BE Civil Engineering (IOE, TU) Engineering Economics (IOE, CE 656) Question Paper 2076 Nepal

Given: $i = 10\%$, $n = 12$ years. Both designs have equal lives, so PW of cost (costs as positive) can be compared directly.

Factors at 10%:

• $(P/F,10\%,6) = 1/1.1^{6} = 0.56447$
• $(P/F,10\%,12) = 1/1.1^{12} = 0.31863$
• $(P/A,10\%,12) = \dfrac{1.1^{12}-1}{0.10\times1.1^{12}} = 6.81369$
• $(A/P,10\%,12) = 1/6.81369 = 0.14676$

(a) Present Worth of Cost

Design X:

$$PW_X = 1{,}800{,}000 + 320{,}000(P/A) + 250{,}000(P/F_6) - 200{,}000(P/F_{12})$$ $$= 1{,}800{,}000 + 320{,}000(6.81369) + 250{,}000(0.56447) - 200{,}000(0.31863)$$ $$= 1{,}800{,}000 + 2{,}180{,}381 + 141{,}118 - 63{,}726 = \mathbf{NRs\ 4{,}057{,}773}$$

Design Y:

$$PW_Y = 2{,}600{,}000 + 210{,}000(6.81369) + 150{,}000(0.56447) - 400{,}000(0.31863)$$ $$= 2{,}600{,}000 + 1{,}430{,}875 + 84{,}671 - 127{,}452 = \mathbf{NRs\ 3{,}988{,}094}$$

Since $PW_Y (3{,}988{,}094) < PW_X (4{,}057{,}773)$, Design Y has the lower present worth of cost and should be selected.

(b) EUAC

$$EUAC_X = 4{,}057{,}773 \times 0.14676 = \mathbf{NRs\ 595{,}519/yr}$$ $$EUAC_Y = 3{,}988{,}094 \times 0.14676 = \mathbf{NRs\ 585{,}293/yr}$$

Design Y is cheaper by about NRs 10,200/yr, confirming Design Y as the economic choice.

Given: MARR $= 12\%$, $n = 5$ years.

(a) IRR of Alternative P

Set PW = 0:

$$-900{,}000 + 280{,}000(P/A,i,5) + 100{,}000(P/F,i,5) = 0$$

Try $i = 18\%$: $(P/A,18\%,5)=3.12717$, $(P/F,18\%,5)=0.43711$

$$-900{,}000 + 280{,}000(3.12717) + 100{,}000(0.43711) = -900{,}000 + 875{,}608 + 43{,}711 = +19{,}319$$

Try $i = 20\%$: $(P/A,20\%,5)=2.99061$, $(P/F,20\%,5)=0.40188$

$$-900{,}000 + 280{,}000(2.99061) + 100{,}000(0.40188) = -900{,}000 + 837{,}371 + 40{,}188 = -22{,}441$$

Interpolate:

$$IRR_P = 18 + 2\times\frac{19{,}319}{19{,}319+22{,}441} = 18 + 2(0.4626) = \mathbf{18.93\%}$$

Since $18.93\% > 12\%$ MARR, P is acceptable on its own.

(b) Incremental analysis (Q − P)

Increment: $\Delta$cost $= 500{,}000$, $\Delta$benefit $= 120{,}000/yr$, $\Delta$salvage $= 100{,}000$.

$$-500{,}000 + 120{,}000(P/A,i,5) + 100{,}000(P/F,i,5) = 0$$

Try $i = 10\%$: $(P/A)=3.79079$, $(P/F)=0.62092$

$$-500{,}000 + 120{,}000(3.79079) + 100{,}000(0.62092) = -500{,}000 + 454{,}895 + 62{,}092 = +16{,}987$$

Try $i = 12\%$: $(P/A)=3.60478$, $(P/F)=0.56743$

$$-500{,}000 + 120{,}000(3.60478) + 100{,}000(0.56743) = -500{,}000 + 432{,}574 + 56{,}743 = -10{,}683$$

Interpolate:

$$\Delta IRR = 10 + 2\times\frac{16{,}987}{16{,}987+10{,}683} = 10 + 2(0.6139) = \mathbf{11.23\%}$$

Since $\Delta IRR (11.23\%) < MARR (12\%)$, the extra investment in Q is not justified.

Recommendation: choose Alternative P.

Given: $i = 8\%$, $n = 25$ years. $(P/A,8\%,25) = \dfrac{1.08^{25}-1}{0.08\times1.08^{25}} = 10.67478$ $(A/P,8\%,25) = 1/10.67478 = 0.09368$

Annualize all amounts (use annual basis):

• Equivalent annual first cost (CR): $60{,}000{,}000 \times 0.09368 = NRs\ 5{,}620{,}800/yr$
• Annual benefits $B = 9{,}500{,}000 + 1{,}300{,}000 = NRs\ 10{,}800{,}000/yr$
• Annual disbenefits $D = NRs\ 800{,}000/yr$
• Annual O&M $= NRs\ 2{,}200{,}000/yr$

Conventional B/C

$$B/C = \frac{B - D}{CR + \text{O\&M}} = \frac{10{,}800{,}000 - 800{,}000}{5{,}620{,}800 + 2{,}200{,}000} = \frac{10{,}000{,}000}{7{,}820{,}800} = \mathbf{1.279}$$

Modified B/C

(O&M subtracted from numerator):

$$B/C_{mod} = \frac{B - D - \text{O\&M}}{CR} = \frac{10{,}800{,}000 - 800{,}000 - 2{,}200{,}000}{5{,}620{,}800} = \frac{7{,}800{,}000}{5{,}620{,}800} = \mathbf{1.388}$$

Both ratios exceed 1.0, so the project is economically justified. (Both methods always give a consistent accept/reject decision relative to 1.0.)

Given: $i = 10\%$. Compare on equivalent annual cost basis (the standard defender–challenger approach using each asset's own economic-life AW).

Factors:

• $(A/P,10\%,4) = \dfrac{0.10\times1.1^{4}}{1.1^{4}-1} = 0.31547$
• $(A/F,10\%,4) = (A/P) - i = 0.31547 - 0.10 = 0.21547$
• $(A/P,10\%,8) = \dfrac{0.10\times1.1^{8}}{1.1^{8}-1} = 0.18744$
• $(A/F,10\%,8) = 0.18744 - 0.10 = 0.08744$

Defender EUAC (over 4 remaining years)

The current market value NRs 1,200,000 is the opportunity cost (initial investment) of keeping it.

$$EUAC_D = 1{,}200{,}000(A/P,10\%,4) + 480{,}000 - 300{,}000(A/F,10\%,4)$$ $$= 1{,}200{,}000(0.31547) + 480{,}000 - 300{,}000(0.21547)$$ $$= 378{,}564 + 480{,}000 - 64{,}641 = \mathbf{NRs\ 793{,}923/yr}$$

Challenger EUAC (over 8-year economic life)

$$EUAC_C = 3{,}000{,}000(A/P,10\%,8) + 250{,}000 - 600{,}000(A/F,10\%,8)$$ $$= 3{,}000{,}000(0.18744) + 250{,}000 - 600{,}000(0.08744)$$ $$= 562{,}320 + 250{,}000 - 52{,}464 = \mathbf{NRs\ 759{,}856/yr}$$

Decision

Since $EUAC_C (759{,}856) < EUAC_D (793{,}923)$, the challenger is cheaper by about NRs 34,000/yr.

Recommendation: replace the defender now with the challenger.

Given: real rate $i = 9\%$, inflation $f = 7\%$, actual-rupee cash flows.

(a) Combined (market) rate

$$i_f = i + f + i\,f = 0.09 + 0.07 + (0.09)(0.07) = 0.16 + 0.0063 = \mathbf{0.1663\ (16.63\%)}$$

(b) PW of actual rupees discounted at the market rate $i_f = 16.63\%$

Discount factors $1/(1.1663)^t$:

• $t=1: 0.85741$
• $t=2: 0.73516$
• $t=3: 0.63031$
• $t=4: 0.54043$

$$PW = 428{,}705 + 441{,}096 + 441{,}217 + 432{,}344 = \mathbf{NRs\ 1{,}743{,}362}$$

(c) Verification via constant rupees and real rate

Convert actual to constant (year-0) rupees by dividing by $(1.07)^t$:

• $t=1: 500{,}000/1.07 = 467{,}290$
• $t=2: 600{,}000/1.1449 = 524{,}064$
• $t=3: 700{,}000/1.225043 = 571{,}408$
• $t=4: 800{,}000/1.310796 = 610{,}317$

Discount these at the real rate $9\%$ $(1/1.09^t)$:

$$PW = \mathbf{NRs\ 1{,}743{,}369}$$

The two PWs agree (small rounding), confirming that discounting actual rupees at the market rate equals discounting constant rupees at the real rate.

Given: $P = 250{,}000$, nominal $r = 12\%/yr$, $m = 12$ compoundings/yr, $n = 5$ yr.

(a) Effective annual rate

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

(b) Future value after 5 years

Total monthly periods $= 5\times12 = 60$, monthly rate $= 1\%$.

$$F = 250{,}000(1.01)^{60} = 250{,}000(1.816697) = \mathbf{NRs\ 454{,}174}$$

(Check via effective annual rate: $250{,}000(1.126825)^5 = 250{,}000(1.816697) = NRs\ 454{,}174$ — consistent.)

Given: $F = 5{,}000{,}000$, $n = 10$, $i = 9\%$.

Factors:

• $(A/F,9\%,10) = \dfrac{0.09}{1.09^{10}-1} = \dfrac{0.09}{1.367364} = 0.065820$
• $(F/A,9\%,10) = 1.367364/0.09 = 15.19293$
• $(F/P,9\%,10) = 1.09^{10} = 2.367364$

(a) Equal annual deposit

$$A = 5{,}000{,}000(A/F,9\%,10) = 5{,}000{,}000(0.065820) = \mathbf{NRs\ 329{,}100/yr}$$

(b) Lump sum today plus NRs 300,000/yr

Future worth of the 10 annual deposits:

$$F_{annual} = 300{,}000(F/A,9\%,10) = 300{,}000(15.19293) = 4{,}557{,}879$$

Remaining needed at year 10: $5{,}000{,}000 - 4{,}557{,}879 = 442{,}121$ Lump sum today:

$$P = 442{,}121/(F/P,9\%,10) = 442{,}121/2.367364 = \mathbf{NRs\ 186{,}752}$$

Given: $B = 4{,}000{,}000$, $S = 400{,}000$, $n = 8$ years.

(a) Straight-line

$$D_{SL} = \frac{B - S}{n} = \frac{4{,}000{,}000 - 400{,}000}{8} = \mathbf{NRs\ 450{,}000/yr}$$

Book value end of year 3:

$$BV_3 = 4{,}000{,}000 - 3(450{,}000) = 4{,}000{,}000 - 1{,}350{,}000 = \mathbf{NRs\ 2{,}650{,}000}$$

(b) Double-declining-balance

DDB rate $= 2/n = 2/8 = 0.25$ (25% of book value each year).

Year 1: $D_1 = 0.25\times4{,}000{,}000 = 1{,}000{,}000$; $BV_1 = 3{,}000{,}000$

Year 2: $D_2 = 0.25\times3{,}000{,}000 = 750{,}000$; $BV_2 = 2{,}250{,}000$

So $D_1 = \mathbf{NRs\ 1{,}000{,}000}$, $D_2 = \mathbf{NRs\ 750{,}000}$, $BV_2 = \mathbf{NRs\ 2{,}250{,}000}$. (Both year amounts are above the salvage floor, so no adjustment is needed.)

Given: base $A_1 = 120{,}000$, arithmetic gradient $G = 15{,}000$, $n = 8$, $i = 10\%$.

Factors at 10%, n = 8:

• $(P/A,10\%,8) = 5.33493$
• $(P/G,10\%,8) = \dfrac{(1.1^{8}-1)/(0.1\cdot1.1^{8}) - 8/1.1^{8}}{0.1} = 16.02867$
• $(A/P,10\%,8) = 0.18744$

Present Worth

$$PW = A_1(P/A) + G(P/G) = 120{,}000(5.33493) + 15{,}000(16.02867)$$ $$= 640{,}192 + 240{,}430 = \mathbf{NRs\ 880{,}622}$$

Equivalent Uniform Annual Cost

$$EUAC = PW\times(A/P,10\%,8) = 880{,}622\times0.18744 = \mathbf{NRs\ 165{,}064/yr}$$

(Check via gradient-to-uniform: $(A/G,10\%,8)=3.00448$; $EUAC = 120{,}000 + 15{,}000(3.00448) = 120{,}000 + 45{,}067 = NRs\ 165{,}067/yr$ — consistent.)

Given: $P = 2{,}500{,}000$, $A = 650{,}000/yr$, $n = 6$, salvage $= 0$.

(a) Simple payback period

$$n_p = \frac{P}{A} = \frac{2{,}500{,}000}{650{,}000} = \mathbf{3.85\ years}$$

(about 3 years and 10 months).

(b) IRR

Set $-2{,}500{,}000 + 650{,}000(P/A,i,6) = 0 \Rightarrow (P/A,i,6) = 3.84615$.

Try $i = 14\%$: $(P/A,14\%,6)=3.88867 \Rightarrow$ PW $= 650{,}000(3.88867)-2{,}500{,}000 = +27{,}636$ Try $i = 16\%$: $(P/A,16\%,6)=3.68474 \Rightarrow$ PW $= 650{,}000(3.68474)-2{,}500{,}000 = -104{,}919$

Interpolate:

$$IRR = 14 + 2\times\frac{27{,}636}{27{,}636+104{,}919} = 14 + 2(0.2085) = \mathbf{14.42\%}$$

Since $IRR (14.42\%) > MARR (14\%)$, the project is (marginally) acceptable.

(a) Time value of money (TVM)

Money available now is worth more than the same amount in the future because present money can be invested to earn interest (and because of inflation and risk). Hence cash flows occurring at different times cannot be added directly; they must be moved to a common point in time using interest factors. A cash-flow diagram (horizontal time line with downward arrows for disbursements and upward arrows for receipts) gives a clear visual of the timing, sign and magnitude of every cash flow, which prevents errors when applying equivalence formulas.

(b) Capital recovery factor

The factor $(A/P,i,n)$ converts a present sum $P$ into an equivalent series of $n$ equal end-of-period payments $A$:

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

(c) Annual loan installment

$P = 800{,}000$, $i = 11\%$, $n = 5$.

$$(A/P,11\%,5) = \frac{0.11(1.11)^{5}}{(1.11)^{5}-1} = \frac{0.11(1.685058)}{0.685058} = \frac{0.185356}{0.685058} = 0.270570$$ $$A = 800{,}000\times0.270570 = \mathbf{NRs\ 216{,}456/yr}$$