Time value of Money(Financial Management)

Q. The following are exercise in future (terminal) values:

a)      At the end of three years, how much is an initial deposit of $100 worth, assuming a compound annual interest rate of (i) 100 percent? (ii) 0 percent?

b)      At the end of five years, how much is an initial $500 deposit followed by five year-end, annual $100 payments worth, assuming a compound annual interest rate of (i) 10 percent? (ii) 5 percent? (iii) 0 percent?

c)      At the end of six years, how much is an initial $500 deposit followed by five year-end, annual $100 payments worth, assuming a compound annual interest rate of (i) 10 percent? (ii) 5 percent? (iii) 0 percent?

d)     At the end of three years, how much is an initial $100 deposit worth, assuming a quarterly compounded annual interest rate of  (i) 100 percent? (ii) 10 percent?

e)      Why your answer to part (d) is differ from those to part (a)?

f)       At the end of 10 years, how much is an initial $100 deposit worth, assuming an annual interest rate of 10 percent compounded (i) annually? (ii) Semiannually? (iii) Continuously?

Solution:

   a) FV_n= P₀(1 + i)

(i) FV₃= $100(2.0)³ =$100(8) = $800

(ii)    FV₃= $100(1.10)³= $100(1.331)= $133.10

(iii)   FV₃= $100(1.0)³=   $100(1)   = $100    

   

   b) FV_n = P₀(1 + i)ⁿ; FVA_n = R[([1 + i]n - 1)/i]

(i)    FV₅ = $500(1.10)⁵ = $500(1.611) = $  805.50

                 FVA₅ = $100[([1.10]⁵ - 1)/(.10)]

                  = $100(6.105) =                                     610.50

                                                                          $1,416.00

(ii)    FV₅ = $500(1.05)⁵ = $500(1.276) = $  638.00

                 FVA₅ = $100[([1.05]⁵ - 1)/(.05)]

                 = $100(5.526) =                                       552.60

                                                                           $1,190.60

(iii)   FV₅ = $500(1.0)⁵ = $500(1) =         $  500.00

                 FVA₅ = $100(5)=                                     500.00

                                                                          $ 1,000.00

   c) FV_n= P0(1 + i)ⁿ; FVAD_n= R[([1 + i]n- 1)/i][1 + i]

(i)    FV₆ = $500(1.10)⁶ = $500(1.772) = $  886.50

                 FVA₅ = $100[([1.10]⁵ - 1)/(.10)]

                  = $100(6.105) (1.10) =                            671.55

                                                                          $1,557.55

(ii)    FV₆ = $500(1.05)⁶ = $500(1.340) = $  670.00

                 FVA₅ = $100[([1.05]⁵ - 1)/(.05)]

                 = $100(5.526) (1.10) =                             580.23

                                                                          $1,250.23

(iii)   FV₆ = $500(1.0)⁶ = $500(1) =         $  500.00

                 FVA₅ = $100(5) =                                    500.00

                                                                          $ 1,000.00

 d)   FV_n = PV₀(1 + [i/m])

(i)          FV3= $100(1 + [1/4])₁₂ = $100(14.552)  = $1,455.20

(ii)         FV3= $100(1 + [.10/4])₁₂= $100(1.345) = $  134.50

e)

The more times a year interest is paid, the greater the future 

value.  It is particularly important when the interest rate is

high, as evidenced by the difference in solutions between

Parts 1.a) (i) and 1.d) (i).

f)   FV_n= PV₀(1 + [i/m])^mn; FV_n = PV0(e)ⁱⁿ

         

       (i)     $100(1 + [.10/1])¹⁰ = $100(2.594) = $259.40

       (ii)    $100(1 + [.10/2])²⁰ = $100(2.653) = $265.30

(iii)        $100(1 + [.10/4])⁴⁰= $100(2.685) = $268.50

   (iv)  $100(2.71828)1 = $271.83