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CHAPTER 5: HOMEWORK EXCERSICES NAME: CAO THI NGOC ANH ID: 421 762 CLASS: VISK2010B SOLUTION Problem 4 (p.196) a. We have a formula : PV = FVn [PVFk,n] And substitute what’s known, $856 = $1,122 [PVF7,n] PVF7,n = $856/ $ 1,122 = 0.7629 n = 4 years b. First, we calculate the correct k. Because compounding is monthly k = knom / 12 = 12%/12 = 1% We have a formula: PV = FVn [PVFk,n] And substitute what’s known, $450.00 = $725.50 [PVF1,n] PVF1,n = 0.6203 n = 48 months = 4years c. First, we calculate the correct k. Because compounding is quarterly k= knom/ 4 = 10% / 4 = 2.5 % We have a formula: PV = FVn [PVFk,n] And substitute what’s known, $5,000 = $6,724.44 [PVF2.5,n] PVF2.5,n = 0.7436 n = 12 quarters = 3 years Problem 7 (p. 196) a. A formula of the future value factor for k and n is : FVF k, n = (1 + k) n (1+k)a+b = (1+k)a (1+k)b Thus, we have a new formula: FVFk,a+b = [FVFk,a] [FVFk,b] And, substitute what’s known on equation FVF 3, 85 = [FVF 3, 60] [FVF 3,25] FVF 3, 85 = 5.8916 x 2.0938 = 12.3359 Then, substitute in the formula: FV = PV [FVF3,85] = $ 11.50 x 12.3359 = $141.863 per bond b. Using equation (a)to calculate : FVF 7, 85 = [FVF 7, 60] [FVF 7, 25] FVF 7, 85 = 57.9464 x 5.4274 = 314.49 Then, substitute in the formula: FV = PV [FVF7, 85] = $ 11.50 x 314.49 = $ 3,616.63 Per pond c. Problem 9 (p. 197) a. The amount of note that Charlie will pay in five years with accumulated interest at 6% : FV = PV [FVF6, 5] = $ 8,000 x 1.3382 = $10,705.6 When a bank has offered to discount the note at 14%, Joe will receive: PV = FV [PVF14, 5] = 10,705 .6 x 0.5194 = $5,560.49 Joe‘s investment at 18% for five year will grow to: FV = PV [FVF18, 5] = 5, 560.49 x 2.2878=$ 12,721.29 Thus, Joe will be ahead: $12, 721.29 - $ 10,705.6 = $ 2,015.7 b. The future value of investment at 4% will be : FV = PV [FVF4, 5] = $5,560.49 x 1.2167 = $6,765.45 And, Joe will be behind: $10,705.6 - $6,765.45 = $3,940.15 Problem 10 (p. 197) We have a formula: PVA = PMT [PVFAk, n] And, substitute given information, 30,000 = 761.80 [PVFAk,60] [PVFAk,60] = 33, 3804 Appendix A table-4 gives k = 1.5% Then, substitute k into EAR formula: EAR = (1 + knom /m) m – 1= (1+ 0.015) 12 – 1 = 0.195 0r 19.5 % Problem 14 (p.197) We have a formula: FVAn = PMT [FVFAk, n] And substitute the given information, $20,000 = $850 [FVFA8,n] FVFA8,n = 23.529 n = 13.75 years = 13 years 9 month Problem 16 (p.197) We have a formula: PVA = PMT [PVFAk, n] $ 10,000 = PMT [PVFA 6,4] PVFA 6, 4 =3.4651, and PMT = $10,000/ 3.4651 = $ 2,885.92 An amortization schedule: Period | Beginning balance | Payment | Interest@6% | Principal reduction | Ending balance | 1 | $ 10,000 | $ 2,885.92 | $600 | $ 2,285.92 | $ 7,714.08 | 2 | $ 7,714.08 | $ 2,885.92 | $ 462.84 | $ 2,423.08 | $ 5,291 | 3 | $ 5,291 | $ 2,885.92 | $ 317.46 | $ 2,568.46 | $ 2,722.54 | 4 | $ 2,722.54 | $ 2,885.92 | $ 163.38 | $ 2,722.54 | $ 0 | Problem 18 (p.197) For monthly compounding, k = knom/ 12 = 6% / 12 = 0.5 % n = 2 years x 12 months / year = 24 months And, we have an equation: FVAn = PMT [FVFA k,n]( 1+k) Substitute given information in the above equation: FVAn= $ 250 x [FVFA 0.5, 24] (1.005) Appendix table -3 gives FVFA0.5, 24 =25.4320, and FVAn= $ 250 x 25.4320 x 1.005 = $ 6,389.79 The amount of money, Joe will have at the end of three years: FV = FVAn [FVF 0.5, 12] = $ 6,389.79 x 1.0617 =$ 6,784.04 Problem 20 (p.197) For monthly compounding, k = knom/ 12 = 12% / 12 = 1 % n = 30years x 12 months / year = 360 months And, we have an equation: PVA = PMT [PVFAk,n] Substitute given information in the above equation: $ 150,000 = PMT [PVFA1, 360] Appendix table -4 gives PVFA1, 360 =97. 2183, and: PMT = $150,000/ 97.2183= $ 1,542.92 An amortization schedule: Months | Beginning balance | Payment | Interest@1% | Principal reduction | Ending balance | 1 | $150,000 | $ 1,542.92 | $1,500 | $42.92 | $149,957.08 | 2 | $149,957.08 | $ 1,542.92 | $ 1,499.57 | $ 43.35 | $ 149,913.73 | 3 | $ 149,913.73 | $ 1,542.92 | $ 1,499.13 | $ 43.79 | $ 149,869.94 | 4 | $ 149,869.94 | $ 1,542.92 | $ 1,498.70 | $ 44.22 | $ 149,825.72 | 5 | $149,825.72 | $ 1,542.92 | $ 1,498.26 | $ 44.66 | $ 149,781.06 | 6 | $ 149,781.06 | $ 1,542.92 | $ 1,497.81 | $ 45.11 | $ 149,735.95 | Problem 28 (p.198) a. The present value of first cash inflow with formula PV = FV [PVFk, n] And, substitute given information on that, PV1 = $20, 000[PVF15, 1] = $20,000 x 0.8696 = $17,392 The present value of cash flow annuity in the five next year with formula: PVA = PMT [PVFAk, n] And, substitute given information on that PVA = PMT [PVFA15, 5] = $ 16,000 x 3.3522= $ 53,635.2 The present value of cash before paying for someone in the present: PV 2= FV [PVF k,n ] = $ 53,635. 2 [PVF15, 1] = $ 53,635.2 x 0.8696 = $ 46,641.16 Net present value of the project is: NPV = PV1 + PV 2 - initial cost = $17,392 + $ 46,641.16 -$ 50,000 = $ 14,033.16