# Chapter 3 Interest and Equivalence

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

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

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 1 2 3 5 \$29,100 4 1 2 3 5 \$5,000 \$8,000 \$6,000

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 1 2 \$1000 \$580 \$540 Yr Interest Balance Repayment Cash Flow 1000 1 80 500 -580 2 40 -540

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”!

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

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.

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

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

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

Repaying a debt 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 Yr Balance at Beginning of Year Interest Balance at the end of Year Interest Payment Principal Payment Total Payment 1 \$5,000.00 \$400.00 \$5,400.00 \$0.00 2 3 4 5 Subtotal \$2,000.00 \$7,000.00

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 Yr Balance at the Beginning of year Interest Balance at the end of yr before payment Interest Payment Principal Payment Total Payment 1 \$5,000.00 \$400.00 \$5,400.00 \$852.28 \$1,252.28 2 \$4,147.72 \$331.82 \$4,479.54 \$920.46 3 \$3,227.25 \$258.18 \$3,485.43 \$994.10 4 \$2,233.15 \$178.65 \$2,411.80 \$1,073.63 5 \$1,159.52 \$92.76 Subtotal \$1,261.41 \$6,261.41

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 Yr Balance at Beginning of year Interest Balance at the end of year Interest Payment Principal Payment Total Payment 1 \$5,000.00 \$400.00 \$5,400.00 \$0.00 2 \$432.00 \$5,832.00 3 \$466.56 \$6,298.56 4 \$503.88 \$6,802.44 5 \$544.20 \$7,346.64 \$2,346.64 Subtotal

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

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

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 Single Payment Compound Amount Formula (Eq. 3-3) (Eq. 3-4) The above notation is read as Find F, given P, at i, over n

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

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? F = P(1+i)n Eq 3-5 F=800 P = F(1+i)-n = 800(1+0.05)-4 Eq. 3-5 = \$658.16 Or 1 2 3 4 P = F×(P/F, i, n) Eq. 3-6 = 800(0.8227) = \$658.16 i=5% P=?

Example 3-5 \$500 were deposited in a saving account, paying 6%, compounded quarterly, for 3 years — i, n, F F=? 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) = \$598.00 1 2 12 11 i=1.5% \$500 “…paying 6%...” means paying annually.

How much can you borrow if you can repay it in the two payments shown
How much can you borrow if you can repay it in the two payments shown? Assume a 12% interest rate Solve for P, using Appendix C (or Eq. 3-6) Year Cash Flow +P 1 2 3 -400 4 5 -600 P = 400(P/F,12%,3) (P/F,12%,5) = 400(0.7118) + 600(0.5674) = P=? 4 1 2 3 5 i=12% 400 600

→ P is smaller at 15% than at 12% → Given future payments, P
With the same two payments shown, if the interest is increased to 15%, Will the value of P be larger or smaller? Solve for P P = 400(P/F,15%,3) (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! Year Cash Flow +P 1 2 3 -400 4 5 -600 4 1 2 3 5 P=? 400 i=15% 600

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

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

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?

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( ) = \$2,676.45

Problem 3-18 (c) n = 20 F = 2000(1+0.06)20 = 2000(F/P, 0.06, 20)
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

Problem 3-18 (e) n = 100 F = 2000(1+0.06)100 = 2000(F/P, 0.06, 100)
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

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?

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( ) = \$14,259.72

Problem 3-19 (c) n = 20 P = 20000(1+0.07)-20 = 20000(P/F, 0.07, 20)
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

Problem 3-19 (*) n = 100 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

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.

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), ia Annual interest rate with compounding ia = (1 + r/m)m – (3-7) m = # of compounding periods in a year m = 4, if compounding quarterly m = 12, if compounding monthly

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

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

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

QUESTION CONTINUES W 1 2 3 4 5 \$1000 Given P, ia, n, find W.

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 ia? (b) Starting with \$50, how much is it worth after 1 yr?

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, ia = ( )52 – 1 = 13, = 1,310,363% Year end worth F = Pia = 50( ) = \$655,181.5

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

Continuous Compounding
Again, F = P(er)n has quantities P, F, r, and n. Given any three, you can find the other one! 1) Given P, r & n, F = Pern. (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).

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!

Example 3-11 Method 2: Use Eq. (3-15)
Solution Method 1: Use Eq. (3-3) or (3-4) (single payment compound interest formula) r = 5%, P = \$2000, n = 2 ia = er – 1 = e0.05 – 1 = F = P(F/P, ia, n) = 2000( )2 = \$2,210.34 Method 2: Use Eq. (3-15) F = P[F/P, r, n] = 2000(e)0.05(2) = \$2,210.34

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

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 1 Annual 0.12 2 Semiannual 0.1236 4 Quarterly 12 Monthly 52 Weekly 365 Daily Continuous

End of Chapter 3