**(i)**
The dummy activity after $C$ is needed because $G$ follows $C$ only but $F$ follows both $B$ and $C$. The dummy activity after $H$ is needed because both $H$ and $I$ directly join the same pair of vertices (events). | B1 | For explaining the relationships between $B,C,F,G$
 | B1 | For identifying $H$ and $I$

**(ii)**

| Activity | Duration | Immediate predecessors |
|----------|----------|------------------------|
| A | 5 | — |
| B | 3 | — |
| C | 4 | — |
| D | 2 | A |
| E | 1 | A |
| F | 3 | B,C |
| G | 5 | C |
| H | 2 | D |
| I | 4 | D |
| J | 3 | E,F |

| B1 | For $A,B,C$ having no predecessors
| B1 | For $D,E,F,G$ correct
| B1 | For $H,I,J$ correct

**(iii)**

Forward pass showing early event times, all correct with critical activities $A, D, I$ identified.

| M1 | For carrying out forward pass
| A1 | For all early event times correct

Backwards pass showing late event times, all correct.

| M1 | For carrying out backwards pass
| A1 | For all late event times correct

Minimum completion time $= 11$ days; Critical activities: $A, D, I$

| B1 | For 11 with units
| B1 | For $A, D, I$ only

**(iv)**
1 day | B1 | For 1 (not 'more than 1')

**(v)**
| B1 | For the first four days correct

| M1 | For a resource histogram with no 'hanging' cells
| A1 | For days five onwards correct

Number of workers required $= 4$ | A1 | For 4 or follow through their histogram if possible

**(vi)**
Delay activity $E$ by 2 days, so the start changes from 5 to 7 (it happens on day 8). This causes activities $H$ and $J$ to shift. $J$ shifts by 1 day and $H$ by 2 days. | B1 | For stating how much to move $E$
 | M1 | For saying that $H$ and $J$(only) will need to shift
 | A1 | For describing how much $H$ and $J$ move

---