Engineering Economics Lec 9 PDF

Engineering Economics Lec 9 SIMPLE INTEREST When the total interest paid or earned is linearly proportional to the principal, the interest rate, and the number of interest periods for which the principal is committed, the interest and the interest rate are said to be simple. Simpl...

Engineering Economics Lec 9 SIMPLE INTEREST When the total interest paid or earned is linearly proportional to the principal, the interest rate, and the number of interest periods for which the principal is committed, the interest and the interest rate are said to be simple. Simple interest is not used frequently in commercial practice. When you deposit money into a bank, the bank pays you interest. When you borrow money from a bank, you pay interest to the bank. Simple interest is Rate of interest is money paid only the percent on the principal. charged or earned. I=P rt  Principal is the Time that the money amount of money is borrowed or borrowed or invested. invested (in years). Formula for SI  Example 1: Finding Interest and Total Payment on a Loan To buy a car, Jessica borrowed $15,000 for 3 years at an annual simple interest rate of 9%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay? First, find the interest she will pay. I = P r  t I = 15,000  0.09  3 I = 4050  Example 1 Continued Jessica will pay $4050 in interest. You can find the total amount A to be repaid on a loan by adding the principal P to the interest I. P+I=A 15,000 + 4050 = A 19,050 = A Jessica will repay a total of $19,050 on her loan.  Exercise 1 To buy a laptop computer, Elaine borrowed $2,000 for 3 years at an annual simple interest rate of 5%. How much interest will she pay if she pays the entire loan off at the end of the third year? What is the total amount that she will repay?  Exercise 1 (contd.) First, find the interest she will pay. I = P r  t I = 2,000  0.05  3 I = 300  Exercise 1 Continued Elaine will pay $300 in interest. You can find the total amount A to be repaid on a loan by adding the principal P to the interest I. P+I=A 2000 + 300 = A 2300 = A Elaine will repay a total of $2300 on her loan. Exercise 2: Determining the Amount of Investment Time Nancy invested $6000 in a bond at a yearly rate of 3%. She earned $450 in interest. How long was the money invested?  Exercise 2 (contd.) Nancy invested $6000 in a bond at a yearly rate of 3%. She earned $450 in interest. How long was the money invested? I = P r  t. 450 = 6000  0.03  t 450 = 180t 2.5 = t The money was invested for 2.5 years, or 2 years and 6 months.  Exercise 3 Bertha deposited $1000 into a retirement account when she was 18. How much will Bertha have in this account after 50 years at a yearly simple interest rate of 7.5%?  Exercise 3 Bertha deposited $1000 into a retirement account when she was 18. How much will Bertha have in this account after 50 years at a yearly simple interest rate of 7.5%? I = P r  t I = 1000  0.075  50 I = 3750. The interest is $3750. Now you can find the total.  Exercise 3 Continued P+I=A 1000 + 3750 = A 4750 = A Bertha will have $4750 in the account after 50 years.  Exercise 4 You borrow PKR 1,500 from your friend for three years at a simple interest rate of 8% per year. How much interest you will pay after three years and what is the of total amount that will be paid by you? Solution The interest I paid is computed first Here, P = `1,500; n = 3 years; i = 8% per year. Thus, SI = (`1,500)(3)(0.08) = `360. Total amount paid by you at the end of three years would be `1,500 + `360 = `1,860.  Practice Questions Smith takes a loan of $10000 from a bank for a period of 1 year. The rate of interest is 10% per annum. Find the interest and the amount he has to pay at the end of a year. John borrowed $50,000 for 3 years at the rate of 3.5% per annum. Find the interest accumulated at the end of 3 years. COMPOUND INTEREST Whenever the interest charge for any interest period e.g. a year, is based on the remaining principal amount plus any accumulated interest charges upto the beginning of that period, the interest is said to be compound interest. Compound interest is interest paid not only on the principal, but also on the interest that has already been earned. The formula for compound interest is below. nt A = P(1 + r ) n A is the final dollar value, P is the principal, r is the rate of interest, t is the number of years, and n is the number of compounding periods per year. CI Formula The table shows some common compounding periods and how many times per year interest is paid for them. Compounding Times per year (n) Periods Annually 1 Semi-annually 2 Quarterly 4 Monthly 12  Example 5: Applying Compound Interest David invested $1800 in a savings account that pays 4.5% interest compounded semi-annually. Find the value of the investment in 12 years. r nt A = P(1 + ) n 0.045 2(12) = 1800(1 + ) t 2 = 1800(1 + 0.0225)24 = 1800(1.0225)24  Example 5 Continued ≈ 1800(1.70576) ≈ 3,070.38 After 12 years, the investment will be worth about $3,070.38. Example 6: Applying Compound Interest You deposit $500 in an account that pays 6% interest compounded annually. What is your balance at the end of 4 years?  Example 6 Beginning Year balance Interest Ending balance I = (500)(0.06)(1) A = 500 + 30 1 $500 = $30 = $530 I = (530)(0.06)(1) A = 530 + 31.80 2 $530 = $31.80 = $561.80 I = (561.80)(0.06) A = 561.80 + 33.71 3 $561.80 (1) = $595.51 = $33.71 I = (595.51)(0.06) A = 595.51 + 35.73 4 $595.51 (1) = $631.24 = $35.73  Example 7 What is the value of an investment of $3,500 after 2 years if it earns 1.5% compounded quarterly?  Practice Questions  If you deposit $4000 into an account paying 6% annual interest compounded quarterly, how much money will be in the account after 5 years?  If you deposit $6500 into an account paying 8% annual interest compounded monthly, how much money will be in the account after 7 years?  How much money would you need to deposit today at 9% annual interest compounded monthly to have $12000 in the account after 6 years? Comparing Simple to Compound Interest 27 Simple Vs Compound Interest 45000.0 40000.0 35000.0 30000.0 25000.0 Simple 20000.0 Compound 15000.0 10000.0 P=1000 5000.0 i=8% 0.0 1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 28 Rule of 72 The time it will take an investment (or debt) to double in value at a given interest rate using compounding interest. 72 = Years to Interest double Rate investment (or debt) Cash Flow Diagrams The graphic representation of each monetary value with time is called a cash flow diagram. The benefits are represented as upward arrows and costs as downward arrows. It is drawn to convert the time stream of monetary value into an equivalent single number. Nomenclature Standard Cash flows  A single payment cash flow can occur at the beginning of the time line (designated as t= 0), at the end of the time line (designated as t= n), or at any time in between.  The uniform series cash flow, consists of a series of equal transactions starting at t = 1 and ending at t= n. The symbol A (representing an annual amount) is typically given to the magnitude of each individual cash flow.  The gradient series cash flow starts with a cash flow (typically given the symbol G) at t= 2 and increases by G each year until t= n, at which time the final cash flow is (n - 1) G. The value of the gradient at t= 1 is zero.  Cash flow diagrams For example, consider a truck that is going to be purchased for $55,000. It will cost $9,500 each year to operate including fuel and maintenance. It will need to have its engine rebuilt in 6 years for a cost of $22,000 and it will be sold at year 9 for $6,000. Cash Flow Diagram At time 0 (beginning of $50,000 is paid out the first year) At EOY 3 $23,000 is paid out At EOY 6 $15,000 is received Every year starting at $2,000 is received EOY 1 The Huge Sheet Printing Company is considering purchasing a new web printing press, and they have determined the following cash items: $530,0 Initial Cost: 00 210,00 Annual Revenue 0 Annual 80,000 Maintenance Rebuild 330,00 Expenses EOY 4 0 Salvage value Present Value Present Value (PV) is the value today (in the present) of a payment that is promised to be made in the future. Present Value is the amount that must be invested today in order to realize a specific amount on a given future date. Example Find PV of $100 received in one year, i=5% PV=$100/(1+.05) = $95.24 What will be the present Value of $100 received in 2 ½ years and an interest rate of 8%. PV = $100 / (1.08)2.5 = $82.50 Examples  Julie borrowed $5000 to purchase furniture for her new house. She can repay the loan in either of the two ways described below (a) Five equal annual installments with interest based on 5% per year. (b) One payment 3 years from now with interest based on 7% per year. Solution  (a) The repayment schedule requires an equivalent annual amount A, which is unknown. P = $5000 i =5% per year n = 5 years A = ?  (b) Repayment requires a single future amount F, which is unknown. P = $5000 i = 7% per year n= 3 years F = ? Example The winner of a sweepstakes prize is given the choice of a onetime payment of $1,000,000 or a guaranteed $80,000 per year for 20 years. If the value of money is 5%, which option should the winner choose? Solution Option 1: P = $1,000,000 Option 2: P = 80,000(P/A, 5%, 20) = $996,960 Choose option 1: take the $1,000,000 now Example  Emma and her husband decide they will buy $1,000 worth of utility stocks beginning one year from now. Since they expect their salaries to increase, they will increase their purchases by $200 per year for the next nine years. What would the present worth of all the stocks be if they yield a uniform dividend rate of 10% throughout the investment period and the price/share remains constant? Solution PW of the base amount ($1,000) is: 1,000(P/A, 10%, 10) = $6,144.57 PW of the gradient is: 200(P/G, 10%, 10) = $4,578.27 Total PW = 6,144.57 + 4,578.27 = $10,722.84 Example Production engineers of a manufacturing firm have proposed a new equipment to increase productivity of a manual gas cutting operation. The initial investment (first cost) is `5,00,000 and the equipment will have a salvage value of `1,00,000 at the end of its expected life of 5 years. Increased productivity will yield an annual revenue of `2,00,000 per year. If the firm’s minimum attractive rate of return is 15%, is the procurement of new equipment economically justified? P.W. = `2,00,000(P/A, 15%, 5) + `1,00,000 (P/F,15%,5) = `2,00,000(3.3522) + `1,00,000(0.4972) = `7,20,160 P.W. = `5,00,000 N.P.W. = `7,20,160 − `5,00,000 N.P.W. = `2,20,160 Since N.P.W. > 0, this equipment is economically justified PRESENT WORTH (P.W.) METHOD (i) If N.P.W. > 0, the investment in the project is economically justified (ii) If N.P.W. < 0, the investment in the project is economically not justified (iii) If N.P.W. = 0, the investment in the project is barely economically justified Question s