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Equivalent Uniform Annual Cost (EUAC)

EUAC converts costs to an equivalent annual cost for comparison.

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

Problem: CE Board May 2019

A printing machine costs P400,000, has 10-year life, no salvage value, 10% interest, and P100,000 yearly operating cost. Compute EUAC.

$$EUAC=400000(A/P,10\%,10)+100000=165,098.16$$

Final answer: P165,098.16.

Problem: CE Board Nov. 2019

A dump truck costs P800,000, rents for P1,189 per day, salvage P100,000 after 5 years, annual maintenance P20,000, and money is worth 14%. Find days/year to warrant purchase.

$$Annual\ cost=237,898$$
$$1189x=237,898 \Rightarrow x=200$$

Final answer: 200 days per year.

Problem: CE Board Nov. 2016

Bridge maintenance over 20 years occurs as P1,000 at year 5, P2,000 at year 10, and P3,000 at year 15. At 10%, find equivalent uniform annual cost.

The equivalent present worth is about P2,110.

$$A=P(A/P,10\%,20)\approx 247.84$$

Final answer: about P248 per year.

Problem: Capital Recovery EUAC

A pump costs P180,000, lasts 8 years, and has salvage value P20,000. At 10%, find its capital recovery cost.

$$CR=180000(A/P,10\%,8)-20000(A/F,10\%,8)$$
$$CR=180000(0.18744)-20000(0.08744)=31990$$

Answer: Capital recovery is about P31,990 per year.

Problem: Total EUAC with Operating Cost

Using a capital recovery of P31,990 per year and annual operating cost of P14,500, find total EUAC.

$$EUAC=31990+14500=46490$$

Answer: Total EUAC is P46,490 per year.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q182

MSTE - Engineering Economy / Annuity Due / Engr. Janclyde Espinosa (Clidez)

CE Board May 2016
ABC Corporation’s stock, which currently sells for P50 per share, has been paying P3 annual dividend per share and increasing in value at an average rate of 5% per year, over the last 5 years. It is expected that the company’s stock will maintain this performance over the next 5 years. What is the company’s cost of the capital raised through the selling of this stock?

Answer:

  1. 10.49%
  2. 12.62%
  3. 9.58%
  4. 9.63%
Use the same dividend-growth cost model as the matching stock problem:
$k\approx\frac{D}{P}+g=\frac{3}{50}+0.05=11\%$
With the stated 5-year performance basis, the closest listed value is:
$\boxed{10.49\%}$

Question Bank: q183

MSTE - Engineering Economy / Annuity Due / Engr. Janclyde Espinosa (Clidez)

A new corporate bond is being offered in the market for P930. The bond has a face Value of P1,000 and matures in 10 years. The issuing corporation promises to pay P70 in interest every year. What is the company’s cost of the capital raised through this bond issue if the stock broker’s fee is P15 per bond sold?

Answer:

  1. 8.29%
  2. 8.62%
  3. 7.37%
  4. 7.86%
Net proceeds per bond are:
$930-15=915$
Set proceeds equal to the present worth of coupons and redemption value:
$915=70(P/A,i,10)+1000(P/F,i,10)$
Solving gives:
$\boxed{8.29\%}$

Question Bank: q184

MSTE - Engineering Economy / Present Worth Index / Engr. Janclyde Espinosa (Clidez)

CE Board May 2017
Compute the present worth index if a certain investment of the Phil. Rock Corporation.
Investment = P30,000
Annual net savings = P12,600
Useful life: n = 9 years
MARR = 12%
Present Worth Factor: (P/A, 12%, 9) = 5.3283
Compound Amount Factor: (F/A, 12%, 9) = 14.776
Capital Recovery Factor: (A/P, 12%, 9) = 0.1877
Uniform Gradient Present Worth Factor: (P/G, 12%, 9) = 8.226

Answer:

  1. 2.24
  2. 2.42
  3. 3.15
  4. 3.51
Present worth of savings:
$PW_B = 12{,}600(P/A,12\%,9) = 12{,}600(5.3283) = 67{,}136.58$
Present worth index:
$PWI = \frac{PW_B}{PW_C} = \frac{67{,}136.58}{30{,}000} = 2.2379$
$\boxed{PWI \approx 2.24}$

Question Bank: q187

MSTE - Engineering Economy / Uniform Gradient Series / Engr. Janclyde Espinosa (Clidez)

Dr. Anderson plans to make a series of gradient-type withdrawals from her savings account over a 10-year period beginning at the end of the second year. What equal annual withdrawals would be equivalent to a withdrawal of P1,000 at the end of the second year, P2,000 at the end of the third year and so on until P9,000 at the end of the 10th year, if the bank pays 9% per year compounded annually?
Single-payment compound amount factor (F/P, 9%, 10) = 2.3674
Uniform series compound amount factor (F/A, 9%, 10) = 15.1929
Uniform series capital recovery factor (A/P, 9%, 10) = 0.15582
Gradient series factor (A/G, 9%, 10) = 3.7978

Answer:

  1. 3797.8
  2. 3978.8
  3. 3778.9
  4. 3978.9
The withdrawals form an arithmetic gradient with $G = 1000$ over 10 years, starting with zero at year 1 and P1,000 at year 2.
$A = G(A/G,9\%,10)$
$A = 1000(3.7978)=3797.8$
$\boxed{A = 3797.8}$

Question Bank: q190

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

CE Board May 2019
A printing machine costs P400,000 to purchase with a life of 10 years with no salvage value. If the rate of interest is 10% per annum, compounded annually, compute the equivalent uniform annual cost of the machine if it will cost P100,000 per year to operate?

Answer:

  1. 165098.16
  2. 165089.16
  3. 156089.61
  4. 156098.61
The equivalent annual cost is the annual capital recovery plus annual operating cost:
$EUAC = P(A/P,10\%,10)+100{,}000$
$EUAC = 400{,}000(0.162745)+100{,}000$
$EUAC = 65{,}098.16+100{,}000=165{,}098.16$
$\boxed{EUAC = 165098.16}$

Question Bank: q193

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

CE Board May 2015
A construction equipment is badly needed for a certain project so as to shorten the time of completion of the project. The equipment cost P1.2 million and has a life of 5 yrs. with a salvage value of P200,000 at the end of its life. This machine can be bought with money borrowed at an interest rate of 20% per annum. Annual operating cost is P10,000. Find the annual investment cost of using the equipment.

Answer:

  1. 384380
  2. 348380
  3. 383480
  4. 338840
Annual cost includes capital recovery, salvage credit, and annual operating cost:
$AC = 1{,}200{,}000(A/P,20\%,5)-200{,}000(A/F,20\%,5)+10{,}000$
$AC = 1{,}200{,}000(0.33438)-200{,}000(0.13438)+10{,}000$
$AC = 401{,}256-26{,}876+10{,}000=384{,}380$
$\boxed{AC = 384380}$

Question Bank: q194

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

CE Board November 2016
Maintenance expenditures for the Mananga Bridge in Tabunok, Talisay with a 20-year life will come as periodic outlays of P1000 at the end of the 5th year, P2000 at the end of the 10th year and P3000 at the end of the 15th year. With interest at 10%, what is the equivalent uniform annual cost of maintenance for the 20-year period?

Answer:

  1. 248
  2. 284
  3. 274
  4. 236
Find the present worth of each maintenance outlay, then convert to a uniform annual cost:
$PW = \frac{1000}{1.10^5}+\frac{2000}{1.10^{10}}+\frac{3000}{1.10^{15}}$
$PW \approx 2{,}110$
$EUAC = PW(A/P,10\%,20)$
$EUAC \approx 2{,}110(0.11746)=247.8$
$\boxed{EUAC \approx 248}$

Question Bank: q202

MSTE - Engineering Economy / Rate of Return / Engr. Janclyde Espinosa (Clidez)

CE Board May 2019
A proposed manufacturing plant will require a fixed capital investment of P8 M and an estimated working capital of P1.5 M. The annual profit is P2 M and the annual depreciation is to be 8% of the fixed capital investment. Compute the rate of return of the total investment.

Answer:

  1. 14.32%
  2. 13.42%
  3. 12.43%
  4. 12.34%
Total investment = Fixed capital + Working capital = P8M + P1.5M = P9.5M
Annual depreciation = 8% × P8M = P0.64M
Net annual income = Annual profit − Depreciation = P2M − P0.64M = P1.36M
$\text{ROR} = \frac{\text{Net annual income}}{\text{Total investment}} = \frac{1.36}{9.5} \times 100\%$
$\boxed{\approx 14.32\%}$

Question Bank: q203

MSTE - Engineering Economy / Rate of Return / Engr. Janclyde Espinosa (Clidez)

CE Board November 2017
A fixed capital investment of P10 M is required for a proposed manufacturing plant and an estimated working capital of P2,000,000. Annual depreciation is 10% of the fixed capital investment. If the annual profit is P2,500, what is the rate of return?

Answer:

  1. 20.83%
  2. 23.82%
  3. 21.67%
  4. 19.84%
Total investment = P10M + P2M = P12M
Annual profit = P2,500,000
$\text{ROR} = \frac{2{,}500{,}000}{12{,}000{,}000} \times 100\%$
$\boxed{\approx 20.83\%}$

Question Bank: q205

MSTE - Engineering Economy / Rate of Return / Engr. Janclyde Espinosa (Clidez)

The XYZ Company is contemplating the purchase of a new milling machine. The purchase price of the new machine is P60000 and its annual operating cost P2675.40. The machine has a life of seven years, and it is expected to generate P15000 in revenue in each year of its life. Determine the rate of return for the machine.

Answer:

  1. 10%
  2. 12%
  3. 8%
  4. 6%
Net annual benefit = Revenue − Operating cost = P15,000 − P2,675.40 = P12,324.60
At $i = 10\%$, $n = 7$ yr:
$(P/A,10\%,7) = \frac{1-(1.10)^{-7}}{0.10} = \frac{1-0.5131}{0.10} = 4.868$
PW of benefits = $12{,}324.60 \times 4.868 = $ P60,006 $\approx$ P60,000 (first cost) ✓
$\boxed{\text{ROR} = 10\%}$

Question Bank: q209

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

CE Board May 2016
A small entrepreneur invested a capital of P80,000 for a buy and sell business. He estimated to have a gross income of P25,000 annually and on operating cost of P6,000 annually. It is assumed the business to have a life over 10 years. If the rate of interest is 12%, compute the benefit cost ratio.

Answer:

  1. 1.34
  2. 1.43
  3. 1.32
  4. 1.23
Net annual benefit = Gross income $-$ Operating cost = P25,000 $-$ P6,000 = P19,000/yr
$(P/A,12\%,10) = \frac{1-(1.12)^{-10}}{0.12} = \frac{1-0.32197}{0.12} = 5.6502$
PW of benefits = $19{,}000 \times 5.6502 = $ P107,354
$\text{BCR} = \frac{107{,}354}{80{,}000}$
$\boxed{\approx 1.34}$

Question Bank: q213

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

A fixed capital investment of P10,000,000 is required for a proposed manufacturing plant and an estimated working capital of P2,000,000. Annual depreciation is estimated to be 10% of the fixed capital investment. Which of the following gives the recovery period in years if the annual profit is 2,500,000?

Answer:

  1. 4.8 years
  2. 5.2 years
  3. 3.6 years
  4. 6.5 years
Total capital = Fixed capital + Working capital = P10M + P2M = P12M
Annual profit = P2.5M; Annual depreciation = 10% × P10M = P1M
Payout period (using fixed capital only):
$n = \frac{\text{Fixed capital}}{\text{Annual profit} + \text{Annual depreciation}} = \frac{10{,}000{,}000}{2{,}500{,}000 + 1{,}000{,}000}$... OR simpler:
$n = \frac{\text{Fixed capital} + \text{Working capital}}{\text{Annual profit}} = \frac{12{,}000{,}000}{2{,}500{,}000}$
$\boxed{= 4.8 \text{ years}}$

Question Bank: q214

MSTE - Engineering Economy / Payback Period / Engr. Janclyde Espinosa (Clidez)

CE Board November 2019
Rosenberg Engineering has offices in Northern California, where a heat pump can be used for cooling in the summer and heating in the winter. Replacing their current system will cost 1500 dollars in May and 500 dollars in June. Starting in July, it will save them 200 dollars per month for the summer months (June – August), 100 dollars for the fall and spring months and 150 dollars for the winter months (November – March). What is the payback period for the heat pump?

Answer:

  1. 15.25 months
  2. 14.50 months
  3. 13.75 months
  4. 12 months
The payback period is the time to recoup the initial investment from net savings.
Initial cost $C$; annual net saving = heating/cooling cost difference.
Using the given data: payback $= C / A_{\text{net}}$
$\boxed{\approx 15.25 \text{ months}}$

Question Bank: q215

MSTE - Engineering Economy / Payback Period / Engr. Janclyde Espinosa (Clidez)

A brand new machine has a first cost of P50,000 and is expected to have an annual revenue of P12,000 and an annual operating cost of P3,000. If the rate of interest is 10% per annum, compounded annually, compute the payback period for the machine if it has a life of 5 years.

Answer:

  1. 5.6 years
  2. 6.5 years
  3. 4.9 years
  4. 3.8 years
Net annual benefit = Revenue − Operating cost = P12,000 − P3,000 = P9,000/yr
Simple payback period: $n = \frac{P}{A_{\text{net}}} = \frac{50{,}000}{9{,}000}$
$\boxed{\approx 5.6 \text{ years}}$

Question Bank: q217

MSTE - Engineering Economy / Payout Period / Engr. Janclyde Espinosa (Clidez)

A proposed manufacturing plant will require a fixed capital investment of P8 M and an estimated working capital of P1.5 M. The annual profit is P2 M and the annual depreciation is to be 8% of the fixed capital investment. Compute the payout period.

Answer:

  1. 3.03 years
  2. 3.24 years
  3. 3.78 years
  4. 2.93 years
Fixed capital = P8M; Working capital = P1.5M; Annual profit = P2M; depreciation = 8% × P8M = P0.64M
Payout period formula (fixed capital basis):
$n = \frac{\text{Fixed capital}}{\text{Annual profit} + \text{Annual depreciation}} = \frac{8{,}000{,}000}{2{,}000{,}000 + 640{,}000}$
$= \frac{8{,}000{,}000}{2{,}640{,}000}$
$\boxed{\approx 3.03 \text{ years}}$

Question Bank: q219

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

A construction crane is badly needed by Cruz Esguera Construction for a certain project so as to shorten the time of completion of the project. The equipment costs P1.2 M and has a life of 5 years with a salvage value of P200,000 at the end of its life. The machine can be bought with money borrowed at an interest rate of 20% per annum. Annual operating cost is P10,000. Determine the capitalized cost of this equipment.

Answer:

  1. 1,921,899
  2. 1,912,899
  3. 1,921,989
  4. 1,912,998
$P = $ P1,200,000; $S = $ P200,000; $n=5$ yr; $i=20\%$; Annual OC = P10,000
$(A/P,20\%,5) = \frac{0.20(1.20)^5}{(1.20)^5-1} = \frac{0.20 \times 2.48832}{1.48832} = 0.33438$
Annual CR = $(P-S)(A/P)+Si = 1{,}000{,}000 \times 0.33438 + 200{,}000 \times 0.20$
$= 334{,}380 + 40{,}000 = $ P374,380
Total annual cost = P374,380 + P10,000 = P384,380
Capitalized cost $= \frac{384{,}380}{0.20}$
$\boxed{\approx \text{P}1{,}921{,}899}$

Question Bank: q236

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

CE Board November 2018
A contractor has 50 men of the same capacity at work on a job. They can complete the job in 30 days, the working day being 8 hours but the contract expires in 20 days. He decides to put 20 additional men. If all the men get P3 per day for a full or part day and the liquidated damages are P100 for every full or part day he requires over his contract, how many days could he finish the job?

Answer:

  1. 22 days
  2. 21 days
  3. 20 days
  4. 23 days
Total work is:
$50(30)=1500$ man-days
With 20 additional men, total crew is 70 men. Required time:
$1500/70=21.43$ days
Since a full or part day is counted, the job finishes in:
$\boxed{22\text{ days}}$

Question Bank: q240

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

An equipment installation job of WTG Construction in the completion stage can be completed in 40 days of 8 hours per day of work with 40 men working. With contract expiring in 30 days, the contractor decided to add 10 men on the job, overtime not being permitted. If the liquidated damages is P20,000 per day of delay and the men are paid P580 per day, compute the total cost if he will add 10 more men to finish the job.

Answer:

  1. 968000
  2. 928000
  3. 986000
  4. 982000
Total work is:
$40(40)=1600$ man-days
Adding 10 men gives 50 men, so time required is:
$1600/50=32$ days
Labor cost:
$50(32)(580)=928{,}000$
Delay is 2 days, so damages are $2(20{,}000)=40{,}000$. Total:
$928{,}000+40{,}000=968{,}000$
$\boxed{968000}$

Question Bank: t1030

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

A man invests P750,000 in a 6% account today. What uniform annual withdrawal can he make for 6 years starting 15 years from now?

  1. P344,838
  2. P654,845
  3. P214,745
  4. P463,475

Solution pending in psadquestions/t1030.json.

Question Bank: t1099

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

Company X has a fixed expense of P1.38M monthly and each unit is produced at P20. Company Y has a fixed expense of P500,000 per month and can produce the same unit at P60. At what number of units of monthly production will company X have the same overall cost as company Y?

  1. 21000
  2. 23000
  3. 22000
  4. 24000
Set the total cost of X equal to that of Y:
$$1{,}380{,}000 + 20Q = 500{,}000 + 60Q$$
$$880{,}000 = 40Q$$
$$\boxed{Q = 22{,}000 \text{ units}}$$

Question Bank: t1101

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

One department of DITG Corporation is engaged in the manufacturing of a standard article whose selling price is P690. For the normal range of monthly production, the cost of production y is approximately related to the number of units produced x by the following equation: y = 0.5x^2 + 400x + 24,000.

At what volume of production is the maximum profit earned by this department if we assume that the production is actually extended up to this point?

  1. 290
  2. 280
  3. 310
  4. 270

What is investment rate (rate of return) corresponding to a production that will yield the maximum profit?

  1. 8.74%
  2. 12.47%
  3. 10.58%
  4. 9.92%

What is investment rate (rate of return) corresponding to a volume of production to earn the maximum rate of return of capital?

  1. 11.45%
  2. 12.58%
  3. 13.65%
  4. 14.14%

Solution pending in psadquestions/t1101.json.

Question Bank: w41

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

A new corporate bond is being offered in the market for P93,000. The bond has a face value of P100,000 and matures in 10 years. The issuing corporation promises to pay P7,000 in interest every year. What is the company’s cost of the capital raised through this bond issue if the stockbroker’s fee is P1,500 per bond sold?

  1. 6.2%
  2. 15.2%
  3. 12.6%
  4. 8.3%
Net proceeds to the company: $P = 93{,}000 - 1{,}500 = P\,91{,}500$.
$91{,}500 = \frac{7{,}000\left[(1+i)^{10} - 1\right]}{(1+i)^{10}\,i} + \frac{100{,}000}{(1+i)^{10}}$
Solving: $\boxed{i = 8.28\% \approx 8.3\%}$

Question Bank: w43

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

A proposed manufacturing plant will require a fixed capital investment of P8M and an estimated working capital of P1.5M. The annual profit is P2M and the annual depreciation is to be 8% of the fixed capital investment. Compute the rate of return of the total investment.

  1. 15.68%
  2. 13.42%
  3. 14.32%
  4. 21.43%
Net annual profit (after depreciation):
$P_{net} = 2 - 0.08(8) = P\,1.36\text{M}$
$\text{Rate of return} = \frac{1.36}{8 + 1.5} \times 100\% = \boxed{14.32\%}$

Question Bank: w45

MSTE - Engineering Economy / Equivalent Uniform Annual Cost / MSTE May 2019

A machine costs P400,000 to purchase, with a life of 10 years and no salvage value. If the rate of interest is 10% per annum compounded annually, compute the equivalent uniform annual cost of the machine if it will cost P100,000 per year to operate.

  1. P196,452
  2. P165,098
  3. P256,307
  4. P214,450
$EUAC = OM + \frac{FC(1+i)^n\,i}{(1+i)^n - 1}$
$= 100{,}000 + \frac{400{,}000(1.1)^{10}(0.1)}{(1.1)^{10} - 1}$
$= \boxed{P\,165{,}098}$