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22. FER - Construction Management#
Contributor(s)
Dr. Deng is a content contribuitor for this section as well as the subject matter expert. Fall 2025 Dr. Gugssa is the subject matter expert.
Readings#
Videos#
Typical “Knowledge” Questions#
1. In a typical construction project schedule, the “critical path” is defined as:#
A. The sequence of activities with the highest total cost.
B. The activities with the greatest number of resources assigned.
C. The path of activities that determines the minimum project duration.
D. The activities most likely to cause safety incidents.
2. A cost-loaded schedule is primarily used for:#
A. Identifying the order in which subcontractors must be hired.
B. Determining time–cost tradeoffs during project acceleration.
C. Tracking planned vs. actual expenditures across the project timeline.
D. Ensuring that equipment is delivered to the site on time.
3. Which type of construction contract places the most financial risk on the contractor?#
A. Cost-plus-fixed-fee
B. Lump sum (fixed price)
C. Guaranteed maximum price
D. Unit price
4. In the context of construction safety management, the “EMR” (Experience Modification Rate) is used by insurers to:#
A. Determine whether a contractor is allowed to bid on public work.
B. Calculate a contractor’s workers’ compensation insurance premium.
C. Measure equipment depreciation rates for tax purposes.
D. Quantify the number of days a worker may miss due to injury.
5. A change order is typically issued when:#
A. The contractor wants to purchase new equipment for future projects.
B. The owner requests a modification to the work after contract award.
C. Weather delays require the project to be extended.
D. The contractor submits their final payment application.
Typical Quantitative Questions#
Problem 1 – CPM Network and Floats#
A small building project has the following activities:
Activity |
Duration (days) |
Immediate Predecessor(s) |
|---|---|---|
A |
4 |
|
B |
3 |
A |
C |
5 |
A |
D |
2 |
B |
E |
6 |
C |
F |
3 |
D, E |
Assume the project starts at day 0.
Tasks:
Compute the earliest start (ES) and earliest finish (EF) for each activity.
Compute the latest start (LS) and latest finish (LF) for each activity.
Identify the project duration and the critical path.
Solution 1#
Forward pass (ES/EF)
A:
ES = 0
EF = ES + 4 = 4
B:
ES = EF(A) = 4
EF = 4 + 3 = 7
C:
ES = EF(A) = 4
EF = 4 + 5 = 9
D:
ES = EF(B) = 7
EF = 7 + 2 = 9
E:
ES = EF(C) = 9
EF = 9 + 6 = 15
F:
ES = max[EF(D), EF(E)] = max[9, 15] = 15
EF = 15 + 3 = 18
Project duration = 18 days
Backward pass (LF/LS)
Start from final activity F: project completion time = 18.
F:
LF = 18
LS = LF − 3 = 15
D: (successor = F)
LF = LS(F) = 15
LS = 15 − 2 = 13
E: (successor = F)
LF = LS(F) = 15
LS = 15 − 6 = 9
B: (successor = D)
LF = LS(D) = 13
LS = 13 − 3 = 10
C: (successor = E)
LF = LS(E) = 9
LS = 9 − 5 = 4
A: (successors = B, C)
LF = min[LS(B), LS(C)] = min[10, 4] = 4
LS = 4 − 4 = 0
Float & critical path
Total float = LS − ES (or LF − EF):
Act |
ES |
EF |
LS |
LF |
Float |
|---|---|---|---|---|---|
A |
0 |
4 |
0 |
4 |
0 |
B |
4 |
7 |
10 |
13 |
6 |
C |
4 |
9 |
4 |
9 |
0 |
D |
7 |
9 |
13 |
15 |
6 |
E |
9 |
15 |
9 |
15 |
0 |
F |
15 |
18 |
15 |
18 |
0 |
Activities with zero float: A – C – E – F
Critical path: A → C → E → F Project duration: 18 days
Problem 2 – Time–Cost Tradeoff (Crashing)#
Consider the critical path found in Problem 1: A–C–E–F. You are given the following normal and crash data for these critical activities:
Activity |
Normal Duration (days) |
Crash Duration (days) |
Normal Cost ($) |
Crash Cost ($) |
|---|---|---|---|---|
A |
4 |
3 |
8,000 |
9,500 |
C |
5 |
3 |
10,000 |
14,000 |
E |
6 |
4 |
12,000 |
15,000 |
F |
3 |
2 |
6,000 |
7,800 |
Compute the crash cost per day for each activity.
Suppose the owner offers a bonus of $3,000 per calendar day the project is finished early (up to 3 days maximum). Which activities would you crash, and by how much, to maximize net benefit?
Solution 2#
Crash cost per day
Crash cost per day = (Crash Cost − Normal Cost) / (Normal Duration − Crash Duration)
A:
ΔCost = 9,500 − 8,000 = 1,500
ΔTime = 4 − 3 = 1 day
Crash cost/day = 1,500 / 1 = $1,500/day
C:
ΔCost = 14,000 − 10,000 = 4,000
ΔTime = 5 − 3 = 2 days
Crash cost/day = 4,000 / 2 = $2,000/day
E:
ΔCost = 15,000 − 12,000 = 3,000
ΔTime = 6 − 4 = 2 days
Crash cost/day = 3,000 / 2 = $1,500/day
F:
ΔCost = 7,800 − 6,000 = 1,800
ΔTime = 3 − 2 = 1 day
Crash cost/day = 1,800 / 1 = $1,800/day
Summary:
Act |
Crash Cost / Day ($/day) |
|---|---|
A |
1,500 |
C |
2,000 |
E |
1,500 |
F |
1,800 |
Decide what to crash (bonus = $3,000/day)
Each day saved yields $3,000. Compare with crash cost/day:
A: cost = 1,500 → net +1,500 per day
E: cost = 1,500 → net +1,500 per day
F: cost = 1,800 → net +1,200 per day
C: cost = 2,000 → net +1,000 per day
Because the project duration is 18 days, and all four are on the critical path, each day we reduce along A–C–E–F reduces project duration by 1 day (until some new critical path emerges, which we’ll ignore for this small example as long as we don’t over-crash).
We can crash up to 3 days total. The most economical order is least crash cost/day first:
Crash A by 1 day:
Cost = 1,500
Bonus = 3,000
Net benefit = +1,500
Crash E by 2 days:
Cost = 2 × 1,500 = 3,000
Bonus = 2 × 3,000 = 6,000
Net benefit = +3,000
Total crash = 3 days, which is the limit.
Total:
Extra cost = 1,500 + 3,000 = 4,500
Bonus = 3 × 3,000 = 9,000
Net benefit = 9,000 − 4,500 = $4,500
Crash 1 day on A and 2 days on E for maximum net benefit under the given assumptions.
Problem 3 – Equipment Production Rate and Cost#
A bulldozer is used to spread and compact fill material. Its effective operating time is 45 minutes per hour (the rest is delays, turning, etc.). Each cycle, it moves 5 cubic yards of material, and a full cycle takes 3 minutes.
The ownership and operating cost of the bulldozer is $180 per machine hour.
Tasks:
Compute the effective production rate in cubic yards per hour.
Compute the cost per cubic yard of placing the fill with this bulldozer.
Solution 3#
Production rate
Effective working time = 45 minutes/hour
Cycle time = 3 minutes/cycle
Cycles per effective hour = 45 / 3 = 15 cycles/hour
Each cycle moves 5 CY: Production = 15 cycles/hour × 5 CY/cycle = 75 CY/hour
Cost per cubic yard
Cost per hour = $180
Production = 75 CY/hour
Cost per CY = 180 / 75 = $2.40 per CY
Production rate = 75 CY/hr, cost ≈ $2.40/CY.
Problem 4 – Labor Productivity and Duration#
A concrete crew can place and finish 250 square feet of slab per hour under normal conditions. A project requires placing 12,000 square feet of slab. The crew works 8-hour days, and productivity is expected to drop to 80% of normal after 6 hours each day due to fatigue and coordination losses.
Assume:
First 6 hours each day are at 100% productivity.
Last 2 hours are at 80% of the nominal rate.
Tasks:
Compute the average daily production (in square feet per day).
Estimate the number of days required to place 12,000 square feet.
Solution 4#
Daily production
Nominal productivity: 250 ft²/hour.
First 6 hours: Production = 6 hr × 250 ft²/hr = 1,500 ft²
Last 2 hours at 80%: Effective rate = 0.8 × 250 = 200 ft²/hr; Production = 2 hr × 200 = 400 ft² Total daily production = 1,500 + 400 = 1,900 ft²/day
Number of days Total work = 12,000 ft² Days = 12,000 / 1,900 ≈ 6.3158 days
So: Need 7 working days (rounding up to whole days). Approximate duration: 7 days.