Project crashing and time-cost trade-off

Why it matters

Projects are often needed earlier than the normal schedule allows — a penalty clause, a plant shutdown window, a market launch. Activities can be speeded up with overtime, extra crews or better equipment, but each day saved costs money, while every day the project runs costs overheads. Crashing finds which activities to speed up, by how much, and the project duration at which the total cost is lowest.

Key ideas

Cost components.

• Direct cost (labour, material, equipment on each activity) rises as activities are shortened.
• Indirect cost (supervision, site overheads, interest, penalties) is roughly proportional to project duration, so it falls as the project is shortened.
• Total cost = direct + indirect. It is U-shaped against duration; its minimum is the optimum duration.

Normal and crash points. Each activity has a normal time tₙ at normal cost Cₙ and a crash time t_c (the shortest technically possible) at crash cost C_c. Between them the direct cost is assumed to vary linearly, so the cost slope is constant. An activity cannot be shortened below t_c.

Crashing procedure.

1. Draw the network with normal times; find the critical path(s), duration and total cost.
2. Among critical activities that can still be shortened, choose the cheapest way to cut one day:
   • one critical path: the critical activity with the lowest cost slope;
   • several parallel critical paths: either an activity common to all of them, or a combination with one activity from each path — whichever has the lowest combined slope.
3. Shorten by the smallest of: the remaining crash limit of the chosen activity(ies), and the amount that makes another path critical (its total float).
4. Recompute durations, critical paths and total cost. Repeat.
5. Stop when the cheapest daily cutting cost exceeds the indirect cost per day (total cost starts rising) — that is the optimum duration. Continue only if a deadline forces it; the crash (minimum) duration is reached when a critical path can no longer be shortened.

Points to remember.

• Crashing non-critical activities adds cost but not speed.
• Shortening one path below a parallel path achieves nothing: all critical paths must be shortened together.
• "Crash everything" gives the minimum duration at the highest direct cost; usually the same minimum duration can be reached much more cheaply by crashing only critical activities.

Formulas

• Cost slope: S = (C_c − Cₙ) / (tₙ − t_c)
• Maximum crash of an activity: Δt_max = tₙ − t_c
• Direct cost after crashing: C_D = Σ Cₙ + Σ Sᵢ · Δtᵢ
• Indirect cost: C_I = c_i · T
• Total cost: C_T = C_D + C_I
• Stop crashing when: cheapest daily crashing cost > c_i

Symbols: Cₙ, C_c = normal and crash cost of an activity (₹); tₙ, t_c = normal and crash time (days); S = cost slope (₹/day); Δtᵢ = days by which activity i is crashed; c_i = indirect cost per day (₹/day); T = project duration (days).

Worked examples

Example 1 (standard). Activities: P (5 days, ₹2000; crash 3 days, ₹2600), Q follows P (8 days, ₹3000; crash 6 days, ₹4000), R runs in parallel with P–Q (4 days, ₹1500; crash 3 days, ₹1700). Reduce the project from 13 to 10 days at least cost.

1. Slopes: P = (2600 − 2000)/(5 − 3) = ₹300/day; Q = (4000 − 3000)/(8 − 6) = ₹500/day; R = 200/1 = ₹200/day.
2. Critical path P–Q = 13 days; R has 9 days of float, so crashing R is useless even though it is cheapest.
3. Crash P by 2 days (its limit): cost 2 × 300 = ₹600, duration 11 days.
4. Crash Q by 1 day: cost ₹500, duration 10 days.
5. Extra direct cost = ₹1100; total direct cost = 6500 + 1100 = ₹7600.

Example 2 (GATE-type optimum duration). Activities (normal / crash days, slope ₹/day): A 4/2, 300; B (after A) 6/4, 150; C (after A) 5/3, 200; D (after B, C) 3/2, 400. Normal direct cost ₹10,000; indirect cost ₹380/day. Find the optimum duration and minimum total cost.

1. Paths: A–B–D = 13, A–C–D = 12. T = 13; total = 10,000 + 13 × 380 = ₹14,940.
2. Day 1: only A–B–D is critical; cheapest is B (150 < 380). Crash B by 1 → both paths 12. Direct 10,150; total 10,150 + 4560 = ₹14,710.
3. Day 2–3: both paths critical. Options: A (common, 300), D (common, 400), B + C (150 + 200 = 350). Crash A by 2 (its limit) at 300/day → T = 10. Direct 10,750; total 10,750 + 3800 = ₹14,550.
4. Day 4: options D (400) or B + C (350). B + C at 350 < 380 → crash each by 1 → T = 9. B is now at its crash limit. Direct 11,100; total 11,100 + 3420 = ₹14,520.
5. Day 5: only D (400) remains as a way to shorten both paths, and 400 > 380, so total cost would rise (to ₹14,540 at 8 days).
6. Optimum duration = 9 days, minimum total cost = ₹14,520. The crash (minimum) duration is 8 days.

Common mistakes

• Crashing the activity with the lowest slope even though it is not critical.
• Shortening one of two parallel critical paths and expecting the project to shrink.
• Crashing an activity by more days than the next path's float allows, so cost is spent with no gain.
• Forgetting indirect cost and therefore crashing all the way to the minimum duration.
• Computing the cost slope as crash cost / crash time instead of the ratio of differences.
• Going below an activity's crash time.

For GATE PI

• Cost slope of an activity; which activity to crash first (quick MCQ or NAT).
• Small networks with two paths: cost of reducing the duration by a given number of days, or the optimum duration with an indirect cost per day.
• Practise building a step table (days cut, activities crashed, cost per day, direct, indirect, total) — it prevents most errors.

Quick check

1. Normal 9 days, ₹4000; crash 6 days, ₹5200. What is the cost slope?
2. Indirect cost is ₹500/day and the cheapest crash option costs ₹650/day. Should you crash further to cut cost?
3. Two parallel critical paths have cheapest slopes ₹200 and ₹250, and a common activity costs ₹420/day. Cheapest way to save one day?
4. Why does crashing a non-critical activity not reduce the project duration?

Answers: 1. ₹400/day. 2. No — total cost would rise by ₹150 per day. 3. Crash the common activity at ₹420/day — cheaper than one activity on each path (200 + 250 = ₹450/day). 4. It already has float; the project length is set by the critical path.