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Simple & Compound Interest

Present Value

Ordinary Annuities

Present Value Annuities

General Annuities & Equivalent Rates

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 UNIT 9  :  MATHEMATICS OF INVESTMENT

 LESSON 4: PRESENT VALUE ANNUITY HOMEWORK QUESTIONS CONT’D

 

Quick Review

Present Value of an  Annuity:

 

Definition 1: A sequence of payments made at regular intervals is called an annuity. 

Definition 2: When we calculate the present values of the sequence of payments made at regular intervals this is called the Present Value of the annuity. 

When a lump sum of money is deposited or borrowed today in order to receive a series of payments in the future, this is a PV annuity

A Present Value annuity has the following properties.

Text Box: ·	When a lump sum of money is deposited or borrowed today in order to receive a series of payments in the future, this is a PV annuity
·	The first payment is always at the end of the first interest period of the annuity.  This will be numbered 1 on your time line.
·	The accumulated sum of the present values of each payment at the end of the annuity’s term is called the present value of the annuity. This will be at the number 0 on your time line.
  
·	This accumulated sum forms a geometric series (see below).
·	A time line is very helpful in illustrating an annuity.
·	You can use either the geometric series formula OR the present value of an annuity formula (see below) to find the present value of an annuity.
 

 

 

 

 

 

 

 

 

 

 

 

 

Text Box: P = principal [amount borrowed                                
       or invested]
n = number of interest periods
i  = interest rate per interest period as  
      as a decimal
A = accumulated amount (due or payable)
Text Box: Formulas:
 

 


                                                                                                                       

 

 

 

 

 

 

Text Box: a = the regular payment of the annuity 
n = number of payments or terms
Text Box: Geometric Series Formula:
 

 

 

 

 

 

 

 

 

Text Box: R = the regular payment of the annuity
n  = the number of payments or terms 
i   = interest rate per interest period
PV = the Present Value of the annuity at the time of the last payment
Text Box: Present Value of an Annuity Formula:
 

 

 

 

 

 

 

 


Homework Questions: (Continued)

 

5.  The Witmer foundation wishes to establish an academic athletic scholarship to be awarded each year for 25 years.  The scholarship will be worth $1500 per year.  How much should be deposited now in a trust fund that pays 6.5%/a, compounded annually?

 

Solution:

 

Interest Period   0          1         2          3                                                                                                                                         23   24     25     

                                                                                                                                                                           

Payment                         1500   1500                                                                                                                                              1500  1500 1500

                                                           

1500(1.065)-1                                                                                                                                                                                                       

1500(1.065)-2                                                                                                                                                                                                                                                                                                                                                       

       .

       .

1500(1.065)-23     

                                                                                                                                                                                   

1500(1.065)-24                                                                                                                                                                              

 

                                                                                                                                                                             

1500(1.065)-25

 

 

 

 

 

Hence $18 296.82 should be deposited now to provide for this scholarship.

 

6.  Mr. I. M. Generus donated $100 000 to minor hockey in his home town.  It is to be paid out over a 10 year period starting one year from now.  How much will be paid out each year if interest is 5.4%/a, compounded annually?

 

Solution:

 

Interest Period   0          1         2          3                                                                                                                                           8      9    10     

                                                                                                                                                                           

Payment                         R          R        R                                                                                                                                          R      R      R

                                                           

        R(1.054)-1                                                                                                                                                                                                     

        R(1.054)-2                                                                                                                                                                                                                                                                                                                                                     

       .

       .

       R(1.054)-8     

                                                                                                                                                                                   

       R(1.054)-9                                                                                                                                                                             

 

                                                                                                                                                                             

      R(1.054)-10

 

This forms the following geometric series:

            R(1.054)-10 + R(1.054)-9 + . . . + R(1.054)-2 + R(1.054)-1        

 

Hence  $13 203.22 will be available each year to minor hockey.

 

7.  Betty won $2 000 000 in a recent lottery.  If she uses the funds to purchase an annuity over 35 years, what monthly payment will she receive if interest is 6%/a, compounded monthly?

 

Solution:

 

Interest Period   0          1         2          3                                                                                                                                         418   419 420     

                                                                                                                                                                           

Payment                         R          R        R                                                                                                                                          R      R      R

                                                           

        R(1.005)-1                                                                                                                                                                                                     

        R(1.005)-2                                                                                                                                                                                                                                                                                                                                                     

       .

       .

       R(1.005)-418     

                                                                                                                                                                                   

       R(1.005)-419                                                                                                                                                                           

 

                                                                                                                                                                             

      R(1.005)-420

 

This forms the following geometric series:

            R(1.007)-420 + R(1.007)-419 + . . . + R(1.007)-2 + R(1.007)-1    

 

Hence Betty will receive a monthly income of $11 403.79 for 35 years from her lottery winnings.

 

8.  Mrs. Peres purchased a car for $19 900 including all taxes.  She wishes to finance the purchase over 5 years.  If interest is 9.6%/a, compounded monthly, what will her monthly payment be?

 

Solution:

 

Interest Period   0          1         2          3                                                                                                                                         58    59    60     

                                                                                                                                                                           

Payment                         R          R        R                                                                                                                                          R     R     R

                                                           

        R(1.008)-1                                                                                                                                                                                                     

        R(1.008)-2                                                                                                                                                                                                                                                                                                                                                     

       .

       .

       R(1.008)-58     

                                                                                                                                                                                   

       R(1.008)-59                                                                                                                                                                            

 

                                                                                                                                                                             

      R(1.008)-60

 

This forms the following geometric series:

            R(1.008)-60 + R(1.008)-59 + . . . + R(1.008)-2 + R(1.008)-1       

 

Hence the monthly payments will be $418.91.

 

9.  Find the purchase price of an annuity that pays $4000 every 6 months for 15 years if interest is 6.6%/a, compounded semi-annually.

 

Solution:

 

Interest Period   0          1         2          3                                                                                                                                         23   24    25     

                                                                                                                                                                           

Payment( 1000’s)           50       50                                                                                                                                                   50    50    50

                                                           

50000(1.076)-1                                                                                                                                                                                                     

50000(1.076)-2                                                                                                                                                                                                                                                                                                                                                     

       .

       .

50000(1.076)-23     

                                                                                                                                                                                   

50000(1.076)-24                                                                                                                                                                            

 

                                                                                                                                                                             

50000(1.076)-25

 

 

 

 

10.  Mr. Cameron purchased a new tractor for his farm for $80 000.  He paid $5000 down and financed the rest over 10 years at 10.2%/a, compounded monthly. Determine his monthly payment and the finance charge.

 

Solution:

 

Interest Period   0          1         2          3                                                                                                                                         118  119  120     

                                                                                                                                                                           

Payment                          R         R        R                                                                                                                                          R      R     R

                                                           

        R(1.0085)-1                                                                                                                                                                                                   

        R(1.0085)-2                                                                                                                                                                                                                                                                                                                                                   

       .

       .

       R(1.0085)-118     

                                                                                                                                                                                   

       R(1.0085)-119                                                                                                                                                                          

 

                                                                                                                                                                             

      R(1.0085)-120

 

 

This forms the following geometric series:

            R(1.0085)-120 + R(1.0085)-119 + . . . + R(1.0085)-2 + R(1.0085)-1        

                       

 

 

 

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