BE Civil Engineering (IOE, TU) Engineering Economics (IOE, CE 656) Question Paper 2080 Nepal

(a) Time value of money, simple vs compound interest

Time value of money (TVM): Money available now is worth more than the same nominal amount in the future because present money can be invested to earn a return. Hence cash flows occurring at different points in time are not directly comparable and must be converted to a common time using an interest rate.

Simple interest — interest is charged only on the original principal:

$$F = P(1 + i\,n)$$

where $P$ = principal, $i$ = interest rate per period, $n$ = number of periods.

Compound interest — interest is charged on principal and on accumulated interest:

$$F = P(1 + i)^n$$

Compound interest produces a larger future amount than simple interest for $n > 1$.

(b) Effective annual rate and lump-sum repayment

Nominal rate $r = 12\% = 0.12$ per year, compounded quarterly, so $m = 4$ and the rate per quarter is

$$i_q = \frac{0.12}{4} = 0.03 = 3\%.$$

(i) Effective annual interest rate:

$$i_{eff} = \left(1 + \frac{r}{m}\right)^m - 1 = (1.03)^4 - 1 = 1.12550881 - 1 = 0.125509$$ $$\boxed{i_{eff} \approx 12.55\%}$$

(ii) Lump-sum amount after 5 years. Over 5 years there are $n = 5 \times 4 = 20$ quarters:

$$F = P(1+i_q)^n = 800{,}000 \,(1.03)^{20}.$$

$(1.03)^{20} = 1.806111$, so

$$F = 800{,}000 \times 1.806111 = 1{,}444{,}889.$$ $$\boxed{F \approx \text{Rs. } 1{,}444{,}889}$$

(c) Equal annual repayments

Use the effective annual rate $i = 0.125509$, $N = 5$ years, $P = 800{,}000$. The capital-recovery factor gives the annual payment $A$:

$$A = P\,\frac{i(1+i)^N}{(1+i)^N - 1}.$$

$(1+i)^5 = (1.125509)^5 = 1.806111$ (this equals $(1.03)^{20}$, confirming consistency).

$$A = 800{,}000 \times \frac{0.125509 \times 1.806111}{1.806111 - 1} = 800{,}000 \times \frac{0.226706}{0.806111}.$$ $$A = 800{,}000 \times 0.281234 = 224{,}987.$$ $$\boxed{A \approx \text{Rs. } 224{,}987 \text{ per year}}$$

Factors at i = 10%

• $(P/A, 10\%, 10) = \dfrac{(1.1)^{10}-1}{0.1(1.1)^{10}} = \dfrac{2.593742-1}{0.259374} = 6.144567$
• $(P/F, 10\%, 5) = (1.1)^{-5} = 0.620921$
• $(P/F, 10\%, 10) = (1.1)^{-10} = 0.385543$
• $(A/P, 10\%, 10) = 0.162745$

Costs are taken as positive outflows; we compute PW of costs (smaller is better).

(a) Present Worth of costs

Design X:

$$PW_X = 2{,}400{,}000 + 380{,}000(P/A) + 300{,}000(P/F,5) - 200{,}000(P/F,10)$$ $$= 2{,}400{,}000 + 380{,}000(6.144567) + 300{,}000(0.620921) - 200{,}000(0.385543)$$ $$= 2{,}400{,}000 + 2{,}334{,}935 + 186{,}276 - 77{,}109$$ $$PW_X = 4{,}844{,}102.$$

Design Y:

$$PW_Y = 3{,}600{,}000 + 250{,}000(6.144567) - 450{,}000(0.385543)$$ $$= 3{,}600{,}000 + 1{,}536{,}142 - 173{,}494$$ $$PW_Y = 4{,}962{,}648.$$

Since $PW_X = \text{Rs. }4{,}844{,}102 < PW_Y = \text{Rs. }4{,}962{,}648$,

$$\boxed{\text{Design X is more economical (lower PW of cost by } \approx \text{Rs. }118{,}546).}$$

(b) EUAC

Multiply each PW of cost by $(A/P, 10\%, 10) = 0.162745$:

$$EUAC_X = 4{,}844{,}102 \times 0.162745 = 788{,}353 \;\text{Rs./yr}$$ $$EUAC_Y = 4{,}962{,}648 \times 0.162745 = 807{,}647 \;\text{Rs./yr}$$ $$\boxed{EUAC_X \approx \text{Rs. }788{,}353/\text{yr} < EUAC_Y \approx \text{Rs. }807{,}647/\text{yr}}$$

The EUAC comparison confirms the PW result: Design X is recommended.

(a) IRR of each machine

For a single uniform series, IRR is the rate where $0 = -P + A(P/A, i, 6)$, i.e. $(P/A, i, 6) = P/A_{annual}$.

Machine A: $(P/A, i, 6) = 1{,}500{,}000 / 420{,}000 = 3.5714.$

• At 17%: $(P/A,17\%,6) = 3.5892$
• At 18%: $(P/A,18\%,6) = 3.4976$

Interpolating for 3.5714:

$$i_A = 17\% + \frac{3.5892 - 3.5714}{3.5892 - 3.4976}\times 1\% = 17\% + \frac{0.0178}{0.0916}\times 1\% = 17.19\%.$$ $$\boxed{IRR_A \approx 17.2\% > 12\% \Rightarrow \text{acceptable.}}$$

Machine B: $(P/A, i, 6) = 2{,}200{,}000 / 560{,}000 = 3.9286.$

• At 13%: $(P/A,13\%,6) = 3.9975$
• At 14%: $(P/A,14\%,6) = 3.8887$

Interpolating for 3.9286:

$$i_B = 13\% + \frac{3.9975 - 3.9286}{3.9975 - 3.8887}\times 1\% = 13\% + \frac{0.0689}{0.1088}\times 1\% = 13.63\%.$$ $$\boxed{IRR_B \approx 13.6\% > 12\% \Rightarrow \text{acceptable.}}$$

Both are individually acceptable, so we cannot simply pick the higher IRR; an incremental analysis is required.

(b) Incremental analysis (B − A)

Incremental IRR: $(P/A, \Delta i, 6) = 700{,}000 / 140{,}000 = 5.000.$

• At 5%: $(P/A,5\%,6) = 5.0757$
• At 6%: $(P/A,6\%,6) = 4.9173$

$$\Delta i = 5\% + \frac{5.0757 - 5.000}{5.0757 - 4.9173}\times 1\% = 5\% + \frac{0.0757}{0.1584}\times 1\% = 5.48\%.$$

Since the incremental return $\Delta i \approx 5.5\% < \text{MARR } 12\%$, the extra investment in B is not justified.

$$\boxed{\text{Select Machine A.}}$$

The extra Rs. 700,000 spent on B earns only about 5.5%, well below the 12% MARR; that capital is better employed elsewhere.

Factor

$$(P/A, 8\%, 30) = \frac{(1.08)^{30}-1}{0.08(1.08)^{30}} = \frac{10.062657-1}{0.805013} = 11.257783.$$

Convention used: O&M treated as a cost in the denominator.

$$\text{B/C} = \frac{PW(\text{benefits})}{\text{Initial cost} + PW(\text{O\&M})}.$$

(a) Conventional B/C ratios

Scheme P:

• PW benefits $= 2{,}200{,}000 \times 11.257783 = 24{,}767{,}123$
• PW costs $= 12{,}000{,}000 + 600{,}000 \times 11.257783 = 12{,}000{,}000 + 6{,}754{,}670 = 18{,}754{,}670$

$$\text{B/C}_P = \frac{24{,}767{,}123}{18{,}754{,}670} = 1.321.$$

Scheme Q:

• PW benefits $= 2{,}900{,}000 \times 11.257783 = 32{,}647{,}571$
• PW costs $= 18{,}000{,}000 + 750{,}000 \times 11.257783 = 18{,}000{,}000 + 8{,}443{,}337 = 26{,}443{,}337$

$$\text{B/C}_Q = \frac{32{,}647{,}571}{26{,}443{,}337} = 1.235.$$

Both exceed 1.0, so both are economically justified.

$$\boxed{\text{B/C}_P \approx 1.32,\quad \text{B/C}_Q \approx 1.24}$$

(b) Incremental B/C (Q − P)

Q is the larger-cost alternative; test the increment $\Delta(Q-P)$:

• $\Delta$ PW benefits $= (2{,}900{,}000 - 2{,}200{,}000)\times 11.257783 = 700{,}000 \times 11.257783 = 7{,}880{,}448$
• $\Delta$ PW costs $= (18{,}000{,}000 - 12{,}000{,}000) + (750{,}000 - 600{,}000)\times 11.257783$

$$= 6{,}000{,}000 + 150{,}000 \times 11.257783 = 6{,}000{,}000 + 1{,}688{,}667 = 7{,}688{,}667.$$ $$\Delta \text{B/C} = \frac{7{,}880{,}448}{7{,}688{,}667} = 1.025.$$

Since $\Delta\text{B/C} = 1.025 > 1.0$, the extra investment in Scheme Q is justified.

$$\boxed{\text{Recommend Scheme Q.}}$$

(Note: although Scheme P has the higher individual B/C ratio, incremental analysis correctly selects Q because the additional spending still returns more than it costs.)

(a) Marginal cost of keeping the defender one more year

Keeping the defender one more year costs:

1. Lost capital + interest on retained capital: by keeping, the company forgoes the current market value of Rs. 1,200,000 and recovers only Rs. 850,000 at year end.
   • Capital loss (decline in value) $= 1{,}200{,}000 - 850{,}000 = 350{,}000$
   • Interest (opportunity cost) on the capital tied up $= 0.15 \times 1{,}200{,}000 = 180{,}000$
2. O&M cost next year $= 520{,}000$

$$\text{Marginal cost} = 350{,}000 + 180{,}000 + 520{,}000 = 1{,}050{,}000.$$ $$\boxed{\text{Cost of keeping defender one more year} \approx \text{Rs. }1{,}050{,}000}$$

(b) EUAC of the challenger

Factors at $i = 15\%$, $N = 8$:

• $(A/P, 15\%, 8) = \dfrac{0.15(1.15)^8}{(1.15)^8 - 1} = \dfrac{0.15 \times 3.059023}{2.059023} = \dfrac{0.458853}{2.059023} = 0.222850$
• $(A/F, 15\%, 8) = (A/P) - i = 0.222850 - 0.15 = 0.072850$

$$EUAC_C = 3{,}000{,}000(A/P) - 600{,}000(A/F) + 300{,}000$$ $$= 3{,}000{,}000(0.222850) - 600{,}000(0.072850) + 300{,}000$$ $$= 668{,}550 - 43{,}710 + 300{,}000 = 924{,}840.$$ $$\boxed{EUAC_C \approx \text{Rs. }924{,}840/\text{yr}}$$

Decision

Compare the defender's marginal cost for the next year (Rs. 1,050,000) with the challenger's minimum EUAC (Rs. 924,840):

$$1{,}050{,}000 > 924{,}840.$$

The cost of keeping the defender one more year exceeds the challenger's annual cost, so

$$\boxed{\text{Replace the defender now with the challenger.}}$$

(a) Straight-line method

Annual depreciation:

$$D = \frac{B - S}{N} = \frac{500{,}000 - 50{,}000}{5} = \frac{450{,}000}{5} = 90{,}000 \text{ per year.}$$

Final book value equals the salvage value, as required.

(b) Double-declining-balance method

DDB rate $= \dfrac{2}{N} = \dfrac{2}{5} = 0.40$ (40% of the beginning book value each year).

Check: book value (Rs. 108,000) has not fallen below salvage (Rs. 50,000), so no adjustment is needed.

$$\boxed{\text{Book value at end of year 3 (DDB)} = \text{Rs. }108{,}000}$$

(a) Real (inflation-free) interest rate

The market rate $i$ combines the real rate $i'$ and inflation $f$ via $1 + i = (1 + i')(1 + f)$. Therefore:

$$i' = \frac{1 + i}{1 + f} - 1 = \frac{1.14}{1.06} - 1 = 1.075472 - 1 = 0.075472.$$ $$\boxed{i' \approx 7.55\%}$$

(b) Present worth in constant rupees

The Rs. 100,000/year are stated in actual (then-current) rupees, so the simplest correct approach is to discount them at the market rate of 14% — this directly gives present worth in today's rupees.

$$(P/A, 14\%, 4) = \frac{(1.14)^4 - 1}{0.14(1.14)^4} = \frac{1.688960 - 1}{0.14 \times 1.688960} = \frac{0.688960}{0.236454} = 2.913712.$$ $$PW = 100{,}000 \times 2.913712 = 291{,}371.$$ $$\boxed{PW \approx \text{Rs. }291{,}371}$$

Verification via the real-rate route: Convert each actual cash flow to constant rupees by dividing by $(1.06)^t$, then discount at the real rate $7.55\%$. Because $1.14 = 1.06 \times 1.075472$, discounting actual rupees at 14% is mathematically identical to discounting constant rupees at 7.55%; both yield $\approx$ Rs. 291,371, confirming the result.

This is a uniform series (base) plus an arithmetic gradient.

• Base amount $A_1 = 200{,}000$ (years 1–6)
• Gradient $G = 30{,}000$ per year

Gradient-to-uniform conversion factor at $i = 10\%$, $N = 6$:

$$(A/G, 10\%, 6) = \frac{1}{i} - \frac{N}{(1+i)^N - 1} = \frac{1}{0.10} - \frac{6}{(1.1)^6 - 1}.$$

$(1.1)^6 = 1.771561$, so $(1.1)^6 - 1 = 0.771561$.

$$(A/G, 10\%, 6) = 10 - \frac{6}{0.771561} = 10 - 7.77645 = 2.22355.$$

Equivalent annual cost of the gradient portion:

$$A_G = G\,(A/G) = 30{,}000 \times 2.22355 = 66{,}707.$$

Total EUAC:

$$EUAC = A_1 + A_G = 200{,}000 + 66{,}707 = 266{,}707.$$ $$\boxed{EUAC \approx \text{Rs. }266{,}707 \text{ per year}}$$

(a) Required annual deposit (sinking-fund)

Given $F = 5{,}000{,}000$, $i = 9\%$, $N = 8$:

$$(A/F, 9\%, 8) = \frac{i}{(1+i)^N - 1} = \frac{0.09}{(1.09)^8 - 1}.$$

$(1.09)^8 = 1.992563$, so $(1.09)^8 - 1 = 0.992563$.

$$(A/F, 9\%, 8) = \frac{0.09}{0.992563} = 0.090674.$$ $$A = F \times (A/F) = 5{,}000{,}000 \times 0.090674 = 453{,}370.$$ $$\boxed{A \approx \text{Rs. }453{,}370 \text{ per year}}$$

(b) Total interest earned

Total deposits made over 8 years:

$$\text{Total deposits} = 8 \times 453{,}370 = 3{,}626{,}960.$$

The fund grows to Rs. 5,000,000, so the interest earned is the difference:

$$\text{Interest} = 5{,}000{,}000 - 3{,}626{,}960 = 1{,}373{,}040.$$ $$\boxed{\text{Total interest earned} \approx \text{Rs. }1{,}373{,}040}$$

Capitalized cost = present worth of perpetual annual cost + present worth of the periodic (every-10-year) repairs.

Annual maintenance (perpetuity):

$$P_1 = \frac{A}{i} = \frac{150{,}000}{0.08} = 1{,}875{,}000.$$

Periodic major repair every 10 years. First convert the Rs. 1,200,000 occurring every 10 years into an equivalent annual amount using the sinking-fund factor:

$$(A/F, 8\%, 10) = \frac{0.08}{(1.08)^{10} - 1} = \frac{0.08}{2.158925 - 1} = \frac{0.08}{1.158925} = 0.069029.$$ $$A_{repair} = 1{,}200{,}000 \times 0.069029 = 82{,}835.$$

Capitalize this equivalent annual amount as a perpetuity:

$$P_2 = \frac{82{,}835}{0.08} = 1{,}035{,}438.$$

Total capitalized cost:

$$CC = P_1 + P_2 = 1{,}875{,}000 + 1{,}035{,}438 = 2{,}910{,}438.$$ $$\boxed{CC \approx \text{Rs. }2{,}910{,}438}$$

(a) Simple payback period

Cumulative undiscounted cash flow:

Recovery occurs during year 4. Fraction of year 4 needed $= 300{,}000 / 1{,}500{,}000 = 0.20$.

$$\boxed{\text{Simple payback} = 3 + 0.20 = 3.2 \text{ years}}$$

(b) Discounted payback at 10%

Discount factors $(P/F, 10\%, t)$: 0.90909, 0.82645, 0.75131, 0.68301, 0.62092.

Recovery occurs during year 4. Fraction $= 972{,}201 / 1{,}024{,}521 = 0.949$.

$$\boxed{\text{Discounted payback} \approx 3 + 0.95 = 3.95 \text{ years}}$$

(c) Net present worth

Using the PVs from the table plus year 5:

• Year 5 PV $= 1{,}500{,}000 \times 0.62092 = 931{,}381$

$$NPW = -4{,}000{,}000 + 909{,}091 + 991{,}736 + 1{,}126{,}972 + 1{,}024{,}521 + 931{,}381.$$

Sum of positive PVs $= 4{,}983{,}701$, so

$$NPW = 4{,}983{,}701 - 4{,}000{,}000 = 983{,}701.$$ $$\boxed{NPW \approx +\text{Rs. }983{,}701 > 0 \Rightarrow \text{the project is ACCEPTABLE.}}$$

The positive NPW confirms the project earns more than the 10% MARR; the discounted payback (3.95 yr) being well within the 5-year horizon supports the same conclusion.