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Notes 6: Annual cash flow analysis Chapter 6: Techniques for cash flow analysis Annual cash flow analysis : Analysis Period = Alternative Lives Analysis Period = Common Multiple of Alt. Lives Analysis Period for a Continuing Requirement Infinite Analysis Period Annual Percentage Rate 1 Techniques for cash flow analysis Present Worth Analysis Annual cash flow analysis Rate of Return Analysis Incremental Analysis Other Techniques: Future Worth Analysis Benefit-Cost Ration Analysis Payback Period Analysis Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 2 Techniques for cash flow analysis Present Worth Analysis: PVA(0)=-RA+A(P/A,i,n)+SA(P/F,i,n) PVB(0)=-RB+B(P/A,i,n)+SB(P/F,i,n) If PVA(0)>PVB(0) => choose A, otherwise => choose B. SA A A A A 0 RA B 1 B 2 B 3 B n SB Annual cash flow analysis : EUACA=RA(A/P,i,n) EUABA=A+SA(A/F,i,n) (EUAB-EUAC)A=A+SA(A/F,i,n)- RA(A/P,i,n) (EUAB-EUAC)B=B+SB(A/F,i,n)- RB(A/P,i,n) (EUAB-EUAC)A (EUAB-EUAC)A (EUAB-EUAC)A (EUAB-EUAC)A 0 RB 1 2 3 n If (EUAB-EUAC)A>(EUAB-EUAC)B (EUAB-EUAC)A => choose A, otherwise => choose B. (EUAB-EUAC)B (EUAB-EUAC)B (EUAB-EUAC)B (EUAB-EUAC)B (EUAB-EUAC)B 0 1 2 3 n 0 1 2 3 n3 Analysis Period Equals to Alternative Lives When the analysis period for an economy study is the same as the useful life for each alternative, we have an ideal situation. There are no difficulties. The economy study can be based on the analysis period. 4 Analysis Period = Common Multiple of Alternative Lives Example Two pumps are being considered for purchase. If interest is 7%, which pump should be bought. $1,500 Assume that Pump B will be replaced after its useful life by the same one Pump Initial Cost End-of-Useful-Life Salvage Value Useful life, in years A $7,000 $1,500 12 B $5,000 $1,000 6 0 1 2 3 4 5 6 7 8 9 10 11 12 $7,000 $1,000 0 1 2 3 4 5 6 7 8 9 $1,000 10 11 12 $5,000 $5,000 replace B EUACA = $7,000 (A/P, 7%, 12) - $1,500 (A/F, 7%, 12) EUACB = $5,000 (A/P, 7%, 6) - $1,000 (A/F, 7%, 6) EUACA = $797 EUACB = $909 Choose Pump A Under the circumstances of identical replacement, it is appropriate to compare the annual cash flows computed for alternatives based on their own different service lives (12 years, 6 years). 5 Analysis Period for a Continuing Requirement Many times the economic analysis is to determine how to provide for a more or less continuing requirement. There is no distinct analysis period. The analysis period is undefined. In case when alternatives were compared based on PW analysis, the least common multiple of alternative lives was found, and present worth for that time is calculated. In case alternatives are compared based on annual cash flow analysis , it is appropriate to compare the annual cash flows computed for alternatives based on their own different service lives. Example Pump Initial Cost End-of-Useful-Life Salvage Value Useful life, in years A $7,000 $1,500 12 B $5,000 $1,000 9 EUACA = $797 EUACB = $684 6 Infinite Analysis Period Motivating Example Consider the following three mutually exclusive alternatives: Alternative Cost Uniform Annual Benefit Useful Life, in years A $100.00 $10.00 Infinity B $150.00 $17.62 20 C $200.00 $55.48 5 Assuming that Alternatives B and C are replaced with identical replacements at the end of their useful lives, and an 8% interest rate, which alternative should be selected? Case 1. We have alternatives with limited (finite) lives in an infinite analysis period situation: If we assume identical replacement (all replacements have identical cost, performance, etc.) then we will obtain the same EUAC for each replacement of the limited-life alternative. The EUAC for the infinite analysis period is therefore equal to the EUAC for the limited life situation. With identical replacement: EUACfor infinite analysis period = EUAC for limited life n EUACB= $150(A/P,8%,20) - $17.62 EUACC= $200(A/P,8%,5) - $55.48 7 Infinite Analysis Period Case 2. Another case occurs when we have an alternative with an infinite life in a problem with an infinite analysis period. In this case, EUACfor infinite analysis period = P (A/P,i, ) + any other annual (costs-benefits) Alternative Cost A $100.00 $10.00 Infinity B $150.00 $17.62 20 C $200.00 $55.48 5 (A/P,i, ) = 1/i Uniform Annual Benefit Useful Life, in years EUACA= $100(A/P,8%, ) - $10.00 EUACB= $150(A/P,8%,20) - $17.62 EUACC= $200(A/P,8%,5) - $55.48 = $100 *0.08 - $10.00 = $-2.00 = $150*0.1019- $17.62 =$-2.34 = $200*2.505 - $55.48 =$-5.38 The most general case for annual cash flow analysis is when the analysis period and the lifetimes of the alternatives of interest are all different. In this case, terminal values at the end of the analysis period become very important. 8 Annual Percentage Rate (APR) Mortgage Broker Advertisement Rate 7.625 7.875 8.125 7.000 7.500 Points 1.00 0.50 2.25 1.00 1.00 Term 30 yr 30 yr 30 yr 15 yr 15 yr Max amt $203,150 $203,150 $600,000 $203,150 $600,000 APR 7.883 8.083 8.399 7.429 7.859 Our focus Rate the base interest rate not including any fees and expenses associated with the mortgage Points percentage of loan amount charged up-front for providing the mortgage Terms the time length to pay off the mortgage Max amt the amount borrowed for the mortgage APR the rate of interest that, if applied to the loan amount with all fees and expenses, would result in a monthly payment of A. All the fees and expenses are added to the loan balance, and the sum is amortized at the stated rate over the stated period. This results in a fixed monthly payment amount, A. 9 Mortgage Broker Advertisement APR Example Example: Base Rate: APR: Term: Loan amount: r = 7.625%, ra = 7.883%, 30 years P = $203,150 Our focus Rate 7.625 7.875 8.125 7.000 7.500 Points 1.00 0.50 2.25 1.00 1.00 Term 30 yr 30 yr 30 yr 15 yr 15 yr Max amt $203,150 $203,150 $600,000 $203,150 $600,000 APR 7.883 8.083 8.399 7.429 7.859 number of periods: n = 12 x 30 rate with fees and expenses: ra = 0.07883/12 Capital Recovery Factor: (A/P,n,ra) = 0.007256 monthly payment: Aa = P (A/P,n,ra) = 360 = 0.006569 = $203,150 * 0.007256 = $1,474.11 We have computed the monthly payment. This amount includes amortized fees and expenses. 10 APR Example (Cont d) By contrast, at 7.625% compounded monthly for 360 month, n=360 r=0.007625/12 P=$203,150 A = P (A/P,n,r) = $1,437.88 we would only need to pay $1,437.88 a month. The difference Aa - A = $1,474.11 - $1,437.88 = $36.23 is because of the amortized fees and expenses. Rate without fees and expenses: Multiplier: Present value of payments: r = 0.07625/12 (P/A,n,r) Pa = Aa (P/A,n,r) = 0.006354167 = 141.2840966 = $208,267 Total initial balance = $208,267 This is the present value of the payments of A each month for 360 months, at 7.625%. We see it exceeds the amount of the loan, because Aa includes the other (amortized) fees and expenses. 11 APR Example (Cont d) Total fees and expenses $208,267 - $203,150 = $5,117 Alternatively, at r=7.625%, with n = 60, the PV of the fees and expenses CFS is $36.23 (P/A,n,r) = $36.23 * 141.2840966 $5,118 Loan fee is 1% of $203,150, $2,032. Other expenses = $5,117 - $2,032 = $3,085 (about 1.52% of the loan amount) If we know the amount of the loan, we can compute the other expenses the broker would charge, although they are not in the ad. 12 Summary, Mortgage Example, 30 year loan Data APR = ra = 7.883% with fees & expenses rate = r = 7.625% without fees & expenses n = 30 12 = 360 payments P = $203,150 is the maximum loan amount Monthly Payment With fees & expenses ra = 0.07883/12, n = 360 (A/P,n,ra) = 0,007256 Aa = P (A/P,n,ra) = $1,474.11 monthly payment with amortized fees and expenses Monthly Payment Without fees & expenses r = 0.07625/12, n = 360 (A/P,n,r) = 0,007078 A = P (A/P,n,r) = $1,437.88 monthly payment w/o fees & expenses 13 Summary, Mortgage Example (Cont d) Monthly difference of two CFS s: Aa - A = $1,474.11- 1,437.88 = $36.23 is the amortized fees and expenses. PV of Monthly difference: At r=7.625%, with n = 60, the PV of the fees and expenses CFS is $36.23 (P/A,n,r) = $36.23 * 141.2840966 $5,118 PV of monthly payments with fees & expenses: Alternatively, at r=7.625% with n = 60, the PV of the monthly payment of Aa=$1,474.11 is Aa (P/A,n,r) = $1,474.11 * 141.2840966 = $208,267 Total fees & expenses: Loan fee: Other expenses: $208,267 - $203,150 = $5,117 is 1% of $203,150 $2,032. $5,117 - $2,032 = $3,085 (about 1.52% of the loan amount). 14 Techniques for cash flow analysis Present Worth Analysis Annual cash flow analysis Rate of Return Analysis Incremental Analysis Other Techniques: Future Worth Analysis Benefit-Cost Ration Analysis Payback Period Analysis Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 15

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