An equipment used in producing goods undergoes a major overhaul now, its output can be increased by 20% production, which translates into additional cash flow of P20,000 at the end of each year for five years. If i=15% per year, how much can we afford to invest to overhaul this machine? Add your answer

ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN:9780190931919
Author:NEWNAN
Publisher:NEWNAN
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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Question
7. Please use the formula to solve the problem.
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
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
An equipment used in producing goods undergoes a
major overhaul now, its output can be increased by
20% production, which translates into additional cash
flow of P20,000 at the end of each year for five years.
If i=15% per year, how much can we afford to invest
to overhaul this machine?
Add your answer
Transcribed Image Text:An equipment used in producing goods undergoes a major overhaul now, its output can be increased by 20% production, which translates into additional cash flow of P20,000 at the end of each year for five years. If i=15% per year, how much can we afford to invest to overhaul this machine? Add your answer
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