BE Civil Engineering (IOE, TU) Construction Project Management (IOE, CE 752) Question Paper 2076 Nepal

(a) Network (Activity-on-Arrow):

            B(6)        D(8)
   (1)--A(4)-->(2)---------->(4)
    \           \             ^   \
     \           +--(dummy)---+    \ F(3)
      \                            v
       +--A--(1)..C(5)-->(3)--E(7)-->(4)-->(5)

Reduced event numbering used for the pass: 1 -> A -> 2; 2 -> B -> 3; 2 -> C -> 4; 3 -> D -> 5; 4 -> E -> 5; 5 -> F -> 6.

(b) Forward pass (EST/EFT), Backward pass (LST/LFT):

Forward pass: EFT = EST + duration; an activity's EST = max EFT of predecessors. EFT(F) = 21 = project duration. Backward pass: start LFT(F) = 21; LST = LFT − duration; LFT of an activity = min LST of successors.

(c) Total float = LST − EST = LFT − EFT:

Activities with zero float lie on the critical path.

Critical path: A → B → D → F, with duration $4+6+8+3 = 21$ days.

Project duration = 21 days. The chain A–C–E–F totals $4+5+7+3 = 19$ days, leaving 2 days of float on C and E.

(a) Expected time $t_e = \dfrac{a+4m+b}{6}$, variance $\sigma^2 = \left(\dfrac{b-a}{6}\right)^2$:

(b) Path durations (using $t_e$):

• 1–2–3–4–5: $6+5+4+6 = 21$ days
• 1–2–4–5: $6+8+6 = 20$ days

The longer path is critical. Critical path: 1 → 2 → 3 → 4 → 5, expected duration $T_e = 21$ days.

(c) Variance along critical path:

$$\sigma^2_{cp} = 1.000 + 1.000 + 1.000 + 1.000 = 4.0 \quad\Rightarrow\quad \sigma_{cp} = \sqrt{4.0} = 2.0 \text{ days}$$

For a target $T_s = 25$ days:

$$z = \frac{T_s - T_e}{\sigma_{cp}} = \frac{25 - 21}{2.0} = 2.0$$

From the standard normal table, $P(z \le 2.0) = 0.9772$.

Probability of completion within 25 days ≈ 0.9772, i.e. about 97.7%.

(a) Normal duration. Paths:

• A–B–D: $6+8+4 = 18$ days
• A–C–E: $6+5+7 = 18$ days

Both paths are critical at 18 days (network has two parallel critical chains sharing common start A).

Direct normal cost $= 2000+3000+1500+1200+2800 = \text{Rs }10{,}500$. Indirect $= 18 \times 600 = \text{Rs }10{,}800$. Total normal project cost $= 10{,}500 + 10{,}800 = \text{Rs }21{,}300$.

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

(c) Step-by-step crashing (both paths must be shortened together; A is common to both).

Step 1 — crash A. A lies on both critical paths, slope = Rs 400/day < indirect Rs 600/day, so crashing is economical. Crash A by its full 2 days.

• New duration $= 16$ days; direct cost $= 10{,}500 + 2(400) = 11{,}300$.
• Indirect $= 16 \times 600 = 9{,}600$. Total $= 11{,}300 + 9{,}600 = \text{Rs }20{,}900$.

Step 2 — check further crashing. A is exhausted. To reduce duration further, BOTH paths must be crashed simultaneously: cheapest on path A–B–D is D (300), cheapest on path A–C–E is E (400). Combined slope $= 300 + 400 = \text{Rs }700/\text{day}$, which exceeds the indirect saving of Rs 600/day. Total cost would rise, so we stop.

Optimum duration = 16 days; minimum total project cost = Rs 20,900 (a saving of Rs 400 over the all-normal schedule).

(a) Excavator production.

Output per hour (loose volume):

$$Q_{loose} = \frac{3600}{t} \times q \times f \times E = \frac{3600}{30} \times 1.2 \times 0.90 \times 0.83$$ $$= 120 \times 1.2 \times 0.90 \times 0.83 = 107.57\ \text{m}^3/\text{hr (loose)}$$

(i) Loose output ≈ 107.57 loose-m³/hr.

Convert to bank volume (swell 25%, so bank $=$ loose$/1.25$):

$$Q_{bank} = \frac{107.57}{1.25} = 86.06\ \text{m}^3/\text{hr (bank)}$$

(ii) Bank output ≈ 86.06 bank-m³/hr.

Production per 8-hour shift $= 86.06 \times 8 = 688.5$ bank-m³. Shifts required $= \dfrac{4500}{688.5} = 6.54 \Rightarrow$ 7 shifts (rounded up).

(b) Straight-line depreciation.

$$D = \frac{C - S}{n} = \frac{5{,}000{,}000 - 500{,}000}{8} = \frac{4{,}500{,}000}{8} = \text{Rs }562{,}500\ \text{per year}$$

Book value after 3 years:

$$BV_3 = C - 3D = 5{,}000{,}000 - 3(562{,}500) = 5{,}000{,}000 - 1{,}687{,}500 = \textbf{Rs } 3{,}312{,}500$$

Annual depreciation = Rs 5,62,500; book value at end of year 3 = Rs 33,12,500.

(c) Two factors reducing job efficiency: (i) poor site management / operator delays, equipment breakdown and repair time; (ii) adverse working conditions — wet/sticky soil, restricted swing angle, bad haul-road grade and weather. (Any two acceptable.)

(a) Early-start bar chart (day 1 … 8):

Day:      1   2   3   4   5   6   7   8
A(4)     [###][###][###]
C(5)     [###][###]
B(6)               [###][###][###]
D(3)               [###][###]
E(4)                                 [###][###]

Daily labour totals (early-start):

Peak demand (early-start) = 9 workers (days 1, 2, 4, 5).

(b) Difference.

• Resource levelling: used when resources are limited/fixed; activities are delayed (even beyond the original duration if necessary) so that demand never exceeds the available resource ceiling — the project duration may increase.
• Resource smoothing: the project duration is held fixed; only non-critical activities are shifted within their float to even out the demand curve and remove sharp peaks — the completion date does not change.

(c) Smoothing using float (duration stays 8 days). Shift C (float 3) to start later and shift D (float 2) so peaks of 9 are broken:

• Move C to days 3–4 (uses 2 days of its 3-day float).
• Move D to days 6–7? D float is 2; B finishes day 6 and D follows C-chain — move D to days 6–7 is not allowed by its float window (latest finish day 7). Instead shift D to days 6–7 within float (LST allows). Re-tabulate:

Reassigned: A: 1–3, C: 3–4, B: 4–6, D: 6–7, E: 7–8.

That worsens day 4, so adopt the balanced shift: move C to days 3–4 only partially is not possible; the practical smoothing is to delay C by 1 day (days 2–3) and D by 2 days (days 6–7):

A: 1–3, C: 2–3, B: 4–6, D: 6–7, E: 7–8.

The smoothed profile removes the day-4/5 double peak; demand is more even and peak stays at 9 but occurs on fewer, more spread-out days, with the high-low oscillation reduced. (The critical activities A, B, E are unchanged, and total duration remains 8 days.)

Key point for marks: only float of non-critical activities (C, D) is consumed; critical chain and the 8-day finish are preserved.

(a) Project planning is the process of identifying the activities required to complete a project, establishing their logical sequence and time durations, and determining the resources, methods and standards needed to achieve the project objectives within the planned time, cost and quality.

Four objectives of construction project planning:

1. To complete the project within the stipulated time.
2. To complete the project within the approved budget/cost.
3. To ensure the required quality and specifications are met.
4. To make the most economical and efficient use of resources (men, material, machinery, money) and to provide a basis for monitoring and control.

(b) Bar chart vs Network technique.

Advantages of the bar chart: (1) Simple to draw and easy to understand, even for non-technical staff; (2) Good visual tool for progress reporting and for showing time-scaled status at a glance.

Limitations of the bar chart: (1) It does not show the inter-relationships/logical dependencies between activities; (2) It cannot identify the critical path or float, so it is poor for analysing the effect of delays and for large/complex projects.

(a) Item-rate vs Lump-sum contract.

Suitable when: Item-rate is best where quantities cannot be accurately known in advance (e.g. earthwork, foundations in uncertain ground). Lump-sum is best where the design and scope are complete and well defined (e.g. a standard building from finalised drawings).

(b) Competitive tendering steps (public work, Nepal — Public Procurement Act/Regulation):

1. Preparation of cost estimate, drawings, specifications and the tender/bidding document.
2. Invitation for bids — public notice/advertisement (and posting on the PPMO/e-GP portal) inviting sealed bids.
3. Issue/sale of bidding documents and a pre-bid meeting / clarifications if required.
4. Submission of sealed bids with bid security before the deadline.
5. Bid opening in public at the appointed time and date.
6. Evaluation of bids — preliminary (responsiveness), technical and financial evaluation to find the lowest substantially responsive evaluated bid.
7. Approval and notification/letter of intent / award to the successful bidder.
8. Submission of performance security and signing of the contract agreement.

Given: BAC = Rs 20,00,000; PV = 12,00,000; EV = 10,00,000; AC = 13,00,000.

(a) Cost performance.

$$CV = EV - AC = 10{,}00{,}000 - 13{,}00{,}000 = -\text{Rs }3{,}00{,}000$$ $$CPI = \frac{EV}{AC} = \frac{10{,}00{,}000}{13{,}00{,}000} = 0.769$$

CV is negative and CPI $< 1$ → the project is over budget (getting Rs 0.77 of work per rupee spent).

(b) Schedule performance.

$$SV = EV - PV = 10{,}00{,}000 - 12{,}00{,}000 = -\text{Rs }2{,}00{,}000$$ $$SPI = \frac{EV}{PV} = \frac{10{,}00{,}000}{12{,}00{,}000} = 0.833$$

SV is negative and SPI $< 1$ → the project is behind schedule (only 83.3% of planned work done).

(c) Estimate at Completion (CPI continues):

$$EAC = \frac{BAC}{CPI} = \frac{20{,}00{,}000}{0.769} = \textbf{Rs } 26{,}00{,}000$$

(Equivalently $EAC = AC + (BAC-EV)/CPI = 13{,}00{,}000 + 10{,}00{,}000/0.769 = 13{,}00{,}000 + 13{,}00{,}000 = 26{,}00{,}000$.)

Interpretation: the project is both over budget and behind schedule; at the current efficiency it is forecast to cost Rs 26,00,000, i.e. Rs 6,00,000 over the original budget.

(a) QA vs QC.

(b) Cost of quality is the total cost of achieving quality plus the cost incurred because quality was not achieved. Its four categories are:

1. Prevention costs — planning, training, QA systems.
2. Appraisal costs — inspection, testing, calibration.
3. Internal failure costs — rework, scrap detected before handover.
4. External failure costs — defects found after handover (warranty repairs, claims, loss of reputation). Conformance cost = prevention + appraisal; non-conformance cost = internal + external failure. Investing in prevention/appraisal usually reduces the larger failure costs.

(c) Three concrete QC tests:

1. Slump test (workability of fresh concrete).
2. Compressive strength test on cubes/cylinders (hardened concrete, typically at 7 and 28 days).
3. Compaction factor / water-cement ratio check or non-destructive rebound hammer test for in-situ strength. (Any three acceptable.)

(a) Four major causes of construction accidents:

1. Falls from height (scaffolds, edges, openings).
2. Struck by / caught between moving plant, falling materials or collapsing excavations.
3. Electrical contact and unsafe temporary wiring.
4. Unsafe acts/conditions: lack of training, poor housekeeping, missing guards and non-use of PPE.

(b) Hierarchy of hazard controls (most → least effective):

1. Elimination — remove the hazard. e.g. prefabricate components at ground level to avoid working at height.
2. Substitution — replace with a less hazardous option. e.g. use a water-based, low-VOC curing compound instead of a toxic solvent.
3. Engineering controls — isolate people from the hazard. e.g. guardrails, edge protection, trench shoring, machine guards.
4. Administrative controls — change the way people work. e.g. permit-to-work, safe-work procedures, training, rotation, signage.
5. PPE — protect the individual. e.g. helmet, safety harness, boots, goggles.

(c) Job-Safety Analysis (JSA) is a systematic procedure that breaks a task into its sequential steps, identifies the hazard associated with each step, and specifies the control measure for each hazard before the work starts.

PPE is the last line of defence because it does nothing to remove or reduce the hazard itself — it only protects the worker if an incident occurs, depends entirely on correct use and fit, and fails the moment it is not worn. Higher controls (elimination, substitution, engineering) act on the hazard at source and protect everyone, so PPE is used only after those have been applied.

(a) Float computation. Total float $TF = L_j - E_i - d$. Free float $FF = E_j - E_i - d$.

(b) Activities with zero total float: P, Q, S (dummy), T. Critical path: 1 → 2 → 3 → 4 → 5 (P–Q–S–T), duration $5+7+0+6 = 18$ days. Project duration = 18 days. Activity R has 3 days of total float (but 0 free float, so the slack is shared with following activities).

(c) Total float vs free float.

• Total float is the maximum time an activity can be delayed without delaying the overall project completion; using it may consume the float of following activities.
• Free float is the time an activity can be delayed without delaying the early start of its succeeding activities (and hence without disturbing anyone downstream). Free float is therefore always ≤ total float; here R has $TF = 3$ but $FF = 0$, meaning any delay of R, while not affecting the project end, would immediately push back its successor's earliest start.