Solutions — Project Planning and Critical Path Analysis

Fluency A(4), B(6)→A, C(3)→A. EST of B and C; min completion time.

EST(A) = 0 (no predecessors)
EST(B) = EST(A) + duration(A) = 0 + 4 = 4
EST(C) = EST(A) + duration(A) = 0 + 4 = 4

B ends at 4+6 = 10. C ends at 4+3 = 7.
Minimum project completion time = 10 days (B is the longest path).

Fluency EST = 8, LST = 12. Float and critical status.

Float = LST − EST = 12 − 8 = 4
Float > 0, so this is not a critical activity. It can be delayed up to 4 time units without affecting the project completion time.

Fluency A(5)—none; B(3)→A; C(7)→A; D(2)→B,C. Forward scan.

EST(A) = 0
EST(B) = 0 + 5 = 5
EST(C) = 0 + 5 = 5
EST(D) = max(EST(B) + dur(B), EST(C) + dur(C)) = max(5+3, 5+7) = max(8, 12) = 12

Minimum project duration = 12 + 2 = 14 days

Fluency Backward scan, LSTs, floats and critical path for Q3 network.

Backward scan (working right to left):
LST(D) = 12 (project finishes at 14, D must start at 12)
LST(C) = LST(D) − dur(C) = 12 − 7 = 5
LST(B) = LST(D) − dur(B) = 12 − 3 = 9
LST(A) = min(LST(B), LST(C)) − dur(A) = min(9, 5) − 5 = 5 − 5 = 0

Floats: A=0−0=0 • B=9−5=4 • C=5−5=0 • D=12−12=0

Critical path: A → C → D (all zero float), duration = 5+7+2 = 14 ✓
Activity B has float 4 and is not critical.

Understanding Construction project: A(4), B(5), C(3)→A, D(6)→B, E(4)→C,D, F(2)→E.

Forward scan:
EST(A)=0, EST(B)=0, EST(C)=4, EST(D)=5
EST(E) = max(4+3, 5+6) = max(7, 11) = 11
EST(F) = 11+4 = 15 → Project ends at 15+2 = 17 days

Backward scan:
LST(F)=15, LST(E)=11, LST(D)=5, LST(C)=8, LST(B)=0, LST(A)=4

Floats:

Activity	EST	LST	Float	Critical?
A	0	4	4	No
B	0	0	0	Yes
C	4	8	4	No
D	5	5	0	Yes
E	11	11	0	Yes
F	15	15	0	Yes

Critical path: B → D → E → F, duration = 5+6+4+2 = 17 days

Understanding Q5 project: finish 2 days early (in 15 days). Which activities to shorten?

To save 2 days, the critical path must be reduced from 17 to 15 days.

Only critical activities matter: B, D, E, F are on the critical path.
Shortening any of these by a combined 2 days will achieve the goal.

Non-critical activities A and C cannot help — they have 4 days of float each. Even if reduced to zero duration, they would not affect the project completion time.

Practical options: reduce B by 2 days (5→3), or D by 2 days (6→4), or any combination totalling 2 days from the critical path. The cheapest or most feasible reduction should be chosen.

Understanding Critical path = 20 days. Activity G has float = 4.

(a) G delayed 3 days: 3 < float (4) → no effect on project. Still completes in 20 days.

(b) G delayed 5 days: 5 > float (4) → the project is delayed by 5 − 4 = 1 day. New completion = 21 days.

(c) Critical activity sped up by 2 days → critical path is now 18 days. Project completes in 18 days. (Note: other paths may now become critical; always check.)

Understanding 8-activity project. Find critical path and minimum duration.

Forward scan:
EST(A)=0, EST(B)=0
EST(C)=3, EST(D)=3
EST(E)=max(0+5, 3+4)=max(5,7)=7
EST(F)=3+2=5
EST(G)=max(7+6, 5+3)=max(13,8)=13
EST(H)=13+4=17 → Project duration = 17+2 = 19 days

Backward scan:
LST(H)=17, LST(G)=13, LST(E)=7, LST(F)=10, LST(C)=3, LST(D)=8, LST(B)=2, LST(A)=0

Floats: A=0, B=2, C=0, D=5, E=0, F=5, G=0, H=0
Critical path: A → C → E → G → H, duration = 3+4+6+4+2 = 19 days ✓

Problem Solving Option X: reduce critical G by 2 days. Option Y: reduce non-critical F by 2 days.

Option X (reduce G from 6 to 4 days):
G is on the critical path. Reducing it by 2 days shortens the critical path by 2 days → project completes 2 days earlier.

Option Y (reduce F from 4 to 2 days):
F is NOT on the critical path and has 3 days of float. Reducing F by 2 days increases its float from 3 to 5. The project completion time is unchanged — no time is saved.

The project manager should choose Option X.
Only reducing critical activities shortens the project. Spending $2000 on Option X saves 2 days; Option Y saves nothing despite having a lower or equal cost.

Problem Solving Two parallel critical paths: Path 1 = A(5)+C(8)+F(7)+H(5)=25; Path 2 = B(6)+D(9)+G(5)+H(5)=25.

(a) Minimum project duration = 25 days (both paths take 25 days).

(b) Reducing C from 8 to 5 days (saving 3 days on Path 1):
New Path 1 duration = 5+5+7+5 = 22 days.
Path 2 is still 25 days.
No — the project still takes 25 days because Path 2 is unchanged. The critical constraint is now Path 2 alone.

(c) Activities to reduce for 22-day completion:
Both paths must be reduced to ≤ 22 days (a reduction of 3 days each).
Most efficient: reduce H by 3 days (from 5 to 2), since H is shared by both paths. This reduces both paths by 3 days simultaneously:
Path 1: 5+8+7+2 = 22 ✓ Path 2: 6+9+5+2 = 22 ✓
Minimum total reduction: reduce H by 3 days.
Alternatively, reduce Path 1 and Path 2 separately (e.g. C by 3 and D by 3), but this requires changes to 2 activities instead of 1.