t_(2,3); L_4 - t_(2,4)}`
`=text{Min}{10 - 6; 10 - 2}`
`=text{Min}{4; 8}`
`=4`
`L_1=L_2 - t_(1,2)``=4 - 4``=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 : `1-2-3-4-5-6-7-8` and critical activities are `A,B,d,E,F,G,H`
The total project time is 32
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 | 4 | 0 | 4 | 0 | 4 | `0+4=4` | `4-4=0` | `0-0=0` | `4-0-4=0` | `4-0-4=0` | Yes |
| 2-3 | 6 | 4 | 10 | 4 | 10 | `4+6=10` | `10-6=4` | `4-4=0` | `10-4-6=0` | `10-4-6=0` | Yes |
| 2-4 | 2 | 4 | 10 | 4 | 10 | `4+2=6` | `10-2=8` | `8-4=4` | `10-4-2=4` | `10-4-2=4` | No |
| 3-4 | 0 | 10 | 10 | 10 | 10 | `10+0=10` | `10-0=10` | `10-10=0` | `10-10-0=0` | `10-10-0=0` | Yes |
| 3-6 | 2 | 10 | 21 | 10 | 21 | `10+2=12` | `21-2=19` | `19-10=9` | `21-10-2=9` | `21-10-2=9` | No |
| 4-5 | 7 | 10 | 17 | 10 | 17 | `10+7=17` | `17-7=10` | `10-10=0` | `17-10-7=0` | `17-10-7=0` | Yes |
| 5-6 | 4 | 17 | 21 | 17 | 21 | `17+4=21` | `21-4=17` | `17-17=0` | `21-17-4=0` | `21-17-4=0` | Yes |
| 6-7 | 8 | 21 | 29 | 21 | 29 | `21+8=29` | `29-8=21` | `21-21=0` | `29-21-8=0` | `29-21-8=0` | Yes |
| 7-8 | 3 | 29 | 32 | 29 | 32 | `29+3=32` | `32-3=29` | `29-29=0` | `32-29-3=0` | `32-29-3=0` | Yes |
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then