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Equivalent Annual Cost Alternatives

EAC compares alternatives by converting each cost stream into an equivalent annual amount.

Problem: CE Board May 2019

A government center compares standard construction and energy-efficient construction for a 40-year life at 10%. Which alternative has lower EAC?

Standard construction: $EAC=250,536+10,226+10,000=270,762$.

Energy-efficient construction: $EAC=255,649+8,692+6,000=270,341$.

Final answer: choose energy-efficient construction because its EAC is lower.

Problem: Deferred Repair EAC

A generating unit has a 2-year warranty. Repair costs of $3,500 per year begin after warranty and continue through a 15-year life. At 7%, find EAC for repair costs.

Convert the deferred repair series to future worth, then annualize over 15 years.

$$A \approx 2,805$$

Final answer: $2,805 per year.

Problem: Compare Two Machines by EAC

Machine A has EAC P95,000 and Machine B has EAC P88,000 for the same service. Which is preferred?

For cost alternatives providing the same service, choose the smaller equivalent annual cost.

Answer: Choose Machine B.

Problem: Different Lives by Annual Cost

Alternative X has present cost P300,000 for 6 years. At 10%, convert it to EAC.

$$EAC=300000(A/P,10\%,6)=300000(0.22961)=68883$$

Answer: EAC is about P68,883 per year.

Problem: Annual Cost with Salvage Credit

Equipment costs P250,000, lasts 5 years, has P40,000 salvage, and costs P18,000 yearly to operate. Use i = 8%.

$$EAC=250000(A/P,8\%,5)-40000(A/F,8\%,5)+18000$$
$$EAC=250000(0.25046)-40000(0.17046)+18000=73897$$

Answer: EAC is about P73,897 per year.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q192

MSTE - Engineering Economy / Equivalent Uniform Annual Cost / Engr. Janclyde Espinosa (Clidez)

CE Board November 2019
A contractor can buy dump trucks for P800,000 each (surplus) or rent them for P1189 per truck per day. The truck has a salvage value of P100,000 at the end of its useful life of 5 years. Annual cost of maintenance is P20,000. If money is worth 14% per annum, determine the number of days per year that a truck must be used to warrant the purchase of the truck.

Answer:

  1. 200
  2. 190
  3. 180
  4. 170
Annual owning cost of one truck:
$EUAC = 800{,}000(A/P,14\%,5)-100{,}000(A/F,14\%,5)+20{,}000$
$EUAC = 800{,}000(0.291284)-100{,}000(0.151284)+20{,}000$
$EUAC \approx 237{,}899$
Set this equal to rental cost:
$1189D = 237{,}899$
$D = 200.08$ days
$\boxed{D \approx 200\text{ days per year}}$

Question Bank: q195

MSTE - Engineering Economy / Equivalent Uniform Annual Cost / Engr. Janclyde Espinosa (Clidez)

CE Board May 2019
Lapu-Lapu City is building a new government center. The civil engineer for walls and ceiling finishes and the mechanical engineer for the heating are evaluating two plans for insulating and heating the building. They have identified construction (SC) and energy- efficient construction (EEC) as the two alternatives. DBP the funding agency uses an interest rate of 10%, and the new government center is expected to have a life of 40 years. Using EAC, which alternative is better?

q195

Answer:

  1. EEC
  2. SC
  3. Both are equal
  4. None of the alternatives are profitable

Solution pending in psadquestions/q195.json.

Question Bank: q196

MSTE - Engineering Economy / Equivalent Uniform Annual Cost / Engr. Janclyde Espinosa (Clidez)

Davao Central Power & Light has just purchased a generating unit that comes with a 2-year warranty. Once the warranty expires, repair costs are expected to average $3500 per year. Find the EAC for the repair costs over the 15-year life of the generating unit. The interest rate is 7%.

Answer:

  1. 2805
  2. 2508
  3. 2580
  4. 2850
Repair costs begin after the 2-year warranty, so use a deferred 13-year annuity over years 3 through 15:
$PW = 3500(P/A,7\%,13)(P/F,7\%,2)$
$PW = 3500(8.35765)(1.07)^{-2}=25{,}549.64$
Convert this present worth to an equivalent annual cost over 15 years:
$EAC = PW(A/P,7\%,15)=25{,}549.64(0.109795)$
$\boxed{EAC \approx 2805}$

Question Bank: q208

MSTE - Engineering Economy / Benefit Cost Ratio / Engr. Janclyde Espinosa (Clidez)

CE Board May 2015
If the benefit ratio is to be 1.8, how much should he spend in a project that provides annual benefits of P120,000 for a period of 5 years without a salvage value. Cost of money is 12%.

Answer:

  1. 240,318
  2. 432,573
  3. 243,308
  4. 423,375
Annual benefit $A = $ P120,000; $n=5$ yr; $i=12\%$; BCR $= 1.8$
PW of benefits: $(P/A,12\%,5) = \frac{1-(1.12)^{-5}}{0.12} = \frac{1-0.56743}{0.12} = 3.6048$
$PW_B = 120{,}000 \times 3.6048 = $ P432,576
$\text{BCR} = \frac{PW_B}{\text{Cost}} \Rightarrow \text{Cost} = \frac{432{,}576}{1.8}$
$\boxed{\approx \text{P}240{,}318}$

Question Bank: q218

MSTE - Engineering Economy / Capitalized Cost / Engr. Janclyde Espinosa (Clidez)

CE Board November 1999
The first cost of a certain equipment is P324,000 and a salvage value of P50,000 at the end of its life of 4 years. Find the capitalized cost if money is worth 6% compounded annually.

Answer:

  1. 1,367,901.15
  2. 1,376,901.15
  3. 1,396,710.15
  4. 1,369,710.15
$P = $ P324,000; $S = $ P50,000; $n = 4$ yr; $i = 6\%$
$(A/P,6\%,4) = \frac{0.06(1.06)^4}{(1.06)^4-1} = \frac{0.075749}{0.262477} = 0.28859$
Annual CR $= (P-S)(A/P) + Si = 274{,}000 \times 0.28859 + 50{,}000 \times 0.06$
$= 79{,}074 + 3{,}000 = $ P82,074
Capitalized Cost $= \frac{CR}{i} = \frac{82{,}074}{0.06}$
$\boxed{\approx \text{P}1{,}367{,}901}$

Question Bank: q220

MSTE - Engineering Economy / Capitalized Cost / Engr. Janclyde Espinosa (Clidez)

A hydroelectric dam has a design life of 90 years. The dam will cost P275 M to build. The electric turbines will last 30 years and cost P45 M. The interest rate is 8%, and all salvage values are zero. What is the EAC (equivalent annual cost) and the capitalized cost?

Answer:

  1. 325.3M
  2. 352.3M
  3. 253.3M
  4. 235.3M
Dam: $P=$ P275M, $n=90$ yr; Turbines: $P=$ P45M, $n=30$ yr; $i=8\%$; SV=0
$(A/P,8\%,90) \approx \frac{0.08(1.08)^{90}}{(1.08)^{90}-1} \approx 0.08001$ (very nearly $i$ for long life)
$(A/P,8\%,30) = \frac{0.08(1.08)^{30}}{(1.08)^{30}-1} = \frac{0.08 \times 10.0627}{9.0627} = 0.08883$
EAC$_{\text{dam}} = 275 \times 0.08001 = $ P22.00M
EAC$_{\text{turbine}} = 45 \times 0.08883 = $ P3.997M
Total EAC = 22.00 + 3.997 = P26.00M
Capitalized cost $= \frac{26.00}{0.08}$
$\boxed{\approx \text{P}325.3 \text{M}}$

Question Bank: t1052

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

A printing machine is bought at P1.5 million and is estimated to have a salvage value of P100,000 after 500,000 copies. The annual cost of renting the space for the business is P80,000, power cost per copy is P1.50, and maintenance and paper cost per copy is P4.00. The expected annual production of the machine is 100,000 copies. Annual interest is 15%.

Determine the annual operation and maintenance cost of the machine.

  1. P680,000
  2. P630,000
  3. P560,000
  4. P720,000

Determine the annual depreciation of the machine.

  1. P207,642
  2. P256,325
  3. P321,478
  4. P187,445

Determine the production cost per copy.

  1. P15.68
  2. P21.54
  3. P10.63
  4. P12.89

Solution pending in psadquestions/t1052.json.

Question Bank: t1077

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

Two bridges A and B are to be compared on the basis of capitalized cost at 5 per cent interest. Bridge A has an estimated life of 25 years, initial cost of P50M, renewal cost of P35M, annual maintenance of P0.5M, repairs every five years amounting to P2M, and salvage value of P5M. Bridge B has an estimated life of 50 years, initial cost of P75M, renewal cost of P75M, annual maintenance cost of P0.1M, repairs every five years amounting to P1M, and salvage value of P10M. The initial cost can be paid out of available funds. All other expenses will be defrayed by sinking funds.

What is the capitalized cost of Bridge A?

  1. P65.24M
  2. P86.73M
  3. P78.97M
  4. P84.25M

What is the capitalized cost of Bridge B?

  1. P84.25M
  2. P78.97M
  3. P65.24M
  4. P86.73M

How much savings is realized by choosing the more economical of the two bridges?

  1. P19.01
  2. P7.76M
  3. P13.73
  4. P5.28M

Solution pending in psadquestions/t1077.json.

Question Bank: w84

MSTE - Engineering Economy / Breakeven / MSTE November 2019

A contractor can buy a dump truck for P800,000 each (surplus) or rent them for P1,189 per truck per day. The truck has a salvage value of P100,000 at the end of its useful life of 5 years. Annual cost of maintenance is P20,000. If money is worth 14% per annum, determine the number of days per year that a truck must be used to warrant the purchase of the truck.

  1. 198
  2. 199
  3. 200
  4. 197
Annual cost of owning (interest on first cost + maintenance + sinking-fund depreciation):
$AC_{own} = FC\,i + OM + \dfrac{(FC - SV)\,i}{(1+i)^n - 1}$
$AC_{own} = 800{,}000(0.14) + 20{,}000 + \dfrac{(800{,}000 - 100{,}000)(0.14)}{(1.14)^5 - 1} = \text{P}237{,}898.50$
Annual cost of renting for $x$ days: $AC_{rent} = 1{,}189x$.
Set equal: $1{,}189x = 237{,}898.5 \Rightarrow x = 200$ days
$\boxed{x = 200\text{ days}}$