To maintain an asphalt road in Davao City, P5,000 will be required at the end of three years and annually thereafter. If the money is worth 8%, determine the capitalized cost of all future maintenance.
To maintain an asphalt road in Davao City, P5,000 will be required at the end of three years and annually thereafter. If the money is worth 8%, determine the capitalized cost of all future maintenance.
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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Question
9. Please use the formula to solve the problem.
![To maintain an asphalt road in Davao City, P5,000 will
be required at the end of three years and annually
thereafter. If the money is worth 8%, determine the
capitalized cost of all future maintenance.
Add your answer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7053b554-3a97-4d68-beed-d147b9a2b76e%2Fd7627206-365b-4cd5-87c4-d291a65d8ccc%2Ftgeg5hh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To maintain an asphalt road in Davao City, P5,000 will
be required at the end of three years and annually
thereafter. If the money is worth 8%, determine the
capitalized cost of all future maintenance.
Add your answer
![Legend:
Cc - Capitalized Cost (Currency)
Fc - First Cost (Currency)
Mc - maintenance Cost (Currency)
Rc - Replacement Cost (Currency); if no Re: Re Fe
Sy- Salvage Value (Curency); if no Sy: Sy = 0
A - Periodic Amount (Currency)
F-Future Value (Currency)
G - Periodic Amount Increment Amount (Currency)
P - Present Value (Currency)
g - Periodic Amount Increment Rate (Persentage)
i-Nominal Interest Rate (Percent)
m - Number of Periods per Year (Number)
r - Effective Interest Rate (Percent)
t-Number of Years (Number)
Formulae:
Compounding Transformation (i, (m₂)→ 1₂ (m₂)): (1 + ) = (1 +)**
Perpetuity (t=00):
Ordinary Annuity (Payment at End of Period):
P=A1-(1+1)-²)
Arithmetic Gradient:
Geometric Gradient:
Capitalized Cost:
n = mt
Annuity Due (Payment at Beginning of Period):
r=-
m
F = P(1+r)"
P = F(1+r)-R
F=A
F=A
= A ((1+r)^²-1)
F=
P=A (1-(1+r)^)(1+r)
= A (¹ + r)² - 1) (1 + r)
T
FA[(1+r)-1]G[-nr + (1+r)" - 1]
T
r2
A[(1+r)"-(1+g)"]
T-9
Rc-Sv
Cc=Fc++ (1+r)"-1
A = Cer](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7053b554-3a97-4d68-beed-d147b9a2b76e%2Fd7627206-365b-4cd5-87c4-d291a65d8ccc%2Ftn85w1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Legend:
Cc - Capitalized Cost (Currency)
Fc - First Cost (Currency)
Mc - maintenance Cost (Currency)
Rc - Replacement Cost (Currency); if no Re: Re Fe
Sy- Salvage Value (Curency); if no Sy: Sy = 0
A - Periodic Amount (Currency)
F-Future Value (Currency)
G - Periodic Amount Increment Amount (Currency)
P - Present Value (Currency)
g - Periodic Amount Increment Rate (Persentage)
i-Nominal Interest Rate (Percent)
m - Number of Periods per Year (Number)
r - Effective Interest Rate (Percent)
t-Number of Years (Number)
Formulae:
Compounding Transformation (i, (m₂)→ 1₂ (m₂)): (1 + ) = (1 +)**
Perpetuity (t=00):
Ordinary Annuity (Payment at End of Period):
P=A1-(1+1)-²)
Arithmetic Gradient:
Geometric Gradient:
Capitalized Cost:
n = mt
Annuity Due (Payment at Beginning of Period):
r=-
m
F = P(1+r)"
P = F(1+r)-R
F=A
F=A
= A ((1+r)^²-1)
F=
P=A (1-(1+r)^)(1+r)
= A (¹ + r)² - 1) (1 + r)
T
FA[(1+r)-1]G[-nr + (1+r)" - 1]
T
r2
A[(1+r)"-(1+g)"]
T-9
Rc-Sv
Cc=Fc++ (1+r)"-1
A = Cer
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