BE Civil Engineering (IOE, TU) Engineering Economics (IOE, CE 656) Question Paper 2079 Nepal

Given: $i_{MARR}=10\%$, $n=10$ yr.

Factor values at 10%, 10 yr:

• $(P/A,10\%,10)=6.1446$
• $(P/F,10\%,10)=0.3855$

(a) Present Worth of Costs

Treat costs as positive outflows; salvage reduces PW.

Pump A:

$$PW_A = 600{,}000 + 120{,}000(6.1446) - 60{,}000(0.3855)$$ $$= 600{,}000 + 737{,}352 - 23{,}130 = \mathbf{Rs\ 1{,}314{,}222}$$

Pump B:

$$PW_B = 900{,}000 + 70{,}000(6.1446) - 120{,}000(0.3855)$$ $$= 900{,}000 + 430{,}122 - 46{,}260 = \mathbf{Rs\ 1{,}283{,}862}$$

Since $PW_B < PW_A$, Pump B has the lower present worth of cost and should be selected.

(b) Incremental Rate of Return (B − A)

Incremental cash flows (B minus A):

• Extra first cost: $900{,}000-600{,}000 = 300{,}000$ (outflow at year 0)
• Annual savings: $120{,}000-70{,}000 = 50{,}000$ per year (inflow, years 1–10)
• Extra salvage: $120{,}000-60{,}000 = 60{,}000$ (inflow at year 10)

Set incremental PW to zero:

$$0 = -300{,}000 + 50{,}000(P/A,i^*,10) + 60{,}000(P/F,i^*,10)$$

Try i = 12%: $(P/A,12\%,10)=5.6502$, $(P/F,12\%,10)=0.3220$

$$PW = -300{,}000 + 50{,}000(5.6502) + 60{,}000(0.3220) = -300{,}000 + 282{,}510 + 19{,}320 = +1{,}830$$

Try i = 13%: $(P/A,13\%,10)=5.4262$, $(P/F,13\%,10)=0.2946$

$$PW = -300{,}000 + 50{,}000(5.4262) + 60{,}000(0.2946) = -300{,}000 + 271{,}310 + 17{,}676 = -11{,}014$$

Interpolating:

$$i^* = 12\% + \frac{1{,}830}{1{,}830+11{,}014}(1\%) \approx 12.14\%$$

Since $i^* = 12.14\% > MARR = 10\%$, the extra investment in B is justified — select Pump B. This confirms part (a).

(c) Agreement condition

PW analysis and (correctly performed) incremental ROR analysis always give the same decision for mutually exclusive alternatives, provided that: (i) the same study period / common service life is used, (ii) the analysis is done on the increment (defender vs. challenger), not on individual project IRRs, and (iii) reinvestment is assumed at the MARR. Ranking by individual IRRs alone can mislead because a higher IRR on a smaller investment may not maximise wealth.

Given: $i=8\%$, $n=20$ yr, $(P/A,8\%,20)=9.8181$.

Convert annual flows to present worth using $(P/A)$. Net annual benefit = benefits − O&M (treating O&M in numerator) — but part (a) uses the conventional ratio with O&M in the denominator as a cost.

(a) Conventional B/C (O&M as annual cost in denominator)

$$B/C = \frac{\text{PW of benefits}}{\text{First cost} + \text{PW of O\&M}}$$

X: $PW_B = 130(9.8181)=1276.35$; $PW_{O\&M}=25(9.8181)=245.45$

$$B/C_X = \frac{1276.35}{800+245.45} = \frac{1276.35}{1045.45}=\mathbf{1.221}$$

Y: $PW_B = 175(9.8181)=1718.17$; $PW_{O\&M}=30(9.8181)=294.54$

$$B/C_Y = \frac{1718.17}{1100+294.54}=\frac{1718.17}{1394.54}=\mathbf{1.232}$$

Z: $PW_B = 225(9.8181)=2209.07$; $PW_{O\&M}=40(9.8181)=392.72$

$$B/C_Z = \frac{2209.07}{1500+392.72}=\frac{2209.07}{1892.72}=\mathbf{1.167}$$

All three exceed 1.0, so each is individually acceptable.

(b) Incremental B/C

Order by increasing total cost (denominator): X (1045.45) < Y (1394.54) < Z (1892.72). Baseline = X (acceptable, B/C > 1).

Y − X:

$$\Delta B = 1718.17-1276.35 = 441.82;\quad \Delta C = 1394.54-1045.45 = 349.09$$ $$\Delta B/C = \frac{441.82}{349.09}=1.266 > 1 \Rightarrow \text{Y dominates X.}$$

Z − Y:

$$\Delta B = 2209.07-1718.17 = 490.90;\quad \Delta C = 1892.72-1394.54 = 498.18$$ $$\Delta B/C = \frac{490.90}{498.18}=0.985 < 1 \Rightarrow \text{extra spending on Z not justified; keep Y.}$$

Recommendation: build Alignment Y.

(c) O&M placement

If O&M is netted in the numerator as a disbenefit, the ratio becomes $\frac{PW_B - PW_{O\&M}}{\text{First cost}}$. This lowers the numerator and the denominator differently, changing the absolute B/C number (it shifts away from 1). However, for incremental analysis the cross-over test ($\Delta B/\Delta C \gtrless 1$) is mathematically equivalent to checking whether $\Delta(B-\text{O\&M}) > \Delta(\text{first cost})$, i.e. whether incremental net PW > 0. Hence the selection (Y) is unchanged; only reported magnitudes differ.

Given: MARR $=12\%$. Factors at 12%: $(A/P,12\%,6)=0.24323$, $(A/F,12\%,6)=0.12323$.

(a) Defender — cost of one more year of service

The defender's "investment" if retained is its current market value Rs 1,500,000 (the opportunity cost — money forgone by not selling now).

Annual cost of keeping one more year:

$$AW_D = \underbrace{1{,}500{,}000(A/P,12\%,1)}_{\text{capital recovery on retained value}} - \underbrace{1{,}100{,}000(A/F,12\%,1)}_{\text{salvage credit}} + 900{,}000$$

For $n=1$: $(A/P,12\%,1)=1.12$ and $(A/F,12\%,1)=1.00$.

$$AW_D = 1{,}500{,}000(1.12) - 1{,}100{,}000(1.00) + 900{,}000$$ $$= 1{,}680{,}000 - 1{,}100{,}000 + 900{,}000 = \mathbf{Rs\ 1{,}480{,}000\ per\ year}$$

(b) Challenger — AW over 6-year economic life

Capital recovery:

$$CR = (5{,}000{,}000 - 800{,}000)(A/P,12\%,6) + 800{,}000(0.12)$$ $$= 4{,}200{,}000(0.24323) + 96{,}000 = 1{,}021{,}566 + 96{,}000 = 1{,}117{,}566$$

(Equivalently $5{,}000{,}000(0.24323) - 800{,}000(0.12323) = 1{,}216{,}150 - 98{,}584 = 1{,}117{,}566$.)

Add operating cost:

$$AW_C = 1{,}117{,}566 + 350{,}000 = \mathbf{Rs\ 1{,}467{,}566\ per\ year}$$

(c) Decision

$$AW_C = 1{,}467{,}566 < AW_D = 1{,}480{,}000.$$

The challenger's equivalent annual cost is lower (by about Rs 12,400/yr), so replace the defender now with the challenger.

Meaning of the defender's first cost: its original purchase price is a sunk cost and is irrelevant to the replacement decision. What matters is the defender's current market value, which represents the opportunity cost (capital tied up) of continuing to own it.

Given: real MARR $i_r = 9\%$, inflation $f = 6\%$, $n=8$ yr. Year-0-rupee cash flow $A_0 = 400{,}000$ each year (constant purchasing power).

(a) Market (nominal) rate

$$i_m = i_r + f + i_r\,f = 0.09 + 0.06 + (0.09)(0.06) = 0.1554$$ $$\boxed{i_m = 15.54\%\ \text{per year}}$$

(b)(i) Real-rate method (constant Rs)

The cash flow expressed in constant year-0 rupees is a level annuity of Rs 400,000. Discount at the real rate 9%:

$$(P/A,9\%,8)=5.5348$$ $$PW = 400{,}000(5.5348) = \mathbf{Rs\ 2{,}213{,}920}$$

(b)(ii) Nominal-rate method (actual Rs)

Actual (then-current) cash flow in year $t$ grows with inflation:

$$A_t = 400{,}000(1.06)^t$$

Discount each at the nominal rate 15.54%. Because the actual flow is a geometric series growing at $g=f=6\%$ discounted at $i_m=15.54\%$:

$$PW = A_1 \frac{1-\left(\frac{1+g}{1+i_m}\right)^{n}}{i_m - g},\quad A_1 = 400{,}000(1.06)=424{,}000$$ $$\frac{1+g}{1+i_m} = \frac{1.06}{1.1554}=0.917432$$ $$0.917432^{8}=0.50637$$ $$PW = 424{,}000 \times \frac{1-0.50637}{0.1554-0.06} = 424{,}000 \times \frac{0.49363}{0.0954}$$ $$= 424{,}000 \times 5.17431 = \mathbf{Rs\ 2{,}193{,}907}$$

The two results agree to within rounding (≈ Rs 2.21 million vs Rs 2.19 million); the small gap is purely from rounding the geometric-series factor and the $(P/A)$ table value. Using more decimals both converge to ≈ Rs 2.20 million.

(c) Why they agree

Deflating an actual-rupee cash flow by $(1+f)^t$ and discounting at the real rate $i_r$ is algebraically identical to discounting the actual cash flow at the nominal rate $i_m=(1+i_r)(1+f)-1$, because $(1+i_m)^t=(1+i_r)^t(1+f)^t$. The inflation factor cancels, so PW is invariant to the rupee basis chosen — provided the rate matches the rupee basis (real with constant Rs, nominal with actual Rs).

Given: Cost $B = 2{,}400{,}000$, salvage $S = 240{,}000$, life $N=5$ yr.

(a) Straight-Line

$$D_{SL} = \frac{B-S}{N} = \frac{2{,}400{,}000 - 240{,}000}{5} = \frac{2{,}160{,}000}{5} = 432{,}000\ \text{per year}$$

(b) Double-Declining-Balance

DDB rate $= \dfrac{2}{N} = \dfrac{2}{5} = 0.40$. Each year $D_t = 0.40\times BV_{t-1}$, but never below salvage Rs 240,000.

Year 5 check: unrestricted DDB would give BV = 311,040 − 124,416 = 186,624, which is below salvage (240,000). Therefore year-5 depreciation is capped at $311{,}040 - 240{,}000 = \mathbf{71{,}040}$, leaving BV = Rs 240,000.

(c) End-of-year-3 book value and comparison

• SL: $BV_3 = \mathbf{Rs\ 1{,}104{,}000}$
• DDB: $BV_3 = \mathbf{Rs\ 518{,}400}$

DDB writes the asset down much faster in the early years (larger early deductions, smaller later ones), giving a far lower year-3 book value. This accelerated pattern is advantageous for early tax shielding, whereas SL spreads the deduction evenly across the life.

Given: nominal $r=12\%$, $m=12$ compoundings/yr, $n=6$ yr.

(a) Effective annual rate

$$i_a = \left(1+\frac{r}{m}\right)^{m} - 1 = (1+0.01)^{12} - 1 = 1.126825 - 1$$ $$\boxed{i_a = 12.6825\% \approx 12.68\%}$$

(b) Future worth at year 6

Use the effective annual rate (12.6825%) with annual cash flows.

Single deposit (compounded 6 yr):

$$F_1 = 50{,}000(1+0.126825)^{6} = 50{,}000(2.04097) = 102{,}049$$

Annual series (6 end-of-year payments):

$$(F/A,12.6825\%,6) = \frac{(1.126825)^{6} - 1}{0.126825} = \frac{2.04097 - 1}{0.126825} = \frac{1.04097}{0.126825} = 8.2079$$ $$F_2 = 30{,}000(8.2079) = 246{,}237$$

Total:

$$F = F_1 + F_2 = 102{,}049 + 246{,}237 = \mathbf{Rs\ 348{,}286}$$

Given: base $A = 80{,}000$ (year 1), gradient $G = 15{,}000$, $n=8$, $i=10\%$.

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

(a) Present worth

$$PW = A(P/A,10\%,8) + G(P/G,10\%,8)$$ $$= 80{,}000(5.3349) + 15{,}000(16.0287)$$ $$= 426{,}792 + 240{,}431 = \mathbf{Rs\ 667{,}223}$$

(b) Equivalent uniform annual cost

$$EUAC = A + G(A/G,10\%,8) = 80{,}000 + 15{,}000(3.0045)$$ $$= 80{,}000 + 45{,}068 = \mathbf{Rs\ 125{,}068\ per\ year}$$

Check: $EUAC = PW(A/P,10\%,8) = 667{,}223(0.18744) = 125{,}064$ — agrees within rounding.

Given: $P=1{,}200{,}000$, $A=320{,}000$/yr, $S=150{,}000$, $n=6$.

Set PW = 0:

$$0 = -1{,}200{,}000 + 320{,}000(P/A,i,6) + 150{,}000(P/F,i,6)$$

Try i = 16%: $(P/A,16\%,6)=3.6847$, $(P/F,16\%,6)=0.4104$

$$PW = -1{,}200{,}000 + 320{,}000(3.6847) + 150{,}000(0.4104)$$ $$= -1{,}200{,}000 + 1{,}179{,}104 + 61{,}560 = +40{,}664$$

Try i = 18%: $(P/A,18\%,6)=3.4976$, $(P/F,18\%,6)=0.3704$

$$PW = -1{,}200{,}000 + 320{,}000(3.4976) + 150{,}000(0.3704)$$ $$= -1{,}200{,}000 + 1{,}119{,}232 + 55{,}560 = -25{,}208$$

Interpolate:

$$i^* = 16\% + \frac{40{,}664}{40{,}664+25{,}208}(2\%) = 16\% + \frac{40{,}664}{65{,}872}(2\%) \approx 17.23\%$$ $$\boxed{IRR \approx 17.2\%}$$

Since $IRR = 17.2\% > MARR = 14\%$, the project is acceptable.

Given: $i=7\%$.

(a) Capitalized cost (perpetual maintenance)

For a perpetual uniform cost, $P = A/i$.

$$CC = \text{First cost} + \frac{A}{i} = 8{,}000{,}000 + \frac{250{,}000}{0.07}$$ $$= 8{,}000{,}000 + 3{,}571{,}429 = \mathbf{Rs\ 11{,}571{,}429}$$

(b) Capital recovery of the culvert

$$CR = P(A/P,7\%,25),\quad (A/P,7\%,25)=0.08581$$ $$CR = 600{,}000(0.08581) = \mathbf{Rs\ 51{,}486\ per\ year}$$

(No salvage, so there is no $(A/F)$ credit term.)

Given: $B=900{,}000$, $S=100{,}000$, $N=4$. Depreciable base $= B-S = 800{,}000$.

Sum of years digits: $\sum = 1+2+3+4 = 10$.

(a) Annual depreciation

$$D_t = (B-S)\times\frac{N - t + 1}{\sum}$$

Check: $320{,}000+240{,}000+160{,}000+80{,}000 = 800{,}000$ ✓ (equals depreciable base).

(b) Book value end of year 2

$$BV_2 = B - (D_1 + D_2) = 900{,}000 - (320{,}000 + 240{,}000)$$ $$= 900{,}000 - 560{,}000 = \mathbf{Rs\ 340{,}000}$$

(a) Required deposit today (P/F)

$$P = F(P/F,9\%,12) = F\,(1+i)^{-n} = 1{,}000{,}000(1.09)^{-12}$$ $$(1.09)^{12} = 2.81266 \Rightarrow (P/F,9\%,12)=0.35553$$ $$P = 1{,}000{,}000(0.35553) = \mathbf{Rs\ 355{,}530}$$

(b) Time to triple at 8%

$$3 = (1.08)^{n} \Rightarrow n = \frac{\ln 3}{\ln 1.08} = \frac{1.098612}{0.076961} = 14.27$$ $$\boxed{n \approx 14.3\ \text{years}}$$

(Rule-of-72 check: $72/8 = 9$ yr to double; tripling sensibly takes longer, ≈ 14 yr.)

(c) PW vs AW

• Present Worth (PW): converts all cash flows to a single equivalent value at time 0. Best for comparing total lifetime wealth impact and for one-time / non-repeating decisions.
• Annual Worth (AW): converts all cash flows to an equivalent uniform end-of-period amount.

When AW is more convenient: when comparing mutually exclusive alternatives with unequal lives, AW automatically reflects a per-year (repeated-service) basis, so there is no need to find a least-common-multiple study period as PW would require. AW is also natural when results are wanted on a per-unit-time basis (e.g., cost per year, per tonne).