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Chapter 3 Interest and Equivalence EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1.

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Presentation on theme: "Chapter 3 Interest and Equivalence EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1."— Presentation transcript:

1 Chapter 3 Interest and Equivalence EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1

2 Learning Objectives Understand the concept of “time value of money” Distinguish between simple and compound interest Understand the concept of “equivalence” of cash flows Solve problems using Single Payment Compound Interest Formulas Slide 2

3 Computing cash flow To purchase a new $30,000 machine, two alternatives: Pay the full price now minus a 3% discount or Pay $5000 now; $8000 at the end of year 1; and $6000 at the end of each of the next 4 years (A total of $37,000!) Why do the second? 4 0 1 23 5 $29,100 4 0 1 23 5 $ 5,000 $ 8,000 $ 6,000 Slide 3

4 Computing Cash flow (cont…) To repay a loan of $1,000 at 8% interest in 2 years Repay half of $1000 plus interest at the end of year 1 Repay the remaining $500 plus interest at the end of yr 2 YrInterestBalanceRepayment Cash Flow 01000 180500 -580 2400500-540 0 1 2 $1000 $580 $540 Slide 4

5 Time value of money Money has purchasing power Money has earning power Money is a valuable asset People are willing to pay some charges (interest) to have money available for their use Money loses value in time! Or money has a time value! The time value lost over time is made up by “interest”! Slide 5

6 Interest Calculations Simple interest (Eq. 3-1) Total interest earned = whereP = Principal (loan) i = Simple annual interest rate n = Number of years whereF = Amount due at the end of n years Interest in computed on just the principal in each year! (Eq. 3-2) Slide 6

7 Simple interest (example) Loan of $10,000 for 5 yrs at simple interest rate of 6% Total interest earned = $10000(6%)(5) = $3000 Amount due at end of loan = $10000 + $3000 = $13000 Although it is “simple”, it is rarely used. Slide 7

8 Compound interest Interest is computed on the unpaid balance (not just the principal), which includes the principal and any unpaid interest from the preceding period Common practice for interest calculation, unless specifically stated otherwise Slide 8

9 Compound interest (example) Loan of $10K for 5 yrs at compound interest rate of 8% Year Balance Beginning of yearInterestBalance at end of year 1 $10,000.00$10000 x 0.08 = $800.00 $10,000 + 800 =10,800.00 2 $10,800.00$10,800 x 0.08 = $864.00$10,800 + 864 =11,664.00 3 $11,664.00$11,664 x 0.08 = $933.12 $11,664 + 933.12 =12,597.12 4 $12,597.12 x 0.08 = $1007.77 $12,597.12 + $1007.77=13,604.89 5 $13,604.89 x 0.08 = $1088.39 $13,604.89 + $ 1088.39=14,693.28 Slide 9

10 Repaying a debt Yr Balance at the Beginning of YearInterest Balance at the end of yr before payment Interest Payment Principal Payment Total Payment 1$5,000.00$400.00$5,400.00$400.00$1,000.00$1,400.00 2$4,000.00$320.00$4,320.00$320.00$1,000.00$1,320.00 3$3,000.00$240.00$3,240.00$240.00$1,000.00$1,240.00 4$2,000.00$160.00$2,160.00$160.00$1,000.00$1,160.00 5$1,000.00$80.00$1,080.00$80.00$1,000.00$1,080.00 Subtotal$1,200.00$5,000.00$6,200.00 Repay of a loan of $5000 in 5 yrs at interest rate of 8% Plan #1: Pay $1000 principal plus interest due Slide 10

11 Repaying a debt Yr Balance at Beginning of YearInterest Balance at the end of Year Interest Payment Principal Payment Total Payment 1$5,000.00$400.00$5,400.00$400.00$0.00$400.00 2$5,000.00$400.00$5,400.00$400.00$0.00$400.00 3$5,000.00$400.00$5,400.00$400.00$0.00$400.00 4$5,000.00$400.00$5,400.00$400.00$0.00$400.00 5$5,000.00$400.00$5,400.00$400.00$5,000.00$5,400.00 Subtotal$2,000.00$5,000.00$7,000.00 Repay of a loan of $5000 in 5 yrs at interest rate of 8% Plan #2: Pay interest due at end of each year and principal at end of 5 years Slide 11

12 Yr Balance at the Beginning of yearInterest Balance at the end of yr before payment Interest Payment Principal Payment Total Payment 1$5,000.00$400.00$5,400.00$400.00$852.28$1,252.28 2$4,147.72$331.82$4,479.54$331.82$920.46$1,252.28 3$3,227.25$258.18$3,485.43$258.18$994.10$1,252.28 4$2,233.15$178.65$2,411.80$178.65$1,073.63$1,252.28 5$1,159.52$92.76$1,252.28$92.76$1,159.52$1,252.28 Subtotal$1,261.41$5,000.00$6,261.41 Repaying a debt Repay of a loan of $5000 in 5 yrs at interest rate of 8% Plan #3: Pay in 5 equal end-of-year payments Slide 12

13 Yr Balance at Beginning of yearInterest Balance at the end of year Interest Payment Principal Payment Total Payment 1$5,000.00$400.00$5,400.00$0.00 2$5,400.00$432.00$5,832.00$0.00 3$5,832.00$466.56$6,298.56$0.00 4$6,298.56$503.88$6,802.44$0.00 5$6,802.44$544.20$7,346.64$2,346.64$5,000.00$7,346.64 Subtotal$2,346.64$5,000.00$7,346.64 Repaying a debt Repay of a loan of $5000 in 5 yrs at interest rate of 8% Plan #4: Pay principal and interest in one payment at end of 5 years Slide 13

14 Equivalence Total payment summary: Plan #1Pay $1K principal plus interest due $6,200.00 Plan #2Pay interest at yr 1-4, & balance at yr 5 $7,000.00 Plan #3Make 5 equal payments at end of yr 1-5 $6,261.41 Plan #4Make one payment = balance at yr 5 $7,346.64 Slide 14

15 Equivalence The 4 repayment plans are equivalent to one another and to $5000 now at 8% interest Using the concept of equivalence, one can convert different types of cash flows at different points of time to an equivalent value at a common reference point Equivalence is dependent on interest rate Slide 15

16 Single payment. Compound interest formulas Notation: i = interest rate per compounding period n = number of compounding periods P = a present sum of money F = a future sum of money (Eq. 3-3) (Eq. 3-4) The above notation is read as Find F, given P, at i, over n Single Payment Compound Amount Formula Slide 16

17 Single payment. Compound interest formulas Solving for P in Eq. 3-3 yields (Eq. 3-5) (Eq. 3-6) The above notation is read as Find P, given F, at i, over n Single Payment Present Worth Formula Slide 17

18 Simple payment. Compound interest formulas (ex.) Wish to have $800 at the end of 4 years, how much should be deposited in an account that pays 5% annually? P = F×(P/F, i, n) Eq. 3-6 = 800(0.8227) = $658.16 P=? F=800 i=5% 0 1 23 4 P = F(1+i) -n = 800(1+0.05) -4 Eq. 3-5 = $658.16 Or F = P(1+i) n Eq 3-5 Slide 18

19 Example 3-5 $500 were deposited in a saving account, paying 6%, compounded quarterly, for 3 years — i, n, F F=? i=1.5% 0 1 2 12 11 i = 6%/4 = 1.5% - quarterly interest rate n = 3 x 4 = 12 quarters F = P(1+i) n = P(F/P, i, n) = 500(1+0.015) 12 = 500(F/P,1.5%,12) = 500(1.196) = $598.00 $500 “…paying 6%...” means paying annually. Slide 19

20 How much can you borrow if you can repay it in the two payments shown? Assume a 12% interest rate P = 400(P/F,12%,3) + 600(P/F,12%,5) = 400(0.7118) + 600(0.5674) = 625.16 YearCash Flow 0+P 10 20 3-400 40 5-600 Solve for P, using Appendix C (or Eq. 3-6) 4 01 23 5 P=? 400 600 i=12% Slide 20

21 With the same two payments shown, if the interest is increased to 15%, Will the value of P be larger or smaller? P = 400(P/F,15%,3) + 600(P/F,15%,5) = 400(0.6575) + 600(0.4972) = 561.32 at 15% P = 625.16 at 12% → P is smaller at 15% than at 12% → Given future payments, P decreases, as i increases! YearCash Flow 0+P 10 20 3-400 40 5-600 Solve for P 600 4 01 23 5 P=? 400 i=15% Slide 21

22 Problem 3-2 Borrow $2000 Repay at end of 3 yrs, with 10% simple interest/yr How much to pay? Solution P = $2000, i = 10%, n = 3 yrs F = P(1+in) = 2000[1+0.1(3)] = $2,600 Slide 22

23 Problem 3-11 Borrow $5M Repay at end of 5 yrs, with 10% compound interest/yr How much to pay? Solution P = $5,000,000, i = 10%, n = 5 yrs F = P(1+i) n = P(F/P, i, n) = 5M(1+0.1) 5 = 5M(F/P, 0.1, 5) = 5M(1.61051) = $8,052,550 Slide 23

24 Problem 3-18 Deposit $2K in an account, & earn 6% interest How much is in the account (a)after 5 years? (b)after 10 years? (c)after 20 years? (d)after 50 years? (e)after 100 years? Slide 24

25 Problem 3-18 Solution P = $2k, and i = 6% (a)n = 5 F = P(1+i) n = P(F/P, i, n) = 2000(1+0.06) 5 = 2000(F/P, 0.06, 5) = 2000(1.338226) = $2,676.45 Slide 25

26 Problem 3-18 Solution (b)n = 10 F = P(1+i) n = P(F/P, i, n) = 2000(1+0.06) 10 = 2000(F/P, 0.06, 10) = $3,581.70 (c)n = 20 F = 2000(1+0.06) 20 = 2000(F/P, 0.06, 20) = $4,414.27 Slide 26

27 Problem 3-18 Solution (d)n = 50 F = 2000(1+0.06) 50 = 2000(F/P, 0.06, 50) = $36,840.31 (e)n = 100 F = 2000(1+0.06) 100 = 2000(F/P, 0.06, 100) = $678,604.17 Slide 27

28 Problem 3-19 Inheritance amount of $20k, interest rate is 7% How much is it worth, if it is received (a)after 5 years? (b)after 10 years? (c)after 20 years? (d)after 50 years? Slide 28

29 Problem 3-19 Solution F = $20k, and i = 7% (a)n = 5 P = F(1+i) -n = F(P/F, i, n) = 20000(1+0.07) -5 = 20000(P/F, 0.07, 5) = 20000(0.712986) = $14,259.72 Slide 29

30 Problem 3-19 Solution (b)n = 10 P = F(1+i) -n = P(F/P, i, n) = 20000(1+0.07) -10 = 20000(P/F, 0.07, 10) = $10,166.99 (c)n = 20 P = 20000(1+0.07) -20 = 20000(P/F, 0.07, 20) = $5,168.38 Slide 30

31 Problem 3-19 Solution (d)n = 50 P = 20000(1+0.07) -50 = 20000(P/F, 0.07, 50) = $678.96 (*)n = 100 P = 20000(1+0.07) -100 = 20000(P/F, 0.07, 100) = $23.05 Slide 31

32 Single Payment Formulas Single payment compound amount formula F = P(1+i) n - given P, i, & n, find F.(3-3) Single payment present worth formula P = F(1+i) -n - given F, i, & n, find P.(3-5) Single payment interest formula i = (F/P) 1/n – 1 - given P, F, & n, find i. Single payment time formula n = log(F/P)/log(1+i) - given P, F, & i, find n. Slide 32

33 Nominal & Effective Interest Nominal interest rate (per year), r Annual interest rate, without compounding, also called APR (annual percentage rate) Effective interest rate (per year), i a Annual interest rate with compounding i a = (1 + r/m) m – 1 (3-7) m = # of compounding periods in a year m = 4, if compounding quarterly m = 12, if compounding monthly Slide 33

34 Nominal and Effective Interest Rate Question: What is the effective interest rate for 12% interest compounded monthly? 34

35 Nominal & Effective Interest Example 3-9 A Bank pays 1.5% interest quarterly What are r and i a ? Solution Nominal annual interest rate, r = 4(1.5%) = 6% Effective interest rate, i a = (1 + 0.015) 4 – 1 = 6.14% Slide 35

36 Nominal and Effective Interest Rate Question: If you deposit $1000 in a bank that pays a 12% interest compounded quarterly what will be the amount you can withdraw in five equal yearly sums? In other words, how much should you withdraw each year? Solution r = 12% and m = 4 36

37 QUESTION CONTINUES W 0 1 2 3 4 5 WWWW $1000 37 Given P, i a, n, find W.

38 Nominal & Effective Interest Example 3-10 Loan shark: “If I give you $50 on Monday, you own me $60 on the following Monday.” (a) What are r and i a ? (b) Starting with $50, how much is it worth after 1 yr? Slide 38

39 Nominal & Effective Interest Solution (a) Weekly interest rate = 60/50 – 1 = 0.2 Nominal annual interest rate, r = 52(0.2) = 10.4 (b)Effective annual interest rate, i a = (1 + 0.2) 52 – 1 = 13,103.63 = 1,310,363% Year end worth F = Pi a = 50(13103.63) = $655,181.5 Slide 39

40 Continuous Compounding For nominal annual interest rate r, compounded continuously, effective annual interest rate is i a = e r – 1 (3-14) Compound amount F = P(e r ) n = P[F/P, r, n] (3-15) Present Worth P = F(e r ) -n = F[P/F, r, n] (3-16) Slide 40

41 Continuous Compounding Again, F = P(e r ) n has quantities P, F, r, and n. Given any three, you can find the other one! 1) Given P, r & n, F = Pe rn.(3-15) 2) Given F, r & n, P = Fe -rn.(3-16) 3) Given P, F, & r, n = (1/r)ln(F/P). 4) Given P, F, & r, r = (1/n)ln(F/P). Slide 41

42 Continuous Compounding Example 3-11 If a bank pays 5% nominal interest, compounded continuously, how much would $2000 be worth after 2 years? We can solve it by two methods! Slide 42

43 Example 3-11 Solution Method 1: Use Eq. (3-3) or (3-4) (single payment compound interest formula) r = 5%, P = $2000, n = 2 i a = e r – 1 = e 0.05 – 1 = 0.051271 F = P(F/P, i a, n) = 2000(1+0.051271) 2 = $2,210.34 Method 2: Use Eq. (3-15) F = P[F/P, r, n] = 2000(e) 0.05(2) = $2,210.34 Slide 43

44 Continuous Compounding Example 3-13 How long will it take for money to double at 10% nominal interest, compounded continuously? Solution This is the case that given P, F, & r, find n. In particular, F = 2P, & r = 10%, n = (1/r)ln(F/P) = (1/0.1)ln2 = (10)(0.693) = 6.93 years Slide 44

45 Nominal and Effective Interest Rate Table below shows effective interest rates for various compounding periods with r = 12% Number of Compounding Periods per year Length of Periods Effective Interest Rate 1Annual0.12 2Semiannual0.1236 4Quarterly0.12551 12Monthly0.12683 52Weekly0.12734 365Daily0.12747 Continuous0.12749 45

46 End of Chapter 3 Slide 46


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