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Capital Recovery

Capital recovery is the annual equivalent cost of owning an asset after considering salvage value.

$$CR=\frac{(FC-SV)(1+i)^n i}{(1+i)^n-1}+SV(i)$$

Problem: CE Board Nov. 2016

A machine costs Php50,000, has 10-year life, and salvage equal to 10% of original value. Find capital recovery at 8%.

Salvage value is P5,000.

$$CR=\frac{(50000-5000)(1.08)^{10}(0.08)}{(1.08)^{10}-1}+5000(0.08)=7,105$$

Final answer: P7,105 per year.

Problem: CE Board Nov. 2016

A machine costs $100,000, has life 15 years, salvage 20% of original cost, and interest 10%. Find capital recovery.

$$CR=12,517.90$$

Final answer: $12,517.90 per year.

Problem: Capital Recovery

A loader costs P1,200,000, has salvage P250,000 after 6 years, and i = 10%. Find capital recovery.

$$CR=1200000(A/P,10\%,6)-250000(A/F,10\%,6)$$
$$CR=1200000(0.22961)-250000(0.12961)=243130$$

Answer: Capital recovery is about P243,130 per year.

Problem: Capital Recovery with No Salvage

Equipment costs P500,000 and is fully depreciated over 5 years. Use i = 8% to find annual capital recovery.

$$CR=500000(A/P,8\%,5)=500000(0.25046)=125230$$

Answer: Annual capital recovery is about P125,230.

Problem: Effect of Salvage on Annual Cost

At 9% for 4 years, what annual credit is produced by a P60,000 salvage value?

$$A_S=60000(A/F,9\%,4)=60000(0.21547)=12928$$

Answer: Salvage reduces annual cost by about P12,928.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q210

MSTE - Engineering Economy / Capital Recovery / Engr. Janclyde Espinosa (Clidez)

CE Board November 2016
A machine which costs Php50,000 when new has a 10-year lifetime and a salvage value equal to 10% of its original value. Determine the capital recovery based upon an interest rate of 8% per year, compounded annually.

Answer:

  1. 7105 per year
  2. 7501 per year
  3. 7150 per year
  4. 7510 per year
$P = $ P50,000; $S = $ P5,000; $n = 10$ yr; $i = 8\%$
$(A/P,8\%,10) = \frac{0.08(1.08)^{10}}{(1.08)^{10}-1} = \frac{0.08 \times 2.1589}{1.1589} = 0.14903$
$CR = (P-S)(A/P) + Si = 45{,}000 \times 0.14903 + 5{,}000 \times 0.08$
$= 6{,}706 + 400$
$\boxed{\approx \text{P}7{,}105 \text{ per year}}$

Question Bank: q211

MSTE - Engineering Economy / Capital Recovery / Engr. Janclyde Espinosa (Clidez)

CE Board November 2016
A machine which costs P100,000 when new has a lifetime of 15 years and a salvage value equal to 20% of its original cost. Determine the capital recovery for this machine if the interest rate is 10% per year, compounded annually.

Answer:

  1. 12517.9
  2. 12715.9
  3. 12175.9
  4. 12571.9
$P = $ P100,000; $S = 20\% \times 100{,}000 = $ P20,000; $n = 15$ yr; $i = 10\%$
$(A/P,10\%,15) = \frac{0.10(1.10)^{15}}{(1.10)^{15}-1} = \frac{0.10 \times 4.1772}{3.1772} = 0.13147$
$CR = (P-S)(A/P) + S \cdot i = 80{,}000 \times 0.13147 + 20{,}000 \times 0.10$
$= 10{,}518 + 2{,}000$
$\boxed{\approx \text{P}12{,}518 \text{ per year}}$

Question Bank: t1091

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Machine cost = P550,000; Life = 12 years; Salvage Value = P120,000. What minimum cash return would the investor demand annually from the operation of this machine if he desires interest annually at the rate of 32% on his investment and accumulates a capital replacement fund by investing annual deposits at 12%?

  1. P125,485
  2. P193,818
  3. P232,587
  4. P157,445
Minimum return = interest demanded on the investment + sinking-fund deposit to replace capital.
Interest: $550{,}000(0.32) = 176{,}000$.
Sinking fund at 12%: $(A/F,12\%,12) = \dfrac{0.12}{1.12^{12}-1} = 0.041437$.
$$(550{,}000 - 120{,}000)(0.041437) = 17{,}818$$
$$\boxed{\text{Annual return} = 176{,}000 + 17{,}818 = \text{P}193{,}818}$$

Question Bank: w44

MSTE - Engineering Economy / Rate of Return / MSTE May 2019

A man bought a government bond which cost P1M and will pay P50,000 interest each year for 20 yrs. The bond will mature at the end of 20 years and he will receive the original P1M. If there is 2% annual inflation during this period, what rate of return will the investor receive after considering the effect of inflation?

  1. 2.94%
  2. 1.82%
  3. 3.14%
  4. 2.41%
Bond yield (nominal): $i = \frac{50{,}000}{1{,}000{,}000} = 5\%$.
Inflation adjustment: $(1 + i) = (1 + R)(1 + h)$
$1.05 = (1 + R)(1.02)$
$\boxed{R = 2.94\%}$