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Chapter 7 Rate of Return Analysis 1

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Chapter Contents Internal Rate of Return Rate of Return Calculations Plot of NPW versus interest rate i Fees or Discounts Examples Incremental Analysis Using Spreadsheet 2Engineering Economics

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Rate of return analysis is the most frequently used exact analysis technique in industry. Major advantages Rate of return is a single figure of merit that is readily understood. Calculation of rate of return is independent from the minimum attractive rate of return (MARR). Rate of Return Analysis 3Engineering Economics

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Internal Rate of Return What is the internal rate of return (IRR)? IRR is the interest rate at which present worth or equivalent uniform annual worth is equal to 0. In other words, the internal rate of return is the interest rate at which the benefits are equivalent to the costs. 4Engineering Economics

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Internal Rate of Return Internal rate of return is commonly used to evaluate the desirability of investments or projects. IRR can be used to rank multiple prospective projects. Because the internal rate of return is a rate quantity, it is an indicator of the efficiency, quality, or yield of an investment. To decide how to proceed, IRR will be compared to preselected minimum attractive rate of return (Chapter 8) 5Engineering Economics

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6 Given a cash flow stream, IRR is the interest rate i which yields a zero NPW (i.e., the benefits are equivalent to the costs), or a zero worth at any point in time. This can be expressed in 5 different ways as follows. NPW = 0 PW of benefits PW of costs = 0 PW of benefits = PW of costs PW of benefits/PW of costs = 1 EUAB EUAC = 0 Internal Rate of Return (IRR) Engineering Economics

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Example A person invests $1000 at the end of each year. If the person would like to have $80,000 in savings at EOY 26 what interest rate should he select? When the compound interest tables are visited the value of i where (F/A, i%, 26)=80 is found as 8%, so i=8% Checking the Tables 26 yrs @ 6%, F/A = 59.156 26 yrs @ 10%, F/A = 109.182 26 yrs @ 8%, F/A = 79.954 7Engineering Economics

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Example A person invests $1000 at the end of each year. If the person would like to have $80,000 in savings at EOY 26 what interest rate should he select? When the compound interest tables are visited the value of i where (F/A, i%, 26)=80 is found as 8%, so i=8% Checking the Tables 26 yrs @ 6%, F/A = 59.156 26 yrs @ 10%, F/A = 109.182 26 yrs @ 8%, F/A = 79.954 8Engineering Economics

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Example EXCEL solution RATE(n, A, P, F, type, guess) rate(26, 1000, 0, -80000) = 8% rate (26, -1000, 0, 80000) = 8% A, P, F must have different signs (+ or )! IRR(value range, guess) value range = the cash flow stream 9Engineering Economics

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Example Cash flows for an investment are shown in the following figure. What is the IRR to obtain these cash flows? YEARCASH FLOW 0($500) 1$100 2$150 3$200 4$250 10Engineering Economics

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EXAMPLE CONTINUES -8.85 YEARCASH FLOW 0($500) 1$100 2$150 3$200 4$250 11Nov. 2, 2011Engineering Economics

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QUESTION CONTINUES -8.85 12Nov. 2, 2011Engineering Economics

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INTERPOLATION 5% 15% X% 30.95 -8.85 5-X 30.95 1039.80 0 13Engineering Economics

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INTERPOLATION 5% 15% X% 30.95 -8.85 5-X 30.95 -1039.80 0 14Engineering Economics

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INTERPOLATION 12% 15% X% 3.350 -8.850 12-X 3.350 -3 12.200 0 15Engineering Economics

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INTERPOLATION 12% 15% X% 3.350 -8.850 12-X 3.350 -3 12.200 0 16Engineering Economics

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EXCEL solution IRR(C1:C5) = 12.83% C1 ~ C5 stores the stream of the 5 cash flows: -500, 100, 150, 200, 250 17Engineering Economics

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Example A student, who will graduate after 4 years, borrows $10,000 per year at 5% interest rate at the beginning of each year. No interest is charged till graduation. If the student makes five equal annual payments after the graduation (end-of-period payments). a) What is each payment after the graduation? b) Calculate IRR of loan? (hint: use cash flow from when the student started borrowing the money to when it is all paid back) c) Is the loan attractive to the student? 18Engineering Economics

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EXAMPLE CONTINUES Year Cash Flow 010,000 1 2 3 40 5(9240) 6 7 8 9 a) b) 19Engineering Economics

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INTERPOLATION: c) Since the rate is low, the loan looks like a good choice. 20Engineering Economics

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INTERPOLATION: c) Since the rate is low, the loan looks like a good choice 21Engineering Economics

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a) pmt(5%, 5, -40000) = $9,238.99 per month. b) irr(g1:g10) = 2.66%. c) Since the rate is low, the loan looks like a good choice. 22Engineering Economics EXCEL Solution

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Plot of NPW versus Interest Rate Borrowing Cases YearCash Flow 0200 1-50 2 3 4 5 :: :: 23 p. 218 Engineering Economics

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Plot of NPW versus Interest Rate Investment Cases YearCash Flow 0-200 150 2 3 4 5 :: :: 24Engineering Economics

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Fees or Discounts Question: Option 1: If a property is financed through a loan provided by a seller, its price is $200,000 with 10% down payment and five annual payments at 10%. Option 2: If a property is financed through the same seller in cash, the seller will accept 10% less. However, the buyer does not have $180,000 in cash. What is the IRR for the loan offered by seller? 25Engineering Economics

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QUESTION CONTINUES YearPay Cash Borrow from Seller 0($180,000)($20,000) 1($47,484) 2 3 4 5 26Engineering Economics

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QUESTION CONTINUES INTERPOLATION: 27Engineering Economics

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QUESTION CONTINUES INTERPOLATION: This is a relatively high rate of interest, so that borrowing from a bank and paying cash to the property owner is better. 28 Engineering Economics

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EXCEL Solution Combined cash flows (difference between options 1 & 2): At time 0:-$160,000 EOY 1-5: $47,484 IRR = rate(5, 47484, -160000) = 14.78% per year IRR = irr(a1:a6) = 14.78% per year 29 Engineering Economics

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Loan and Investments are Everywhere Question: A student will decide whether to buy weekly parking permit or summer semester parking permit from USF. The former costs $16 weekly; the latter costs $100 due May 17 th 2010; in both cases the duration is 12 weeks. Assuming that the student pay the weekly fee on every Monday: a) What is the rate of return for buying the weekly permit? b) Is weekly parking attractive to student? *Total 12 weeks 30Engineering Economics

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QUESTION CONTINUES WeekWeeklySemester May 170($16)($100) May 241($16) May 312($16) June 73($16) June 144($16) June 215($16) June 286($16) July 57($16) July 128($16) July 199($16) July 2610($16) August 211($16) a) To find IRR%, set cash flows equal in PW terms 100 = 16 16 (P/A,i%,11) (P/A,i%,11) = (100 - 16) / 16 (P/A,i%,11) = 5.25 Looking in the table for the above value: IRR = 15% b) Nominal interest rate for 52 weeks IRR 15%/week or 15*52 = 780%/yr Since the rate is high, paying the semester fee looks like a good choice. 31 Effective annual interest = (1+.15)^52 1 = 1432% ! Engineering Economics

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EXCEL Solution Combined cash flows (difference between the 2 options): At time 0:-$84 EOM 1~11: $16 IRR = rate(11, 16, -84) = 14.92% per month IRR = irr(C1: C12) = 14.92% per month 32Engineering Economics

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Rate Of Return Calculations Question: There are two options for an equipment: Buy or Lease for 24 months. The equipment might be either leased for $2000 per month or bought for $30,000. If the plan is to buy the equipment, the salvage value of the equipment at EOM 24 is $3,000. What is the IRR or cost of the lease? 33Engineering Economics

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QUESTION CONTINUES MonthBuy OptionLease Option 0($30,000)($2,000) 1-23($2,000) 24$3,0000 To find IRR%, set cash flows of Buy and Lease options equal in PW terms 34Engineering Economics

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35Engineering Economics Try i = 5% 28000 3000(0.3101) 20000(13.489) = 91.7 Try i = 4% 28000 3000(0.3477) 2000(14.148) = -1339.1 i 5% per month

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Combined cash flows (difference of buy & lease): At time 0:-$28000 EOM 1~23: $2000 EOM 24: $3000 IRR = irr(e1:e25) = 4.97% per month EXCEL Solution 36Engineering Economics

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When there are two alternatives, rate of return analysis is often performed by computing the incremental rate of return, IRR, on the difference between the two alternatives. Incremental Analysis 37Engineering Economics

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Incremental Analysis The cash flow for the difference between alternatives is calculated by taking the higher initial-cost alternative minus the lower initial-cost alternative. The following decision path is made for incremental rate of return (IRR) on difference between alternatives: Two -Alternative Situations Decision IRRMARRChoose the higher-cost alternative IRR