Mathematics of Finance The formulae
1. A = P(1 + i.n) – simple interest
2. A = P(1 + i)n – compound interest formula 3. A = P(1-i.n) – depreciation straight line
4. A = P(1 – i)n – compound decrease – dimishing balance
5. P = - present value of annuity 6. A = - future value of annuity
7. A = - future value of annuity – one instalment does not earn interest Use for sinking fund.
8. Bn = - m is remaining period or payments – all payments equal.
9. P = - d is deferred period A. Calculation Problems
1. Use the formula A = P(1 + i.n) to find 1.1 A if P = R5000 ; i = 10% pa; n= 6 years 1.2 P if A = R10000 ; i= 9% pa; n = 5 years 1.3 i if A = R20000; P = R10000 ; n = 10years 1.4 n if A = R15000; P = R10000; i = 7 years 2. Use the formula A = P(1 + i)n to find
2.1 A if P = R5000 ; i = 10% pa compounded monthly; n= 6 years 2.2 P if A = R10000 ; i= 9% pa compounded quarterly; n = 5 years
2.3 i if A = R20000; P = R10000 ; n = 10years; interest compounded monthly 2.4 n if A = R15000; P = R10000; i = 7 years; interest compounded quarterly 2.5 A if P = R60000 ; i= 12%pa compounded daily; n = 5 years
3. Use the formula A = to find
3.1 A if x = R3500; i = 10% pa compounded monthly; n = 10 years. 3.2 x if A = R500000; i = 13% pa compounded monthly; n = 8 years
3.3 n if A = R600000; i = 12%; x = R10000 ; interest compounded monthly.
4. Use the formula A = to find
4.1 A if x = R3500; i = 10% pa compounded monthly; n = 10 years. 4.2 x if A = R500000; i = 13% pa compounded monthly; n = 8 years
4.3 n if A = R600000; i = 12%; x = R10000 ; interest compounded monthly.
5. Use the formula P =
5.1 P if x = R2000; i = 8% pa compounded monthly and n = 10 years 5.2 P if x = R50000; i = 10% pa compounded annually; n = 6 years 5.3 x if P = R2 200000; i = 9% pa compounded monthly; n = 20 years 5.4 n if P = R1000000; i = 8,3% pa compounded monthly; n = 18 years.
6. Determine the effective rate of interest for the year if the nominal rate of interest is 6.1 15% pa compounded monthly 6.2 21% pa compounded monthly 6.3 12% pa compounded daily 6.4 16% pa compounded daily 6.5 24% pa compounded quarterly
7. Determine the nominal rate of interest pa if the effective rate of interest is 7.1 14% pa; compounded monthly
7.2 15,4 % pa ; compounded monthly 7.3 16,8% pa compounded daily 7.4 19,2% pa compounded quarterly
8. Determine the effective rate of interest for 6 months if
8.1 the nominal rate of interest is 13,6% pa compounded monthly 8.2 the nominal rate of interest is 11,4% pa compounded monthly.
B. Problems on Mathematics of Finance
1. Mrs Khumalo wanted to invest a certain amount of money to ensure that her two children , ages 3 years and 5 years and whose birthday fall on the same day, are guaranteed a sum of R300000 each when they each reach 19. She was given an interest rate of 9,6% compounded monthly for the given period. Determine the amount she must invest now in order to achieve her objective. 2. Mr Zee saved a certain amount so that on his 40th birthday he would receive R200000 and on his 50th birthday he would receive R600000. He is 24 years old when he did his investment at a rate of interest of 15% pa compounded monthly. What must he invest now?
3. Present value of Annuity Problems
3.1 Mr Y took out a loan for a certain amount. He paid this loan off in 10 equal monthly instalments of R1000 each at an interest rate of 12% pa compounded annually. Determine the amount of the loan taken if the first payment starts at the end of the first year.
First Principles: - P = 1 000 (1,12)-1 + 1 000 (1,12)-2 + 1 000 (1,12)-3 + … + 1 000 (1,12)-10 P = 1 000 (1,12)-1 [ 1 + (1,12)-1 + (1,12)-2 +…+ (1,12)-10] 1 + (1,12)-1 + (1,12)-2 + … + (1,12)- 9 = a[rn – 1] = [(1,12) –1] 10 – 1 r – 1 (1,12) -1 – 1 P = 1000 (1,12)-1 [(1,12)-10 – 1] (1,12) -1 – 1 = 1000 [(1,12)-10 – 1] 1,12[(1,12) -1 – 1] = 1000 [(1,12)-10 – 1] 1 – 1,12 = 1000 [1 – (1,12)-10 ] 0,12 = R5 650,22 P = x [1 – (1 + i)-n ] i
3.2 A homeowner takes a loan of R400 000. The interest is 9,8% p.a.
compounded monthly. The loan is to be repaid monthly in 240 instalments with the first payment starts at the end of the first month.
3.2.1 Determine the monthly instalment. 3.2.2 Determine total amount paid. 3.2.3 Calculate the actual interest.
3.2.4 If inflation is calculated at a rate of 9% p.a. compounded annually determine the new value of the house after 10 years.
3.2.1 Discussion
Basically all loans are deferred payments. No one takes out a loan and pay an amount immediately. Suppose the borrower wants to pay R5 000 immediately then that person should take out a loan for R395 000. The formula is for one deferred payment. P= x [1 – (1 + i)-n ] i 400 000 = x [1 – (1 + 9,8 ÷1200)-240 ] 9,8÷ 1200
x = R3807,24 (instalment is actually R3808, not less) 3.2.2 A = 3808 x 240 = R913920 3.2.3 i = x . n – P = 3808 x 240 – 400 000 = R513 920 3.2.4 New Value = 400000(1 + 9÷100)10 = R946 946
3.3 A homeowner takes a loan of R950 000. The interest is 8,9% p.a.
compounded monthly. The loan is to be repaid monthly in 240 instalments with the first payment starts at the end of the first month.
3.3.1 Determine the monthly instalment. 3.3.2 Determine total amount paid. 3.3.3 Calculate the actual interest.
3.3.4 If inflation is calculated at a rate of 8% p.a. compounded annually determine the new value of the house in 20 years time.
3.4 A small business enterprise decided to take a loan for R200 000 at a low interest of 6% p.a. compounded monthly for 5 years.
3.4.1 Determine the monthly instalment.
3.4.2 Determine the total interest paid on this loan.
3.4.3 Determine the new monthly instalment if the first payment is deferred for 6 months and the total interest paid.
Discussion
Let us consider a simpler problem:-
A loan of R100 000 was to be repaid in yearly instalments at a rate of 10% p.a compounded annually.
Two cases:-
(a) Payment to commence 1 year after the loan was awarded. (b) Payment to commence 3 years after the loan is awarded. Determine the instalment in each case.
(a) Standard formula:- P= x [1 – (1 + i)-n ] i 100 000 = x [1 – (1 + 10 ÷100)-10 ] 10÷ 100 x = R16 275
(b) Let us go back to first principles
P = x (1,10)-3 + x (1,10)-4+ x (1,10)-5 + … + x (1,10)-12 = x(1,10)-2. (1,10)-1 [1 +(1,12)-1 + (1,12)-2 +…+ (1,12)-10] The only difference is the additional factor of (1,10)-2 P = (1,1)-2.x [1-(1+i)-10]
i x = R19693 (c) New formula:
P = (1+i)-d.x [1-(1+i)-n] ‘d’is deferred period = deferred per. -1 i
3.5 A homeowner takes a home loan of R750 000. This amount is to be repaid in monthly instalments over 20 years at a rate of 9.6% p.a. compounded monthly. Calculate the monthly instalments if the first payment is made at the end of the 6 months.
3.6 A shopkeeper takes a loan and pays it off in 12 equal yearly instalments of R30 000 starting at the end of the 4th year at an interest rate of 10% p.a. compounded annually. How much can she borrow?
Hint: Remember one postponed payment (standard) is (1 + i)-1. In this case
we start with (1 + i)-4. P = 30 000 (1 + i)-4 + 30 000 (1 + i)-5 + … + 30 000 (1 + i)-15
3.7 A company takes out a loan for R3 000 000 at an interest rate of 7,5% pa compounded monthly for 10 years.
3.7.1 Determine the monthly instalment.
3.7.2 How long will it take to pay of the loan if they double the instalment ? 3.8 Jojo invests R900000 with a bank that agreed to give him a return of 18% pa
compounded monthly. How much will Jojo receive every month for 10 years ? 3.9 Patience invested R600000 with a bank that agreed to give her 15% pa
compounded monthly on this investment.
What will her balance be after 5 years if she is given a monthly annuity of R6000 ?
4. Future Value of Annuity
4.1 Shirley invested R5000 per month starting on 1 May 2000 earning interest at a rate of 17% pa compounded monthly. What can Shirley expect on 30 April 2012?
4.2 Trevor invested R20 000 per annum starting on 1 July 2003 earning interest at a rate of 15,8%pa compounded monthly. What is the total amount that Trevor received on 30 June 2011?
4.3 Dudu wanted to become a millionaire in 8 years time. The bank was very happy with her commitment to save and offered her a whopping 20% pa return compounded monthly. She invested her money at the end of each month starting on 31 August 2006. At what date can she expect to be a millionaire? What must her monthly investment be to achieve this objective? 4.4 Randy took out an insurance policy which was fixed at R500 per month. The
guaranteed interest rate was 12% pa compounded annually. What will Randy’s policy be worth in 30 years time?
5. General Problems
5.1 A loan of R750 000 was taken by a homeowner. Interest was to be paid at 9.5%p.a. compounded monthly. The repayment period is 20 years.
5.1 Calculate the monthly instalment .
5.2 Calculate the balance on the loan after (a) 5 years (b) 10 years
5,2 ABC Incorporated invested a certain amount in a bank earning interest at a
rate of 12,6% p.a compounded monthly. They intend withdrawing R9 000 per month for 10 years.
How much must they invest now to ensure that this withdrawal can be made ? 5.3 Mrs Wise invested a certain amount in a bank earning interest at a rate of 15% p.a compounded monthly. She intend withdrawing R15 000 per month for 10 years.
How much must they invest now to ensure that this withdrawal can be made ? 5.4 A father decided to buy a house for his family for R800000. He agreed to pay
monthly instalments of R10000 on a loan which incurred interest at a rate of 14% pa compounded monthly. The first payment was made at the end of the first month.
5.4.1 Show that the loan would be paid off in 234 months.
5.4.2 Suppose the father encountered unexpected problems and was unable to pay any instalments at the end of the 120th; 121st; 122nd; 123rd months. At the end of the 124th month he increased his payment so as to still pay off the loan in 234 months by 111 equal monthly payments. Calculate the value of this new instalment.
5.5 A company purchased a building for R5000000. They took out a loan for this amount at an interest rate of 6% per annum compounded monthly.
They paid a fixed amount of R40000 per month. 5.5.1 How many payments must be made? 5.5.2 Determine the final payment.
5.5.2 What will the balance be after 36 months?
5.6 Sinking Fund
5.6.1 A company purchased a new vehicle for R250 000. The company wished to replace the vehicle in 5 years time. The rate of depreciation is 10% p.a on diminishing balance. It is envisaged that a new vehicle will appreciate in value at a rate of 11% p.a.
Calculate
(a) the residual value of the vehicle in 5 years time. (b) the price of the new vehicle in 5 years time.
(c) the value that the sinking fund must attain in 5 years time.
(d) the monthly payments into the sinking fund ( payment commenced 1 month after the sinking fund is set up) at a rate of 13,6%p.a compounded monthly.
5.6.2. A borehole machine presently cost R325 000. It depreciates in value at a rate of
12 % on a reducing balance. A new borehole machine will be required in 5 years time. The cost of the borehole machine is expected to grow at a rate of 10% p.a. compound interest. Determine
(a) scrap value of old borehole machine. (b) cost of new borehole machine
(c) and hence the value that the sinking fund must attain.
(d) equal monthly instalments which will start in 6 months time and finish in 5 years time. Interest is at 11,2% p.a. compounded monthly.
5.6.3 A car presently cost R150 000. It depreciates at a rate of 10% p.a. on a reducing balance. A new car will be needed in 6 years time. The cost of the new car is expected to grow by 8% p.a. compounded annually. Determine:-
(a) the scrap value of the car to the nearest R. (b) the cost of the new car to the nearest R.
(c) hence the value that the sinking fund must attain.
(d) the equal monthly instalments which will start in 6 months time and finish in 5 years time.
Interest is calculated at 14% p.a. compounded monthly
6. Suppose an amount of R3000 is withdrawn every 6 months starting in one year’s time and finish 6 months before the purchase.
HIRE PURCHASE
1. Mr X purchase a car to the value of R168 000. He takes out a loan for this amount.
He repays this loan at a rate of 12% p.a. compounded monthly over 48 months.
Calculate his monthly instalments.
Rather than take out a bank loan the purchaser signed a H.P
agreement because the H.P charges were 9%p.a. simple interest. Find 1.1 the new instalments.
1.2 nominal rate of interest 1.3 effective rate of interest
2. Appliances at home cost a new homeowner R90 000. The owner pays a deposit of 20% and pays the balance over 36 months in equal
monthly instalments.
2.1 Find the instalments if the borrower borrows the outstanding amount and agrees to pay interest at a rate of 11% p.a. compounded monthly .
2.2 Suppose the homeowner borrows the outstanding balance and agrees to pay the H.P at a rate of 10% p.a. simple interest. Find
2.2.1 new instalments.
2.2.2 nominal rate of interest. 2.2.3 effective rate of interest.
3. Student loans are being offered by a leading banking institution to tertiary students. It offers loans of up to R80000. The interest is fixed at 10% of the amount.
A further condition is that this loan be paid over 10 months. Student A takes a loan of R30 000.
Student B takes a loan of R80 000.
3.1 Determine the monthly repayment for student A and B. 3.2 Calculate the nominal and effective rate of interest p.a.
3.3 Suppose a parent was offered these loans and paid at 12% p.a. compounded monthly over one year. Calculate the monthly instalment and the actual interest paid on these loans. Comment on these two options and how would you advise.
