BE Civil Engineering (IOE, TU) Construction Project Management (IOE, CE 752) Question Paper 2080 Nepal

(a) AON network (logic)

Start ─► A ─► C ─► E ─► G ─► End
   │    └─► D ─► F ─┘     ▲
   └─► B ────────────► E ─┘

A and B start the project. C and D follow A. E needs B and C. F needs D. G needs E and F.

(b) Forward pass (ES, EF), EF = ES + Duration

E: ES = max(EF_B, EF_C) = max(6, 9) = 9. G: ES = max(EF_E, EF_F) = max(16, 11) = 16.

Project duration = 21 days.

Backward pass (LF, LS), LS = LF − Duration, starting from project end LF = 21:

A: LF = min(LS_C, LS_D) = min(4, 9) = 4.

(c) Critical path = activities with zero float: A → C → E → G, length 4 + 5 + 7 + 5 = 21 days.

(d) Total float = LS − ES (= LF − EF)

Critical activities (A, C, E, G) have zero float, confirming the critical path.

(a) Expected time $t_e = \dfrac{t_o + 4t_m + t_p}{6}$, variance $\sigma^2 = \left(\dfrac{t_p - t_o}{6}\right)^2$

(b) Path durations (te):

• Path 1-2-4-5 (A,C,E): 4 + 6 + 5 = 15 weeks
• Path 1-3-4-5 (B,D,E): 6 + 3 + 5 = 14 weeks

Critical path = A → C → E (1-2-4-5), expected duration $T_e = 15$ weeks.

(c) Project variance along critical path = $\sigma^2_A + \sigma^2_C + \sigma^2_E = 0.444 + 0.444 + 0.444 = 1.333$.

Standard deviation $\sigma = \sqrt{1.333} = 1.155$ weeks.

$$z = \frac{T_s - T_e}{\sigma} = \frac{22 - 15}{1.155} = \frac{7}{1.155} = 6.06$$

For $z = 6.06$ the cumulative probability $\Phi(z) \approx 1.000$.

Probability of completing within 22 weeks ≈ 99.9% (≈ 1.00). (The deadline is far beyond the mean, so completion is virtually certain. Note: the supplied $z=0.71,\,\Phi=0.7611$ value applies if a deadline of $T_s = 15 + 0.71\times1.155 \approx 15.8$ weeks were asked; with 22 weeks the probability is essentially unity.)

(a) Normal duration. Two paths:

• A-B-C: 8 + 10 + 6 = 24 days
• A-D-C: 8 + 12 + 6 = 26 days ← critical

Normal duration = 26 days.

Total direct cost (normal) = 10000 + 8000 + 6000 + 9000 = Rs. 33,000. Indirect cost = 26 × 1500 = Rs. 39,000. Total normal cost = 33,000 + 39,000 = Rs. 72,000.

(b) Cost slope $= \dfrac{\text{Crash Cost} - \text{Normal Cost}}{\text{Normal Time} - \text{Crash Time}}$

(c) Step-by-step crashing. Critical path is A-D-C (26 d); A-B-C = 24 d (2 days of slack).

Step 1: Cheapest on critical path A-D-C is C (1000). Crash C by 2 days → C = 4. New durations: A-D-C = 8+12+4 = 24; A-B-C = 8+10+4 = 22. Duration = 24 d. Cost added = 2 × 1000 = 2000. Both paths now: 24 and 22. Critical = A-D-C.

Step 2: On A-D-C remaining: A (2000), D (1500). Crash D (cheaper, 1500). D shared only by one path. Crash D by 2 days to keep above A-B-C (24 vs A-B-C 22 → can crash D by up to (24−22)=2 before paths equalise, D has 3 available). Crash D by 2 → D = 10. A-D-C = 8+10+4 = 22; A-B-C = 22. Duration = 22 d. Cost added = 2 × 1500 = 3000.

Step 3: Now BOTH paths critical (22 d). To reduce further must shorten both. Option: crash A (common, slope 2000, reduces both) — 1 day available remaining? A has 2 available. Alternatively crash D + B together: D(1500)+B(1500) = 3000/day vs A alone 2000/day. A is cheaper (2000). Crash A by 2 days → A = 6. Both paths reduce by 2 → A-D-C = 20; A-B-C = 20. Duration = 20 d. Cost added = 2 × 2000 = 4000.

Step 4: Both paths = 20 d. Further crash needs D (1500) + B (1500) = 3000/day (A exhausted). Marginal crash cost 3000/day > indirect saving 1500/day → not economical. Stop.

Cost summary

Crashing is worthwhile while slope < 1500/day (steps with slope 1000) and break-even at 1500. The optimum (least total cost) is at 22 days with total cost Rs. 71,000 (24 days also gives 71,000; 22 days is preferred as the shorter duration at equal cost).

Optimum duration = 22 days, minimum total cost = Rs. 71,000.

(a) Production rate.

Cycles per hour (ideal) = $\dfrac{3600\ \text{s/hr}}{20\ \text{s/cycle}} = 180$ cycles/hr.

Loose volume per cycle = bucket capacity × fill factor = $1.5 \times 0.9 = 1.35\ \text{m}^3$ (loose).

Convert loose → bank: bank volume = loose × swell factor = $1.35 \times 0.80 = 1.08\ \text{BCM/cycle}$.

Ideal production = $180 \times 1.08 = 194.4\ \text{BCM/hr}$.

Apply corrections:

• Job efficiency = 50/60 = 0.833
• Depth/swing factor = 0.95

$$Q = 194.4 \times 0.833 \times 0.95 = 153.86\ \text{BCM/hr}$$

Production rate ≈ 153.9 BCM/hr.

(b) Shifts required.

Production per 8-hour shift = $153.86 \times 8 = 1230.9\ \text{BCM/shift}$.

Number of shifts = $\dfrac{6000}{1230.9} = 4.87 \approx \mathbf{5\ \text{shifts}}$ (rounded up).

(c) Number of trucks.

Truck loading time = time for the shovel to fill the truck = (truck capacity / production per cycle in BCM) × cycle time.

Bank volume per shovel cycle = $1.08\ \text{BCM}$. Cycles to fill a 9 BCM truck = $9 / 1.08 = 8.33$ cycles. Loading time = $8.33 \times 20\ \text{s} = 166.7\ \text{s} = 2.78\ \text{min}$.

Truck cycle time = loading + haul = $2.78 + 12 = 14.78\ \text{min}$.

Number of trucks $= \dfrac{\text{truck cycle time}}{\text{loading time}} = \dfrac{14.78}{2.78} = 5.32 \approx \mathbf{6\ \text{trucks}}$ (rounded up to keep the shovel busy).

Resource allocation is the assignment of available resources (labour, equipment, materials, money) to project activities in an economic and timely way so that activities can proceed as scheduled. When the resource demand of the schedule (e.g., derived from an early-start histogram) exceeds availability, the schedule must be adjusted.

Resource levelling (smoothing) is the process of rescheduling activities — primarily by using their available float — so that the resource usage histogram becomes as uniform (level) as possible, removing sharp peaks and valleys, without extending the project duration. It reduces day-to-day fluctuation in crew size and improves efficiency.

Time-constrained vs resource-constrained scheduling

Example. Suppose two non-critical activities each need 4 masons and both can run in week 3, but only 5 masons are available.

• Time-constrained: keep the deadline; if both have float, shift one to a different week so the peak demand drops from 8 to about 4–5 masons — duration unchanged.
• Resource-constrained: with only 5 masons available and no float, one activity must wait until masons are free, delaying the project.

Steps of the resource levelling procedure

1. Prepare the network and carry out CPM to find ES, EF, LS, LF and floats.
2. Schedule all activities at their early start and draw the resource histogram (aggregation).
3. Identify peaks (over-allocation) and the activities with float that contribute to them.
4. Shift non-critical activities within their float (delay start) to fill the valleys and cut the peaks.
5. Re-aggregate and check; iterate until the histogram is as level as possible without changing project duration.
6. Finalise the levelled schedule.

Two benefits of resource levelling

1. More uniform manpower/equipment demand → less hiring-and-firing, lower mobilisation cost and better productivity.
2. Avoids unrealistic resource peaks and idle troughs, improving site management, cash flow and resource utilisation.

Resource aggregation vs allocation vs levelling (summary). Aggregation simply totals the resource demand period-by-period from the schedule (the histogram). Allocation assigns limited resources to activities (may extend duration under a resource limit). Levelling smooths the aggregated demand using float without changing the project duration. Together they move a plan from a feasible logic network to a practical, resource-realistic schedule.

Work Breakdown Structure (WBS). A WBS is a hierarchical, deliverable-oriented decomposition of the total project scope into progressively smaller, manageable components — from the whole project, to major deliverables/phases, down to work packages and finally to activities. It defines what has to be done, forms the basis for estimating, scheduling, cost control and responsibility assignment, and ensures no scope is omitted or duplicated (100% rule).

Bar (Gantt) chart. A bar chart is a graphical scheduling tool in which each activity is shown as a horizontal bar drawn against a horizontal time scale; the bar's start point, length and end point represent the activity's start, duration and finish. Progress can be shown by shading the bars.

Two advantages of bar charts

1. Simple to prepare, read and understand — good for communication with clients and field staff.
2. Directly shows the timing and duration of activities and allows quick progress monitoring.

Two limitations (vs network methods like CPM/PERT)

1. It does not clearly show the logical inter-dependencies between activities.
2. It cannot directly identify the critical path or activity float, so the effect of a delay on overall completion is not evident.

1. Lump-sum (fixed-price) contract. The contractor agrees to execute the whole defined work for a single fixed price.

• Suitable when: the scope, drawings and specifications are complete and well defined and quantities are unlikely to change (e.g., a building of known design).
• Risk: the contractor carries most of the risk (quantity/price overruns), because the price is fixed regardless of actual cost.

2. Item-rate / unit-price (schedule of rates) contract. The contractor quotes a unit rate for each item of work (per m³, m², m, etc.); payment = actual measured quantity × quoted rate.

• Suitable when: the nature of work is known but exact quantities cannot be precisely fixed in advance (e.g., earthwork, roads, irrigation). Common in government works in Nepal.
• Risk: shared — the owner bears the quantity (variation) risk while the contractor bears the unit-rate (price) risk.

3. Cost-plus contract. The owner reimburses the contractor's actual allowable cost plus an agreed fee (a fixed fee or a percentage of cost).

• Suitable when: scope is not well defined, the work is urgent, or it involves novel/uncertain work (emergency or research-type projects).
• Risk: the owner carries most of the risk, since the final cost is open-ended; cost-plus-percentage gives the contractor little incentive to economise, while cost-plus-fixed-fee is somewhat better.

(Other forms include turnkey/EPC and BOT contracts.)

(a) Cumulative cost (BCWS / planned value)

Total project cost = Rs. 1,600,000.

S-curve: plotting cumulative cost against time gives an S (ogive) shape — a gentle slope at the start (M1, low expenditure), a steep middle (M2–M3, peak activity) and flattening near the end (M4). The characteristic flat–steep–flat profile is the S-curve.

Cum cost (Rs '000)
1600 |                         ____●  (M4)
1300 |                  ___●           
 700 |          __●  (M2)
 200 |   ●  (M1)
     +----+----+----+----+--- time (months)
        1    2    3    4

(b) Earned-value analysis at end of Month 2.

Given: ACWP (actual cost) = Rs. 800,000; BCWP (earned value) = Rs. 650,000.

Cost Variance: $CV = BCWP - ACWP = 650{,}000 - 800{,}000 = -150{,}000$.

$$CPI = \frac{BCWP}{ACWP} = \frac{650{,}000}{800{,}000} = 0.8125$$

CV = −Rs. 150,000 (negative) and CPI = 0.81 (< 1) ⇒ the project is OVER BUDGET — it is costing more than the value of work actually performed.

Quality Assurance (QA) vs Quality Control (QC)

In short, QA builds quality into the process; QC verifies quality in the product.

Four common causes of construction accidents and corresponding safety measures

(Other measures: site induction & PPE, machine guarding, good housekeeping, and an emergency/first-aid plan.)

Given: $ES = 8$, $EF = 14$, $LS = 11$, $LF = 17$, duration $D = 6$, and the successor's earliest start $ES_{succ} = 14$.

Total float $= LS - ES = 11 - 8 = 3$ days (also $LF - EF = 17 - 14 = 3$). → Maximum delay possible without delaying project completion.

Free float $= ES_{succ} - EF = 14 - 14 = 0$ days. → Delay possible without delaying the early start of any following activity. Here it is 0, so any delay of this activity will push the successor.

Independent float $= ES_{succ} - LF_{pred} - D$. Taking the latest finish of the predecessor as the activity's own $LS$ reference, independent float $= \max[0,\ ES_{succ} - (LS) - D]$ when the activity starts as late as allowed by its predecessor. Using the standard relation Independent float $= ES_{succ} - D - LF_{pred}$; with $LF_{pred}$ assumed $= ES = 8$ (activity at its own ES governed by predecessor early finish): $= 14 - 6 - 8 = 0$ days. → Float available even if preceding activities take all their float and following activities start as early as possible; here 0.

Results: Total float = 3 days, Free float = 0 days, Independent float = 0 days.

Why a critical activity has zero total float. On the critical path $LS = ES$ and $LF = EF$, so total float $= LS - ES = 0$. A critical activity has no spare time — any delay to it directly delays the project completion. The longest path through the network (the critical path) therefore consists entirely of zero-float activities and fixes the project duration.

(a) Straight-line depreciation.

Annual depreciation $= \dfrac{P - S}{n} = \dfrac{6{,}000{,}000 - 1{,}000{,}000}{5} = \dfrac{5{,}000{,}000}{5} = \textbf{Rs. 1,000,000 / year}$.

Hourly depreciation $= \dfrac{1{,}000{,}000}{2{,}000\ \text{hr}} = \textbf{Rs. 500 / hr}$.

(b) Owning cost — investment (interest + insurance + tax).

Average investment $= \dfrac{P + S}{2} = \dfrac{6{,}000{,}000 + 1{,}000{,}000}{2} = \text{Rs. } 3{,}500{,}000$.

Annual investment cost $= 12\% \times 3{,}500{,}000 = \text{Rs. } 420{,}000$/year.

Hourly investment cost $= \dfrac{420{,}000}{2{,}000} = \text{Rs. } 210$/hr.

Owning cost = depreciation + investment cost = $500 + 210 = \text{Rs. } 710$/hr.

Operating cost = Rs. 1,200/hr (given).

$$\text{Total owning \& operating cost} = 710 + 1200 = \textbf{Rs. 1,910 / hr}$$