Simple And Compound Interest (2) PDF

CHAPTER 02 SIMPLE INTEREST, CASH FLOW DIAGRAMS AND COMPOUND INTEREST CE214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE INSTRUCTOR SIMPLE INTEREST SIMPLE INTEREST ORDINARY SIMPLE INTEREST EXACT SIMPLE INTEREST 12 month...

CHAPTER 02 SIMPLE INTEREST, CASH FLOW DIAGRAMS AND COMPOUND INTEREST CE214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE INSTRUCTOR SIMPLE INTEREST SIMPLE INTEREST ORDINARY SIMPLE INTEREST EXACT SIMPLE INTEREST 12 months of 30 days 365 days in a year Ordinary year 360 days in a year 366 days in a year Leap year except for century years and not divisible by 400. CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SIMPLE INTEREST CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SIMPLE INTEREST Simple Interest The total amount F to be repaid is equal to the sum of the principal and the total interest and is I = Pin given by the formula: Let: F = P + I = P(1 + in) I = Total interest earned by the principal P = Principal or present worth n = Number of interest periods i = interest rate per interest period CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.1 If a man borrowed money from his girlfriend with simple rate of 12%, determine the present worth of P74, 900.00, which is due at the end of seven months. F = P ( 1 + i n) Given: i = 0.12 F = 74, 900 n = 7 /12 Solution: F = P ( 1 + i n) 74, 900 = P [1 + 0.12 P = 70,000 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.2 Clara has invested P10,000, part at 5% and the remainder at 10% simple interest. How much is invested at higher rate if the annual interest from this investment is P950.00, CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO. 2 Given: Solution: I = Pi n 950 = X (0.05)(1) + (10,000 – X)(0.10) (1) X = 1000 (1st investment) 10, 000 10,000 - X = 9,000 (2nd investment) X 10,000 - X i = 0.05 i = 0.10 P = 9,000 I = P950.00 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.3 Determine the accumulated amount using exact simple interest on P10,000 for the period from January 20, 1990 to November 28 of the same year at 15% interest rate. CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.3 January - 31 - 20 = 11 F = P ( 1 + i n) February - 28 312 March - 31 P = 10, 000 i = 0.15 n= 365 April - 30 May - 31 F = 10,000 [ 1 + (0.15)] June - 30 July - 31 August - 31 F = 11, 282.19 September - 30 October - 31 November - 28 312 days CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE  SITUATION NO.4 Determine the ordinary simple interest on P 10,000 for 9 months and 10 days if the rate of interest is 12%. Solution: Based on a banker’s year, 9 months and 10 days = 9(30) +10 = 280 days ( 360 ) 280 Ordinary simple interest = P10,000 (0.12)( Ordinary simple interest = P 933.33 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.5 A man borrows P10,000 from a loan firm. The rate of simple interest is 15%, but the interest is to be deducted from the loan at the time the money is borrowed. At the end of one year he has to pay back P10,000. What is the actual rate of interest? Given: F = 10,000 n=1 Solution: P = 10,000 - 0.15(10,000) 10,000 = 8500( 1 + i (1)) P = 8500 i = 0.1765 F = P ( 1 + i n) i = 17.65% CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE CASH-FLOW DIAGRAM Cash-flow diagram is simply a graphical representation of cash flows on a time scale. Cash-flow diagram for economic analysis problem is analogous to that of free body diagram for mechanics problems. Receipt - positive cash flow or cash inflow Disbursement - negative cash flow or cash outflow CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE  CASH-FLOW DIAGRAM Example: A loan of P100 at a simple interest of 10% will become P150 after 5 years. Viewpoint of the lender: Viewpoint of the borrower: P150 P100 0 1 2 3 4 5 0 1 2 3 4 5 P100 P150 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE COMPOUND INTEREST Compound Interest interest is computed every end of each interest period and the interest earned for that period is added to the principal. CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE COMPOUND INTEREST Fn n, period 0 1 2 3 n P F1 F2 F3 F Let: F = accumulated amount or future worth m = number of interest F1 = P (1 + i) 2 period P = Principal or present worth n = total number or F2 = P (1 + i) (1 + i)=P (1 + i) 3 i = interest rate per interest period 2 (1 + i) interest period F3 = P (1 + i) =P (1 + i) = m (t) Fn = P (1 + i) n r = nominal interest rate t = number of years of investment CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE COMPOUND INTEREST F = P (1 + i) n m(t) F= P Let: F = accumulated amount or future worth m = number of interest period per year P = Principal or present worth n = total number or r = nominal interest rate interest period i = interest rate per interest period = m (t) t = number of years of investment CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE COMPOUND INTEREST 6% compounded monthly for 10 years Nominal m = 12 Number of interest years, t rate, r 0.06 i = n = 12 (10) compounded annually = 1 12 compounded semi - annually = 2 compounded quarterly = 4 compounded bi - monthly = 6 compounded monthly = 12 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE COMPOUND INTEREST CONTINUOUS COMPOUNDING - m approaches infinity rt F = Pe CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE COMPOUND INTEREST EFFECTIVE RATE OF INTEREST - the actual rate of interest on the principal for one year Interest earned in one year E.R. = principal at the beginning of the year CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.1 If the sum of P12,000 is deposited in an account earning interest rate of 9% per year compounded quarterly, what will it become after 1 year? F=P (1 + i) n Given: P = 12,000 i = = 0.09 4 n = 4 (1) CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.1 Solution: F=P (1 + i) n 4(1) F = P [1 + F = 13, 117 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.2 In the previous problem, what is the effective rate? F=P (1 + i) n F = 13, 117 Given: 0.09 P = 12,000 i = = 4 n = 4 (1) CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.2 Solution: F-P E.R. = (100%) P 13117 - 12000 E.R. = (100%) 12000 E.R. = 9.31% E.R. = 0.0931 or 9.31% CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.3 How many years are required for P1,000 increase to P2,000 if invested at 9% per year compounded continuously? rt F = Pe Given: P = 1,000 F = 2,000 r = 0.09 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.3 Solution: rt F = Pe 0.09 t 2000 = 1000 e t = 7.7 years CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.4 When compounded bi-monthly, P150,000 becomes P223,183 after 5 years. What is the nominal rate of interest? F = P (1 + i) n Given: P = 150,000 F = 223,138 t=5 m=6 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.4 Solution: F=P (1 + i) n m(t) F =P 223,138 = 150,000 [1 + r = 0.08 or 8% CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.5 What payment X ten years from now is equivalent to a payment of P1,000 six years from now, if interest is 15% compounded monthly? X 0 1 2 3 4 5 6 7 8 9 10 1000 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.5 Solution: F = P (1 + i) n F = P (1 + i)n where : X = F 12(10) F X = 408.844 [1 + ] P= (1 + i)n 1000 X = 1815.35 P= 12(6) = 408.844 [1 + ] CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE SITUATION NO.5 What payment X ten years from now is equivalent to a payment of P1,000 six years from X now, if interest is 15% compounded monthly? 0 1 2 3 4 5 6 7 8 9 10 F = P (1 + i)n where : X = F 1000 12(4) X = 1000 [1 + ] X = 1815.35 CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE  END OF PRESENTATION CHAPTER O2 SIMPLE INTEREST, CASH-FLOW DIAGRAMS AND COMPOUND INTEREST INSTRUCTOR CE 214 - ENGINEERING ECONOMICS ENGR. CHRISTINE JOYCE C. ROSETE