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

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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

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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? $29, $ 5,000 $ 8,000 $ 6,000 Slide 3

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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 $1000 $580 $540 Slide 4

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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

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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

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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 = $ $3000 = $13000 Although it is “simple”, it is rarely used. Slide 7

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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

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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 = $ $10, =10, $10,800.00$10,800 x 0.08 = $864.00$10, =11, $11,664.00$11,664 x 0.08 = $ $11, =12, $12, x 0.08 = $ $12, $ =13, $13, x 0.08 = $ $13, $ =14, Slide 9

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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, $4,000.00$320.00$4,320.00$320.00$1,000.00$1, $3,000.00$240.00$3,240.00$240.00$1,000.00$1, $2,000.00$160.00$2,160.00$160.00$1,000.00$1, $1,000.00$80.00$1,080.00$80.00$1,000.00$1, Subtotal$1,200.00$5,000.00$6, Repay of a loan of $5000 in 5 yrs at interest rate of 8% Plan #1: Pay $1000 principal plus interest due Slide 10

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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$0.00$ $5,000.00$400.00$5,400.00$400.00$0.00$ $5,000.00$400.00$5,400.00$400.00$0.00$ $5,000.00$400.00$5,400.00$400.00$0.00$ $5,000.00$400.00$5,400.00$400.00$5,000.00$5, Subtotal$2,000.00$5,000.00$7, 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

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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, $4,147.72$331.82$4,479.54$331.82$920.46$1, $3,227.25$258.18$3,485.43$258.18$994.10$1, $2,233.15$178.65$2,411.80$178.65$1,073.63$1, $1,159.52$92.76$1,252.28$92.76$1,159.52$1, Subtotal$1,261.41$5,000.00$6, 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

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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, Subtotal$2,346.64$5,000.00$7, 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

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

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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

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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

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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

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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) = $ P=? F=800 i=5% P = F(1+i) -n = 800(1+0.05) -4 Eq. 3-5 = $ Or F = P(1+i) n Eq 3-5 Slide 18

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Example 3-5 $500 were deposited in a saving account, paying 6%, compounded quarterly, for 3 years — i, n, F F=? i=1.5% 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( ) 12 = 500(F/P,1.5%,12) = 500(1.196) = $ $500 “…paying 6%...” means paying annually. Slide 19

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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) = YearCash Flow 0+P Solve for P, using Appendix C (or Eq. 3-6) P=? i=12% Slide 20

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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) = at 15% P = at 12% → P is smaller at 15% than at 12% → Given future payments, P decreases, as i increases! YearCash Flow 0+P Solve for P P=? 400 i=15% Slide 21

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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

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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( ) = $8,052,550 Slide 23

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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

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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( ) = $2, Slide 25

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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, (c)n = 20 F = 2000(1+0.06) 20 = 2000(F/P, 0.06, 20) = $4, Slide 26

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

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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

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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( ) = $14, Slide 29

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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, (c)n = 20 P = 20000(1+0.07) -20 = 20000(P/F, 0.07, 20) = $5, Slide 30

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

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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

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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

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

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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 = ( ) 4 – 1 = 6.14% Slide 35

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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

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QUESTION CONTINUES W WWWW $ Given P, i a, n, find W.

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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

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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 = ( ) 52 – 1 = 13, = 1,310,363% Year end worth F = Pi a = 50( ) = $655,181.5 Slide 39

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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

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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

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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

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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 = F = P(F/P, i a, n) = 2000( ) 2 = $2, Method 2: Use Eq. (3-15) F = P[F/P, r, n] = 2000(e) 0.05(2) = $2, Slide 43

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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

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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 2Semiannual Quarterly Monthly Weekly Daily Continuous

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End of Chapter 3 Slide 46

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