Business Mathematics Lecture Note #9 Chapter 5

1 Business Mathematics Lecture Note #9 Chapter 5

Financial Mathematics 1. Arithmetic and Geometric Sequences and Series 2. Simple Interest, Compound Interest and Annual Percentage Rates 3. Depreciation 4. NPV and IRR 5. Annuities, Debt Repayments, Sinking Funds 6. Interest Rates and Price of Bonds

Arithmetic and Geometric Sequences and Series Sequence : A list of numbers which follow a definite pattern or rule. 1. Arithmetic Sequence : Each term, after the first, is obtained by adding a constant, d, to the previous term, where d is called the common difference 2. Geometric Sequence : Each term, after the first, is obtained by multiplying the previous term by a constant, r, where r is called the common ratio

Arithmetic and Geometric Sequences and Series Series : The sum of the terms of a sequence. Finite series: the sum of a finite number of terms of a sequence Infinite series: the sum of an infinite umber of terms of a sequence 1. Arithmetic Series (Arithmetic Progression, AP) : the sum of the terms of an arithmetic sequence. 2. Geometric Series (Geometric Progression, GP) : the sum of the terms of a geometric sequence.

Arithmetic Sequences and Series

Geometric Sequences and Series

Application of Arithmetic and Geometric Series A manufacturer produces 1,200 computers in the first week. But after week 1, it increases production by: i) scheme I: 80 computers each week Ii) scheme II: 5% each week. (a) Find out the production quantity in week 20 under each scheme. (b) Find out the total production quantity over the first 20 weeks under each scheme. (c) Find the week in which the production quantity reaches 8,000 or more for the first time under each scheme.

Application of Arithmetic and Geometric Series

Assignment #2 Problems 1, 9, 10, 11 of Progress Exercises 5.1 Due on 2013/05/02 (Thursday)

Simple Interest

Compound Interest

How compounding is carried out (when annual interest rate i %) The next slide demonstrates.how interest is calculated at the end of each year.interest earned is added to the principal.principal at the start of next year = (principal + interest) from previous year 21

The table below will be filled in, row by row..to demonstrate the idea of compounding annually at an interest rate i % Amount at start of year = principal Interest earned during year Amount at end of Year = principal + interest Year 1 P 0 ip 0 P 0 + ip 0 = P 0 (1+ i) = P 1 Year 2 P 1 ip 1 P 1 + ip 1 = P 1 (1+ i) = P 2 Year 3 P 2 ip 2 P 2 + ip 2 = P 2 (1+ i) = P 3 In general, at the end of year t. Year t P t-1 ip t-1 P t-1 + ip t-1 = P t-1 (1+ i) = P t 22

Compound interest formula (II) principal + interest P 0 + ip 0 = P 0 (1+ i) = P 1 Next year P 1 + ip 1 = P 1 (1+ i) = P 2 Next year P 2 + ip 2 = P 2 (1+ i) = P 3 In general. BUT, in terms of P 0 so P 1 = P 0 (1+ i) But P 1 (1+ i) = P 0 (1+ i) (1+ i) so P 2 = P 0 (1+ i) 2 But P 2 (1+ i) = P 0 (1+ i) 2 (1+ i) so P 3 = P 0 (1+ i) 3 P 4 = P 0 (1+ i) 4..and so on.p t = P 0 (1+ i) t 23

Worked Example 5.5 (see text) Calculate the amount owed on a loan of 1000 at the end of three years, interest compounded annually, rate of 8% you will need.. P t = P 0 (1+ i) t..the compound interest formula Method Substitute the values given into the compound interest formula t = 3 years P 0 = 1000 8 i = = 0.08 100 Calculations P 3 3 P0 ( 1i) 3 1000(1 0.08) 3 1000(1.08) 1000(1.2597120) 1259.712 24

Terminilogy: present value; future value In the compound interest formula; P t = P 0 (1+ i) t P t is called the future value of P 0 at the end of t years when interest at i% is compounded annually P 0 is called the present value of P t when discounted at i% annually see following examples 25

The present value formula is deduced from the compound interest formula as follows: P P (1 i) t 0 t P (1 t i) t P 0 (1 (1 i) i) t t (1 P t i) t P 0 P 0 (1 P t i) t 26

Worked Example 5.6 (a)(i) 5000 is invested at an interest rate of 8% for three years You will need P t = P 0 (1+ i) t..the compound interest formula Method Substitute the values given into the compound interest formula t = 3 years P 0 = 5000 8 i = = 0.08 100 Calculations P 3 3 P0 ( 1i) 3 5000(1 0.08) 3 5000(1.08) 5000(1.2597120) 6298.5 27

Revise terminilogy: present value; future value In the compound interest formula; P t = P 0 (1+ i) t future value present value In Worked Example 5.6 P t = 6298.5 is called the future value of P 0 = 5000 at the end of 3 years when invested at 8% compounded annually P 0 = 5000 is called the present value of P t = 6298.5 when discounted at 8% annually for 3 years 28

Worked Example 5.6(b)(i) Present value calculations ( 6298.5 discounted at 8% annually for three years) P 0..the present value formula will be required Method Substitute the values given into the present value formula t = 3 years P t = 6298.5 8 i = = 0.08 100 P (1 t i) t Calculations P 0 P 3 ( 1 i) 3 6298.5 3 (1 0.08) 6298.5 3 (1.08) 5000 29

Worked Example 5.6 (b)(ii) Present value calculations ( 15,000 discounted at 8% annually for three years) P 0..the present value formula will be required Method Substitute the values given into the present value formula t = 3 years P t = 15,000 8 i = = 0.08 100 P (1 t i) t Calculations P 0 P 3 ( 1 i) 3 15000 3 (1 0.08) 15000 3 (1.08) 11907.48 30

How to compound twice annually (rate = i % pa) P P P (1 i) t t P 0 0 i 1 2 At each compoumding t 2t..compounding once annually..compounding twice annually Two compoundings necessary in 1 year use the annual rate, i, divided by 2 2 x t compoundings necessary in t years 31

How to compound three times annually (rate = i% pa) P P P (1 i) t t P 0 0 i 1 3 t 3t..compounding once annually..compounding three times annually At each compoumding use the annual, I, rate divided by 3 Three compoundings necessary in 1 year 3 x t compoundings necessary in t years 32

How to compound m times annually (rate = i% pa) P t P P (1 i) t 0 P0 1 i m t mt..compounding once annually..compounding m times annually At each compoumding use the annual rate,i, divided by m m compoundings necessary in 1 year m x t compoundings necessary in t years 33

Compounding continuously P P (1 i) P t t P t 0 P0 1 t i m P 1 mt i m m compounding once annually compounding m times annually t 0 rearranging i e t Pt P0 P0 e it Pt P0 e it 1 i m m e i as m 34

Worked Example 5.8 (a). 5000 is invested at an interest rate of 8% for three years compounded semiannually you will need the formula.. P t P 1 mt Method Substitute the values given in the question into the compound interest formula above m = 2 t = 3 years P 0 = 5000 8 i = = 0.08 100 i Calculations 0 m3 m i P3 P0 1 m 23 0.08 P 3 50001 2 6 5000(1 0.04) 6 5000(1.04) 5000(1.2653190) 6326.595 35

Worked Example 5.8 (c)(i). 5000 is invested at an interest rate of 8% for three years compounded monthly you will need the formula.. mt P t P 1 i m Method Substitute the values given into the compound interest formula above m = 12 t = 3 years P 0 = 5000 8 i = = 0.08 100 Calculations 0 m i 3 P3 P0 1 P 3 m 50001 5000(1.270237) 6351.185 0.08 12 123 36

Worked Example 5.8 (c)(ii) 5000 is invested at an interest rate of 8% for three years compounded daily (assume 365 days per year) you will need the formula... P t Method Substitute the values given into the compound interest formula above m = 365 t = 3 years P 0 = 5000 P0 1 8 i = = 0.08 100 i m mt Calculations P P 3 3 P0 1 1095 5000(1.0002192) 5000(1.2712157) 6356.079 i m 50001 m3 0.08 365 3653 37

Worked Example 5.9 5000 is invested at an interest rate of 8% for three years compounded continuously you will need the formula... it Pt P0 e Method Substitute the values given into the compound interest formula above t = 3 years P 0 = 5000 8 i = = 0.08 100 Calculations P 3 5000e P e 0 i3 P3 5000e 0.24 5000(1.2712492) 6356.246 0.083 38

How much do you gain when interest is compounded more than once annually? Review results in Worked Examples 5.6, 5.7 and 5.9 5000 is invested at a nominal interest rate of 8% for three years but compounded at various intervals annually. The future value at the end of 3 years was calculated: 6298.560 compounded once annually 6326.595 compounded twice annually 6351.185 compounded monthly 6356.079 compounded daily 6356.246 compounded continuously 39

How much do you gain when interest is compounded more than once annually? Review results in Worked Examples 5.6, 5.7 and 5.9 5000 is invested at a nominal interest rate of 8% for three years but compounded at various intervals annually 6298.560 one conversion period 6326.595 2 conversion periods 6351.185 12 conversion periods 6356.079 365 conversion periods 6356.246 infinete conversion periods (continuous) 40

How much do you gain by compounding more than once annually? Conversion periods/year Amount at end of 3 years 1 6298.560 Difference over annual compounding 2 6326.595 6326.595-6298.560 = 28.035 12 6351.185 6351.185-6298.560 = 52.625 365 6356.079 6356.079-6298.560 = 57.519 Infinitely many (continuous) 6356.246 6356.246-6298.560 = 57.686 41

How do we make comparisons when different conversions periods are used? Use Annual Percentage Rates: APR What is the APR? The APR is the interest rate, compounded annually that yeilds an amount P t the same amount P t would be yeilded when any other method of compounding is used, for example.. 42

Annual Percentage Rates: APR P t calculated using the APR rate annually is the same as P t calculated by any other method mt P t P0 1 P t P 0 e it i m P P (1 APR) t 0 t 43

Calculate the APR when interest is compounded m times annually P t P0 1 i m mt compounding m times annually at a nominal rate of i % p.a. P P (1 APR) t 0 t compounding once annually at APR% p.a. But P t is the same whcihever method is used, hence P 1 t 0 ( 1 APR ) P0 i m mt Next slide 44

Calculate the APR when interest is compounded m times annually But P t is the same whcihever method is used, hence P 1 t 0 ( 1 APR ) P0 i m mt t i ( 1 APR ) 1 m m (1 APR ) 1 APR i m m mt i 1 1 m 45

Calculate the APR when interest is compounded continuously But P t is the same whcihever method is used, hence t P0 ( 1 APR) 0 P e it (1 APR) e t it i (1 APR) e APR e i 1 46

Calculate the APR: Progress Exercises 5.4 no 11(a) P t is the same whcihever method is used, hence 3 0.06 5500(1 APR ) 55001 2 23 3 0.06 (1 APR ) 1 2 2 0.06 ( 1 APR ) 1 2 23 APR 2 1 0.03 1 0. 0609 47

Calculate the APR Progress Exercises 5.4 no 11(d) But P t is the same whcihever method is used, hence P 3 0 (1 APR) P 0 e 0.063 (1 APR) 3 e 0.063 ( 1 APR) e 0.06 APR e 0.06 1 0.06184 48

Assignment #3 Problems 4 and 6 of Progress Exercises 5.2 Problems 5, 6, 15 of Progress Exercises 5.3 Problems 5, 6 of Progress Exercises 5.4 Due on 2013/05/07 (Tuesday)

Depreciation Depreciation: allowance made for the wear and tear of equipment during the production process which involves reduction of the asset value. There are two depreciation methods: 1. Straight-line depreciation subtracts equal amount from the original asset value each year. This is the converse of simple interest. 2. Reducing-balance depreciation subtracts equal rate from the asset value of the previous year. This is the converse of compound interest.

Depreciation

Worked Example 5.11

Worked Example 5.12

NPV and IRR NPV and IRR are the two techniques which are used to appraise investment projects (or investment alternatives). More specifically, NPV and IRR are used to determine whether to invest in a certain investment project, or are used to select one or a few among many investment alternatives. NPV(Net Present Value) IRR(Internal Rate of Return)

Net Present Value(NPV) NPV is the sum of the present values of several future cash flows discounted at a given rate. NPV uses present values to appraise the profitability of investment projects. While calculating NPV, a given discount rate (i) is used to convert all future cash flows into present values. Each Cash flow is either cash inflow or cash outflow. Cash inflow is a return from the investment. Cash outflow is a cost or money to be invested.

Net Present Value(NPV)

Calculating NPV Year (t) Cash flow 0-400,000 1-400,000 1 120,000 0.9259 111,111 2 130,000 0.8573 111,454 3 140,000 0.7938 111,137 4 150,000 0.7350 110,254 43,956

Internal Rate of Return(IRR) In the previous example of calculating NPV, discount rate of 8% was used. As a result, NPV=$43,956 was obtained. The value of NPV, however, changes as the discount rate changes. If the discount rate increases, the NPV decreases. If the discount rate is increased to 12.6555%, the NPV becomes zero. If the discount rate is further increased to 15%, the NPV would result in a negative value. (See Table 5.4 on page 232) IRR is the discount rate at which the NPV is zero. For the previous example, the IRR is 12.6555%

Internal Rate of Return(IRR) Decision rule for using IRR: Invest in the project, if IRR > market rate of interest Do not invest in the project, if IRR < market rate of interest

(1) Graphical method Calculating IRR Calculate NPV s for several different discount rates so that NPV s range from positive to negative values. Then, plot the points of (discount rate, NPV) on a graph where the horizontal axis represents discount rate and the vertical axis represents value of NPV. Then connect the points to get a curve. The value of the discount rate of the point at which the curve crosses the horizontal axis is the IRR.

Calculating IRR

Comparison of NPV and IRR When comparing the profitability of two or more projects, the most profitable project would be the project with the largest NPV which would be the project with the largest IRR. Advantage of using NPV: results are given in cash terms Disadvantage of using NPV: results change when discount rate is changed Advantage of using IRR: results are independent of any external rates of interest Disadvantage of using IRR: does not differentiate between the scale of projects. Higher IRR with smaller NPV due to small scale of the project.

Comparison of NPV and IRR Project I (discount rate = 10%) Project II (discount rate = 10%) year Cash flow Discount factor PV year Cash flow Discount factor 0-100,000 1-100,000 0-10 1-10 1 120,000 109,091 1 50 45 NPV of Project I = 9,091 NPV of Project II = 35 PV

Compound Interest for Fixed Deposits at Regular Intervals of Time

Compound Interest for Fixed Deposits at Regular Intervals of Time Worked Example 5.15 New members of a golf club are admitted at the start of each year and pay a joining fee of $2,000. Henceforth members pay the annual fee of $400, which is due at the end of each year. How much does the club earn from a new member over the first 10 years, assuming annual compounding at an annual interest rate of 5.5%.

Compound Interest for Fixed Deposits at Regular Intervals of Time

Annuities

Worked Example 5.16 Annuities To provide for future education, a family considers various methods of saving. Assume saving will continue for a period of 10 years at an interest rate of 7.5% per annum. (a) Calculate the value of the fund at the end of year 10 when equal deposit of $2,000 is made at the end of each year. (b) How much should be deposited each year if the final value of the fund is $40,000? (c) How much should be deposited each month if the final value of the fund is $40,000?

Annuities

Debt Repayment We say a loan is amortized if both principal and interest are to be paid back by a series of equal payments made at equal intervals of time assuming a fixed rate of interest throughout. Mortgage is an amortized loan used to purchase a real estate (house or building) by offering the real estate to be purchased as a collateral. Mortgage repayment is one type of debt repayment(or loan repayment).

Debt Repayment

Sinking Funds

Assignment #4 Problems 3 of Progress Exercises 5.5 Problems 2, 3, 5 of Progress Exercises 5.6 Due on 2013/05/09 (Thursday)