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Capitalized Cost and Perpetuity

Capitalized cost represents the present worth of a perpetual or very long-life service.

$$Capitalized\ Cost=FC+\frac{FC-SV}{(1+i)^n-1}+\frac{OC}{i}$$
$$P_{perpetuity}=\frac{A}{i}$$

Problem: CE Board Nov. 1999

First cost is P324,000 with salvage P50,000 at end of 4-year life. Find capitalized cost if money is worth 6% compounded annually.

$$CC=324,000+\frac{324,000-50,000}{(1.06)^4-1}=1,367,901.15$$

Final answer: P1,367,901.15.

Problem: CE Board May 2016

Find the annual payment needed to retire P70,000,000 in bonds over 50 years at 6%. Treat 50 years as nearly infinite.

$$A=Pi=70,000,000(0.06)=4,200,000$$

Final answer: P4,200,000 per year.

Problem: Perpetual Income Stream

Viewing fees bring $800K in the first year and increase by $50K per year. At 10% and perpetual life, find equivalent annual worth.

$$EAW=800+\frac{50}{0.10}=1,300\text{ K}$$

Final answer: $1,300K per year.

Problem: Capitalized Maintenance Cost

A facility requires P90,000 per year forever. At 6%, find the capitalized cost of maintenance.

$$P=\frac{A}{i}=\frac{90000}{0.06}=1500000$$

Answer: The capitalized maintenance cost is P1,500,000.

Problem: Perpetual Replacement Fund

A component costing P250,000 must be replaced every 10 years forever. At 8%, find the capitalized replacement cost.

$$A=250000(A/F,8\%,10)=250000(0.06903)=17258$$
$$P=\frac{A}{i}=\frac{17258}{0.08}=215725$$

Answer: Capitalized replacement cost is about P215,725.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q221

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

CE Board May 2019
The first cost of an electric rebar bender is P324,000 and a salvage value of P50,000 at the end of its life for 4 years. Money is worth 6% annually. If there is no salvage value and the annual maintenance cost is P18,000, find the capitalized cost of perpetual service.

Answer:

  1. 624000
  2. 642000
  3. 637000
  4. 673000
For perpetual service with no salvage value and annual maintenance cost = P18,000:
Capitalized cost = First cost + $\frac{\text{Annual cost}}{i}$
$= 324{,}000 + \frac{18{,}000}{0.06}$
$= 324{,}000 + 300{,}000$
$\boxed{= \text{P}624{,}000}$

Question Bank: q225

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

The National Monuments Commission is considering the imposition of viewing fees for several monuments. At one monument, these fees would bring in P800 K for the first year, and this should increase by P50 K per year as the number of tourist increases. If the government uses an interest rate of 10% and perpetual life is assumed, what is the EAW for the income stream?

Answer:

  1. 1300K
  2. 1200K
  3. 1400K
  4. 1100K
First year income $A_1 = $ P800K; annual increase $G = $ P50K; $i = 10\%$; perpetual.
For a gradient perpetuity: EAW $= A_1 + \dfrac{G}{i}$
$= 800{,}000 + \dfrac{50{,}000}{0.10} = 800{,}000 + 500{,}000$
$\boxed{= \text{P}1{,}300{,}000}$

Question Bank: q241

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

CE Board November 2021
An electric replacement pump is being considered for purchase. It is capable of providing 200 hp. The pertinent data are as follows:
Cost = P3200
Maintenance Cost per year = P50 Electric efficiency = 0.85
Life expectancy = 14 years
The pump is used for 400 hours per year and the cost of electricity is P0.04 per kilowatt-hour
(1 horsepower = 0.746 KW). Assuming the pump will have no salvage value, what will be the monthly cost?

Answer:

  1. 238.2
  2. 283.2
  3. 223.8
  4. 232.8

Solution pending in psadquestions/q241.json.

Question Bank: w42

MSTE - Engineering Economy / Capitalized Cost / MSTE May 2019

An electric rebar bender has a first cost of P324,000 and a salvage value of P50,000 at the end of its life of 4 years. Money is worth 6% annually. If there is no salvage value and the annual maintenance cost is P18,000, find the capitalized cost of perpetual service.

  1. P856,000
  2. P714,000
  3. P624,000
  4. P936,000
Capitalized cost for perpetual service (no salvage):
$K = FC + \frac{OM}{i} = 324{,}000 + \frac{18{,}000}{0.06} = \boxed{P\,624{,}000}$