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Chapter 6: Annual Cash Flow Analysis

Chapter 6: Annual Cash Flow Analysis. Annual Percentage Rate. 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

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Chapter 6: Annual Cash Flow Analysis

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  1. Chapter 6: Annual Cash Flow Analysis Annual Percentage Rate 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

  2. Techniques for Cash Flow Analysis • Present Worth Analysis • Annual Cash Flow Analysis • Rate of Return Analysis • Other Techniques: • Future Worth Analysis • Benefit-Cost Ration Analysis • Payback Period Analysis • Chapter 5 • Chapter 6 • Chapter 7 • Chapter 9

  3. “present” P-Pattern 1 2 3 n “future” F-Pattern 1 2 3 n “annual” A-Pattern 1 2 3 n “gradient” G-Pattern 1 2 3 n Cash Flow Diagrams

  4. Equivalence of Cash Flow Patterns

  5. Annual Equivalent Cost Comparison • Incomes are converted to an A-pattern. • Costs are converted to an A-pattern. • The costs are subtracted from the incomes to determine the ANEV. • Mutually Exclusive Alternatives – choose the one with highest ANEV • Independent Alternatives – choose all with positive ANEV ANEV: Annual Net Equivalent Value

  6. 50k 50k 50k 50k 50k 50k 50k 1 2 3 4 5 6 7 P Example: A new circuit board component insertion tool will save $50,000 in production costs each year and will have a life of seven years. What is the highest price that can be justified for the tool using a 12% interest rate? Solution:

  7. 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. EUAC Equivalent Uniform Annualized Cost EUAB Equivalent Uniform Annualized Benefit

  8. Techniques for Cash Flow Analysis SA A A A A 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. 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) If (EUAB-EUAC)A>(EUAB-EUAC)B => choose A, otherwise => choose B. 0 1 2 3 … n RA B B B B SB 0 1 2 3 … n RB (EUAB-EUAC)A (EUAB-EUAC)A (EUAB-EUAC)B (EUAB-EUAC)B (EUAB-EUAC)A (EUAB-EUAC)A (EUAB-EUAC)A (EUAB-EUAC)B (EUAB-EUAC)B (EUAB-EUAC)B 0 1 2 3 … n 0 1 2 3 … n

  9. Analysis Period = Common Multiple of Alternative Lives Example Two pumps are being considered for purchase. If interest is 7%, which pump should be bought. Assume that Pump B will be replaced after its useful life by the same one 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 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). $1,500 0 1 2 3 4 5 6 7 8 9 10 11 12 $7,000 $1,000 $1,000 0 1 2 3 4 5 6 7 8 9 10 11 12 $5,000 $5,000 replace B Choose Pump A

  10. Example: A new circuit board component insertion tool is needed. Which should you buy? Solution: The EUAC is calculated for each: JACO: Cheepo: JACO

  11. 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 EUACA = $797 EUACB = $684

  12. Infinite Analysis Period Motivating Example Consider the following three mutually exclusive alternatives: 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

  13. 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) (A/P,i,) = i EUACA= $100(A/P,8%,) - $10.00 = $100 *0.08 - $10.00 = $-2.00 EUACB= $150(A/P,8%,20) - $17.62 = $150*0.1019- $17.62 =$-2.34 EUACC= $200(A/P,8%,5) - $55.48 = $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.

  14. Annual Percentage Rate (APR) • 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.”

  15. APR Example Example: Base Rate: r = 7.625%, APR: ra = 7.883%, Term: 30 years Loan amount: P = $203,150 number of periods: n = 12 x 30 = 360 rate with fees and expenses: ra = 0.07883/12 = 0.006569 Capital Recovery Factor: (A/P,n,ra) = 0.007256 monthly payment: Aa = P (A/P,n,ra) = $203,150 * 0.007256 = $1,474.11 We have computed the monthly payment. This amount includes amortized fees and expenses.

  16. 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: r = 0.07625/12 = 0.006354167 Multiplier: (P/A,n,r) = 141.2840966 Present value of payments: Pa = Aa (P/A,n,r) = $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.

  17. 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.

  18. 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

  19. 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 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: $208,267 - $203,150 = $5,117 Loan fee: is 1% of $203,150  $2,032. Other expenses: $5,117 - $2,032 = $3,085 (about 1.52% of the loan amount).

  20. Techniques for Cash Flow Analysis • Present Worth Analysis • Annual Cash Flow Analysis • Rate of Return Analysis • Other Techniques: • Future Worth Analysis • Benefit-Cost Ration Analysis • Payback Period Analysis • Chapter 5 • Chapter 6 • Chapter 7 • Chapter 9

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