BE Civil Engineering (IOE, TU) Construction Project Management (IOE, CE 752) Question Paper 2078 Nepal

Definition. A construction project is a temporary, unique endeavour undertaken to create a defined physical facility (building, road, dam, etc.) within agreed constraints of scope, time, cost and quality, using a one-time organisation of resources that is disbanded on completion. Unlike repetitive manufacturing, each project has a definite start and finish, a single deliverable, and progressive elaboration of detail.

Four phases of the project life cycle.

1. Conception / Initiation (Feasibility). Need is identified; pre-feasibility and feasibility studies, site selection, preliminary cost-benefit and financial appraisal, and the go/no-go decision are made. Effort and cost are low but the influence on final cost is highest here.
2. Planning & Design (Definition). Detailed engineering design, drawings, specifications, BOQ, master schedule, budget, procurement strategy and contract documents are prepared.
3. Execution / Construction (Implementation). Mobilisation, procurement, actual construction, monitoring and control of progress, cost and quality. Resource consumption peaks here.
4. Commissioning & Closeout (Termination). Testing, commissioning, defect liability/handover, final accounts, demobilisation, and project review/lessons learned.

Cost &        Execution
Manpower      ____
  ^          /    \
  |  Plan  _/      \_ Closeout
  | _/                \__
  +---------------------------> Time
  Conception

WBS for a multi-storey commercial building (3 levels).

1.0 Commercial Building (Level 1 - total project)
  1.1 Substructure (Level 2)
      1.1.1 Site clearing & earthwork (Level 3)
      1.1.2 Foundation & footings
      1.1.3 Basement / raft, waterproofing
  1.2 Superstructure
      1.2.1 RCC columns, beams, slabs
      1.2.2 Masonry & blockwork
      1.2.3 Staircases & lift cores
  1.3 Finishing
      1.3.1 Plastering & painting
      1.3.2 Flooring & tiling
      1.3.3 Doors, windows & glazing
  1.4 MEP Services
      1.4.1 Electrical & lighting
      1.4.2 Plumbing & sanitary
      1.4.3 HVAC & fire-fighting
  1.5 External & Commissioning
      1.5.1 Site development & landscaping
      1.5.2 Testing & commissioning

How the WBS links to project functions.

• Scheduling: the lowest-level work packages become the activities of the bar chart / CPM network; their logical sequencing and durations build the master schedule.
• Budgeting: each work package is costed (labour + material + plant + overhead), and rolling-up the costs gives the project budget and the cost baseline / S-curve for earned-value control.
• Responsibility assignment: every work package is assigned an owner via a Responsibility Assignment Matrix (RACI), eliminating gaps and overlaps and forming the basis of accountability and progress reporting.

Thus the WBS is the single backbone that integrates scope, time, cost and organisation.

Step 1 – Network logic and times. Critical path is A→B→D→E.

• A: ES 0, finishes day 3 → occupies days 1–3
• B: starts day 4, dur 4 → days 4–7
• C: starts day 4, dur 2 → days 4–5 (then waits for E)
• D: starts day 8, dur 3 → days 8–10
• E: starts day 11, dur 2 → days 11–12

Project duration = 12 days. C has float: C must finish before E (day 11), C earliest 4–5, latest 9–10, so total float of C = 10 − 5 = 5 days.

(a) Early-start bar chart.

Day:   1  2  3 | 4  5  6  7 | 8  9 10 |11 12
A     [=======]
B                 [==========]
C                 [====]
D                            [=======]
E                                     [====]

(b) Early-start resource histogram (workers/day).

Workers
 9 |    ##
 6 |    ##
 5 |    ##                 ##
 4 |## ##                 ##
 3 |## ## ##              ##
 2 |## ## ## ##           ##
   +---------------------------
    1-3 4-5 6-7 8-10 11-12

Peak demand (early start) = 9 workers (days 4–5), caused by B and C overlapping.

(c) Resource levelling using C's float. C has 5 days of float, so shift C to start later (e.g. days 8–9) instead of days 4–5, removing the B+C overlap while still finishing before E (day 11).

New profile:

Moving C fully off B reduces the peak from 9 to where C now overlaps D (6+2 = 8). To do better, schedule C on days 6–7 (overlapping the tail of B: 3+6 = 9 — no gain) — so the best non-overlap placement is days 8–9 giving peak 8. Project duration is unchanged at 12 days.

New peak = 8 workers (a smoother profile with no increase in project duration).

(a) Network (AOA).

        A(4)        C(5)
   (1)------->(2)--------->(4)
    |          |            |
  B(6)       E(8)         F(5)
    |          |            |
    v          v            v
   (3)        (5)          (6)
    |          |            ^
    +--D(3)-->(4)           |
               (5)--G(4)----+

Logic: node 4 fed by C(2–4) and D(3–4); node 6 fed by F(4–6) and G(5–6).

(b) Forward pass — earliest event times (ET).

• $ET_1 = 0$
• $ET_2 = ET_1 + 4 = 4$
• $ET_3 = ET_1 + 6 = 6$
• $ET_4 = \max(ET_2+5,\;ET_3+3) = \max(9,\;9) = 9$
• $ET_5 = ET_2 + 8 = 12$
• $ET_6 = \max(ET_4+5,\;ET_5+4) = \max(14,\;16) = 16$

Project duration = 16 weeks.

Backward pass — latest event times (LT) (start $LT_6 = 16$):

• $LT_6 = 16$
• $LT_5 = LT_6 - 4 = 12$
• $LT_4 = LT_6 - 5 = 11$
• $LT_3 = LT_4 - 3 = 8$
• $LT_2 = \min(LT_4-5,\;LT_5-8) = \min(6,\;4) = 4$
• $LT_1 = \min(LT_2-4,\;LT_3-6) = \min(0,\;2) = 0$

(c) Activity times and floats. $ES = ET_i,\;EF = ES+d,\;LF = LT_j,\;LS = LF-d,\;TF = LS-ES,\;FF = ET_j - EF.$

(d) Critical path. Activities with zero total float: A → E → G, i.e. path 1–2–5–6.

$$\text{Critical Path} = A \to E \to G,\qquad \text{Project Duration} = 4 + 8 + 4 = \mathbf{16\ weeks.}$$

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

(b) Paths and critical path. Node 4 is fed by C (2–4) and D (3–4).

• Path 1: 1–2–4–5 = A + C + E = 4 + 6 + 5 = 15 days
• Path 2: 1–3–4–5 = B + D + E = 6 + 3 + 5 = 14 days

The longer path is A–C–E (1–2–4–5) → expected project duration $T_e = 15$ days.

Variance of critical path: $\sigma^2_{cp} = \sigma^2_A + \sigma^2_C + \sigma^2_E = 0.444 + 0.444 + 1.000 = 1.889.$

Standard deviation $\sigma_{cp} = \sqrt{1.889} = 1.374$ days.

(c) Probability of completion in $T_s = 17$ days.

$$z = \frac{T_s - T_e}{\sigma_{cp}} = \frac{17 - 15}{1.374} = 1.455 \approx 1.46.$$

From the standard normal table, $\Phi(1.46) \approx 0.9279.$

Probability ≈ 0.928, i.e. about 93%.

(d) Time for 90% confidence. For 90% probability, $z_{0.90} = 1.28$:

$$T_s = T_e + z\,\sigma_{cp} = 15 + 1.28 \times 1.374 = 15 + 1.759 = 16.76 \approx \mathbf{16.8\ days.}$$

So there is a 90% chance of finishing within about 16.8 days, and roughly a 93% chance of finishing within 17 days.

Types of construction contract (by method of payment).

1. Lump-sum (fixed-price) contract — contractor quotes a single fixed price for the whole defined scope.
   • Advantage: cost certainty for the owner; minimal financial supervision.
   • Disadvantage: risky for the contractor if scope is ill-defined; costly variations.
2. Item-rate / unit-price (BOQ) contract — payment by measured quantity × quoted unit rate.
   • Advantage: fair when quantities are uncertain; easy to handle variations.
   • Disadvantage: final cost is unknown until measurement; risk of unbalanced bidding.
3. Cost-plus contract (cost + fixed fee / + percentage / + fee with GMP) — owner reimburses actual cost plus a fee.
   • Advantage: allows an early start when scope is unclear; flexible for urgent/complex work.
   • Disadvantage: little incentive to economise (especially cost-plus-percentage); needs heavy auditing.
4. Percentage-rate contract — contractor quotes a percentage above/below the engineer's estimated rates.
   • Advantage: simple comparison of bids.
   • Disadvantage: contractor cannot optimise individual rates.
5. Turnkey / EPC contract — single party handles engineering, procurement and construction, delivering a ready-to-use facility.
   • Advantage: single-point responsibility; fast.
   • Disadvantage: higher price; reduced owner control over design.

Competitive tendering process in Nepal (per the Public Procurement Act/Regulations).

1. Procurement planning & cost estimate — prepare master procurement plan and the engineer's estimate.
2. Preparation of bidding documents — ITB, conditions of contract, specifications, drawings, BOQ.
3. Invitation for bids (IFB) — notice published (national/international, e-GP/PPMO portal) inviting sealed bids.
4. Issue/sale of documents & pre-bid meeting — clarifications and addenda issued.
5. Submission of bids — bids submitted with bid security by the deadline.
6. Bid opening — opened publicly at the stated time in presence of bidders.
7. Evaluation — responsiveness check, then technical and financial evaluation; identify the substantially responsive lowest evaluated bid.
8. Award & notification — issue Letter of Intent / Letter of Acceptance to the successful bidder.
9. Contract agreement — winner submits performance security, then signs the contract; work order issued.

Earnest money / Bid security. A refundable guarantee (commonly 2–3% of estimated cost, as cash deposit or bank guarantee) submitted with the bid. It (i) deters frivolous/non-serious bids, and (ii) is forfeited if a bidder withdraws during validity or fails to sign the contract / furnish performance security after award. It is returned to unsuccessful bidders after award.

Performance security. A guarantee (typically 5% of contract amount, as a bank guarantee valid through the defect-liability period) furnished by the successful bidder before signing. It secures the owner against the contractor's failure to perform; it can be invoked to recover losses on default and is released after satisfactory completion of obligations.

CPM vs PERT.

Where CPM suits: (i) construction of a building or road where durations are reliably known from past data; (ii) maintenance/overhaul of plant with repetitive tasks; cost-optimisation (crashing) studies.

Where PERT suits: (i) R&D or first-of-a-kind projects (new dam technology, prototype) with uncertain durations; (ii) projects needing the probability of meeting a deadline, e.g. event-driven or aerospace/defence work.

Given: bucket capacity $q = 0.9\ \text{m}^3$, fill factor $K = 0.85$, cycle time $t = 24\ \text{s}$, efficiency $= 50/60$ working hour, swell factor $= 1.25$.

Step 1 – Cycles per hour (theoretical).

$$N = \frac{3600\ \text{s/hr}}{24\ \text{s}} = 150\ \text{cycles/hr}.$$

Step 2 – Volume moved per cycle (loose).

$$q \times K = 0.9 \times 0.85 = 0.765\ \text{m}^3/\text{cycle (loose)}.$$

Step 3 – Loose production with efficiency.

$$Q_{loose} = q\,K \times N \times \text{eff} = 0.765 \times 150 \times \frac{50}{60}.$$ $$Q_{loose} = 0.765 \times 150 \times 0.8333 = 95.6\ \text{m}^3/\text{hr}.$$

(a) Loose production ≈ 95.6 m³/hr (LCM).

Step 4 – Convert to bank (in-situ) volume. Bank volume = loose volume ÷ swell factor:

$$Q_{bank} = \frac{95.6}{1.25} = 76.5\ \text{m}^3/\text{hr}.$$

(b) Bank (in-situ) production ≈ 76.5 m³/hr (BCM).

Quality Assurance (QA). A process-oriented, proactive set of planned and systematic activities (procedures, standards, audits, training, documentation) put in place to provide confidence that quality requirements will be met. QA prevents defects by getting the process right. Example: approved method statements, ITPs (inspection & test plans), supplier pre-qualification, internal audits.

Quality Control (QC). A product-oriented, reactive set of operational activities (inspection, sampling, testing, measurement) used to verify that the output actually meets the specification, and to detect and correct defects. Example: concrete cube tests, slump tests, dimensional checks, rejection of non-conforming material.

Distinction.

Four essential elements of a site safety management plan.

1. Safety policy & responsibilities — written policy, clear roles, a designated safety officer.
2. Hazard identification & risk assessment (HIRA) — identify hazards (falls, excavation collapse, electrocution, lifting) and control them by the hierarchy of controls.
3. Personal protective equipment (PPE) & engineering controls — helmets, harnesses, guardrails, barricades, scaffolding inspection.
4. Training, emergency preparedness & incident reporting — toolbox talks, induction, first-aid/fire response plan, and accident recording with corrective action.

Step 1 – Cost slope $= \dfrac{\text{crash cost} - \text{normal cost}}{\text{normal time} - \text{crash time}}$ (Rs/day).

Total normal direct cost $= 800+600+1000+1200+500 =$ Rs. 4100.

Step 2 – Normal duration & critical path. Node 4 fed by C(2–4) and D(3–4); E(4–5) is common to all paths.

• Path 1: A–C–E = 8+10+5 = 23 d
• Path 2: B–D–E = 6+12+5 = 23 d

Both paths are critical → Normal duration = 23 days.

Total cost at normal = direct + indirect = $4100 + 400\times23 = 4100 + 9200 =$ Rs. 13,300.

Step 3 – Crashing (since both paths critical, shorten the common activity E first, then one activity on each path).

• 23→22 (1 day): crash E (common, cheapest, slope 150). Added direct = 150.
• 22→21 (1 day): crash E again (E now fully crashed at 3 d). Added direct = 150.
• 21→20 (1 day): E exhausted, so crash one activity on each critical path: cheapest is C (100) on path 1 and D (250) on path 2 → combined 350/day.

Step 4 – Cost table.

Beyond 20 days, the next crash step would cost ≥ 350/day in direct cost while saving only 400/day indirect (marginal benefit shrinking and remaining cheap options exhausted); total cost reaches its minimum at 20 days.

Result. Cost slopes as above; normal duration = 23 days (Rs. 13,300); optimum duration = 20 days with minimum total cost = Rs. 12,750.

Given (Rs. lakh): $PV = 50,\ EV = 42,\ AC = 48,\ BAC = 120.$

Step 1 – Variances.

$$\text{Cost Variance } CV = EV - AC = 42 - 48 = -6\ \text{lakh}.$$ $$\text{Schedule Variance } SV = EV - PV = 42 - 50 = -8\ \text{lakh}.$$

Both are negative → over budget and behind schedule.

Step 2 – Performance indices.

$$CPI = \frac{EV}{AC} = \frac{42}{48} = 0.875.$$ $$SPI = \frac{EV}{PV} = \frac{42}{50} = 0.840.$$

Both $< 1$ → cost and schedule efficiency below plan.

Step 3 – Estimate at Completion (assuming current cost performance continues).

$$EAC = \frac{BAC}{CPI} = \frac{120}{0.875} = 137.14\ \text{lakh}.$$

Estimate to Complete $ETC = EAC - AC = 137.14 - 48 = 89.14$ lakh. Variance at Completion $VAC = BAC - EAC = 120 - 137.14 = -17.14$ lakh (projected overrun).

Step 4 – Interpretation.

• $CV = -6$ lakh and $CPI = 0.875$: the project is over budget — it gets only Rs. 0.875 of value per rupee spent.
• $SV = -8$ lakh and $SPI = 0.84$: the project is behind schedule — only 84% of planned work is done.
• At the current rate the project will finish at about Rs. 137 lakh against a budget of Rs. 120 lakh, an overrun of ≈ Rs. 17 lakh.

Conclusion: the project is unhealthy on both cost and time and requires corrective action (recovery plan, resource/scope review, tighter cost control).

Given: Purchase price $P =$ Rs. 60,00,000; salvage $S =$ Rs. 8,00,000; life $n = 5$ years $= 10{,}000$ working hours.

(a) Annual straight-line depreciation.

$$D_{annual} = \frac{P - S}{n} = \frac{60{,}00{,}000 - 8{,}00{,}000}{5} = \frac{52{,}00{,}000}{5} = \textbf{Rs. 10,40,000 per year.}$$

(b) Hourly ownership depreciation.

$$D_{hour} = \frac{P - S}{\text{total working hours}} = \frac{52{,}00{,}000}{10{,}000} = \textbf{Rs. 520 per hour.}$$

(Equivalently, Rs. 10,40,000/yr ÷ 2,000 hr/yr = Rs. 520/hr.)

(c) Other components of total owning & operating cost.

• Owning (ownership) costs: depreciation, interest on investment, insurance, taxes/registration, and storage.
• Operating costs: fuel, lubricants & filters, tyres (or tracks/undercarriage), repairs & maintenance, and operator's wages.

Total hourly cost = (sum of ownership costs per hour) + (sum of operating costs per hour); this rate is the basis for equipment hire rates and unit-rate analysis.