530solution-CH9 - 9-32 —m m Cost includin land $500,000 $900,000 $2,200,000 6 44 m Annual Income A $70,000 $105,000(“ 0.4 l Salvae Value

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Unformatted text preview: 9-32 —m m Cost includin land $500,000 $900,000 $2,200,000 6- 44 m Annual Income A $70,000 $105,000 (“+0.4 l Salvae Value $200,000 $300,000 BIC Ratio Analysis - PW of Salvage Value = F PIF, 8%, 20 = 0.2145F $42,900 Iimm— $835,650 $2,114,200 PW of Benefit = A (PIA, 3%, $687.260 $1 330,890 20 = 9.818A - - Benefit/PW of Cost Incremental BIC Ratio Anal!“3 5 Stories Rather than 2 10 Stories Rather than — Stories 2 Stories A PW of Cost $835,650 - $457,100 $2,114,200 - $457,100 =51.657.100 APW of Benefii $1,030,890 - $687,260 E$3.3.2.tfii;11052)$687.260 = $343,630 = 1 I 113100 = APW oi $343,6301$378.550 $11,812:.1solst 557,100 Benefits/APW of Costs = 0.91 = - _ ' ' ' ‘ncrement. >1 Desnable “”"“°”"‘*”“ increment. Re' 5 stories. With 03/130 = 0.91, the increment of 5 stories rather than 2 stories is undesirable. The 10 stories rather than 2 stories is desirable. Conclusion: Choose the 10-May alternative. 9-33 Note that the three alternatives have been rearranged below in order of increasing cost. _ mm:— Uniform Annual Benefit Salva : Value Comute BIC Ratio A First Cost A Uniform Annual $60 Benefit A Salvae Value Compute AB/AC value Beneflt- Cost Ratio Computations: Alternative A: BIG [$140 (PIA, 10%. 6)]f[$560 - $40 (P/F, 10%, 6)] $130 (4.355)]I($560 - $40 (05645)] Alternative B: BIC = [$100 (PIA, 10% 6)]/$340 = 1 I‘ .28 Alternative C: BIC = [$40 (PIA. 10%. (SH/$120 = 1.45 Incremental Mews“: B- C ABIAC = [$60 (PM. 10%. 6)]/$220 = 1.19 B- C is a desirable increment. A- B AB/AC = [$40 (PIA, 10%, 6)l[$220 - $40 (P/F, 10%. 6)] = 0.88 A- B is an undesirable increment. Conclusion: Choose B. The solution may be checked by Net Present Worth or Rate of Return NPW Solution NPWA = $140 (PIA, 10%, 6) + $40 (PIF, 10%, 6) - $560 = $140 (4.355) + $40 (0.5645) - $560 = +$72.28 NPWB = $100 (PIA, 10%, 6) - $340 = +$95.50 Nch = $40 (PIA. 10%. 6) - 5120 = +$54.2o Select B. Rate of Return Solution _ Benefit De > 10% ACOBt B. Re'ect C. cision Select B. 9-44 (a) Net Future Worth 6.6% < 10% Re'ect A. Illlll Illlll NFWA = $18.8 (FIA, 10%, 5) - $75 (FIP, 10%, 5) NFWa = $13.9 (FM. 10%. 5) - $50 (PIP. 10%. 5) NFWc = $4.5 (FM, 10%, 5) - $15 (FIP. 10%, 5) NFWD = $23.5 (FM, 10%. 5) - $90 (FIP, 10%, 5) II II II I! 9:5??? 2238 1 Select B. (b) Incremental Analysis Computed Uniform Annua! $3.96 $13.19 $19.78 3 Cost UAC Conclusion: Select B. (c) Payback Period Payback A = $753183 = 4,0 Paybacka = $508133 = 3.6 Paybackc = 315/545 = 3.3 (_ Paybacko = $905233 = 33 To minimize Payback. select 0. 9-45 Alternative A A = $28.8 ssossossossossossovo NFWA = $28.8 (FM. 12%, 12) - $50 (NP, 12%, 2) (PIA, 12%, 12) z $23.3 (24.133) - $50 (0.5917) (24.133) = 013.94 Alternative B A = $39.6 $150 $150 *Fw NFWB = $39.6 (FM, 12%, 12) - $150 (HP, 12%, 6) - $150 (HP, 12%, 12) = $39.6 (24.133) - $150 [1.974 + 3.896] = +$75.17 Alternative C A = $39.6 E $1i0 $110 3110 VW NFWc = $39.6 (FIA. 12%, 12) - $110 (FIP, 12%. 4) - $110 (FIP. 12%, 8) — $110 F/P.12%.12) = $39.6 (24.133) - $110 [1.574 + 2.476 + 3.896] = $81.61 Choose Altmtive g because it maximizes Future Worth. (b) Solve by Benefit-Cost ratio analysis With neither input nor output fixed. incremental analysis is required. Alternative c- Altematlve A For the Increment c- A: PW of Cost = $60 PW of Benefits = $10.8 (PIA, 12%, 4) + $50 (P/F, 12%, 2) = $10.3 (3.037) + $50 (0.7972) = $72.66 Twelve years is a suitable analysis period for Alternatives B and C- For the Increment B- c lgnori the tential drfficz‘ ; ' - . cash 33w: p0 "‘3 s'gna'ed by 3 sign changes in the e- c PW 01 Cost = $40 + $150 (P/Fl 12%, 6) = $40 + $150 (0.5066) = $115.99 PW of Benefits = $110 (PIF, 12%, 4) + $110 (PIF. 12%, 8) = $110 (0.6355) + $110 (0.4039) = $114.33 AB/AC = PW of Benefits/PW of Cost = $114.33]$115.99 < 1 The increment is undesirable and therefore Alternative C is preferred over Alternative B. Alternative Analysis of the Increment B- c An examination of the B— C cash flow suggests there is an external investment of money at the end of Year 4. Using an external interest rate (say 12%) the +$110 at Year 4 becomes: +$110 (FIP, 12%, 2): $110 (1.254) = $137.94 at the end of Yr. 6. The altered cash flow becomes: '51 50 +$1 37.94 For the altered B- C cash flow: PW of Cost = $40 + $12.06 (WP, 12%, 6) = $40 + $12.06 (0.5066) = $46.11 PW of Benefits = $110 (HF. 12%, 8) = $110 (0.4039) = $44.43 ABIAC = PW of Benefitle of Cost = $44.43/$46-11 < 1 The increment is Alternative B. Solutions for part (b): Qhoose Alternative Q. (C) (d) Payback Period = $50!$28.8 = 1.74 yr = 5150/5395: 3.79 yr = $1501$39.6= 2.78 yr To minimize the Payback Period, choose Alternative A. Alternative A: Payback Alternative B: Payback Alternative C: Payback Payback period is the time required to recover the investment. Here we have three alternatives that have rates of return varying from 10% to 16.4%. Thus each generates uniform annual benefits in excess of the cost. during the life of the alternative. From this is must follow that the alternative with a 2-year life has a payback period less than 2 years. The alternative with a 4-year life has a payback period less than 4 years, and the alternative with a 6-year life has a payback period less than 6 years. Thus we see that the shorter-lived asset automatically has an advantage over longer-lived alternatives in a situation like this. While Alternative A takes the shortest amount of time to recover its investment. Alternative C is best for long-term economic efficiency. ...
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This note was uploaded on 04/07/2008 for the course IMSE 530 taught by Professor Chang during the Fall '07 term at Kansas State University.

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530solution-CH9 - 9-32 —m m Cost includin land $500,000 $900,000 $2,200,000 6 44 m Annual Income A $70,000 $105,000(“ 0.4 l Salvae Value

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