Engineering Economics EGN3615

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HW-7 * Read Newnan Chap 7. Work problems: 12th Ed. pp. 207ff: 6-1, 6-3, 6-8, 6-10, 6-21, 6-22, 6-37, 6-49, 6-54 Instructions Do all your work below; use this as a text file or use the Excel functions if you need them. List each problem number and do the work here. Save this file to your computer as you work. When you are finished, go back to the Assignments page where you found this and open the Assignment. You will find a section called “Assignment Materials.” Near the bottom of that section is “Attach File” and a button to Browse for Local File. Again do your work, save it, and them use the above to attach it. Do your work here hem. open the on is “Attach File” 12th Ed. pp. 211ff: 6-1, 6-3, 6-8, 6-10, 6-21, 6-22, 6-37, 6-49, 6-54 6-1 E = $6,000− $1,500 (A/G, 12%, 4)= $6,000 − $1,500 (1.359) = $3,962 6-3 ALTERNATIVE p n int/yr Salv EUAC Eff int EUAC First, compute A: A = ($20,000 − $4,000) (A/P, 4%, 10) + $4,000 (0.04) = $16,000 (0.1233) + $160= $2,132.80 per semiannual period Now, compute the equivalent uniform annual cost: EUAC = $2,132.80 (F/A, 4%, 2)= $2,132.80 (2.040)= $4,350.91 6-8 A = F[(er −1)/(em − 1)] = $5 × 106 [(e0.15 − 1)/(e(0.15)(40) − 1)] = $5 × 106 [0.161834/402.42879] = $2,011 6-10 Solution at back of book is not correct. 1 2 3 4 5 6 1.4 1.5 1.6 1.7 1.8 1.9 7 8 9 10 npv= EUAC= Or = 2 2.1 2.2 2.3 $9.94 ($1.76) 1.4 + .1 * (A/G,0.12,10) 1.7585 6-21 P = $1,000 (P/A, 9%, 5) + $3,500 (P/F, 9%, 4) + $1,500 (P/A, 9%, 5) (P/F, 9%, 5) + $3,500 (P/F, 9%, 8) = $1,000 (3.890) + $3,500 (0.7084) + $1,500 (3.890) (0.6499) + $3,500 (0.5019) = $11,918 Equivalent Uniform Annual Amount = $11,918 (A/P, 9%, 10) = $1,857 6-22 This problem is much harder than it looks! EUAC = {$600 (P/A, 8%, 5) + $100 (P/G, 8%, 5) + [$900 (P/A, 8%, 5) − $100 (P/G, 8%, 5)][(P/F, 8%, 5)]}{(A/P, 8%, 10)} = {$600 (3.993) + $100 (7.372) + [$900 (3.993) − $100 (7.372)][0.6806]}{0.1490} = $756.49 PVa(0) PVb(0) PV(0) EUAC 3133 1944.1339 0.149 5077.1339 756.4929511 6-37 EUAC Comparison Gravity Plan Initial Investment: = $2.8 million (A/P, 10%, 40) = $2.8 million (0.1023) = $286,400 Annual Operation and maintenance = $10,000 Annual Cost = $296,400 Pumping Plan Initial Investment: = $1.4 million (A/P, 10%, 20) = $1.4 million (0.1023) = $143,200 Additional investment in 10th year: = $200,000 (P/F, 10%, 10) (A/P, 10%, 40) = $200,000 (0.3855) (0.1023) = $7,890 Annual Operation and maintenance = $25,000 Power Cost: = $50,000 for 40 years = $50,000 Additional Power Cost in last 30 years: = $50,000 (F/A, 10%, 30) (A/F, 10%, 40) = $50,000 (164.494) (0.00226) = $18,590 Annual Cost = $244,680 Select the Pumping Plan. 6-49 Annual Cost of Diesel Fuel = [$80,000km/(16 km/liter)] × $0.88/liter = $4,400 Annual Cost of Gasoline = [$80,000km/(11 km/liter)] × $0.92/liter = $6,691 EUACdiesel = ($24,000 − $4,000) (A/P, 6%, 5) + $4,000 (0.06) + $4,400 fuel + $900 repairs + $1,000 insurance = $20,000 (0.2374) + $240 + $6,300 = $11,288 EUACgasoline = ($19,000 − $6,000) (A/P, 6%, 4) + $6,000 (0.06) + $6,691 fuel + $700 repairs + $1,000 insurance = $12,503 The diesel taxi is more economical. 6-54 It is important to note that the customary “identical replacement” assumption is not applicable here. Alternative A EUAB – EUAC = $1,500 − $5,000 (A/P, 15%, 10) = $1,500 − $5,000 (0.1993) PS solution Interest= 0.1 Years Grav 0 2800000 10 Ann yr 1 to 10 yr 11-40 EUAC 10000 $296,326.36 = +$504 Alternative B EUAB – EUAC = $6,000 (P/A, 15%, 5) (A/P, 15%, 10) − $18,000 (A/P, 15%, 10) = +$421 Choose A. Check solution using NPW: Alternative A: NPW = $1,500 (P/A, 15%, 10) − $5,000 = +$2,529 Alternative B: NPW = $6,000 (P/A, 15%, 5) − $18,000 = +$2,112 -20000 5 0.08 4000 0.0816 (annually) $4,350.62 21753.08211 10 0.04 Alternative way npv= pmt= 600 700 800 900 1000 900 800 700 600 500 $5,076.75 ($756.59) Pump $1,400,000.00 One-time $200,000.00 One-time $25,000.00 $50,000.00 $100,000.00 $77,108.66 $244,476.27 $307,228.36 $363,448.36 $2,392,261.64 $244,631.27 PV for 200k for 10 yrs PV for 25,000 for 40 yrs PV of 50,000 for 10 yrs PV for yrs 11-40 which are then moved to yr 0 Total PV 12th Ed. pp. 207ff: 6-1, 6-3, 6-8, 6-10, 6-21, 6-22, 6-37, 6-49, 6-54 Classroom session 10-15-15 6-1 Int== Year 0.12 0 1 2 3 4 6000 4500 3000 1500 using PV ($5,357.14) ($3,587.37) ($2,135.34) ($953.28) Note: When seeking EUAB, EUAC or EU the best method will be to use P Using Gradient 6000 6000 6000 6000 0 -1500 -3000 -4500 PW (NPV)= $12,033.13 PW (PV) ($12,033.13) EUAB= $3,961.72 using PMT -1500(A/G,.12,4) 6000 -2038.5 3961.5 EUAW=EUAB 6-22 Int= 0 1 2 3 4 5 6 7 8 9 10 PV= 600 700 800 900 1000 900 800 700 600 500 $5,076.75 0.08 using PV ($555.56) ($600.14) ($635.07) ($661.53) 661.5269 ($680.58) ($567.15) ($466.79) ($378.19) ($300.15) ($231.60) ($5,076.75) $756.59 -5 -4 -3 -2 -1 0 1 2 3 4 600 700 800 900 1000 900 800 700 600 500 When seeking EUAB, EUAC or EUAW the best method will be to use PMT(). This was calulated using the (1+i )^n method. It shows that the result is the same as the red numbers. The red numbers uses PV(). Note: To use PMT you must have a value at either the beginning of the time range (0, or in this case, -6), or at the end (4 or 10 depending on which set of years you are using. You cannot use the 0 time when it is in the middle of the range. NOTE: At the bottom of this window are labeled tabs. TAB: HW-6 is this sheet with the new homework HW - 6 Problems due the day BEFORE the next class. * Read Newnan Chap. 5, 6 12th Ed. pp. 173ff: 5-1, 5-2, 5-3, 5-5, 5-9, 5-15, 5-32, 5-38, 5-37, 5-43, 5-61, 5-83 Instructions Do all your work below; use this a text file or use the Excel functions if you need them. List each problem number and do the work here. Save this file to your computer as you work. When you are finished, go back to the Assignments page where you found this and open the Assignment. You will find a section called “Assignment Materials.” Near the bottom of that section is “Attach File” and a button Again do your work, save it, and them use the above to attach it. Do your work here on is “Attach File” and a button to Browse for Local File. 12th Ed. pp. 173ff: 5-1, 5-2, 5-3, 5-5, 5-9, 5-15, 5-32, 5-38, 5-37, 5-43, 5-61, 5-83 On some I added notes--PS Also see the Recitation Excel Files on this assignment when posted. 5-1 NPW at start of year on 1/1/2018 is same as NWP at end of year 2017 NPW12/31/2017= –$14,000 NPW12/31/2015= –$14,000 (P/F, 9%, 2) = –$14,000 (0.8417) = –$11,784 0,09 ($11.783,52) 5-2 P = $18,000 (P/A, 9%, 10) + $245,000 (P/F, 9%, 10) = $18,000 (6.418) + $245,000 (0.4224) = $219,012 P= ($219.008,49) 0,09 ($115.517,84) 0,09 ($103.490,65) 5-3 P = the first cost = $980,000 ($98.000,00) F = the salvage value = $20,000 0,12 AB = the annual benefit = $200,000 $1.130.044,61 Remember our convention of the costs being negative and the benefits being 0,12 positive. Also, remember the P occurs at time = 0. $113.004,46 NPW = –P + AB (P/A, 12%, 10) + F (P/F, 12%, 10) = –$980,000 + $200,000 (5.650) + $20,000 (0.3220) = $156,440 Therefore, purchase the machine, as NPW is positive. PS note: You can also do it these two ways. Doulble-click on the two cells to see how I did it. Note the slightly different answer due to rounding in the tables used by the author.: NPV= $156.484,07 $156.484,07 5-5 P = ?, n = 36 months, i = 15%/12mo = 1.25%/month, A = $500 P = $500 (P/A, 1.25%, 36) = $500 (28.847) = $14,424 0,0125 ($14.423,63) 5-9 NPW = -$600,000 - $100,000(P/A,15%,6) - $5,000(P/G,15%,6) + $300,000(P/A,15%,6) -$15,000(P/G,15%,6) + $100,000(P/F,15%,6) NPW = -$600,000 + ($300,000 - $100,000)(P/A,15%,6) + (-$5,000 - $15,000)(P/G,15%,6) + $100,000(P/F,15%,6) NPW = -$600,000 + $200,000(3.784) - $20,000(7.937) + $100,000(0.4323) NPW = $41,290 Given a positive NPW, the plan should be accepted. 5-32 36 Compute the PW of Cost for a 25-year analysis period. Note that in both cases the annual maintenance is $100,000 per year after 25 years. Thus after 25 years all costs are identical. Single Stage Construction PW of Cost = $22,400,000 + $100,000 (P/A, 4%, 25) = $22,400,000 + $100,000 (15.622) = $23,962,000 Two Stage Construction PW of Cost = $14,200,000 + $75,000 (P/A, 4%, 25) + $12,600,000 (P/F, 4%, 25) = $14,200,000 + $75,000 (15.622) + $12,600,000 (0.3751) = $20,098,000 Choose two stage construction. 5-37 NPW for 10 years of alternative A (identical equipment replace at year 5): NPW(A) = -$4,500 + ($500 - $4,500)(P/F,6%,5) - $300(P/A,6%,10) + $500(P/F,6%,10) NPW(A) = -$4,500 - $4,000(0.7473) - $300(7.360) + $500(0.5584) = -$9,418 Alternatively, we could write this showing the two lives as: NPW(A) = -$4,500 - $300(P/A,6%,5) + $500(P/F,6%,5) - $4,500(P/F,6%,5) - $300(P/A,6%,5)(P/F,6%,5) + $500(P/F,6%,5)*(P/F,6%,5) If you notice the second 3 terms are the same as the first three except for the 5 year time difference. Therefore, we could rewrite this as: NPW(A) = [-$4,500 - $300(P/A,6%,5) + $500(P/F,6%,5)][1 + (P/F,6%,5)] NPW(A) = [-$4,500 - $300(4.212) + $500(0.7473)][1 + (0.7473)] = -$9,418 NPW for 10 years of alternative B: NWP(B) = -$5,500 - $400(P/A,6%,10) = -$5,500 - $400(7.360) = -8,444 Alternative B is the better of the two, yielding a lower present worth of the costs. 5-38 To equate the lives we will replace alternative A at the end of year 10 with an identical process. PW(A) = $18M [1 + (P/F, 9%, 10)] + $5M (P/A, 9%, 20) − $4M[(P/F, 9%, 10) + (P/F, 9%, 20)] PW(A) = $18M [1 + (0.4224)] + $5M (9.129) − $4M[(0.4224) + (0.1784)] PW(A) = $68.8M PW(B) =$25M + $3M (P/A, 9%, 20) − $6M (P/F, 9%, 20) PW(B) =$25M + $3M (9.129) − $6M (0.1784) PW(B) = $51.3M PW(A) - PW(B) = $69.8M - $51.3M = $17.5M Select Alternate B which saves $17.5M 5-43 To equate the lives we will replace alternative A at the end of year 2 with an identical process. PW(A) = $200,000 [1 + (P/F, 2%, 8)] + $10,000 (P/A, 2%, 16) − $25,000 [(P/F, 2%, 8) + (P/F, 2%, 16)] PW(A) = $200,000 [1 + (.8535)] + $10,000 (13.578) − $25,000 [(.8535) + (0.7284)] PW(A) = $466,933 PW(B) =$250,000 + $15,000 (P/A, 2%, 16) − $40,000 (P/F, 2%, 16) PW(B) =$250,000 + $15,000 (13.578) − $40,000 (0.7284)= $424,534 Select Alternate B. 5-61 Compute the present worth of the cash flow: PW = ($725,000 + $143,000) + $260,000 (P/F, 9%, 25) + $12,000 (P/A, 9%, 40) - $32,000 (P/F, 9%, 40) PW = $868,000 + $260,000 (0.1160) + $12,000 (10.757) - $32,000 (0.0318) PW = $1,026,226.40 Compute the equivalent annual worth of this present value over 40 years: AW = 1,026,226.40 (A/P, 9%, 40) = 1,026,226.40 (0.0930) = $95,439.06 Capitalized cost = $95,439.06/0.09 = $1,060,434 5-83 Assuming monthly compounding, we use a per period interest rate of 1.5% Details below int costs 1 2 3 4 5 6 7 8 Sum: Or: 30000 50000 210000 530000 620000 460000 275000 95000 $2.106.599,68 2 14000 10 18000 10 245000 10 200000 10 20000 Using PV() e how I did it. 500 Using PV() 0,015 PW 29557 48533 200827 499358 575521 420689 247782 84333 $2.106.599,68 HW-9 Work problems: 12th Ed. 7-2, 7-6, 7-14, 7-18, 7-46. NEXT EXAM DATE 10-27-2015. It will cover Chaps 5, 6 and 7 You may use a COMPUTER, and the textbook. No notes or homework. Instructions Do all your work below; use this as a text file or use the Excel functions if you need them. List each problem number and do the work here. Save this file to your computer as you work. When you are finished, go back to the Assignments page where you found this and open the Assignment. You will find a section called “Assignment Materials.” Near the bottom of that section is “Attach File” and a button to Browse for Local File. Again do your work, save it to the computer, and then attach it to the Assignment. Do your work here hem. open the on is “Attach File” 12th Ed. 7-2, 7-6, 7-14, 7-18, 7-46. 7-2 (F/A, i, 35) = 10^6/5800 = = 172.414 and is very close to 8% from tables. (Exact = 8.003%) PS alternative A -5800 Using Rate() N 35 7-6 PS Alternative $9,375 = $325 (P/A, i%, 36) (P/A, i%, 36) = $9,375/$325 = 28.846 From compound interest tables, i = 1.25% Nominal Interest Rate = 1.25 × 12 = 15% Effective Interest Rate = (1 + 0.0125)12 − 1 = 16.08% PV A N Rate()= Nominal Effective -9375 325 36 1,25% 15,003% =12*H17 16,079% =(1+H17)^12-1 7-14 The easiest solution is to solve one cycle of the repeating diagram: PS alternatives: 0 1 80 -120 IRR()+ 50% Rate() 50% Note that to use Rate() the period is 1 and you use the 2nd value as the pmt and the first as pv. Another: $120 = $80 (F/P, i%, 1) $120 = $80 (1 + i) (1 + i) = $120/$80 = 1.50 i* = 0.50 = 50% Alternative Solution: EUAB = EUAC $80 = [$200 (P/F, i%, 2) + $200 (P/F, i%, 4) + $200 (P/F, i%, 6)] (A/P, i%, 6) Try i = 50% $80 = [$200 (0.4444) + $200 (0.1975) + $200 (0.0878)] (0.5481) = $79.99 Therefore i* = 50% 7-18 Year 0 1 2 3 4 5 6 Using IIR() Cash flow 12500 -1000 -2000 -3000 -4000 -5000 -6000 13,18% $12,500 = $1,000 (P/A, i%, 6) + $1,000 (P/G, i%, 6) at 12%, $1,000 (4.111) + $1,000 (8.930) = $13,040 at 15%, $1,000 (3.784) + $1,000 (7.937) = $11,720 i* = 12% + (3%) ((13,040 – 12,500)/(13,040 −11,720)) = 13.23% 7-46 0 1 2 3 4 5 6 80 -120 80 -120 80 -120 50% ($0,00) PS alternatives PW of Benefits – PW of Cost = $0 $15,000 (P/F, i%, 4) − $9,000 − $80 (P/A, i%, 4) = $0 Try i = 12% $15,000 (0.6355) − $9,000 − $80 (3.037) = +$289.54 Try i = 15% $15,000 (0.5718) − $9,000 − $80 (2.855) = −$651.40 Performing Linear Interpolation: i* = 12% + (3%) [289.54/(289.54 + 651.40)] = 12.9% Price Ann Taxes N Fut Val Using Rate() -9000 -80 4 15000 12,88% years EOY 0 1 2 3 4 -9000 -9000 -80 -80 -80 -80 -80 -80 14920 =15000-80 12,88% =IRR(C5:C9) Future 1000000 8,00% =(1+H17)^12-1 use Rate() the nd you use the 2nd pmt and the first 0,499999 =Int found by Goal seek 50% HW-8 * Read Newnan Chap 7. Work problems: 12th Ed. pp. 245ff: 7-1, 7-3, 7-4, 7-7, 7-15, 7-30, 7-32, 7-42, 7-44, 7-48, 7-54, 7-70. NEXT EXAM DATE 10-27-2015. It will cover Chaps 5, 6 and 7 You may use a COMPUTER, and the textbook. No notes or homework. Instructions Do all your work below; use this as a text file or use the Excel functions if you need them. List each problem number and do the work here. Save this file to your computer as you work. When you are finished, go back to the Assignments page where you found this and open the Assignment. You will find a section called “Assignment Materials.” Near the bottom of that section is “Attach File” and a button to Browse for Local File. Again do your work, save it to the computer, and then attach it to the Assignment. Do your work here 4, 7-70. hem. open the on is “Attach File” 12th Ed. pp. 249ff: 7-1, 7-3, 7-4, 7-7, 7-15, 7-30, 7-32, 7-42, 7-44, 7-48, 7-54, 7-70. 7-1 (F/A, i, 25) = 7 5 10 40 (F/A, 3.5%, 25) = 38.950 and (F/A, 4%, 25) = 41.646 bound the value. So we can interpolate: i = 3.5% + (0.5%)􁉂 􁉂􁉂􁉂􁉂􁉂.􁉂􁉂􁉂 􁉂􁉂.􁉂􁉂􁉂􁉂􁉂􁉂.􁉂􁉂􁉂􁉂 = 3.69% 7-3 We can solve for the rate of return directly: F = P (1 + i)n $20,000 = $5,000 (1 + i)3 (1 + i) = ($20,000/$5,000)1/3 = 1.587 i* = 0.587 Rate of Return = 58.7% 7-4 F = $5, P = $1, n = 5 F = P (1 + i)^n $5 = $1 (1 + i)^5 (1 + i) = 5^0.20 = 1.38 i* = 38% 7-7 $5,000 = $1,500 (P/A, i%, 5) (P/A, i%, 5) = $5,000/$1,500 = 3.333 Performing Linear Interpolation: (P/A, i% 5) i 3.352 15% 3.127 18% i = 15% + (3%)((3.333 − 3.352)/(3.127 − 3.352))= 15.25% 7-15 Solve for the rate that yields a NPW of zero (ignore salvage value initially): NPW = 0 = $14,000 (P/A, i%, 30) - $250,000 (P/A, i%, 30) = $250,000/$14,000 = 17.857 At i = 3.5%, P/A = 18.392; At i = 4%, P/A = 17.292. i = 3.5% + 0.5% [(17.857 – 18.392)/(17.292 – 18.392)] = 3.7% Including salvage value of $200,000: NPW = 0 = $14,000 (P/A, i%, 30) + $200,000 (P/F, i%, 30) - $250,000 Try i = 5%: NPW = $14,000 (15.372) + $200,000 (0.2314) - $250,000 = $11,488 Try i = 6%: NPW = $14,000 (13.765) + $200,000 (0.1741) - $250,000 = -$22,470 Performing Linear Interpolation: NPW i $11,488 5% ($22,470) 6% i = 5% + (1%)((0 − 11,488)/(-22,470 − 11,488))= 5.34% 7-30 PIC PW of Benefits – PW of Cost = $0 $30 (P/A, i%, 27) + $1,000 (P/F, i%, 27) − $875 = $0 Try i = 3 ½% $30 (17.285) + $1,000 (0.3950) − $875 = $38.55 >$0 Try i = 4% $30 (16.330) + $1,000 (0.3468) − $875 = −$38.30 < $0 i* = 3.75% Nominal rate of return = 2 (3.75%) = 7.5% 7-32 a) Interest paid quarterly = $1,000 (0.06/4) = $15 NPW = 0 = $15 (P/A, i%, 40) + $1,000 (P/F, i%, 40) - $900 Try i = 1.75% NPW = $15 (28.594) + $1,000 (0.4996) - $900 = $28.50 Try i = 2.00% NPW = $15 (27.355) + $1,000 (0.4529) - $900= -$36.80 Interpolating we get: i = 1.859%/qtr Annual effective rate = (1 + 0.0186)4 - 1 = 7.65% b) NPW = 0 = $15 (P/A, i%, 20) + $950 (P/F, i%, 20) - $900 Try i = 1.75% NPW = $15 (16.753) + $950 (0.7068) - $900 = $22.80 Try i = 2.00% NPW = $15 (16.351) + $950 (0.6730) - $900= -$15.40 Interpolating we get: i = 1.899%/qtr Annual effective rate = (1 + 0.01899)4 - 1 = 7.82% 7-42 a) The foregone cash rebate is like a hidden finance charge. You pay $12,000 for the car but receive a car only worth $12,000 – $3,000 = $9,000. The monthly payments = 12000/(4*12) = $250 for 48 months. NPW = 0 = 9000 – 250 (P/A, i, 48), so (P/A, i%, 48) = 36.0 and interpolating i = 1% + (0.25%)*(37.974 - 36.0)/(37.974 35.932) = 1.242%, so r = (12) (1.242%) = 14.90% and ia = (1 + 0.01242)12–1 = 0.15965 or 15.97%. 1.24% using Rate() b) Worth of car = Cost – Rebate = $18,000 – $3,000 = $15,000. The monthly payments = 18000/(4*12) = $375 for 48 months. NPW = 0 = 15,000 – 375 (P/A, i%, 48), so, (P/A, i%, 48) = 40.0 and interpolating i = 0.75% + (0.25%)(40.185 - 40.0)/(40.185 37.974) = 0.771%, so r = (12)(0.771%) = 9.65% and ia = (1 + 0.00771)12–1 = 0.0965 or 9.65%. c) Worth of car = Cost – Rebate = $24,000 – $3,000 = $21,000. The monthly payments = 24000/(4*12) = $500 for 48 months. NPW = 0 = 21000 – 500 (P/A, i%, 48), so, (P/A, i%, 48) = 42.0 and interpolating i = 0.50% + (0.25%)(42.580 - 42.0)/(42.580 - 40.185) = 0.561%, so r = (12)(0.561%) = 6.73% and ia = (1 + 0.00561)12–1 = 0.0694 or 6.94%. 7-44 The amount of cash paid will be $75,000 – $50,000 = $25,000 with $50,000 financed, so, the monthly payments will be 50,000 (A/P, 8%, 4) = (50,000) (0.3019) = $15,095. The reduction in cost if one pays entirely in cash is $75,000 x 0.10 = $7,500, so, a 100% cash payment would be $75,000 − $7,500 = $67,500 (true value of equipment). Year Pay Cash Borrow from Incremental Manufacturer Difference 0 -67,500 -25,000 -42,500 1 -15095 15095 2 -15095 15095 3 -15095 15095 4 -15095 15095 NPW = 0 = –42500 + 15095 (P/A, IRR, 4), so (P/A, IRR, 4) = 2.816. Interpolating IRR = 15% + (3%)(2.855 - 2.816)/(2.855 - 2.690) = 15.72%. Using the IRR() function: 7-48 Total Annual Revenues = $500 (12 months) (4 apt.) = $24,000 Annual Revenues – Expenses = $24,000 − $8,000 = $16,000 To find Internal Rate of Return the Net Present Worth must be $0. NPW = $16,000 (P/A, i*, 5) + $160,000 (P/F, i*, 5) − $140,000 At i = 12%, NPW = $8,464 At i = 15%, NPW = −$6,816 15.69% IRR= 12% + (3%) [$8,464/($8,464 + $6,816)]= 13.7% 7-54 The problem requires an estimate for n, the expected life of the infant. Seventy or seventy-five years might be the range of reasonable estimates. Here we will use 71 years. The purchase of a $1,000 life subscription avoids the series of beginning-of-year payments of $64.50. Based on 71 beginning-of-year payments, $1,000 – $64.50 = $64.50 (P/A, i%, 70) (P/A, i%, 70) = $935.50/$64.50 = 14.50 6% < i*< 8%. By Calculator: i* = 6.83% 6.83% =RATE(75,-64.5,935.5) 7-70 Using incremental analysis, computed the internal rate of return for the difference between the two alternatives. Note: Internal Rate of Return (IRR) equals the interest rate that makes the PW of costs minus the PW of Benefits equal to zero. $9,000 – $3,000 (P/A, i*, 7) − $1,200 (P/F, i*, 8) = $0 Try i = 25%: $9,000 − $3,000 (3.161) − $1,200 (0.1678) = −$684.36< $0 Try i = 30%: $9,000 − $3,000 (2.802) − $1,200 (0.1226) = $446.88> $0 i* = 25% + (5%) [$684.36/($446.88 + $684.36)]= 28.0% (actual value is 27.9%) The contractor should choose Alternative A and lease because 28% > 15% MARR. OR: $9,000 – $3,000 (P/A, i*, 7) − $1,200 (P/F, i*, 8) = $0 Amount years Init -9000 9000 Ann 3000 7 ($8,832.41) Future 1200 8 ($167.59) Int 0.278988 PV Goal seek: ($0.00) Chapter 7 problems worked in class 7-6 Price Down month N Loan Rate= Nominal Two ways effective interest 12375 3000 -325 36 9375 1.250% 15.003% Pmt N PV /mo P/A= 28.84615 0.0125 0.1607858 Using Effect() 0.1607858 Using (1+int)^12) - 1 7-14 0 1 2 3 4 5 6 0.499999 =Int found by Goal seek 80 -120 80 -120 80 -120 50% ($0.00) 50% 7-42 Price n pmt PV Rate= Effective Case a 12000 48 3000 -250 9000 1.241% 15.96% In words: Case b Case c 0% interest!! 18000 24000 48 48 Discount for cash 3000 3000 -375 -500 15000 21000 =RATE(B30,B32,B33) 0.770% 0.559% 9.64% 6.91% =RATE(n,pmt,Price-discount) years EOY 7-46 0 1 2 3 4 -9000 -9000 -80 -80 -80 -80 -80 -80 14920 =15000-80 12.88128% =IRR(C5:C9) int -9000 -79.2079 -78.4237 -77.6472 14337.83 5102.548 0.01 -9000 ($79.21) ($78.42) ($77.65) $14,337.83 Years Cash 0 1 2 3 4 -9000 -80 -80 -80 14920 INT= 0.01 -9000
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Explanation & Answer

Hi there!This assignment has been completed in its entirety. All 5 questions (including the extra credit) have been answered, fully and completely. Moreover, the answers provided are correct. Please review this document and let me know if you have any questions :)Thanks again!Selenica

Exam 2 Engineering Economics
Instructions:
There are five (5) questions on this exam, one on each tab at the bottom of the window.
They are all similar to the problems you have worked for homework. They are not exactly the
same!
Problem 5 has an extra credit section!
Show your work to get partial credit; this may be in the cell with the answer.
You are encouraged to use Excel functions wherever they will work.
You may use the Interest Tables in your book although I discourage their use.
If you do your work on paper and only show the answer then it will be right or wrong; no partial
credit.
I suggest that you look at each problem and first do the ones that are easiest for you.

Prob 1

Here are your steps:
a)
b)
c)
d)
e)

Create a cashflow table for each machine.
Calculate the IRR for the ...


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