CHAPTER 2.0 MONEY-TIME RELATIONSHIP 1 CASH FLOW DIAGRAM 2 TIME VALUE OF MONEY 3 EQUIVALENCE 4 SINGLE PAYMENT COMPOUND I NTEREST 5 UNIFORM SERIES COMPOUND I NTEREST 6 ARITHMETIC AND GEOMETRIC GRADIENT Cash Flow Diagrams Cash flow diagrams (CFD) summarize the costs and benefits of projects Example: Time Period Size of Cash Flow 0 (today) 1 2 3 4 5 A CFD illustrates the size, sign, and timing of individual cash flows Periods may be months, quarters, years, etc. Receive $100 (positive CF) Pay $100 (negative CF) Positive CF of $100 Negative CF of $150 Negative CF of $150 Positive CF of $50 Tomorrow COMMENTS: • The end of one period is the beginning of the next one 100 100 50 • Arrows point up for revenues or benefits, down for costs • One person’s payment (cash outflow w. neg. sign) is another person’s receipt (cash inflow w. pos. sign) It is essential to use only one perspective in any CFD . 0 Today 1 100 2 3 4 150 150 5 Categories of Cash Flows Category Description First Cost Expense to build or to buy and install Operations and Maintenance (O&M) Annual expense, which often includes electricity, labor, minor repairs, etc Salvage Value Receipt at project termination for sale or transfer of the equipment (can be salvage lost) Revenues Annual receipts due to sale of products or services Overhaul Major capital expenditure that occurs during the asset’s life Prepaid Expenses Annual expenses, such as leases and insurance payments, that must be pain in advance Timing of Cash Flows Time could start at:◦ At the beginning of analysis ◦ When the project is expected to be approved ◦ When design is started ◦ When the material are first ordered ◦ When construction begins ◦ When use begin ◦ The first cost Cash Flow Diagram Practice 1 Ernie’s Earthmoving is considering the purchase of a piece of heavy equipment. What is the cash flow diagram if the following cash flow are anticipated? ◦ First cost: RM129K ◦ O & M Cost:RM30K per year ◦ Overhaul Cost: RM 35K in year 3 ◦ Salvage value: RM40K after 5 years Cash Flow Diagram Practice 2 Rather than purchasing the heavy equipment, as in Example 2, Ernie’s Earthmoving is planning on leasing it. The lease payments will be RM25K per year. Ernie’s Earthmoving is responsible for the overhaul and O & M costs. What is the cash flow diagram? Time Value of Money Question: Would you prefer $100 today or $100 after 1 year? There is a time value of money. Money is a valuable asset, and people would pay to have money available for use. The charge for its use is called interest rate. Interest The cost of money. More specifically, it is a cost to the borrower and an earning to the lender, above and beyond the initial sum borrowed or loaned. Interest Rate A percentage periodically applied to a sum of money to determine the amount of interest to be added to that sum. Equivalence Money is a valuable commodity. People will pay you to use your money. You will have to pay someone to use his or her money. For example, leaving RM100 in the bank for one year may generate an interest of RM6 at the end of the year. That is, the bank has paid you 6% interest to use your money. In the same vane, the bank will turn around and lend the money to another individual and charge 10% interest. The borrower will have to pay back RM110 at the end of one year. This means that your RM100 today is worth RM106 one year from now because of the interest earned on the principal. With the same token, RM100 lent by the bank is worth RM110 to the bank one year from now. We can also say that, for you, RM100 today is equivalent to RM106 one year from now. This principle is called as THEORY OF EQUIVALENCE. Theory of Equivalence In economics, because of the time value of money, a certain amount at present is equal to different future amounts given at an interest rate. Conversely, an amount of dollars invested now will be worth more when the principal and its accumulated interest are received in the future. The total interest earned or charged is linearly proportional to the initial amount of the loan (principal), the interest rate and the number of interest periods for which the principal is committed. Simple Interest When applied, total interest “i” may be found by i = ( P ) ( N ) ( i ), where ◦ P = principal amount lent or borrowed ◦ N = number of interest periods ( e.g., years ) ◦ i = interest rate per interest period Compounding In some financial investments such as fixed deposit (FD) which involves a deposit of a certain amount of money, known as a principal, at a given time, a specified interest rate compounded periodically, and a specified period of time; F = P(1+i)n Compound Interest The practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn from the initial sum. Compound interest is by far the most commonly used system in the real world Economic Equivalence It exists between individual cash flows and/or pattern of cash flows that have the save value. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal. Example 1 An investor deposits RM1000 into FD account that pays 10% a year, compounded annually, for 1 year. F = P(1+i)n F = 1000(1+0.10)1 = RM1100 If deposited for 2 years F = 1000(1+0.10)2 = RM1210 If deposited for 5 years at 6%interest rate F = 1000(1+0.06)5 = RM1338 Practice 3 If you invest RM4000 in a bank at an interest rate of 6.25% per year, how much money would you have at the end of three years? Annual Percentage Rate (APR) Interest is most frequently quoted by financial institutions as an APR. However, compounding often occurs more than once annually, and the APR does not account for the effect of this more frequent compounding. This situation leads to the distinction between nominal and effective interest. Nominal Interest. A stated rate of interest for a given period (usually a year). Effective Interest. The actual rate of interest, which accounts for the interest amount accumulated over a given period. Example 2 If RM100 is invested in a bank at a rate of 12% per year to b compounded monthly, what would be the effective interest rate? Interest rate per period (month) = 12%/12month = 1%/month 1 year contains 12 compounding period F = P(1+i)n = 100(1+0.01)12 = RM112.68 This is equivalent to applying an interest rate of 12.68% and compounding on a yearly basis. F = 100(1+0.1268)1 = RM112.68 In other words, the effective interest rate is 12.68% if the interest rate 12%/year and is compounding monthly. The 12% /year rate is called a nominal interest rate Effective and Nominal i = (1+j/m)m – 1 i = effective interest rate per period j = nominal interest rate period m = the number of compounding sub periods Practice 4 If RM100 invested in a bank at interest rate of 12% per year to be compounded quarterly, what would be the effective interest rate? Equivalence When an organization is indifferent as to whether it has a present sum of money now or the assurance of some other money (or a series of sums of money) in future, we say that the present sum of money is equivalent to the future sum or series of sums. Equivalence Given the choice of these two plans which would you choose? Year 1 2 3 4 5 Total Plan 1 $1400 1320 1240 1160 1080 $6200 Plan 2 $400 400 400 400 5400 $7000 Theory of Equivalence Determine a single equivalent value at a point in time for plan 1 Determine a single equivalent value at a point in time for plan 2 Both at the same interest rate Judge the relative attractiveness of the two alternatives from the comparable equivalent values Interest Rate Formulas P = present value of an amount of money i = annual interest rate n = number of annual periods F = future value at the end of the period A = annual amount, in series of equal payment or receipts, made at the end of each year Single Payment Compound Interest General Notation ◦ F = P(1+i)n Functional notation ◦ F = P(F/P, i, n) ◦ Read as follows – to find a future sum F, given a present sum, P at an interest rate i per interest rate period and n interest periods ◦ Called compound interest factor for single payment Refer to table to get the factor value Example 3 If RM1000 is invested for 5 years, interest rate compounded annually, the future value is obtain by multiplying RM1000 with the F/P for a single payment at the given interest rates: at i = 6%, F = RM1000(1.338) = RM1338 at i = 10%, F = 1000(1.611) = RM1611 at i = 14%, F = 1000(1.925) = RM1925 Single Payment Present Value Factor General Notation (P/F, i, n) is called single payment present value factor 1 F P F n n 1 i 1 i Functional notation ◦ P = F(P/F, i, n) ◦ Read as follows – to find a present sum P, given a future sum, F at an interest rate i per interest rate period and n interest periods Refer to table to get the factor value Example 4 What is a present value of RM1000 due 5 years from now, discounted at an interest rate of 10% compounded annually? P = RM1000(P/F, 10%, 5) = 1000(0.6209) P = RM620.90 Example 5 Question: You borrowed RM1000 at 12% compounded annually. The loan is paid back after 5 years. How much should be repaid? Solution: Using the compound interest tables for 12% and five periods, the value of the (F/P, 12%, 5) factor is found to be 1.7623. F = P(F/P, 12%, 5) = 1000(1+0.12)5 = 1762.30 Therefore the amount to be repaid equals to RM1762.30 Practice 5 If you invest RM1000 in a bank at an interest rate of 6.25% per year, how much money would you have at the end of three years? Practice 7 & 8 Practice 7 If $500 were deposited in a bank savings account, how much would be in the account three years hence if the bank paid 6% interest compounded annually? Practice 8 Suppose the bank changed their interest policy in Example A to 6% interest, compounded quarterly. For this situation, how much money would be in the account at the end of 3 years, assuming a $500 deposit now? Uniform Series Compound Interest Formulas Uniform Series Compound Interest F is required when A, i , n is known Functional Notation: F = A(F/A, i%, n) General Equation: 1 i n 1 F A i Uniform Series Compound Amount/Sinking Fund Factors A is required when F, i%, n is known Functional Notation: A = F(A/F, i%, n) General Equation i A F n 1 i 1 Uniform Series Compound Interest Formulas Capital Recovery Factor ◦ A is required when P, i%, n is known ◦ Functional Notation: A = P (A/P, i%, n) ◦ General Equation i 1 i n A P n 1 i 1 Uniform Series Present Value Factor ◦ P is required when A, i%, n is known ◦ P = A(P/A, i, n) ◦ General Equation 1 i n 1 P A n i 1 i Uniform Series Compound Interest It is often required to know the present value of a series of equal payments (or receipt) over a number of years at a given interest rate. Example: What is a present value of an investment which will generate RM1000 per year over the next 3 years if the interest rate is 10% compounded annually? P = 1000[(P/F, 10, 1) + (P/F, 10, 2) + (P/F, 10, 3)] P A A A 1 i 1 1 i 2 1 i 3 P = 1000[(0.9091) + (0.8265) + (0.7513)] = RM2486.90 OR P = 1000*(P/A, 10, 3) = 1000*(2.4869) = RM2486.90 P/A is called uniform series present value factor Practice 9 If you want to invest sufficient money in the bank such that at an interest rate of 8%, you will receive RM20,000 per year for the next 10 years, how much money should you invest in the bank now? Practice 10 If you take a home loan improvement loan of RM10,000 to be paid over a five year period, what would be the yearly payment if the interest rate is 12% per year? Practice 11 You wish to deposit a single sum of money in a savings account so that five equal annual withdrawals of RM2000 can be made before depleting the fund. If the first withdrawal is to occur 1 year after the deposit and the fund pays interest at a rate of 12% compounded annually, how much should be deposited? Practice 12 If you deposited RM10,000 at the end of each year, how much would you accumulate at the end of 5 years at an interest rate of 6% Practice 13 If you need RM 100,000 ten years from now for your son’s education, how much should you invest at the end of each year at an interest rate of 8% Practice 14 Jim Hayes read that out west, a parcel of land could be purchased for $1000 cash. Jim decided to save a uniform amount at the end of each month so that he would have the required $1000 at the end of one year. The local credit union pays 6% interest, compounded monthly. How much would Jim have to deposit each month? Arithmetic and Geometric Gradient In applications, the annuity cash flow pattern is not the only type of pattern encountered Two other types of end of period patterns are common ◦ The Linear or arithmetic gradient ◦ The geometric (% per period) gradient Arithmetic Gradient Factors An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a constant amount over n time periods. A linear gradient is always comprised of TWO components: Arithmetic Gradient Factors •The Two Components are: •The Gradient component •The base annuity component •The objective is to find a closed form expression for the Present Worth of an arithmetic gradient Linear Gradient Example A1+(n-1)G Assume the following: A1+(n-2)G A1+2G A1+G 0 1 2 3 n-1 N This represents a positive, increasing arithmetic gradient Example: Linear Gradient • Typical Negative, Increasing Gradient: G=$50 The Base Annuity = $1500 Example: Linear Gradient • Desire to find the Present Worth of this cash flow The Base Annuity = $1500 Arithmetic Gradient Factors • The “G” amount is the constant arithmetic change from one time period to the next. •The “G” amount may be positive or negative! •The present worth point is always one time period to the left of the first cash flow in the series or, •Two periods to the left of the first gradient cash flow! Derivation: Gradient Component Only Focus Only on the gradient Component (n-1)G “0” G (n-2)G +2G G Removed Base annuity 0 1 2 3 n-1 N Present Worth Point… The Present worth point of a linear gradient is always: ◦2 periods to the left of the “1G” point or, ◦1 period to the left of the very first cash flow in the gradient series. DO NOT FORGET THIS! Present Worth Point… $700 $600 $500 $400 $300 $200 $100 X 0 1 2 3 4 5 The Present Worth Point of the Gradient 6 7 Gradient Component $600 •The Gradient Component $500 $400 $300 $200 $100 $0 X0 1 2 3 4 The Present Worth Point of the Gradient 5 6 7 Present Worth Point… •PW of the Base Annuity is at t = 0 •PWBASE Annuity=$100(P/A,i%,7) Base Annuity – A = $100 X 0 1 2 3 4 The Present Worth Point of the Gradient 5 6 7 Gradient Series 1 1 ni 1 i n P G 2 i or P = G(P/G, i, n) Where; (P/G, I, n) is called gradient series, present worth factor Uniform Series Equivalent to the Gradient series A = G(A/G, i, n) Where; (A/G, i, n) is called gradient to uniform series conversion factor Practice 15 A man has purchased a new automobile. He wishes to set aside enough money in a bank account to pay the maintenance on the car for the first five years. It has been estimated that the maintenance cost of an automobile is as follows: Year Maintenance Cost 1 $120 2 150 3 180 4 210 5 240 Assume the maintenance costs occur at the end of each year and that the bank pays 5% interest. How much should the car owner deposit in the bank now? Practice 16 On a certain piece of machinery, it is estimated that the maintenance expenses will be as follows: Year Maintenance Cost 1 $100 2 200 3 300 4 400 What is the equivalent uniform annual maintenance cost for the machinery if 6% interest is used? Practice 17 A textile mill in India installed a number of new looms. It is expected that the initial maintenances costs and expenses for repairs with high but then decline for several years. The projected cost is: Year Maintenance and Repair Costs (rupees) 1 24000 2 18000 3 12000 4 6000 What is the projected equivalent annual maintenance and repair cost if interest is 10%? Practice 18 Maintenance costs for a particular production machine increase by $1000/year over the 5 year life of the equipment. The initial maintenance cost is $3000. Using an interest rate of 8% compounded annually, determine the present worth equivalent for the maintenance costs. Geometric Gradients • An arithmetic (linear) gradient changes by a fixed dollar amount each time period. •A GEOMETRIC gradient changes by a fixed percentage each time period. •We define a UNIFORM RATE OF CHANGE (%) for each time period •Define “g” as the constant rate of change in decimal form by which amounts increase or decrease from one period to the next Geometric Series 1 1 g n 1 i n P A ig nA1 P 1 i ig i g Or P = A1(P/A1, i, g, n) Where, (P/A1, i,g,n) is called geometric series present value factor Practice 19 A company is considering purchasing a new machine tool. In addition to the initial purchase and installation costs, company management is concerned about the machine’s maintenance cost. The maintenance cost of the machine tool are expected to be $1000 at the end of the first year of the machine’s life and increase 8% per year thereafter. The expected life of the machine tool is 15 years. Company management would like to endow a maintenance fund for expected costs. If the endowment account earns 10% per year compounded annually, how much money must be initially deposited in the account? Geometric Series 1 i n 1 g n F A1 ig F nA1 1 i n 1 ig ig Or F = A1(F/A1, I, g, n) Where, (F/A1, i,g,n) is called geometric series future value factor Practice 20 You receive an annual bonus and deposits it in a savings account that pays 8% compounded annually. The size of the bonus increases by 10% each year; the initial deposit was RM500. Determine how much will be in the fund immediately after tenth deposit.