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Problem 2 An investment is under consideration. If the total annual payments to the investment of $10,000/year is made unifor

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Answer #1

Total Investment after n payments have been made:

Bn=A(1+n)n+{(P/i)*[(1 + i)n – 1]}

Where Bn= balance at the end of period

A= Initial Payment at the beginning

P = Principal,

i = nominal annual interest rate in percentage terms, and

n = number of compounding periods.

1).   Payments are made at the end of each year with yearly interest compounding.

P= $10000

i = 8%

A= 0 ( as payments are made at the end of year 1)

n= 10

Bn= 0 + [10000/0.08]*[(1 + 0.08)10– 1]

= 125000*1.1589

= $ 144862.5

Total annual payments= $ 10000*10: = $100000

Accumulated Interest = Accumulated amount of interest & principal - total annual payments

= $ 144862.5 - $ 100000

= $ 44862.5

2).   Payments are made at the end of each week with weekly interest compounding.

P= $10000/52: P= $ 192.308

i = 8%/52 : i=0.0015

A= 0 ( as payments are made at the end of year 1)

n= 52 weeks* 10 years: n= 520

Bn= 0 + [192.308/0.0015]*[(1 + 0.0015)520– 1]

= 128205.33*1.1801

= $ 151295

Total annual payments= $ 192.308*52*10: = $100000

Accumulated Interest = Accumulated amount of interest & principal - total annual payments

= $ 151295 - $ 100000

= $ 51295

3). Payments are made Continously Throughout the year.

Formula= CF*( ert-1)/(er-1)

CF= Cash flows

R= rate

T= time

= 10000* ( e0.08*10-1)/( e0.08​​​​​​-1)

= 10000* ( 2.22554-1)/(1.08329-1)

= $ 147141.31

Total annual payments= $ 10000*10: = $100000

Accumulated Interest = Accumulated amount of interest & principal - total annual payments

= $ 147141.31 - $ 100000

= $ 47141

So, Weekly compounding interest pays the highest accumulated interest.

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