BE Civil Engineering (IOE, TU) Engineering Economics (IOE, CE 656) Question Paper 2078 Nepal

Factors at $i=8\%$, $n=30$:

$$(P/A,8\%,30)=\frac{1-(1.08)^{-30}}{0.08}=11.2578$$ $$(P/F,8\%,30)=(1.08)^{-30}=0.09938$$

Work in present worth (PW). Benefits are converted to PW; costs are the PW of first cost + PW of O&M, with salvage treated as a cost reduction.

Proposal X

PW of benefits $=3{,}200{,}000\times11.2578 = \text{Rs. }36{,}024{,}960$

PW of costs $=18{,}000{,}000 + 900{,}000\times11.2578 - 2{,}000{,}000\times0.09938$ $=18{,}000{,}000 + 10{,}132{,}020 - 198{,}760 = \text{Rs. }27{,}933{,}260$

$$(B/C)_X=\frac{36{,}024{,}960}{27{,}933{,}260}=1.290$$

Proposal Y

Annual benefit $=4{,}100{,}000+600{,}000=4{,}700{,}000$

PW of benefits $=4{,}700{,}000\times11.2578 = \text{Rs. }52{,}911{,}660$

PW of costs $=26{,}000{,}000 + 700{,}000\times11.2578 - 4{,}000{,}000\times0.09938$ $=26{,}000{,}000 + 7{,}880{,}460 - 397{,}520 = \text{Rs. }33{,}482{,}940$

$$(B/C)_Y=\frac{52{,}911{,}660}{33{,}482{,}940}=1.580$$

(a) Both ratios exceed 1.0, so both proposals are individually acceptable.

(b) Incremental analysis (Y − X, since Y has the larger PW of cost):

$$\Delta B = 52{,}911{,}660-36{,}024{,}960 = \text{Rs. }16{,}886{,}700$$ $$\Delta C = 33{,}482{,}940-27{,}933{,}260 = \text{Rs. }5{,}549{,}680$$ $$\Delta(B/C)=\frac{16{,}886{,}700}{5{,}549{,}680}=3.043$$

Since $\Delta(B/C)=3.04 > 1.0$, the extra investment in Y is justified. Recommend Proposal Y (Detention Basin).

(c) A high individual B/C ratio only means a project returns more than each rupee spent; it does not test whether the additional money spent on a larger alternative is itself worthwhile. Ranking by individual B/C can favour a small, cheap project that leaves available, productive capital unused. The incremental ratio asks the correct question — "does the next block of spending earn at least the MARR?" — so for mutually exclusive choices the alternative justified by incremental analysis maximises net present worth.

(a) IRR of each machine — set PW $=0$:

$$-P + A(P/A,i,8) + S(P/F,i,8)=0$$

Machine M1 ($P=4{,}000{,}000,\;A=1{,}150{,}000,\;S=500{,}000$):

At $i=25\%$: $(P/A,25\%,8)=3.3289,\;(P/F,25\%,8)=0.16777$ $PW=-4{,}000{,}000+1{,}150{,}000(3.3289)+500{,}000(0.16777)=-4{,}000{,}000+3{,}828{,}235+83{,}885=-87{,}880$

At $i=24\%$: $(P/A,24\%,8)=3.4212,\;(P/F,24\%,8)=0.17891$ $PW=-4{,}000{,}000+1{,}150{,}000(3.4212)+500{,}000(0.17891)=-4{,}000{,}000+3{,}934{,}380+89{,}455=+23{,}835$

Interpolate: $IRR_{M1}=24\%+\dfrac{23{,}835}{23{,}835+87{,}880}\times1\% = 24\%+0.21\% \approx \mathbf{24.2\%}$

Machine M2 ($P=6{,}200{,}000,\;A=1{,}620{,}000,\;S=800{,}000$):

At $i=22\%$: $(P/A,22\%,8)=3.6193,\;(P/F,22\%,8)=0.20376$ $PW=-6{,}200{,}000+1{,}620{,}000(3.6193)+800{,}000(0.20376)=-6{,}200{,}000+5{,}863{,}266+163{,}008=-173{,}726$

At $i=20\%$: $(P/A,20\%,8)=3.8372,\;(P/F,20\%,8)=0.23257$ $PW=-6{,}200{,}000+1{,}620{,}000(3.8372)+800{,}000(0.23257)=-6{,}200{,}000+6{,}216{,}264+186{,}056=+202{,}320$

Interpolate: $IRR_{M2}=20\%+\dfrac{202{,}320}{202{,}320+173{,}726}\times2\% = 20\%+1.08\% \approx \mathbf{21.1\%}$

(b) Incremental IRR on (M2 − M1):

$\Delta P = 2{,}200{,}000,\;\Delta A = 470{,}000,\;\Delta S = 300{,}000$

Set $-2{,}200{,}000+470{,}000(P/A,i,8)+300{,}000(P/F,i,8)=0$.

At $i=14\%$: $(P/A)=4.6389,\;(P/F)=0.35056$ $PW=-2{,}200{,}000+470{,}000(4.6389)+300{,}000(0.35056)=-2{,}200{,}000+2{,}180{,}283+105{,}168=+85{,}451$

At $i=16\%$: $(P/A)=4.3436,\;(P/F)=0.30503$ $PW=-2{,}200{,}000+470{,}000(4.3436)+300{,}000(0.30503)=-2{,}200{,}000+2{,}041{,}492+91{,}509=-66{,}999$

Interpolate: $\Delta IRR=14\%+\dfrac{85{,}451}{85{,}451+66{,}999}\times2\%=14\%+1.12\%\approx 15.1\%$

Since $\Delta IRR = 15.1\% > MARR = 12\%$, the extra Rs. 2,200,000 invested in M2 earns more than the MARR. Recommend Machine M2.

(c) Annual-worth check at $i=12\%$ $((A/P,12\%,8)=0.20130,\;(A/F,12\%,8)=0.08130)$:

$AW_{M1}=-4{,}000{,}000(0.20130)+1{,}150{,}000+500{,}000(0.08130)=-805{,}200+1{,}150{,}000+40{,}650=\text{Rs. }385{,}450$

$AW_{M2}=-6{,}200{,}000(0.20130)+1{,}620{,}000+800{,}000(0.08130)=-1{,}248{,}060+1{,}620{,}000+65{,}040=\text{Rs. }436{,}980$

Since $AW_{M2} > AW_{M1} > 0$, Machine M2 is confirmed as the best choice, consistent with the incremental IRR result.

All costs are positive numbers; we compare EUAC (lower is better).

(a) Challenger EUAC, $n=6$, $i=10\%$:

Factors: $(A/P,10\%,6)=0.22961,\;(A/F,10\%,6)=0.12961$

$$EUAC_C = 1{,}400{,}000(A/P)+180{,}000-300{,}000(A/F)$$ $$= 1{,}400{,}000(0.22961)+180{,}000-300{,}000(0.12961)$$ $$= 321{,}454+180{,}000-38{,}883 = \text{Rs. }462{,}571$$

(b) Defender EUAC, $n=4$, $i=10\%$:

The operating cost is a gradient: base $A=320{,}000$, gradient $G=60{,}000$.

Factors: $(A/P,10\%,4)=0.31547,\;(A/F,10\%,4)=0.21547,\;(A/G,10\%,4)=1.3812$

Equivalent annual operating cost $=320{,}000+60{,}000(A/G,10\%,4)=320{,}000+60{,}000(1.3812)=320{,}000+82{,}872=402{,}872$

$$EUAC_D = 600{,}000(A/P)+402{,}872-150{,}000(A/F)$$ $$= 600{,}000(0.31547)+402{,}872-150{,}000(0.21547)$$ $$= 189{,}282+402{,}872-32{,}321 = \text{Rs. }559{,}833$$

(c) Decision:

$$EUAC_C = \text{Rs. }462{,}571 < EUAC_D = \text{Rs. }559{,}833$$

The challenger has the lower equivalent uniform annual cost by about Rs. 97,262 per year. The rising operating-cost gradient makes the defender progressively expensive, and its low salvage cannot offset that. Replace the defender now with the challenger. (Note the trade-in value of Rs. 600,000 is correctly used as the defender's first cost — its original purchase price is a sunk cost and is irrelevant.)

Let $f=$ inflation $=0.07$, real rate $i'=0.06$.

(a) Market (combined) interest rate:

$$i = i' + f + i'\!\cdot\! f = 0.06 + 0.07 + (0.06)(0.07) = 0.1342 = \mathbf{13.42\%}$$

(b) Convert constant rupees to actual rupees, then discount at $i=13.42\%$.

Actual rupees in year $t$: $CF_t^{actual}=CF_t^{constant}\times(1.07)^t$. Discount: $\div(1.1342)^t$.

A neat shortcut: discounting actual rupees at the market rate is algebraically identical to discounting constant rupees at the real rate, because $\dfrac{(1.07)^t}{(1.1342)^t}=\dfrac{1}{(1.06)^t}$. We carry the long form here for transparency.

Sum of PW $\approx 1{,}887{,}159+1{,}780{,}720+2{,}100{,}272+1{,}981{,}386+2{,}243{,}243 = \mathbf{Rs.\;9{,}992{,}780}$

(c) Direct discount of constant rupees at the real MARR $i'=6\%$ $((P/F,6\%,t))$:

Sum $\approx \mathbf{Rs.\;9{,}987{,}839}$.

Comment: The two present worths agree (to within small rounding, $\approx$ Rs. 9.99 million). This confirms the fundamental rule: discount actual rupees at the market rate, or equivalently discount constant rupees at the real rate — never mix an actual-rupee cash flow with a real rate (or vice-versa). The tiny discrepancy is purely from rounding the tabulated factors.

Cost $B=9{,}000{,}000$, salvage $S=1{,}000{,}000$, life $N=5$, depreciable base $B-S=8{,}000{,}000$.

(a) Straight-line: $D=\dfrac{B-S}{N}=\dfrac{8{,}000{,}000}{5}=\text{Rs. }1{,}600{,}000$ each year.

(b) Sum-of-years'-digits: $\sum =1+2+3+4+5=15$. $D_t=(B-S)\times\dfrac{N-t+1}{15}$.

(Total depreciation $=8{,}000{,}000$ ✓)

(c) Double-declining-balance: rate $d=\dfrac{2}{N}=\dfrac{2}{5}=0.40$. Charge $=0.40\times$ beginning book value, but stop so book value does not drop below salvage Rs. 1,000,000.

In year 5 the unadjusted DDB charge (466,560) would push book value to 699,840, below salvage. So depreciation is limited to $1{,}166{,}400-1{,}000{,}000=166{,}400$ to land exactly on salvage. Total $=3{,}600{,}000+2{,}160{,}000+1{,}296{,}000+777{,}600+166{,}400=8{,}000{,}000$ ✓

(d) Year-2 depreciation: SL = Rs. 1,600,000; SOYD = Rs. 2,133,333; DDB = Rs. 2,160,000. DDB gives the largest year-2 charge (Rs. 2,160,000). A firm may prefer an accelerated method (DDB or SOYD) because larger early depreciation defers taxable income to later years, improving early-year after-tax cash flow and present worth of tax savings.

Nominal rate $r=9\%$/yr, $m=12$ compounding periods/yr, so monthly rate $=0.09/12=0.0075$.

(a) Effective annual rate:

$$i_{eff}=\left(1+\frac{0.09}{12}\right)^{12}-1=(1.0075)^{12}-1=1.093807-1=0.093807$$ $$\boxed{i_{eff}=9.381\%}$$

(b) Future value after 6 years ($n=6\times12=72$ months):

$$F=P(1+0.0075)^{72}=500{,}000\,(1.0075)^{72}$$ $$(1.0075)^{72}=1.712553 \Rightarrow F=500{,}000\times1.712553=\mathbf{Rs.\;856{,}277}$$

(Equivalently $F=500{,}000(1.093807)^6=856{,}277$, confirming the effective-rate result.)

(a) Equal annual installment (capital recovery):

$$A=P\,(A/P,i,n)=P\cdot\frac{i(1+i)^n}{(1+i)^n-1}$$ $$(1.11)^{10}=2.839421,\quad (A/P,11\%,10)=\frac{0.11\times2.839421}{2.839421-1}=\frac{0.312336}{1.839421}=0.169801$$ $$A=3{,}000{,}000\times0.169801=\mathbf{Rs.\;509{,}403\text{ per year}}$$

(b) First payment split:

• Interest in year 1 $=0.11\times3{,}000{,}000=\text{Rs. }330{,}000$
• Principal repaid in year 1 $=509{,}403-330{,}000=\mathbf{Rs.\;179{,}403}$

So the first installment of Rs. 509,403 consists of Rs. 330,000 interest and Rs. 179,403 principal, leaving a balance of Rs. 2,820,597.

This is a base annuity $A=400{,}000$ plus an arithmetic gradient $G=50{,}000$, $n=8$, $i=10\%$.

Factors at $10\%, n=8$: $(P/A)=5.3349,\;(P/G)=16.0287,\;(A/G)=3.0045$.

(a) Present worth:

$$P = A(P/A,10\%,8)+G(P/G,10\%,8)$$ $$= 400{,}000(5.3349)+50{,}000(16.0287)$$ $$= 2{,}133{,}960 + 801{,}435 = \mathbf{Rs.\;2{,}935{,}395}$$

(b) Equivalent uniform annual cost:

$$A_{eq} = A + G(A/G,10\%,8) = 400{,}000 + 50{,}000(3.0045) = 400{,}000+150{,}225 = \mathbf{Rs.\;550{,}225/\text{yr}}$$

Check: $A_{eq}(P/A)=550{,}225\times5.3349=2{,}935{,}396$ ✓ (matches part a).

Equal lives (10 yr), so compare PW of total cost directly. Factors at $9\%, n=10$: $(P/A)=6.4177,\;(P/F)=0.42241$.

Pump A:

$$PW_A = 700{,}000 + 240{,}000(6.4177) - 60{,}000(0.42241)$$ $$= 700{,}000 + 1{,}540{,}248 - 25{,}345 = \text{Rs. }2{,}214{,}903$$

Pump B:

$$PW_B = 1{,}050{,}000 + 150{,}000(6.4177) - 120{,}000(0.42241)$$ $$= 1{,}050{,}000 + 962{,}655 - 50{,}689 = \text{Rs. }1{,}961{,}966$$

Since $PW_B = \text{Rs. }1{,}961{,}966 < PW_A = \text{Rs. }2{,}214{,}903$, Pump B is more economical (lower present worth of cost by about Rs. 252,937). Its higher first cost is outweighed by lower operating cost and higher salvage.

(a) Simple payback:

$$n_{pb}=\frac{\text{First cost}}{\text{Annual cash flow}}=\frac{5{,}000{,}000}{1{,}100{,}000}=4.55\text{ years} \approx \mathbf{4.5\text{ years}}$$

(b) Discounted payback at $i=10\%$ — find $n$ where cumulative discounted cash flow reaches Rs. 5,000,000.

The cumulative discounted cash flow crosses Rs. 5,000,000 during year 7. Fraction into year 7 $=\dfrac{5{,}000{,}000-4{,}790{,}786}{564{,}474}=\dfrac{209{,}214}{564{,}474}=0.371$.

$$\text{Discounted payback}=6+0.371\approx \mathbf{6.37\text{ years}}$$

(c) Limitation of simple payback: it ignores the time value of money (treats early and late rupees as equal) and ignores all cash flows occurring after the payback point, so it does not measure overall profitability. Here the simple payback (4.5 yr) looks attractive but the discounted payback (6.37 yr) reveals the project recovers cost much later once interest is charged.

Declining-balance: book value $BV_t = B(1-d)^t$, with $B=1{,}200{,}000$, $d=0.20$.

(a) Book value at end of year 3:

$$BV_3 = 1{,}200{,}000(1-0.20)^3 = 1{,}200{,}000(0.8)^3 = 1{,}200{,}000\times0.512 = \mathbf{Rs.\;614{,}400}$$

(b) Depreciation in year 3 $=BV_2-BV_3$:

$$BV_2 = 1{,}200{,}000(0.8)^2 = 1{,}200{,}000\times0.64 = 768{,}000$$ $$D_3 = 768{,}000 - 614{,}400 = \mathbf{Rs.\;153{,}600}$$

(Equivalently $D_3 = d\times BV_2 = 0.20\times768{,}000 = 153{,}600$ ✓.)