BE Civil Engineering (IOE, TU) Engineering Economics (IOE, CE 656) Question Paper 2077 Nepal

(a) Time value of money (2 marks)

The time value of money (TVM) is the principle that a given amount of money available today is worth more than the same amount in the future, because money on hand can be invested to earn a return (interest) over time. Hence equal cash amounts occurring at different points in time are not directly comparable; they must be moved to a common point using an interest rate.

Reasons a rupee today is worth more than a rupee a year from now:

• It can be invested immediately to earn interest, growing to more than NPR 1.
• Inflation erodes future purchasing power.
• Future receipts carry uncertainty/risk, while money in hand is certain.

(b) Future worth of single deposit (3 marks)

Given $P = 250{,}000$, $i = 12\% = 0.12$, $n = 8$.

Using the single-payment compound-amount factor:

$$F = P(1+i)^n = P\,(F/P,\,12\%,\,8)$$ $$(1.12)^8 = 2.47596$$ $$F = 250{,}000 \times 2.47596 = 618{,}990.79$$

F = NPR 618,990.79

(c) Uniform series — present and future worth (5 marks)

Given $A = 80{,}000$, $i = 10\% = 0.10$, $n = 15$.

Cash-flow diagram (year-end deposits):

 A   A   A   ...   A
 |   |   |         |
 1   2   3   ...  15

Present worth (uniform-series present-worth factor):

$$P = A\,\frac{(1+i)^n - 1}{i(1+i)^n} = A\,(P/A,\,10\%,\,15)$$ $$(1.10)^{15} = 4.177248$$ $$(P/A,10\%,15) = \frac{4.177248 - 1}{0.10 \times 4.177248} = \frac{3.177248}{0.417725} = 7.60608$$ $$P = 80{,}000 \times 7.60608 = 608{,}486.36$$

P = NPR 608,486.36

Future worth (uniform-series compound-amount factor):

$$F = A\,\frac{(1+i)^n - 1}{i} = A\,(F/A,\,10\%,\,15)$$ $$(F/A,10\%,15) = \frac{4.177248 - 1}{0.10} = 31.77248$$ $$F = 80{,}000 \times 31.77248 = 2{,}541{,}798.54$$

F = NPR 2,541,798.54

(a) Validity conditions for PW comparison (2 marks)

• Alternatives must be compared over an equal service life (here both are 12 years, so the equal-life condition is satisfied). If lives differ, use a common multiple of lives or the study-period/annual-worth approach.
• A single, consistent interest rate (MARR) must be applied to all alternatives.
• All relevant cash flows (first cost, recurring costs, salvage) must be included and placed at correct points in time.

(b) Present worth of cost (5 marks)

Factors at $i=9\%$, $n=12$:

$$(1.09)^{12} = 2.812665$$ $$(P/A,9\%,12) = \frac{2.812665 - 1}{0.09 \times 2.812665} = \frac{1.812665}{0.253140} = 7.16073$$ $$(P/F,9\%,12) = \frac{1}{2.812665} = 0.355535$$

Design X:

$$PW_X = 4{,}000{,}000 + 520{,}000(7.16073) - 400{,}000(0.355535)$$ $$= 4{,}000{,}000 + 3{,}723{,}580 - 142{,}214 = 7{,}581{,}366$$

Design Y:

$$PW_Y = 6{,}500{,}000 + 300{,}000(7.16073) - 900{,}000(0.355535)$$ $$= 6{,}500{,}000 + 2{,}148{,}219 - 319{,}982 = 8{,}328{,}237$$

Since these are costs, the smaller PW is preferred:

$$PW_X = \text{NPR } 7{,}581{,}366 < PW_Y = \text{NPR } 8{,}328{,}237$$

Select Design X.

(c) EUAC of Design X (3 marks)

$$(A/P,9\%,12) = \frac{1}{(P/A,9\%,12)} = \frac{1}{7.16073} = 0.139651$$ $$EUAC_X = PW_X \times (A/P,9\%,12) = 7{,}581{,}366 \times 0.139651 = 1{,}058{,}749$$

EUAC of Design X = NPR 1,058,749 per year.

Interpretation: choosing Design X commits the municipality to an equivalent uniform cost of about NPR 1.06 million per year for 12 years, which is the lowest among the alternatives; this annual figure is convenient for budgeting and for comparison against any future alternative with a different life.

(a) Definition and decision rule (2 marks)

The internal rate of return (IRR) is the interest rate $i^*$ that makes the net present worth of a project's cash flows equal to zero (equivalently, the rate at which PW of benefits equals PW of costs):

$$NPW(i^*) = 0.$$

Decision rule for a single project: accept if $i^* \ge \text{MARR}$; reject if $i^* < \text{MARR}$.

(b) IRR by interpolation (5 marks)

The IRR satisfies:

$$-1{,}200{,}000 + 220{,}000\,(P/A,\,i^*,\,10) = 0 \;\Rightarrow\; (P/A,i^*,10) = \frac{1{,}200{,}000}{220{,}000} = 5.4545$$

Trial at 12%: $(P/A,12\%,10) = 5.65022$

$$NPW_{12\%} = -1{,}200{,}000 + 220{,}000(5.65022) = +43{,}049$$

Trial at 13%: $(P/A,13\%,10) = 5.42624$

$$NPW_{13\%} = -1{,}200{,}000 + 220{,}000(5.42624) = -6{,}226$$

The IRR lies between 12% and 13%. Linear interpolation:

$$i^* = 12\% + (13\% - 12\%)\times\frac{43{,}049}{43{,}049 + 6{,}226} = 12\% + 1\%\times\frac{43{,}049}{49{,}275}$$ $$i^* = 12\% + 0.8736\% = 12.87\%$$

IRR ≈ 12.87% per annum.

(c) Acceptability and limitation (3 marks)

Since $IRR = 12.87\% > \text{MARR} = 11\%$, the investment is acceptable — it earns more than the contractor's minimum required rate.

Limitation of IRR for competing projects: when ranking mutually exclusive alternatives, the project with the highest IRR is not necessarily the best, because IRR ignores the scale of investment. A small project may have a very high IRR but contribute little net value, while a larger project with a slightly lower IRR may add more total worth. Correct selection requires incremental IRR analysis (analysing the IRR of the extra investment $\Delta$) or use of the PW/AW method. IRR can also yield multiple roots for non-conventional cash flows.

(a) B/C method (2 marks)

The benefit-cost (B/C) ratio is the ratio of the equivalent worth of benefits (to the public) to the equivalent worth of costs (to the sponsor/government), both expressed at the same point in time and same discount rate. It is the standard tool for evaluating public projects, where the goal is overall social welfare rather than private profit.

Decision criterion: accept the project if $B/C \ge 1$; reject if $B/C < 1$. A ratio of exactly 1 means benefits just equal costs.

(b) B/C calculations (6 marks)

Factor at $i=8\%$, $n=20$:

$$(1.08)^{20} = 4.660957$$ $$(P/A,8\%,20) = \frac{4.660957 - 1}{0.08 \times 4.660957} = \frac{3.660957}{0.372877} = 9.81815$$

Present worth of items:

$$PW_{\text{benefits}} = 900{,}000 \times 9.81815 = 8{,}836{,}333$$ $$PW_{\text{O\&M}} = 150{,}000 \times 9.81815 = 1{,}472{,}722$$ $$PW_{\text{initial cost}} = 5{,}000{,}000$$

Conventional B/C (O&M treated as a cost in the denominator):

$$B/C = \frac{PW_{\text{benefits}}}{PW_{\text{initial}} + PW_{\text{O\&M}}} = \frac{8{,}836{,}333}{5{,}000{,}000 + 1{,}472{,}722} = \frac{8{,}836{,}333}{6{,}472{,}722} = 1.365$$

Modified B/C (O&M netted out of benefits in the numerator):

$$B/C_{\text{mod}} = \frac{PW_{\text{benefits}} - PW_{\text{O\&M}}}{PW_{\text{initial}}} = \frac{8{,}836{,}333 - 1{,}472{,}722}{5{,}000{,}000} = \frac{7{,}363{,}611}{5{,}000{,}000} = 1.473$$

Conventional B/C = 1.365 and Modified B/C = 1.473; both exceed 1, so the project is economically justified.

(Note: the two ratios differ in value but always give the same accept/reject conclusion.)

(c) Conventional vs modified B/C (2 marks)

• Conventional B/C: annual O&M (and other recurring operating disbursements) are placed in the denominator as part of total cost. $B/C = \dfrac{\text{PW benefits}}{\text{PW (capital + O\&M)}}$.
• Modified B/C: recurring O&M is subtracted from benefits in the numerator, leaving only capital (and salvage) in the denominator. $B/C_{\text{mod}} = \dfrac{\text{PW (benefits} - \text{O\&M)}}{\text{PW capital}}$.

Both yield identical accept/reject decisions, but their numerical magnitudes differ, so the convention used must be stated when ranking projects.

(a) Depreciation and the three methods (3 marks)

Depreciation is the systematic allocation of the cost of a tangible asset (less salvage) over its useful life, representing the loss in value due to use, wear, age and obsolescence. It is a non-cash expense used for cost accounting and tax purposes.

• Straight-line (SL): equal depreciation each year, $D = (C - S)/N$.
• Sum-of-years-digits (SOYD): an accelerated method; year-$t$ charge $= \dfrac{N-t+1}{\sum_{k=1}^{N}k}\,(C-S)$, giving larger charges early.
• Double-declining-balance (DDB): accelerated method applying a fixed rate $2/N$ to the current book value each year; ignores salvage in the rate (book value not allowed to drop below salvage).

Given $C = 2{,}400{,}000$, $S = 200{,}000$, $N = 8$, so $C-S = 2{,}200{,}000$.

(b) Year-3 depreciation and book value (6 marks)

Straight-line:

$$D_t = \frac{2{,}200{,}000}{8} = 275{,}000 \text{ per year}$$ $$D_3 = \text{NPR } 275{,}000$$ $$BV_3 = C - 3D = 2{,}400{,}000 - 3(275{,}000) = \text{NPR } 1{,}575{,}000$$

Sum-of-years-digits: $\sum k = \frac{8(9)}{2} = 36$.

$$D_3 = \frac{8-3+1}{36}(2{,}200{,}000) = \frac{6}{36}(2{,}200{,}000) = \text{NPR } 366{,}667$$

Cumulative depreciation through year 3 uses digits $8,7,6$:

$$\text{Cum}_3 = \frac{8+7+6}{36}(2{,}200{,}000) = \frac{21}{36}(2{,}200{,}000) = 1{,}283{,}333$$ $$BV_3 = 2{,}400{,}000 - 1{,}283{,}333 = \text{NPR } 1{,}116{,}667$$

Double-declining-balance: rate $= 2/8 = 0.25$.

$$D_3 = \text{NPR } 337{,}500, \qquad BV_3 = \text{NPR } 1{,}012{,}500$$

(Each ending BV exceeds the NPR 200,000 salvage, so no adjustment is needed.)

Summary:

(c) Most rapid early write-off (1 mark)

DDB writes off the most value in the earliest years (lowest $BV_3$ here). A firm may prefer accelerated depreciation because larger early depreciation expenses reduce taxable income sooner, deferring tax payments and improving early-year cash flow (a time-value benefit).

(a) Nominal vs effective rate (2 marks)

• The nominal annual rate $r$ is the stated yearly rate that ignores the effect of compounding within the year (it is simply the periodic rate multiplied by the number of periods).
• The effective annual rate $i_{\text{eff}}$ is the actual rate earned/paid in one year once intra-year compounding is taken into account. Whenever compounding occurs more than once per year, $i_{\text{eff}} > r$.

(b) Effective annual rate (3 marks)

Given nominal $r = 12\%$, compounding $m = 12$ times/year, periodic rate $= 0.12/12 = 0.01$ (1% per month).

$$i_{\text{eff}} = \left(1 + \frac{r}{m}\right)^m - 1 = (1 + 0.01)^{12} - 1$$ $$(1.01)^{12} = 1.126825$$ $$i_{\text{eff}} = 1.126825 - 1 = 0.126825$$

Effective annual interest rate = 12.68% per annum.

(a) Accumulated amount, quarterly compounding (3 marks)

Nominal $r = 9\%$, $m = 4$, so periodic rate $= 0.09/4 = 0.0225$; number of periods $n = 5 \times 4 = 20$.

$$F = P\left(1 + \frac{r}{m}\right)^{mn} = 50{,}000\,(1.0225)^{20}$$ $$(1.0225)^{20} = 1.560509$$ $$F = 50{,}000 \times 1.560509 = 78{,}025.46$$

F = NPR 78,025.46.

(b) Additional interest vs annual compounding (2 marks)

With annual compounding at 9% for 5 years:

$$F_{\text{annual}} = 50{,}000\,(1.09)^5 = 50{,}000 \times 1.538624 = 76{,}931.20$$

Additional interest from quarterly compounding:

$$\Delta = 78{,}025.46 - 76{,}931.20 = 1{,}094.26$$

Additional interest ≈ NPR 1,094.26, earned purely because more frequent (quarterly) compounding produces a higher effective rate.

Arithmetic-gradient present worth (5 marks)

The series is a base annuity $A_1 = 20{,}000$ plus an arithmetic gradient $G = 3{,}000$, $n = 6$, $i = 10\%$.

Cash flow (year-end, NPR):

Year:  1      2      3      4      5      6
Cost:  20,000 23,000 26,000 29,000 32,000 35,000

Present worth $= A_1(P/A,10\%,6) + G(P/G,10\%,6)$.

Factors at $i=10\%$, $n=6$ (with $(1.10)^6 = 1.771561$):

$$(P/A,10\%,6) = \frac{1.771561 - 1}{0.10 \times 1.771561} = \frac{0.771561}{0.177156} = 4.35526$$ $$(P/G,10\%,6) = \frac{(1.10)^6 - 1}{0.10^2(1.10)^6} - \frac{6}{0.10(1.10)^6} = \frac{0.771561}{0.0177156} - \frac{6}{0.177156}$$ $$= 43.5526 - 33.8684 = 9.6842$$

Therefore:

$$P = 20{,}000(4.35526) + 3{,}000(9.6842)$$ $$= 87{,}105.20 + 29{,}052.53 = 116{,}157.73$$

Present worth of maintenance costs = NPR 116,157.73.

(a) Then-current (actual) future cost (2 marks)

Inflated future price grows at the inflation rate $f = 7\%$:

$$C_5 = C_0(1+f)^5 = 100{,}000\,(1.07)^5$$ $$(1.07)^5 = 1.402552$$ $$C_5 = 100{,}000 \times 1.402552 = 140{,}255.17$$

Then-current cost in 5 years ≈ NPR 140,255.17.

(b) Combined (market) interest rate (3 marks)

The combined rate $i_c$ links the real rate $i'$ and the inflation rate $f$ by:

$$i_c = i' + f + i'\,f$$ $$i_c = 0.10 + 0.07 + (0.10)(0.07) = 0.10 + 0.07 + 0.007 = 0.177$$

Combined (market) interest rate = 17.7% per annum.

Use: when cash flows are expressed in then-current (actual, inflated) rupees, they must be discounted at the combined/market rate (17.7%) to obtain present worth. Equivalently, if cash flows are stated in constant (real) rupees, the real rate (10%) is used. Mixing a real rate with actual-rupee cash flows (or vice versa) gives an incorrect equivalence.

(a) Defender and challenger (1 mark)

In replacement analysis, the defender is the asset currently owned and in service; the challenger is the proposed new (candidate) asset being considered to replace it.

(b) EUAC of the challenger (4 marks)

Given $C = 800{,}000$, $S = 100{,}000$, $n = 6$, $i = 10\%$, annual O&M $= 60{,}000$.

Factors with $(1.10)^6 = 1.771561$:

$$(A/P,10\%,6) = \frac{i(1+i)^n}{(1+i)^n - 1} = \frac{0.10 \times 1.771561}{0.771561} = 0.229607$$ $$(A/F,10\%,6) = \frac{i}{(1+i)^n - 1} = \frac{0.10}{0.771561} = 0.129607$$

EUAC (capital recovery + O&M − salvage annuity):

$$EUAC = C\,(A/P,10\%,6) - S\,(A/F,10\%,6) + \text{O\&M}$$ $$= 800{,}000(0.229607) - 100{,}000(0.129607) + 60{,}000$$ $$= 183{,}685.6 - 12{,}960.7 + 60{,}000$$ $$= 230{,}724.9$$

EUAC of the challenger ≈ NPR 230,725 per year.

The challenger should replace the defender if this EUAC is lower than the defender's marginal annual cost of continued ownership.

(Attempt any two; 2.5 marks each.)

(a) Payback period method

The payback period is the time required for the cumulative net cash inflows of a project to recover its initial investment. For uniform annual returns:

$$\text{Payback} = \frac{\text{Initial investment}}{\text{Annual net cash flow}}.$$

A project is favoured if its payback is shorter than a preset maximum. Main limitation: the simple payback method ignores the time value of money and ignores all cash flows after the payback point, so it can favour short-term, lower-value projects. (A discounted payback partially fixes the first defect but not the second.)

(b) Sunk cost

A sunk cost is an expenditure already incurred in the past that cannot be recovered or altered by any current or future decision (e.g., money already spent on a study or on the original purchase of an old machine). Because economic decisions compare future differences between alternatives, a sunk cost is the same regardless of which alternative is chosen and therefore must be excluded from the analysis. Including it leads to the "sunk-cost fallacy" of throwing good money after bad.

(c) MARR (minimum attractive rate of return)

MARR is the lowest rate of return an organisation is willing to accept on an investment, given the risk and the available alternatives; it serves as the hurdle rate (the $i$ used in PW/AW and the benchmark for IRR). Factors influencing MARR:

• Cost of capital (interest on borrowed funds and required return on equity).
• Risk of the project (higher risk → higher MARR).
• Opportunity cost — returns available on other competing investments.
• Availability of funds (capital rationing) and overall business/market conditions and inflation.