BE Civil Engineering (IOE, TU) Construction Project Management (IOE, CE 752) Question Paper 2079 Nepal

(a) Network logic (Activity-on-Node):

        +--> C --+
        |        v
START --+--> A --+--> D --> F --+
        |                       v
        +--> B --------> E -----+--> G --> END
                          ^
                          |
                  (E follows B and C)
              (G follows E and F)

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

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

Project duration = EF of G = 18 days.

Backward pass (LF, LS), LS = LF − Duration: Set LF of G = 18.

(c) Total float (TF = LS − ES) and Free float (FF = ES of successor − EF):

(d) Critical path = activities with TF = 0 forming a continuous chain: A → C → E → G.

Check: 4 + 5 + 7 + 2 = 18 days = project duration. ✔

Project duration = 18 days; Critical path: A–C–E–G.

(a) PERT formulas: $t_e = \dfrac{t_o + 4t_m + t_p}{6}$, $\sigma^2 = \left(\dfrac{t_p - t_o}{6}\right)^2$.

(b) Path lengths using $t_e$:

• A–C–E–F = 4 + 3 + 5 + 5 = 17 weeks
• A–D–F = 4 + 7 + 5 = 16 weeks
• B–E–F = 6 + 5 + 5 = 16 weeks

Longest = A–C–E–F = 17 weeks. Critical path = A–C–E–F, expected duration $T_e$ = 17 weeks.

(c) Variance along critical path:

$$\sigma^2_{cp} = 0.444 + 0.444 + 1.000 + 1.778 = 3.667 \text{ weeks}^2$$ $$\sigma_{cp} = \sqrt{3.667} = 1.915 \text{ weeks}$$

Standard normal variate for $T_s = 26$ weeks:

$$Z = \frac{T_s - T_e}{\sigma_{cp}} = \frac{26 - 17}{1.915} = \frac{9}{1.915} = 4.70$$

Since $Z = 4.70$ is far in the right tail, $P(T \le 26) \approx \Phi(4.70) \approx 1.000$.

Probability ≈ 1.00 (≈ 100%); completing within 26 weeks is virtually certain. (For comparison, $Z = 1.0$ corresponds to a completion target of $17 + 1.915 = 18.9$ weeks with probability 0.8413.)

(a) Cost slope = $\dfrac{\text{Crash cost} - \text{Normal cost}}{\text{Normal time} - \text{Crash time}}$ (Rs/day):

Normal duration = 8 + 6 + 10 + 5 = 29 days (single path, so the whole chain is critical).

Normal direct cost = 10,000 + 8,000 + 15,000 + 6,000 = Rs 39,000. Normal indirect cost = 29 × 2,000 = Rs 58,000. Normal total cost = Rs 97,000.

(b) Step-by-step crashing. Indirect cost = Rs 2,000/day. Crash the activity with the lowest slope first; crashing is worthwhile while slope ≤ indirect saving (2,000/day).

A, B, C all have slope = 2,000/day = indirect cost; D has slope 3,000 > 2,000 (never worth crashing).

Crashing A, B or C costs exactly Rs 2,000/day but saves Rs 2,000/day of indirect cost ⇒ net total cost unchanged, while duration reduces. So the optimum is a range; the shortest duration at no extra total cost is obtained by fully crashing A, B and C.

Crash A by 2 (→6), B by 2 (→4), C by 3 (→7): total reduction = 7 days.

• New duration = 29 − 7 = 22 days.
• Added direct cost = 7 × 2,000 = Rs 14,000 ⇒ direct cost = 39,000 + 14,000 = Rs 53,000.
• Indirect cost = 22 × 2,000 = Rs 44,000.
• Total cost = 53,000 + 44,000 = Rs 97,000 (same as normal).

Crashing D further (slope 3,000 > 2,000) would raise total cost by 1,000, so stop.

Optimum: any duration from 29 down to 22 days gives the minimum total cost of Rs 97,000; the least crashed duration achieving this minimum is the normal 29 days, and the shortest duration at minimum cost is 22 days. The recommended optimum duration is 22 days at total cost Rs 97,000 (shortest time at least cost).

(a) Hourly production (loose then bank).

Volume of soil moved per cycle (loose) = bucket capacity × fill factor = 1.5 × 0.85 = 1.275 LCM per cycle.

Ideal cycle time = 20 s ⇒ ideal cycles per hour = 3600/20 = 180.

Apply efficiency (50 min/hr ⇒ factor 50/60 = 0.833) and swing-depth factor 0.90. Effective cycles per hour = 180 × 0.833 × 0.90 = 135 cycles/hr.

Loose production = 135 × 1.275 = 172.1 LCM/hr.

Convert loose to bank: $V_{bank} = \dfrac{V_{loose}}{1 + \text{swell}} = \dfrac{172.1}{1.25}$.

Bank production = 137.7 BCM/hr (≈ 138 BCM/hr).

(b) Loading a truck.

Volume per pass (loose) = 1.275 LCM. Truck heaped capacity = 9 LCM.

Number of passes = 9 / 1.275 = 7.06 ⇒ round up to 8 passes (you cannot half-fill the last bucket; truck carries 8 × 1.275 = 10.2 LCM but is filled to its 9 LCM rated heaped capacity, last pass partial).

Using the standard rounding-up convention, 8 passes are needed.

Actual time per cycle = 20 s ÷ (efficiency × swing factor not applied per single observed cycle); using the effective rate, time per cycle = 3600/135 = 26.67 s.

Time to load one truck = 8 × 26.67 s = 213.3 s = 3.56 minutes (≈ 3 min 34 s) including efficiency.

Using ideal cycle time only: 8 × 20 = 160 s = 2.67 minutes.

Answers: (a) ≈ 138 BCM/hr; (b) 8 passes, ≈ 3.6 min per truck (effective) or 2.7 min (ideal).

(a) Resource smoothing vs resource levelling.

(b)(i) Early-start labour histogram.

Schedule: P days 0–2 (3 days, 4/day); Q days 0–1 (2 days, 6/day); R days 3–6 (4 days, 3/day); S days 3–4 (2 days, 5/day).

Labour
10 | ## ##
 8 |          ## ##
 6 |
 4 |       ##
 3 |                ## ##
   +--------------------------
     0  1  2  3  4  5  6   day

Peak daily labour demand = 10 labourers (days 0–1).

(b)(ii) Levelling within float. The peak (10) occurs on days 0–1 because P (critical) and Q overlap. Q has total float = 2, so shift Q later. Move Q to start on day 2 (days 2–3):

That worsens day 3. Instead shift Q to start day 1 is not enough; the better move is to delay S (float 3) and keep Q early but stagger. Try Q on day 2–3 makes day 3 worse. Re-examine: shift Q by exactly its float to days 2–3 conflicts with R+S on day 3.

Best feasible levelling: delay S by its float to days 5–6 and delay Q by 1 day to days 1–2:

Peak still 10. Because P (4) + Q (6) = 10 must overlap at least one day (Q float only 2, P spans days 0–2), the peak of P+Q cannot drop below 10 on any shared day unless Q is delayed past P. Q max start = day 2 (float 2). On day 2, P still active (4) and Q (6) ⇒ 10.

Conclusion: the minimum achievable peak by smoothing within float is 10 labourers, but the demand is made more uniform (eliminating the deep trough). To reduce the peak below 10 one must accept resource levelling that extends the project (delay Q beyond its float to start after day 2), giving a new peak of 6.

New levelled peak (within float) = 10; achievable peak only if Q delayed beyond float = 6 with project extension.

Bar chart (Gantt chart): A bar chart is a graphical project-scheduling tool in which each activity is represented by a horizontal bar drawn against a common time scale. The bar's left end marks the activity start, its length represents the duration, and its right end the finish. Introduced by Henry L. Gantt, it shows what work is planned and when, and progress can be shown by shading the bar.

Three advantages:

1. Simple and visual — easily understood by site staff, clients and non-technical stakeholders.
2. Easy progress monitoring — actual vs planned progress can be marked directly on the bars.
3. Quick to prepare for small projects and useful as a communication/reporting tool.

Three limitations:

1. Does not show interdependencies clearly — logical relationships between activities are not explicit.
2. No critical path / float information — cannot identify which activities control the project duration or have spare time.
3. Poor for analysis of changes — the effect of a delay in one activity on the rest of the project is not directly visible, making updating large projects difficult.

(Network techniques such as CPM/PERT overcome these by explicitly modelling logic, floats and the critical path.)

1. Lump-sum (fixed-price) contract: The contractor agrees to execute the whole specified work for a single fixed price, irrespective of the actual quantities of work done. Detailed drawings and specifications must be complete before tendering. Risk of quantity/price variation lies mainly with the contractor.

• Best for: Building works (e.g., a residential or office block) where the scope and quantities are well defined before tender.

2. Item-rate / unit-price (schedule of rates) contract: The contractor quotes a unit rate for each item in the Bill of Quantities (BOQ); payment = actual measured quantity × quoted rate. Quantities are provisional and re-measured on site.

• Best for: Road, irrigation, water-supply and earthwork projects where exact quantities cannot be fixed in advance (most government works under the Public Procurement Act use this form).

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) for overheads and profit.

• Best for: Emergency works, disaster repair, or projects with undefined scope where work must begin before design is complete (e.g., post-earthquake reconstruction or urgent landslide stabilization).

Key contrast: Lump-sum fixes price (owner low risk on cost, high on scope clarity); item-rate fixes rates (risk shared, quantity risk on owner); cost-plus fixes fee only (cost risk largely on owner, used when uncertainty is high).

Quality Assurance (QA): A set of planned and systematic activities (procedures, audits, documentation, training) carried out to provide confidence that the project will meet quality requirements. QA is process-oriented and proactive — it prevents defects (e.g., approved method statements, calibrated equipment, qualified personnel, inspection plans).

Quality Control (QC): The operational techniques and inspections/tests used to verify that the actual output meets the specified requirements. QC is product-oriented and reactive/detective — it identifies defects in the finished or in-progress work (e.g., testing samples, measuring dimensions).

Four common concrete QC tests:

1. Slump test (workability of fresh concrete).
2. Compressive strength test on cubes/cylinders (hardened concrete, e.g., at 7 and 28 days).
3. Concrete temperature / fresh density test of fresh concrete.
4. Water–cement ratio check and/or rebound hammer / core test (non-destructive in-situ strength of hardened concrete).

Importance of safety management: Construction is among the most accident-prone industries. Effective safety management (a) protects the lives and health of workers and the public, (b) reduces direct and indirect costs from accidents (compensation, lost time, equipment damage, project delay), (c) ensures legal compliance (Labour Act and safety regulations), (d) improves productivity and worker morale, and (e) protects the contractor's reputation and avoids litigation. Safety is achieved through hazard identification, risk assessment, provision of PPE, training, and a safety culture.

Four common hazards and control measures:

(Other valid hazards: dust/noise → masks and ear protection; machinery contact → guards and trained operators.)

Given: Initial cost $P$ = Rs 24,00,000; Salvage value $S$ = Rs 4,00,000; Life $n$ = 8 years. Depreciable amount = $P - S$ = 24,00,000 − 4,00,000 = Rs 20,00,000.

(a) Straight-line annual depreciation:

$$D = \frac{P - S}{n} = \frac{20,00,000}{8} = \textbf{Rs 2,50,000 per year.}$$

(b) Book value at end of year 3 (straight-line): Accumulated depreciation in 3 years = 3 × 2,50,000 = Rs 7,50,000.

$$BV_3 = P - 3D = 24,00,000 - 7,50,000 = \textbf{Rs 16,50,000.}$$

(c) Sum-of-years-digits, first-year depreciation: Sum of digits = $\dfrac{n(n+1)}{2} = \dfrac{8 \times 9}{2} = 36$.

First-year factor = $\dfrac{n}{36} = \dfrac{8}{36}$.

$$D_1 = \frac{8}{36} \times (P - S) = \frac{8}{36} \times 20,00,000 = \textbf{Rs 4,44,444 (approx).}$$

Summary: (a) Rs 2,50,000/yr; (b) Rs 16,50,000; (c) Rs 4,44,444.

(a) Work Breakdown Structure (WBS): A WBS is a hierarchical, deliverable-oriented decomposition of the total project scope into progressively smaller and more manageable components (work packages). The top level is the whole project; each lower level breaks the work into sub-deliverables until the lowest level (work package) is small enough to be estimated, scheduled, assigned and controlled.

Two benefits:

1. Complete scope definition — ensures no part of the work is omitted or duplicated (it defines 100% of the scope).
2. Basis for estimating and control — each work package can be costed, scheduled, assigned to a responsible party, and used for progress measurement. (Other valid: clarifies responsibility, aids communication, foundation for the project network.)

(b) Four phases of the construction project life cycle (in sequence):

1. Conception / Initiation phase — feasibility study, defining objectives and need.
2. Planning & Design phase — detailed design, drawings, specifications, scheduling, budgeting, tendering.
3. Execution / Construction phase — actual construction, procurement, resource deployment, monitoring and control.
4. Commissioning / Closeout (handover) phase — testing, inspection, handover to client, final accounts and documentation.