LESSON 2.5 — Resource Planning and Crashing
A. Standard Map
| Topic | Governing Source | Exam Focus |
|---|---|---|
| Resource leveling vs resource smoothing | ch05-part01; standard CPM | Which extends duration; which holds duration fixed |
| Resource histogram | Standard project management practice | Visual representation of resource usage over time |
| Crashing — cost slope formula | CPM cost-time trade-off theory | Formula; lowest cost slope crashed first |
| Time–cost trade-off curve | CPM theory | All-normal, all-crash, optimal points |
| Direct vs indirect cost | Standard project economics | Direct increases with crashing; indirect decreases |
B. Why It’s Used
Resource planning and crashing bridge project scheduling with cost management — a core construction management competency. GATE AR includes crashing NATs requiring multi-step cost slope calculations and sequential decision-making. UPSC-CPWD tests conceptual distinctions (smoothing vs leveling; which cost rises or falls). The cost slope formula and its application rule (lowest slope crashed first) appear in MCQs and NATs across every exam cycle.
C. Mechanism in Words
Resource Leveling / Smoothing:
1. A project schedule is created from the network; resource requirements are computed per time period.
2. The resulting resource histogram shows peaks (over-allocation) and valleys (under-utilisation).
3. Non-critical activities are rescheduled within their available float to flatten the histogram.
4. If float is insufficient to achieve uniform allocation, the project end date may be extended (leveling) or a constrained best-fit result is accepted (smoothing).
Project Crashing:
1. The project duration is computed at normal activity durations.
2. Each activity has a maximum compression (crash time) achievable at additional cost (crash cost).
3. Crashing the critical path reduces project duration; each day saved reduces indirect costs (overhead).
4. The process continues as long as the cost saving from shorter duration (indirect cost reduction) outweighs the crashing cost (direct cost increase).
5. The optimal project duration is the point of minimum total cost.
D. Core Concept Explanations
C1. Resource Leveling vs Resource Smoothing
| Property | Resource Smoothing | Resource Leveling |
|---|---|---|
| Project duration | FIXED — cannot be extended | MAY be extended |
| Method | Reschedule non-critical activities within available float | Reschedule activities; extend project end date if necessary |
| Primary constraint | Time (deadline is non-negotiable) | Resources (labour continuity or equipment limits matter more) |
| Float usage | Uses existing float; if insufficient, peaks remain | May exhaust float AND extend the schedule |
| Application | Contractually fixed completion date | Continuous crew employment; equipment mobilisation costs |
Exam Anchor: “Resource Smoothing = Duration Fixed; Resource Leveling = Duration May Extend.” If a question asks which technique may extend project duration, the answer is leveling, not smoothing.
Resource histogram: A bar chart plotting resource quantity (persons, equipment units) on the Y-axis against time on the X-axis. Peaks represent over-allocation (more resources than available); valleys represent idle periods. The objective of both techniques is to flatten this histogram.
C2. Project Crashing — Terminology and Cost Slope Formula
| Term | Definition |
|---|---|
| Normal Time (N_t) | Activity duration under normal resource allocation |
| Normal Cost (N_c) | Activity cost at normal duration |
| Crash Time (C_t) | Minimum achievable activity duration (maximum resources) |
| Crash Cost (C_c) | Activity cost at crash time. Always ≥ Normal Cost. |
| Cost Slope | Rate of additional cost per unit time reduction. = (C_c − N_c) / (N_t − C_t) |
| Direct Cost | Cost that INCREASES when activities are crashed (overtime, extra equipment, double shifts) |
| Indirect Cost | Cost that DECREASES with shorter project duration (overhead, site facilities, supervision, penalties for late completion avoided) |
| Optimum Duration | Project duration at which Total Cost (Direct + Indirect) is minimised |
Cost Slope formula:
$$text{Cost Slope} = frac{C_c – N_c}{N_t – C_t}$$
Crash sequence rule: Always crash the critical activity with the LOWEST cost slope first. This minimises the additional direct cost per day saved.
After each crash step: Recompute the network. The critical path may change — a previously non-critical path may become critical. If two paths are simultaneously critical, BOTH must be crashed together to reduce overall duration.
C3. Time–Cost Trade-Off Curve
Three reference points define the curve:
| Point | Duration | Cost |
|---|---|---|
| All-normal | Longest (all activities at normal time) | Lowest direct cost; highest indirect cost |
| All-crash | Shortest (all activities at crash time) | Highest direct cost; lowest indirect cost |
| Optimum | Between the two | Minimum total cost |
The curve shape: as project duration decreases from all-normal, total cost initially falls (indirect savings > crash cost), reaches a minimum at the optimum duration, then rises (crash cost > indirect savings for further reduction).
Exam Trap: “Crashing reduces total project cost.” This is ONLY true down to the optimum duration. Beyond the optimum, crashing INCREASES total cost. Total cost has a minimum — it does not decrease indefinitely as duration falls.
E. Worked Problems — Crashing
CRASHING PROBLEM 2.5-1 (Series Network — Single Critical Path)
Network: A → B → C (series; all activities critical)
| Activity | Normal Time (days) | Normal Cost (₹) | Crash Time (days) | Crash Cost (₹) | Cost Slope (₹/day) |
|---|---|---|---|---|---|
| A | 6 | 6,000 | 4 | 9,000 | (9000−6000)/(6−4) = 1,500 |
| B | 8 | 10,000 | 5 | 13,000 | (13000−10000)/(8−5) = 1,000 |
| C | 4 | 8,000 | 2 | 10,000 | (10000−8000)/(4−2) = 1,000 |
Normal project duration = 6+8+4 = 18 days
Normal direct cost = 6,000 + 10,000 + 8,000 = ₹24,000
Indirect cost = ₹2,000/day
Crashing sequence (crash lowest cost slope first — B and C tied at ₹1,000/day; crash either):
Step 1 — Crash C by 1 day (cost slope = ₹1,000):
Duration = 17 days.
Direct cost change: +₹1,000.
Indirect cost saving: 1 day × ₹2,000 = −₹2,000.
Net change: −₹1,000 (total cost decreases → economical).
Step 2 — Crash C by 1 more day (C now at crash limit = 2 days):
Duration = 16 days. Direct +₹1,000; Indirect −₹2,000. Net: −₹1,000. Still economical.
C is now fully crashed. Next cheapest = B or A (both ₹1,000/day vs ₹1,500/day).
Step 3 — Crash B by 1 day:
Duration = 15 days. Direct +₹1,000; Indirect −₹2,000. Net: −₹1,000. Economical.
Steps 4 & 5 — Crash B by 2 more days (B at crash limit = 5 days):
Each step: net −₹1,000. Duration: 14, 13 days.
Now B and C are fully crashed. Only A remains (cost slope ₹1,500/day vs indirect saving ₹2,000/day).
Step 6 — Crash A by 1 day:
Duration = 12 days. Direct +₹1,500; Indirect −₹2,000. Net: −₹500. Still economical.
Step 7 — Crash A by 1 more day (A at crash limit = 4 days):
Duration = 11 days. Direct +₹1,500; Indirect −₹2,000. Net: −₹500. Economical.
Summary table:
| Duration (days) | Direct Cost (₹) | Indirect Cost (₹) | Total Cost (₹) |
|---|---|---|---|
| 18 (all-normal) | 24,000 | 36,000 | 60,000 |
| 16 | 26,000 | 32,000 | 58,000 |
| 13 | 29,000 | 26,000 | 55,000 |
| 11 (all-crash) | 31,000 | 22,000 | 53,000 |
In this network, because indirect cost (₹2,000/day) exceeds cost slope of all activities, ALL crashing is economical. The optimum = all-crash at 11 days (total cost ₹53,000).
CRASHING PROBLEM 2.5-2 (Parallel Paths — Both Paths Must Be Crashed)
Network: Start → End via two parallel paths.
– Path 1: A(5) → C(4) → End [duration = 9 days]
– Path 2: B(7) → D(2) → End [duration = 9 days]
Both paths are critical (both = 9 days).
| Activity | Normal | Crash | Normal Cost (₹) | Crash Cost (₹) | Cost Slope (₹/day) |
|---|---|---|---|---|---|
| A | 5 | 3 | 500 | 700 | (700−500)/(5−3) = 100 |
| B | 7 | 5 | 600 | 1,200 | (1200−600)/(7−5) = 300 |
| C | 4 | 2 | 400 | 600 | (600−400)/(4−2) = 100 |
| D | 2 | 1 | 200 | 350 | (350−200)/(2−1) = 150 |
Indirect cost = ₹300/day. Normal total cost = (500+600+400+200) + 9×300 = ₹1,700 + ₹2,700 = ₹4,400.
To reduce duration by 1 day, BOTH paths must be shortened by 1 day simultaneously.
Cheapest combination of one activity from each path:
| Combination | Cost per Day Saved |
|---|---|
| A (Path 1) + B (Path 2) | 100 + 300 = 400 |
| A (Path 1) + D (Path 2) | 100 + 150 = 250 ← cheapest |
| C (Path 1) + B (Path 2) | 100 + 300 = 400 |
| C (Path 1) + D (Path 2) | 100 + 150 = 250 ← tied |
Indirect saving per day = ₹300. Combined crash cost = ₹250/day < ₹300 saving → economical.
Step 1 — Crash A and D by 1 day each:
– A: 5→4; D: 2→1 (D now at crash limit)
– Duration: 8 days.
– Path 1 = 4+4 = 8; Path 2 = 7+1 = 8 → both still critical ✓
– Direct cost: +₹100 (A) + ₹150 (D) = +₹250; Indirect: −₹300. Net: −₹50.
Step 2 — D is now at crash limit. Must find new combination.
– A (Path 1) + B (Path 2): 100 + 300 = ₹400/day vs ₹300 saving → NOT economical.
– C (Path 1) + B (Path 2): 100 + 300 = ₹400/day → NOT economical.
No further economical crashing is possible.
Optimum duration = 8 days.
| Duration | Direct Cost (₹) | Indirect Cost (₹) | Total Cost (₹) |
|---|---|---|---|
| 9 (all-normal) | 1,700 | 2,700 | 4,400 |
| 8 (optimum) | 1,950 | 2,400 | 4,350 |
Key lesson from Problem 2: When two paths are simultaneously critical, individual activity crashing on only ONE path is insufficient — you must crash one activity on EACH critical path simultaneously. The cheapest pair of activities (one from each path) is selected.
F. Design Criteria
| Decision | Criterion | Rule |
|---|---|---|
| Which activity to crash first | Cost slope = (Cc − Nc)/(Nt − Ct) | Crash the activity with the LOWEST cost slope first |
| When to stop crashing | Marginal analysis | Stop when cost slope > indirect cost per day saved |
| Resource smoothing | Available float | Reschedule within float; project duration fixed |
| Resource leveling | Resource constraint vs time constraint | May extend duration; resource continuity prioritised |
| Optimum project duration | Minimum total cost | Total cost = Direct cost + Indirect cost |
Code Comparisons
| Parameter | Resource Smoothing | Resource Leveling |
|---|---|---|
| Project duration | FIXED | May extend |
| Method | Uses existing float | May exhaust float AND extend schedule |
| Primary constraint | Time (deadline fixed) | Resources (labour/equipment continuity) |
| Application | Contractual deadlines | Crew continuity / equipment demobilisation costs |
| Cost Component | Behaviour when crashing | Components |
|---|---|---|
| Direct cost | INCREASES | Labour, materials, overtime, extra equipment |
| Indirect cost | DECREASES | Overhead, site facilities, supervision, penalty avoidance |
| Total cost | Decreases to optimum, then increases | Minimum at optimum duration |
G. Application Zones
| Context | Technique | Reason |
|---|---|---|
| Contractually fixed deadline | Resource smoothing | Duration cannot change; work redistributed within float only |
| Continuous crew deployment required | Resource leveling | Labour cost control; no idle periods; schedule may slip |
| Client demands early completion | Crashing | Pay more direct cost to save indirect cost until optimum |
| Two simultaneously critical paths | Crash one activity on each path | Crashing only one path is wasteful — the other path still governs duration |
| Budget-limited project | Optimum duration analysis | Minimum total cost point, not minimum duration point |
H. Common Confusions
| Confusion | Correct Distinction |
|---|---|
| Resource smoothing extends project duration | Only resource leveling may extend the project duration. Resource smoothing holds duration fixed and uses existing float. |
| Crashing always reduces total project cost | Crashing reduces total cost ONLY down to the optimum. Beyond the optimum, crash costs exceed indirect savings and total cost increases. |
| Crash the longest activity first | Always crash the critical activity with the LOWEST COST SLOPE first, not the longest duration. Duration is irrelevant to the crashing sequence. |
| Crashing non-critical activities helps | Crashing a non-critical activity only increases direct cost without shortening the project (the critical path is not shortened). Only crash critical activities. |
| After crashing, the critical path stays the same | After each crash step, recompute the network. The critical path MAY change — a non-critical path may become critical after a formerly critical activity is shortened. |
| Both direct and indirect costs increase when crashing | Direct cost INCREASES when crashing (more resources, overtime). Indirect cost DECREASES (overhead, supervision spread over shorter duration). They move in opposite directions. |
I. Compare & Contrast
| Pair | Item 1 | Item 2 | Key Difference |
|---|---|---|---|
| Resource techniques | Smoothing | Leveling | Smoothing: duration fixed, use existing float; Leveling: duration may extend |
| Cost types | Direct cost | Indirect cost | Direct: rises with crashing (more resources); Indirect: falls (less overhead per day) |
| Crashing points | Normal duration | Crash duration | Normal: cheapest direct cost; Crash: fastest; Optimum: minimum total cost |
| Crash selection | Lowest cost slope | Highest crash cost | Always crash LOWEST cost slope — not the most expensive activity |
| Dual critical paths | Crash one path | Crash both paths | With two simultaneous critical paths: MUST crash one activity on EACH path |
J. Memory Hooks
- Smoothing = Same schedule: S for Smooth, S for Same duration
- Leveling = Lets schedule slip: L for Level, L for Longer possible duration
- Direct cost UP when crashing: “Crashing the schedule = crashing the budget upward”
- Indirect cost DOWN when crashing: “Less time on site = less daily overhead”
- Cost slope = ₹/day saved: divide the cost difference by the time difference — it’s a RATE
- Crash LOWEST slope first: cheapest per day = most economical first
- Stop when slope > indirect saving: “The crash costs more than what it saves”
- Two critical paths → crash BOTH: “Both paths are holding the project hostage simultaneously”
K. Revision Ladder
| Level | Task | Test Yourself |
|---|---|---|
| L1 — Recall | State: smoothing vs leveling difference; cost slope formula; which cost rises/falls with crashing; crashing stop condition | Write the cost slope formula and both resource management distinctions from memory |
| L2 — Apply | Compute cost slopes for 3 activities; determine crash sequence; calculate total cost at 2 different durations; identify the optimum | Work through Problems 2.5-1 and 2.5-2 without checking solutions |
| L3 — Analyse | In a three-path network where one path becomes critical after a first crash step, determine the new cheapest crashing strategy; explain why the optimum is usually not the all-crash point; justify why crashing non-critical activities is wasteful | Rebuild Problem 2.5-2 with indirect cost = ₹200/day — does the optimum shift? Why? |
L. Exam Traps
| Trap | Incorrect Assumption | Correct Answer |
|---|---|---|
| T25 | “Resource smoothing and resource leveling are the same” | Smoothing = duration fixed; peaks reduced within float. Leveling = duration MAY extend. They are different operations. |
| T26 | “Crash the activity with the highest crash cost first” | Cash the activity with the LOWEST COST SLOPE first. Crash cost alone does not determine sequence — cost per unit time saved (cost slope) does. |
| T27 | “Crashing continues until all-crash duration” | Crashing stops when further crashing is NOT economical (crash cost slope > indirect cost per day). The optimum is usually not the all-crash point. |
| T28 | “Cost slope = crash cost − normal cost” | Cost slope = (C_c − N_c) / (N_t − C_t) — the cost difference DIVIDED by the time difference. The numerator alone is the total extra cost, not the per-day rate. |
| T29 | “In a two-critical-path network, crashing one activity shortens the project” | When two paths are simultaneously critical, ONE activity on EACH path must be crashed. Crashing only one path is wasteful — the uncrashed path still governs duration. |
M. Answer-Writing Cues
For resource management questions:
“Resource smoothing redistributes work within available float to reduce peak demand without extending the project end date. Resource leveling may extend the project duration if the available float is insufficient to eliminate resource peaks — it prioritises uniform resource utilisation over meeting the original deadline.”
For crashing questions:
“To optimally reduce project duration, the critical activity with the lowest cost slope is crashed first. The cost slope = (crash cost − normal cost) / (normal duration − crash time), representing the additional direct cost per unit time saved. Crashing continues as long as the indirect cost saving per day exceeds the cost slope of the activity being crashed. The optimum duration minimises total project cost (direct + indirect).”
N. PYQ Integration (2007–2026 verified)
| Topic | Exam Appearance | Question Pattern |
|---|---|---|
| Resource smoothing vs leveling | GATE, UPSC-CPWD | MCQ: “Which technique may extend project duration?” → Leveling |
| Cost slope formula | GATE (NAT), UPSC-CPWD | NAT: compute cost slope given normal and crash data |
| Crash sequence (lowest cost slope first) | GATE, UPSC-CPWD | MCQ: given a table of activities, which is crashed first? |
| Optimal crashing stop condition | GATE | MCQ: “Crashing is stopped when…” → cost slope > indirect cost per day |
| Direct vs indirect cost behaviour | GATE, UPSC-CPWD | MCQ: “Which cost increases when activities are crashed?” → Direct cost |
| Time–cost trade-off curve shape | UPSC-CPWD | MCQ: describe the shape; where is the minimum total cost? |
| 4 activities; duration, relationship and cost parameters given; indirect cost ₹5,000/week; complete by 12 weeks; total project cost = ? | GATE AR 2025 Q46 | NAT: Ans: 164,000 |
O. Mini-Check — Lesson 2.5 (5 Questions)
Q1 (MCQ): Which resource management technique may extend the project completion date?
(A) Resource smoothing (B) Gantt chart leveling (C) Resource leveling (D) Critical path compression
A1: (C) Resource leveling. Resource leveling may extend the project duration when available float is insufficient to achieve uniform resource allocation. Resource smoothing keeps the duration FIXED and only redistributes work within existing float.
Q2 (NAT): Activity M has a normal duration of 8 days (normal cost ₹4,000) and can be crashed to 5 days at a cost of ₹5,500. Calculate the cost slope.
A2:
Cost Slope = (C_c − N_c) / (N_t − C_t) = (5,500 − 4,000) / (8 − 5) = 1,500 / 3 = ₹500 per day
Q3 (MCQ): A project has three critical activities with cost slopes of ₹400/day, ₹700/day, and ₹250/day. The indirect cost rate is ₹600/day. Which activity should be crashed first, and is it economical?
(A) Cost slope ₹250/day; economical (B) Cost slope ₹700/day; not economical (C) Cost slope ₹400/day; economical (D) Cost slope ₹250/day; not economical
A3: (A) The activity with the LOWEST cost slope (₹250/day) is crashed first. Since ₹250 < ₹600 (indirect cost saving per day), crashing is economical — each day reduced saves a net ₹350.
Q4 (MCQ): What happens to direct and indirect project costs as activities are crashed (project duration decreases)?
(A) Both direct and indirect costs increase
(B) Direct cost decreases; indirect cost increases
(C) Direct cost increases; indirect cost decreases
(D) Both decrease until the optimum, then both increase
A4: (C). Direct cost INCREASES when crashing (more resources, overtime). Indirect cost DECREASES as the project duration shortens (less overhead, supervision, and facilities per day). They move in opposite directions — this opposing behaviour creates the total cost minimum at the optimum duration.
Q5 (NAT): A two-activity series network (P→Q) has normal duration 14 days and indirect cost ₹1,500/day. Activity P (normal 8 days, crash 6 days) has a cost slope of ₹800/day. Activity Q (normal 6 days, crash 4 days) has a cost slope of ₹600/day. Normal direct costs: P = ₹8,000, Q = ₹6,000. What is the total project cost (direct + indirect) at the optimum duration?
A5:
– Crash Q first (lower cost slope ₹600 < indirect saving ₹1,500 → economical).
– Crash Q by 1 day: duration = 13 days. Direct +600; Indirect −1500. Net: −900. ✓
– Crash Q by 1 more day (Q at crash limit, 4 days): duration = 12 days. Net: −900. ✓
– Now crash P (cost slope ₹800 < ₹1,500 → still economical).
– Crash P by 1 day: duration = 11 days. Net: −700. ✓
– Crash P by 1 more day (P at crash limit, 6 days): duration = 10 days. Net: −700. ✓
– All activities at crash limit; further crashing impossible.
Optimum = all-crash at 10 days (since every step is economical):
– Direct cost = (8,000 + 2×800) + (6,000 + 2×600) = 9,600 + 7,200 = ₹16,800
– Indirect cost = 10 × 1,500 = ₹15,000
– Total cost = ₹31,800
Normal total cost: (8,000 + 6,000) + 14 × 1,500 = 14,000 + 21,000 = ₹35,000. Crashing saves ₹3,200.