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Present Worth, Net Present Value, and Net Benefit Value

Present worth moves all project cash flows to time zero. NPV is benefits minus costs on a present-worth basis.

Problem: Milling Machine NPV

A machine costs P60,000, operating cost P2,675.40 per year, revenue P15,000 per year, life 7 years, and interest 8%. Find NPV.

$$A=15000-2675.40=12,324.60$$
$$P=12324.60(P/A,8\%,7)=64,166.42$$
$$NPV=64,166.42-60,000=4,166.43$$

Final answer: P4,166.43.

Problem: Benefit, Disbenefit, and NBV

A public works project has present-worth benefits of $75 M, costs of $55 M, and disbenefits of $15 M. Determine BCR treatments and NBV.

$$BCR=\frac{75-15}{55}=1.09$$
$$BCR=\frac{75}{55+15}=1.07$$
$$NBV=75-15-55=5$$

Final answer: project is economically acceptable.

Problem: Quick Fix vs Quality Fix

A facility may use quick-fix A or quality-fix B. At 10%, compare net present savings. Quick fix costs $7,000 with decreasing savings; quality fix costs $12,000 with $4,300 annual savings for 4 years.

Quick-fix net present savings is $1,301.26. Quality-fix net present savings is $1,630.46.

Final answer: choose quality-fix B by the larger net present savings.

Problem: NPV Acceptance Check

A project costs P120,000 now and returns P45,000 per year for 4 years. Use i = 12% and decide by NPV.

$$NPV=-120000+45000(P/A,12\%,4)$$
$$NPV=-120000+45000(3.0373)=16679$$

Answer: NPV is positive, so accept the project.

Problem: Present Worth with Salvage

A machine costs P200,000, saves P55,000 yearly for 5 years, and has P30,000 salvage. Find PW at 10%.

$$PW=-200000+55000(P/A,10\%,5)+30000(P/F,10\%,5)$$
$$PW=-200000+55000(3.7908)+30000(0.6209)=27119$$

Answer: Present worth is about P27,119.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q191

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

CE Board November 2017
Talisay City is considering a new P50000 street cleaner. The new machine will operate at a savings of P600 per day compared to the present equipment. Assume the MARR is 12%, and the machine life is 10 years with zero resale value at that time. How many days per year must the machine be used to justify the investment?

Answer:

  1. 15
  2. 12
  3. 13
  4. 14
Annual savings must at least equal the annual capital recovery:
$600D = 50{,}000(A/P,12\%,10)$
$600D = 50{,}000(0.17698)=8{,}849$
$D = \frac{8{,}849}{600}=14.75$ days
Round up to justify the investment.
$\boxed{D \approx 15\text{ days per year}}$

Question Bank: q197

MSTE - Engineering Economy / Present Worth and Net Present and Benefit Value / Engr. Janclyde Espinosa (Clidez)

The XYZ Company is contemplating the purchase of a new milling machine. The purchase price of the new machine is P60,000 and its annual operating cost is P2675.40. The machine has a life of 7 years and is expected to generate P15,000 in revenues in each year of its life. What is the net present value of the investment in this machine if the interest rate is 8% per year?

Answer:

  1. 4166.43
  2. 4164.36
  3. 4134.63
  4. 4146.63
Annual net revenue is:
$A = 15{,}000 - 2675.40 = 12{,}324.60$
Net present value:
$NPV = -60{,}000 + A(P/A,8\%,7)$
$NPV = -60{,}000 + 12{,}324.60(5.20637)$
$NPV = 4{,}166.43$
$\boxed{NPV = 4166.43}$

Question Bank: q200

MSTE - Engineering Economy / Present Worth and Net Present and Benefit Value / Engr. Janclyde Espinosa (Clidez)

A public works project is proposed that the total present-worth benefits of P75 M and total present worth cost of P55 M. In deliberating this proposal, some members of the town council have suggested that the project has a total present-worth disbenefit of P15 M, other members feel the P15 M should be treated as a cost.

Determine the benefit cost ratio if the P15 M is treated as a disbenefit.

  1. 1.09
  2. 1.23
  3. 1.32
  4. 1.90

Determine the benefit cost ratio if the P15 M is treated as a cost.

  1. 1.07
  2. 1.27
  3. 1.23
  4. 1.15

Determine the net benefit value.

  1. 5
  2. 6
  3. 7
  4. 4

Part 1.

P15M as disbenefit (reduces benefits):
$\text{BCR} = \frac{B - D}{C} = \frac{75 - 15}{55} = \frac{60}{55}$
$\boxed{\approx 1.09}$

Part 2.

P15M as additional cost:
$\text{BCR} = \frac{B}{C + D} = \frac{75}{55 + 15} = \frac{75}{70}$
$\boxed{\approx 1.07}$

Part 3.

Net benefit = $B - C - D = 75 - 55 - 15$
$\boxed{= \text{P5 M}}$

Question Bank: q207

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

Grampian Manufacturing Company is attempting the best sized milling machine for their production shop. Five alternative sizes are available. Grampian has a budget of 250,000 dollars and MARR = 15%. Which size machine should they purchase? Assume n = 100 years and that the ultimate salvage value is zero for each machine.

q207

Answer:

  1. Delux
  2. Super Delux
  3. Economy
  4. Regular and Super, which are equal

Solution pending in psadquestions/q207.json.

Question Bank: t1022

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

A man deposited P5,000 on the date his son celebrated his 1st birthday. If money is worth 10% compounded semi-annually, what is the maximum amount the son can withdraw on his 18th birthday?

  1. P28,451
  2. P32,145
  3. P26,267
  4. P29,411
$P = $ P5,000; $i = 10\%$ compounded semi-annually = 5%/period
From 1st to 18th birthday = 17 years = 34 semi-annual periods
$F = P(1+i)^n = 5{,}000(1.05)^{34}$
$(1.05)^{34} \approx 5.2533$
$F = 5{,}000 \times 5.2533$
$\boxed{\approx \text{P}26{,}267}$

Question Bank: t2017

MSTE - Engineering Economy / Simple Interest / BEMz

Mr. Almagro made a money market of P1 000 000 for 30 days at 7.5% per year. If the withholding tax is 20%, what is the net interest that he will receive at the end of the month?

  1. P3 000
  2. P4 000
  3. P6 000
  4. P5 000
Use ordinary simple interest with a 360-day year.
Gross interest:
$I=Prt=1{,}000{,}000(0.075)\left(\frac{30}{360}\right)=\text{P}6{,}250$
Withholding tax $=20\%$ of the interest:
$T=0.20(6{,}250)=\text{P}1{,}250$
Net interest:
$I_{net}=6{,}250-1{,}250$
$\boxed{\text{P}5{,}000}$