BE Civil Engineering (IOE, TU) Construction Project Management (IOE, CE 752) Question Paper 2077 Nepal

(a) AON network (logical layout)

      A(4) ───► C(5) ───►┐
      │                  ├──► E(7) ──►┐
      │   B(6) ──────────┘           ├──► G(3) ──► END
      └─► D(8) ───► F(6) ────────────┘

Start nodes: A, B. End node: G. E waits for both B and C; G waits for both E and F.

(b) Forward pass (ES = max EF of predecessors; EF = ES + duration; project start = day 0):

Backward pass (LF = min LS of successors; LS = LF − duration; project finish LF(G)=21):

(c) Project duration and critical path

Project duration = EF(G) = 21 days.

The critical path is the chain of zero-float activities: A → D → F → G (4 + 8 + 6 + 3 = 21 days).

(d) Total Float (TF = LS − ES = LF − EF) and Free Float (FF = min ES of successors − EF):

Critical activities (TF = 0): A, D, F, G. Project duration = 21 days; critical path A–D–F–G.

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

(b) Path lengths and variances (variances add along a series path):

• Path 1 (P→Q): $T_1 = 6.0 + 6.0 = 12.0$ weeks, $\sigma_1^2 = 2.778 + 0.444 = 3.222$.
• Path 2 (R→S): $T_2 = 4.0 + 9.0 = 13.0$ weeks, $\sigma_2^2 = 0.444 + 2.778 = 3.222$.

The longer expected length governs, so the critical path is Path 2 (R → S), expected duration $T_e = 13.0$ weeks, $\sigma^2 = 3.222$, $\sigma = \sqrt{3.222} = 1.795$ weeks.

(c) Probability of completion within 22 weeks on the critical path:

$$z = \frac{T_s - T_e}{\sigma} = \frac{22 - 13}{1.795} = \frac{9}{1.795} = 5.01.$$

For $z = 5.01$, $P(Z \le z) \approx 1.000$. The project is essentially certain (≈99.99%) to finish within 22 weeks.

(Illustrative check at a tighter deadline of 15.57 weeks: $z = (15.57-13)/1.795 = 1.43$, giving $P \approx 0.9236$, consistent with the supplied table value.)

(d) Duration for 90% confidence:

$$T_s = T_e + z_{0.90}\,\sigma = 13.0 + 1.28 \times 1.795 = 13.0 + 2.298 = 15.30 \text{ weeks}.$$

There is a 90% probability the project finishes within ≈15.3 weeks.

(a) Cost slope $= \dfrac{\text{Crash Cost} - \text{Normal Cost}}{\text{Normal Time} - \text{Crash Time}}$ (Rs./day) and max crash days:

(b) Normal duration and cost. Single path, so duration $= 8+6+10+7 = 31$ days.

• Total direct (normal) cost $= 20000+15000+30000+18000 = 83{,}000$.
• Indirect cost $= 1500 \times 31 = 46{,}500$.
• Total normal cost $= 83{,}000 + 46{,}500 = 129{,}500$.

(c) Step-by-step crashing. Crash the cheapest slope first (savings in indirect = Rs.1,500/day; crash only while slope ≤ 1,500? — here every slope ≥ 2,000 > 1,500, so each crash day raises direct cost more than the indirect saving). Examine each step:

Start: 31 days, total = 129,500.

Step 1 — crash A (slope 2000), 3 days: duration 31→28. Direct +2000×3 = +6,000 → direct 89,000. Indirect = 1500×28 = 42,000. Total = 89,000 + 42,000 = 131,000 (net +1,500/day, increases).

Because the lowest available slope (2,000) already exceeds the indirect rate (1,500), every crash day costs more than it saves. Tabulating each cumulative step confirms the trend:

Total cost rises monotonically as we crash, because the minimum cost slope (Rs.2,000/day) is greater than the indirect cost rate (Rs.1,500/day).

(d) Optimum. The minimum total cost occurs at the normal duration of 31 days with minimum total cost Rs. 129,500. No crashing is economical here; crashing would only be justified if a contractual deadline shorter than 31 days were imposed (then crash cheapest slopes first: A, then B/C/D). If, say, a 28-day deadline is mandated, crash A by 3 days at the least extra cost, giving Rs.131,000.

(a) Excavator productivity.

Effective volume per cycle (loose) $= 1.2 \times 0.85 = 1.02$ loose m³/cycle.

Cycles per 60-min hour at theoretical rate $= 3600/22 = 163.6$ cycles/h; with job efficiency $50/60$: Effective cycles/h $= 163.6 \times (50/60) = 136.4$ cycles/h.

• Loose production $= 1.02 \times 136.4 = 139.1$ loose m³/h.
• Bank production $= 139.1 \times 0.80 = 111.3$ bank m³/h.

(b) Passes and truck load time.

Passes to fill one truck $= 9.0 / 1.02 = 8.82 \Rightarrow$ 9 passes (round up; truck carries $9 \times 1.02 = 9.18 \approx 9.0$ loose m³).

Loading time per truck (use 9 passes) $=$ passes × cycle time $= 9 \times 22 = 198$ s $= 3.30$ min. (If efficiency-adjusted: at 136.4 cycles/h, 9 passes take $9/136.4 \times 60 = 3.96$ min. Use the gross loading-spot value $\approx 3.3$ min plus efficiency in the balance below.)

Using the effective rate, time the excavator needs to load one 9.0-loose-m³ truck $= 9.0/139.1 \times 60 = 3.88$ min.

(c) Number of trucks for continuous excavator work.

Truck total cycle = haul/dump/return/spot + loading = $18 + 3.88 = 21.88$ min. Number of trucks $N = \dfrac{\text{truck cycle time}}{\text{loading time}} = \dfrac{21.88}{3.88} = 5.64 \Rightarrow$ 6 trucks to avoid excavator idling (round up).

(d) System production with only 4 trucks (bottleneck = trucks).

With 4 trucks < 6 required, the trucks are the bottleneck. Each truck completes one full cycle in 21.88 min and delivers 9.0 loose m³.

• Truck fleet delivery rate $= 4 \times \dfrac{9.0}{21.88} \times 60 = 4 \times 24.68 = 98.7$ loose m³/h.
• In bank measure $= 98.7 \times 0.80 = 79.0$ bank m³/h.

Since 98.7 loose m³/h < excavator capacity 139.1 loose m³/h, the trucks are the bottleneck; the excavator works only $98.7/139.1 = 71\%$ of the time. System output ≈ 79 bank m³/h.

(a) Types of construction contracts.

(b) Public tendering process (Public Procurement Act / PPMO procedures):

1. Need identification, cost estimate, and budget assurance.
2. Preparation of bidding documents (specifications, BOQ, conditions of contract).
3. Invitation for Bids (public notice / e-GP, national or international competitive bidding).
4. Issue/sale of bidding documents; pre-bid meeting and clarifications.
5. Submission of sealed bids by deadline (with bid security).
6. Public bid opening, recording of read-out prices.
7. Evaluation: preliminary (responsiveness), technical, then financial; correction of arithmetic errors; determination of lowest substantially responsive evaluated bid.
8. Recommendation and approval by the competent authority.
9. Notification/Letter of Intent and Letter of Acceptance.
10. Signing of contract agreement after submission of performance security.

(c) Securities and retention:

• (i) Bid security (bid bond): guarantees the bidder will not withdraw and will sign the contract; typically 2–3% of the estimated/bid amount (often a fixed amount specified in documents), valid through bid validity + a margin. Forfeited if bidder defaults.
• (ii) Performance security (performance bond): guarantees due performance of the contract; typically 5% of the contract price (8% for abnormally low bids), valid until completion/defect-liability.
• (iii) Retention money: a percentage withheld from each interim payment to cover defects; typically 5% deducted per bill up to a ceiling of about 5% of contract value, released half at completion and half after the defect-liability period.

(d) The Engineer, variations and EOT.

• The Engineer administers the contract: supervises work, certifies payments, issues instructions, inspects/approves materials and quality, and makes fair determinations between owner and contractor.
• Variation order: the Engineer may instruct changes in quantity, quality or scope within contract limits; the contractor executes and is valued at BOQ rates (or new rates negotiated for new items / quantities beyond a threshold, e.g. ±15–25%). A written variation order and approval are required.
• Extension of Time (EOT): when delay is caused by employer-risk events (late possession of site, variations, exceptional weather, force majeure), the contractor submits a claim with notice within the contract time-bar, supported by records and programme analysis. The Engineer assesses entitlement and grants a fair EOT, revising the completion date so liquidated damages do not apply to the excused delay; associated cost/compensation is decided separately per the contract.

Work Breakdown Structure (WBS): a hierarchical, deliverable-oriented decomposition of the total project scope into progressively smaller, manageable components (levels: project → major deliverables/phases → work packages → activities). The lowest level (work package) is small enough to estimate cost, duration and assign responsibility.

Two advantages of preparing a WBS before scheduling:

1. Ensures the entire scope is captured — nothing is omitted or double-counted — giving a complete activity list as the basis for estimating and scheduling.
2. Clarifies responsibility, cost accounts and control points (each work package can be assigned, budgeted and monitored), enabling integrated cost–schedule control.

Major limitation of a bar (Gantt) chart overcome by CPM: A bar chart does not explicitly show the logical interdependencies between activities (which task must precede which) and therefore cannot identify the critical path or float. A CPM network shows these dependencies, computes critical activities and slack, and reveals the effect of a delay in any activity on the project completion date — information a simple bar chart cannot convey.

Resource smoothing (time-limited scheduling): the project completion date is fixed; activities are shifted only within their available float to even out (reduce peaks in) resource demand. The project duration does not change; resource peaks are reduced but a resource ceiling is not guaranteed.

Resource leveling (resource-limited scheduling): the resource availability is fixed (a hard limit); activities are rescheduled so demand never exceeds the limit, even if this extends the project duration. Here meeting the resource constraint takes priority over the deadline.

Example (using float): Suppose activities X (non-critical, float = 3 days, needs 4 masons) and Y (critical, needs 5 masons) both currently run on days 1–3, giving a peak demand of 9 masons. Because X has 3 days of float, it can be delayed to start after Y finishes. Daily demand then becomes 5 masons (Y) followed by 4 masons (X) — peak reduced from 9 to 5 — without delaying the project, since X was moved only within its float. This is smoothing; if even 5 masons exceeded the limit, leveling would push critical work later and lengthen the project.

Given: PV = 8,00,000; EV = 7,20,000; AC = 8,40,000; BAC = 30,00,000.

Schedule Variance: $SV = EV - PV = 7{,}20{,}000 - 8{,}00{,}000 = -80{,}000$ (negative ⇒ behind schedule).

Cost Variance: $CV = EV - AC = 7{,}20{,}000 - 8{,}40{,}000 = -1{,}20{,}000$ (negative ⇒ over budget).

Schedule Performance Index: $SPI = \dfrac{EV}{PV} = \dfrac{7{,}20{,}000}{8{,}00{,}000} = 0.90$ (< 1 ⇒ behind schedule).

Cost Performance Index: $CPI = \dfrac{EV}{AC} = \dfrac{7{,}20{,}000}{8{,}40{,}000} = 0.857$ (< 1 ⇒ cost overrun; getting Rs.0.857 of value per rupee spent).

Estimate at Completion (cost performance continues):

$$EAC = \frac{BAC}{CPI} = \frac{30{,}00{,}000}{0.857} = \textbf{Rs. } 35{,}00{,}000.$$

(Equivalently $EAC = AC + (BAC-EV)/CPI = 8{,}40{,}000 + (30{,}00{,}000-7{,}20{,}000)/0.857 = 8{,}40{,}000 + 26{,}60{,}000 = 35{,}00{,}000$.)

Comment: Both indices are below 1, so the project is behind schedule and over budget. If the present cost trend continues, the project will overrun its budget by about Rs. 5,00,000 (EAC 35 lakh vs BAC 30 lakh). Corrective action on productivity/cost control and schedule recovery is needed.

Given: Purchase price $P = 75{,}00{,}000$; salvage $S = 15{,}00{,}000$; life $n = 6$ yr; usage = 1,800 h/yr; interest rate = 12%.

(a) Straight-line depreciation:

$$D = \frac{P - S}{n} = \frac{75{,}00{,}000 - 15{,}00{,}000}{6} = \frac{60{,}00{,}000}{6} = \textbf{Rs. } 10{,}00{,}000\text{/year}.$$

(b) Average annual investment and interest:

$$\text{Avg. investment} = \frac{(P - S)(n+1)}{2n} + S = \frac{60{,}00{,}000 \times 7}{12} + 15{,}00{,}000 = 35{,}00{,}000 + 15{,}00{,}000 = 50{,}00{,}000.$$

Interest cost $= 0.12 \times 50{,}00{,}000 = \textbf{Rs. } 6{,}00{,}000\text{/year}.$

(c) Ownership cost per hour: Annual ownership cost $= D + \text{interest} + \text{taxes/insurance/storage}$

$$= 10{,}00{,}000 + 6{,}00{,}000 + 2{,}00{,}000 = 18{,}00{,}000 \text{ /year}.$$

Per operating hour:

$$\frac{18{,}00{,}000}{1{,}800} = \textbf{Rs. } 1{,}000 \text{ per hour}.$$

(a) QA vs QC:

• Quality Assurance (QA): process-oriented, proactive set of planned activities/systems that build confidence quality requirements will be met (e.g., establishing a Quality Management Plan, approving mix-design procedures, training, document control, audits). It prevents defects.
• Quality Control (QC): product-oriented, reactive inspection and testing of the actual output to detect non-conformance (e.g., testing concrete cube strength, slump test, checking reinforcement cover before pour). It detects defects.

QA defines and assures the system; QC verifies the product against the standard.

(b) Four common accident causes and preventive measures:

(Other valid pairs: fire/explosion → fire extinguishers & hot-work permits; manual-handling injury → mechanical aids & training.)

(c) Job Safety Analysis (JSA): a systematic procedure that breaks a specific job/task into its sequential steps, identifies the potential hazards associated with each step, and specifies the controls/safe procedures to eliminate or reduce each hazard. It is done before starting the task to anticipate risks, communicate safe methods to workers, comply with safety regulations, and prevent accidents — turning hazard identification into preventive action.

(a) Dummy activity (AOA): a fictitious activity of zero duration and zero resources, shown as a dashed arrow, that maintains correct logic in an arrow diagram. Its two purposes are:

1. Logic (dependency) dummy: to correctly represent dependencies when an activity depends on only some of the activities that another activity also depends on (avoiding wrong shared dependencies).
2. Uniqueness (numbering/grammar) dummy: to give two parallel activities that share the same start and end nodes their own unique identity (so no two activities have the same head and tail node pair).

(b) AOA network for the table.

Let nodes be numbered events. Logic required: C needs only A; D needs both A and B; E needs C and D.

           A            C
   (1) ─────────► (2) ─────────► (4)
    │              :               │
    │ B            : dummy(0)      │ E
    │              ▼               ▼
   (1)──B──► (3) ─────────────► (4) ───► (5)
                    D

Clearer description of nodes/arrows:

• Node 1 = project start.
• A: 1 → 2. B: 1 → 3.
• A finishes at node 2; B finishes at node 3.
• A dummy is drawn from node 2 → node 3 so that D (starting at node 3) depends on both A and B, while C (starting at node 2) depends on A only. Without the dummy, merging A and B at one node would wrongly make C depend on B as well.
• C: 2 → 4. D: 3 → 4. Both C and D end at node 4 (merge before E).
• E: 4 → 5 (project end).

Thus the dependency dummy 2 → 3 is required so that C depends on A only while D depends on A and B; C and D then merge at node 4 to release E.