ENGINEERING ECONOMIC ENHANCED EBOOK
ENGINEERING ECONOMIC ENHANCED EBOOK
14th Edition
ISBN: 9780190931940
Author: NEWNAN
Publisher: OXF
Question
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Chapter 4, Problem 125P
To determine

i.

Whether given statement is true or not” Waiting for 5 years to start saving doubles what you deposit each year”.

Expert Solution
Check Mark

Answer to Problem 125P

The statement is false.

Explanation of Solution

Given:

Retirement period from now: 40 years

Amount required: $ 2 M

Interest Rate = 10%.

Calculation:

If savings are done for amount S immediately:

$2000000=S×FVIFA(10%,40)$2000000=S×443.5926S=$4518.83

If I start saving after 5 years later with annual saving being S, then:

$2000000=S×FVIFA(10%,35)$2000000=S×271.0244S=$7379.41.

Conclusion:

Since $7379.41 is much less than the double of $4518.33 (which is equal to $9036.66), the statement is false.

To determine

ii.

Whether required multiplier is higher or lower than 10% rate.

Expert Solution
Check Mark

Answer to Problem 125P

Required multiplier is higher.

Explanation of Solution

Given:

Interest Rate = 12%

Retirement period from now: 40 years

Amount required: $ 2 M.

Calculation:

When interest rate is 12%

Required multiplier (FVIFA factor) =

FVIFA(12%,40)=767.0914>442.5926(Higher).

Conclusion:

Required multiplier is higher.

To determine

iii.

Interest Rate.

Expert Solution
Check Mark

Answer to Problem 125P

Interest rate is 0% or 14.78%.

Explanation of Solution

Given:

Savings are done of double amount after 5 years

Retirement period from now: 40 years

Amount required: $ 2 M.

Calculation:

Let required interest rate per year be R%

$2000000FVIFA(R%,40)=2×$2000000FVIFA(R%,40)

FVIFA(R%,40)=2×FVIFA(R%,35)

[(1+R)401]R=2×[(1+R)351]R[(1+R)401]=2×[(1+R)351]

(1+R)401=2×[(1+R)352]2×[(1+R)35]=(1+R)40+1

If 1+R = Z, then,

2Z35=Z40+1Z402Z35+1=0

Solving this polynomial equation,

Z=1orZ=1.1478(theonlytworealroots)1+R=1or1+R=1.1478R=0orR=0.1478.

Conclusion:

R = 0% or R = 14.78%.

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Chapter 4 Solutions

ENGINEERING ECONOMIC ENHANCED EBOOK

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