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Engineering Economics 390. Courtesy of Atre, Sundar V Associate Professor, Oregon State University.

### Transcript of Lecture 7

ENGR 390 Lecture 7: Review

Winter 2007

Effective Annual Interest Rate

ia = (1 + r / M ) 1Mr = nominal interest rate per year ia = effective annual interest rate M = number of interest periods per year

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ENGR 390 Lecture 7: Review

Winter 2007

Interest Rates EFFECTIVE INTEREST RATE

i = [1 + r / CK ]C 1C = number of interest periods per payment period K = number of payment periods per year r/K = nominal interest rate per payment period

Continuous Compounding

i = [1 + r / CK ]C 1where CK = number of compounding periods per year continuous compounding => C

i = lim[( 1 + r / CK ) C 1] = ( e r )1 / K 1

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ENGR 390 Lecture 7: Review

Winter 2007

TimingEqual Payment Series (Annuity) Equal cash flows Equal time between cash flows First cash flow at end of first period Last cash flow at end of last periodF? 0 1 2 3 N A F = A(F/A,i,N) P = A(P/A,i,N) 0 P? 1 2 3 N A

TimingUniform (Linear) gradient of amount G 0 at end of Period 1 G at end of Period 2 2G at end of Period 3 (N-1)G at end of Period NP? 0 1 2 G P = G(P/G,i,N) 3 N

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ENGR 390 Lecture 7: Review

Winter 2007

TimingGeometric gradient of amounts A1 and g A1 at end of Period 1 A1 * (1 + g) at end of Period 2 A1 * (1+ g) ^ 2 at end of Period 3

Complex Cash FlowsComplex Cash Flows Separate complex cash flows into component cash flows in order to use the standard formulas. Remember: You can only combine cash flows if they occur at the same point in time.

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ENGR 390 Lecture 7: Review

Winter 2007

Loans When we calculate the annual payment of a loan (A), the payment is actually composed of interest and payment on principal. The mechanics are best shown through an example.

Problem 1You borrow \$1,000 to help pay for rent, food, and books. It is to be repaid in 3 equal, annual payments starting one year from now. Interest on the loan is 12% per year, compounded annually. Determine the amount of the loan payments (A), the corresponding principal (P) and interest (I) amounts in each payment and the remaining balance (B) after each payment

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ENGR 390 Lecture 7: Review

Winter 2007

SolutionGiven: P = \$1,000; i = 12%; N= 3 Find: A, I1, I2, I3 ; P1, P2, P3 ; B1, B2, B3 A = \$1,000(A/P, 12%, 3) = \$416.35 B0 = P = \$1,000 I1 = B0i = \$1,000*0.12 = \$120 P1 = A I1 = \$416.35 - \$120 = \$296.35 B1 = B0 P1 = \$1000 - \$296.35 = \$703.65 I2 = B1i =\$703.65*0.12 = \$84.44 P2 = = A I2 = \$416.35 - \$84.44 = \$331.91 B2 = B1 P2 = \$703.65 - \$331.91 = \$371.74

Method 1: GeneralizingFor the nth paymentBn = P (A Pi) (F/A, i, n) In = (Bn-1)i Pn = A - In

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ENGR 390 Lecture 7: Review

Winter 2007

ApproachesLoan problems can be worked two ways: 1. Create a Table 2. Work problems in the same manner weve been using formulae

Create a TableWith the following column headings: Year or payment number Payment size Interest payment Cumulative interest payment Principal payment Remaining principal

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ENGR 390 Lecture 7: Review

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Solution# 1 2 3 Pmt Size 416.35 416.35 416.35 Interest Cumul. Payment Interest 120.00 84.44 44.61 120.00 204.44 249.05 Principal Payment 296.35 331.91 371.74 Remaining Balance 703.65 371.74 0.00

Problem 2You wish to payoff the loan at the end of 2 years after making your second loan payment. How much do you owe?

371.74

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ENGR 390 Lecture 7: Review

Winter 2007

Method 2: Remaining BalanceBn = A(P/A, i, N-n) In = (Bn-1)i = A(P/A, i, N-n+1)i Pn = A(P/F, i, N-n+1)

Problem 3A student borrowed \$5,000, which she will repay in 30 equal monthly installments. After making her 25th payment, she desires to pay the remainder of the loan in a single payment. At 12% per year, compounded monthly, what is the amount of the payment?

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ENGR 390 Lecture 7: Review

Winter 2007

SolutionGiven: P = \$5,000; i = 1%; N = 30 monthsDIAGRAM: \$ 5 000 1 0 \$A 2 3 30

Find B25 = A(P/A,i, N-n) A = \$5,000(A/P,1%,30) = \$193.74 After 25 payments: N-n = 30-25 = 5 B25 = \$193.74(P/A,1%,5) = \$939.64

B25 = ?

Problem 4A company has obtained a \$10,000 loan at an interest rate of 15% per year, compounded annually. The loan requires \$500 payments at the end of each of the next 3 years (starting one year from now). Determine how much must be paid 4 years from now in order to payoff the loan.

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ENGR 390 Lecture 7: Review

Winter 2007

SolutionGiven: P = \$10,000; i = 15%; A = \$500 Find: B4 B0 = P = \$10,000 I1 = B0i = \$10,000*0.15 = \$1500 B1 = \$10,000 + \$1,500 - \$500 = \$11,000 I2 = B1i =\$11,000*0.15 = \$1,650 B2 = \$11,000 + \$1,650 - \$500 = \$12,150 I3 = B2i = \$12,150*0.15 = \$1,822.5 B3 = B2 + I3 - \$500 = \$13,472.5 I4 = B3i = \$13,472.5*0.15 = 2020.85 B4 = B3 + I4 = \$15,493.375

Alternative SolutionBn = P (A Pi) (F/A, i, n) + A B4 = \$10,000 (50010,000*0.15)(F/A,15%,4) + A B4 = \$10,000 + \$5000 + \$500 = \$15,500

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ENGR 390 Lecture 7: Review

Winter 2007

Perspective Lender is indifferent between loan payments and investing loan amount at interest rate. Payments = Interest Paid +(Beginning Ending Principal) PW of Loans = PW of Payments

HW 1 Problem 14.7 a What is the amount accumulated by \$7000 in 8 years at 9% compounded annually?

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ENGR 390 Lecture 7: Review

Winter 2007

HW 1 Problem 1

HW 1 Problem 24.18 b. What is the future worth of \$2,000 at the end of each year for 10 years at 8.25% compounded annually?

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ENGR 390 Lecture 7: Review

Winter 2007

HW 1 Problem 2

HW 1 Problem 34.26. An individual deposits an annual bonus into a savings account that pays 6% interest, compounded annually. The size of the bonus increases by \$1,000 each year, and the initial bonus amount was \$3,000. Determine how much will be in the account immediately after the 5th deposit.

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ENGR 390 Lecture 7: Review

Winter 2007

HW 1 Problem 3

HW 1 Problem 44.30a An oil well is expected to produce 100,000 barrels of oil during its first year. Its subsequent production is expected to decrease by 10% over the previous years production. The oil well has a reserve of 1,000,000 barrels. If the price per barrel remains steady at \$30, what is the present worth of the anticipated stream of revenue at an annual interest of 12% compounded annually over the next 7 years?

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ENGR 390 Lecture 7: Review

Winter 2007

HW 1 Problem 4

HW 1 Problem 54.53. You have \$10,000 available for investment in stock. You are looking for a growth stock whose value is \$35,000 over 5 years. What kind of growth rate are you looking for?

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ENGR 390 Lecture 7: Review

Winter 2007

HW 1 Problem 5

HW 2 Problem 1A store offers a credit card that charges 0.95% per month, compounded monthly. What is the nominal interest rate for this credit card? What is the effective annual interest rate?

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ENGR 390 Lecture 7: Review

Winter 2007

HW 2 Problem 1

HW 2 Problem 25.6. A company is advertising a 24-month lease of a car for \$520 payable at the beginning of each month. There is a \$2,500 down payment, plus a \$500 refundable deposit. Alternatively, it offers a 24-month lease with a single up-front payment of \$12,780 plus a \$500 refundable security deposit. If the interest rate is 6% compounded monthly, which lease is preferred?

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ENGR 390 Lecture 7: Review

Winter 2007

HW 2 Problem 2

HW 2 Problem 35.20a. What is the future worth of \$3,000 at the end of each 6 month period for 10 years at 6% compounded semi-annually?

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ENGR 390 Lecture 7: Review

Winter 2007

HW 2 Problem 3

HW 2 Problem 45.27. What is the amount A, such that you will be able to withdraw the amounts shown in the cash flow diagram if the interest rate is 8% compounded quarterly?

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ENGR 390 Lecture 7: Review

Winter 2007

HW 2 Problem 4

HW 2 Problem 55.34. Sam plans to retire in 15 years. He deposits equal amounts quarterly in a bank that gives him an interest of 8% compounded quarterly. If he want to withdraw \$25,000 semi-annually over the 5 years of his retirement how much should he deposit quarterly? The first withdrawal occurs at the end of 6 months after his retirement.

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ENGR 390 Lecture 7: Review

Winter 2007

HW 2 Problem 5

HW 2 Problem 5 (continued)

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