Question 4:
4 | 1 | 3 | 2 | 2
4 | (a) | So that E and F
do not start at the same vertex (event )
and also end at the same vertex (event ) | B1
[1] | 2.4 | Uniqueness (E and F)
To make graph simple
Ignore references to durations or start/finish times
4 | (b) | Minimum project completion time
= 9 (minutes) | M1
A1
[2] | 1.1
1.1 | Forward pass seen with at most one independent error
     
0 4 4 7 7 9
9 from correct forward pass seen
SC B1 for 9 without forward pass seen
4 | (c) | Backward pass seen with at most one
independent error
Floatij = LFTj – ESTi - durationij
A B C D E F
2 0 0 1 0 1 | M1
A1
B1
[3] | 1.1
1.1
1.1 | May be implied from A mark achieved
     
0 4 4 7 8 9
Float correct for non-critical activities: A = 2, D = 1, F = 1
Critical activities B, C, E with 0 float
 |  |  |  |  | 
4 | (d) | No change (A is still not critical) | B1
[1] | 1.1 | No effect, none
4 | (e) | If m < 2 then D becomes critical instead of C
and the minimum project completion time is 8
(minutes)
If 2  m < 3 then C is still critical and minimum
project completion time is 6+m (minutes) | B1
B1
[2] | 1.1
1.1 | m < 2  8 or m  2  8
Or from Max{8, 6+m}
6+m
Note: m need not be an integer, so listing time for e.g. m = 1, m = 2
scores B0, B0