3.1 Project Scheduling with CPM and PERT: Critical Path, Float, Probability & Crashing

3.1 Project Scheduling with CPM and PERT

The specification names "PERT/CPM/CCPM: risk analysis, cost, scope, and time; Gantt charts" under project management. This section covers the time calculations (CPM, PERT, and crashing). The next section covers Gantt charts, critical chain, cost, scope, and risk.

CPM (Critical Path Method) was developed at DuPont in the late 1950s with single, deterministic activity durations and a focus on time-cost tradeoffs. PERT (Program Evaluation and Review Technique) was developed for the U.S. Navy's Polaris program at about the same time and uses three time estimates to model uncertainty.

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1. The Example Project

Activities are drawn as nodes (activity-on-node) with arrows for precedence. The PERT estimates are optimistic ($a$), most likely ($m$), and pessimistic ($b$), in weeks.

$t_e = \frac{a + 4m + b}{6}, \qquad \sigma^2 = \left(\frac{b - a}{6}\right)^2$

For activity B: $t_e = (3 + 20 + 13)/6 = 6$ weeks and $\sigma^2 = (10/6)^2 = 2.778$. The long pessimistic tail (13 weeks) pulls the expected time above the most likely value of 5.

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2. Forward and Backward Passes

Forward pass (earliest times): $ES$ = the largest $EF$ of all predecessors (0 for start activities), and $EF = ES + t$.

Backward pass (latest times): $LF$ = the smallest $LS$ of all successors (the project finish for end activities), and $LS = LF - t$.

Floats:

• Total float $TF = LS - ES = LF - EF$: how long an activity can slip without delaying the project.
• Free float $FF = \min(ES \text{ of successors}) - EF$: how long it can slip without delaying any successor.

Key steps: E cannot start until both B (EF 6) and D (EF 7) finish, so $ES_E = 7$. F waits for C (EF 10) and E (EF 12), so $ES_F = 12$ and the project takes 14 weeks. On the backward pass, A must finish by the earlier of $LS_C = 6$ and $LS_D = 4$, so $LF_A = 4$.

The critical path is the zero-float chain A–D–E–F (4 + 3 + 5 + 2 = 14). The other paths are B–E–F (13 weeks) and A–C–F (12 weeks), so B has 1 week of float and C has 2.

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3. PERT Completion Probability

PERT treats the project duration as approximately normal, with mean equal to the critical path length and variance equal to the sum of the critical activities' variances:

$T_e = 14, \quad \sigma_T^2 = 0.444 + 0.111 + 1.778 + 0 = 2.333, \quad \sigma_T = 1.528 \text{ weeks}$

The probability of finishing within 16 weeks is:

$z = \frac{16 - 14}{1.528} = 1.31, \qquad P(T \le 16) = \Phi(1.31) \approx 0.905$

The probability of finishing within 15 weeks is $\Phi(0.65) \approx 0.74$.

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4. Crashing: Time–Cost Tradeoff

Crashing shortens activities by spending more (overtime, extra crews, expedited delivery). Each activity has a normal time and cost and a crash time and cost:

$\text{Cost slope} = \frac{\text{Crash cost} - \text{Normal cost}}{\text{Normal time} - \text{Crash time}}$

Goal: finish in 12 weeks at minimum added cost.

1. Only critical activities matter. Of A, D, E, and F, the cheapest is A at $1,000/week. Crash A by 1 week. Paths become A–D–E–F 13, B–E–F 13, and A–C–F 11. Two paths are now critical.
2. Both critical paths must be shortened together. Options are E alone (shared by both, $1,500) or D plus B ($1,500 + $1,000 = $2,500). Crash E by 1 week for $1,500. Paths become 12, 12, and 11.

The total added cost is $2,500 to reach 12 weeks. C is the cheapest activity ($800) but crashing it is useless, because C is not on a critical path. If indirect costs (supervision, rented equipment, penalties) are $2,000 per week, each week saved is worth $2,000. Both weeks are worth buying: the first costs $1,000 and the second $1,500.