Chapter+5+exercises+solutions+_9th+Edition_

6. Rate X per half year: X = A1.01) R − 1 = 0.061520151 Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 1 2 3 4 5 2500.00 2500.00 2500.00 2500.00 1842.92 615.20 499.25 376.16 245.50 106.81 1884.80 2000.75 2123.84 2254.50 1736.11 10 000.00 8 115.20 6 114.45 3 990.61 1 736.11 0 7. Monthly payment 1 = BY 333 4 9V 7777|6.66U = $717.15 Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 1 2 3 717.15 717.15 717.15 96.00 93.52 91.02 621.15 623.23 626.13 24 000.00 23,378.85 22,755.22 22,757.72 8. X = A1.025) E R ⁄ − 1 = 0.004123915 1 = FB 333 4 ML6 777777|Z = $725.08 Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 1 2 3 4 5 6 725.08 725.08 725.08 725.08 725.08 725.08 379.40 377.97 376.54 375.11 373.66 372.21 345.68 347.11 348.54 349.97 351.42 352.87 92 000.00 91 654.32 91 307.21 90 958.67 90 608.70 90 257.28 89 904.41 Total interest = 180A725.08) ̶ 92 000 = $38 514.40 9. X = A1.02) E R ⁄ − 1 = 0.00330589 1 = G\3 333 4 966 777777|Z = $1946.27 Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 1 2 3 4 5 6 1946.27 1946.27 1946.27 1946.27 1946.27 1946.27 1223.18 1220.79 1218.39 1215.98 1213.57 1211.15 723.09 725.48 727.88 730.29 732.70 735.12 370 000.00 369,276.91 368,851.43 367,823.55 367,093.26 366,360.56 366,358.14 4374.56 The amount of the principal repaid in the first 6 months is $4374.56

1 2 3 4 5 500.00 500.00 500.00 500.00 117.38 15.57 11.94 8.28 4.59 0.87 484.43 488.06 491.72 495.41 116.51 2076.13 1591.70 1103.64 611.92 116.51 0 4. Debt at the end of 2 months = 1500A1.0125) B = $1537.73. Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 1 2 3 4 5 6 7 8 9 200.00 200.00 200.00 200.00 200.00 200.00 200.00 200.00 26.95 19.22 16.96 14.67 12.36 10.01 7.64 5.23 2.80 0.33 180.78 183.04 185.33 187.64 189.99 192.36 194.77 197.20 26.62 1537.73 1356.95 1173.91 988.59 800.94 610.96 418.59 223.83 26.62 0 5. Monthly rate X E = A1.04) E R ⁄ − 1 = 0.006558197 a) Monthly payment @ = EP3333 4 966 777777|Z M = $1373.79 b) Weekly payment = t Y = $343.45 c) Weekly rate X B = A1.04) E BR ⁄ − 1 = 0.001509627 Find i such that 180 000 = 343.45‘ b 7|c 5 ‘ b 7|c 5 = 542.0937545 E^AEjc 5 ) kl c 5 = 542.0937545 1 − A1 + X B ) ^b = 0.7911886245 A1 + X B ) ^b = 0.208813755 −i logA1 + X B ) = log 0.208813755 i = 1038.33212 weeks i = 19 years 51 weeks Adebt paid off in 19 yrs, 51 wks) d) The Gibsons will save a lot of interest payment $343.45 weekly. 6. In the vth payment: Interest = 1 − A1 + i) ^AB3^wjE) Principal = A1 + X) ^AB3^wjE) Find v such that 1 − A1 + X) ^AB3jwjE) = A1 + X) ^AB3^wjE) A1 + X) ^AB3jwjE) = E B A1 + X) AB3jwjE) = 2 AE.E2) 5M AEjc) x = 2 A1 + X) w = 9.410759001

i = f 881.27 weeks Concludng payment = 160 000A1 + X B ) PPB − 307.73D PPE 77777 |c 5 A1 + X B ) = $82.49 Total interest = A881 × 307.73 + 82.49) − 160 000 = $111 192.62 Mortgage B: X E per month = A1.035) E R ⁄ − 1 = 0.005750039 Monthly payment = ER3 333 4 5U6 777777|Z M = $1230.90 Equivalent semi-annual payment if you double the monthly payment every 6 months is 1230.90D R T |c M + 1230.90 = $8723.28 Find i such that 8723.28‘ b 7|3.3G2 = 160 000 E^AE.3G2) kl 3.3G2 = 18.3417467 A1.035) ^b = 0.358039636 i = ±b3.G2P3GFRGR ±bAE.3G2) i = 29.85665665 half years Concluding payment at the end of 15 years: 160 000A1 + X E ) EP3 − 8723.28D BF 7777 |3.3G2 A1.035) − 1230.90D 2 T |c M A1 + X E ) = 449 086.99 − 441 595.79 − 6261.48 = $1229.72 Total interest = ²A179 + 29) × 1230.90 + 1229.72³ − 160 000 = $97 256.92 Mortgage B has lower interest charges. 17. The lender will advance $152 000, but the monthly payment is calculated on The basis of a $160 000 loan and rate 3.3P EB per month. Monthly payment 1 = ER3333 4 ML6 777777|6.6L M5 ⁄ = $1529.05 The true monthly interest rate X = ´ M5 EB is the solution of the equation 1529.05‘ EP3 77777 |c = 152 000 or ‘ EP3 77777 |c = 99.4081 Applying the method of interpolation, we have: ‘ EP3 |c h EB { 104.6406 8% } µ E% = 2.BGB2 R.3Y\B 6.0472 5.2325 99.4081 h EB ¶ 1% ¶ = f 0.87%

7. Quarterly payment 1 = E23 333 4 V6 |Z = $4315.19 At the end of 8 years ] = 150 00(1.02) GB − 4315.19D GB |3.3B = 282 681.09 − 190 848.04 = $91 833.05 Principal repaid = 150 000 − 91 833.05 = $58 166.95 Couple ° sequity = 58 166.95 + 50 000 = $108 166.95 8. X = (1.0325) E R ⁄ − 1 = 0.005344740 1 = BER 333 4 5U6 |Z = $1599.49 After 5 years: ] = 216 000(1 + X) R3 − 1599.49D R3 |c = 297 409.17 − 112 791.02 = $184 618.15 Principal repaid = 216 000 − 184 618.15 = $31 381.85 Owner ° sequity = 31 381.85 + 110 000 = $141 381.85 9. Monthly payment 1 = RP333 4 ML6 |6.66±K = $689.71 At the end of 9 years: ] = 68 000(1.0075) E3P − 689.71D E3P |3.33\2 = 152 369.44 − 114 135.43 = $38 261.01 Principal repaid = 68 000 − 38 261.01 = $29 738.99 Buyer’s equity = 29 738.99 + 12 000 = $41 738.99 Seller’s equity = $38 261.01 10. X = (1.02) E G ⁄ − 1 = 0.006622710 Outstanding balance at the end of 1 year: ] = 10 000(1 + X) EB − 200D EB |c = 10 824.32 − 2489.38 = $8334.94 11. ] = 500‘ Y |3.E3 = $15 849.33 12. Let @ be the original amount of the loan. At the end of 3 years; @(1.04) R − 802D R |3.3Y = 17 630 1.265319019@ − 5319.65 = 17 630 1.265319019@ = 22 949.65 @ = $18 137.44 13. Outstanding balance after 5 installments = 1000‘ E2 |3.32 = $10 379.66 Interest in the 6th installment = 10 379.66(0.05) = $518.98 Principal in the 6th installment = 1000 − 518.98 = $418.02 14. Monthly rate X = (1.025) E R ⁄ − 1 = 0.004123915 Extra amount = 500‘ R3 |c = $26 528.38 15. (a) For retrospective method, must first calculate the loan amount. We use the “P&Q” formula with P = 900, Q = ̶ 15, n = 60 and i = 0.0045 A = 900 ‘ R3 |3.33Y2 ̶ 15 4 V6| 77777 6.66UK ^R3(E.33Y2) kV6 3.33Y2 = $25 068.15 B 24 = 25 068.15(1.0045) 24 ̶ [ 900 D BY |3.33Y2 ̶ 15 ² 5U| 77777 6.66UK ^BY 3.33Y2 ] = $9444.68 For the prospective method, must first calculate the 25 th payment. R 25 = P + ( n ̶ 1) Q = 900 + (24) ( ̶ 15) = 540

6. X E = (1.04) E R ⁄ − 1 = 0.007363123 Monthly payment = B3 333 4 M56 |Z M = $241.29 Outstanding balance after 2 years = 20 000(1 + i E ) BY − 241.29D BY |c M = 23 397.17 − 6249.46 = $17 147.71 a) X B = (1.045) E R ⁄ − 1 = 0.007363123 Find i such that 17 147.71 = 241.29‘ b |c 5 ‘ b |c 5 = 71.06690758 1 − (1 + X B ) ^b = 0.523273648 (1 + X B ) ^b = 0.476726352 i = − mno 3.Y\R\BRG2B mno(Ejc 5 ) i = 100.9811525 months Duration (left) of the loan is 8 years and 5 months. Final payment @ = 17 147.71(1 + X B ) E3E − 241.29D E33 |c 5 (1 + X B ) = 35 974.69 − 35 737.93 = $236.76 b) X G = (1.035) E R ⁄ − 1 = 0.005750039 Find i such that 17 147.71 = 241.29‘ b |c 9 ‘ b |c 9 = 71.06680758 1 − (1 + X G ) ^b = 0.4086363049 (1 + X G ) ^b = 0.591363049 i = − mno 3.2FEGRG3YF mno(Ejc 9 ) i = 91.62268024 months Duration (left) of the loan is 7 years and 8 months. Final payment @ = 17 147.71(1 + X G ) FB − 241.29D FE |c 9 (1 + X G ) = 29 059.72 − 38 909.32 = $150.40 7. a) Semi-annual payment = E 233 33 4 56 |6.65±K = $98 507.60 Total of payments in 2010 is 2 × 98 507.60 = $197 015.20 b) Outstanding balance on December 31, 2019: 1 500 000(1.0275) G − 98 507.60D G |3.3B\2 = $ 1323 460.15 Outstanding balance on December 31, 2020: 1 500 000(1.0275) 2 − 98 507.60D 2 |3.3G\2 = $1 197 527.16 Principal repaid in 2020 = 1 323 460.15 − 1 197 527.16 = $125 932.99 Interest paid in 2020 = 2 × 98 507.60 − 125 932.99 = $71 082.21 The interest deduction on the 2020 tax form will be $71 082.21 c) Outstanding balance on January 1, 2022: 1 500 000(1.0275) \ − 98 507.6D \ |3.3B\2 = $1 064 572.63 Capital gain = (700 000 + 1 064 572.63) − 1 700 000 = $64572.63

EXERCISE 5.3 Concept Questions 1. The initial rate on the loan, i 1 , is used in both methods. 2. a) i 1 is used to determine the 3 months’ interest penalty; b) both rates are used to determine IRD , as IRD = i 1 - i 2. 3. The penalty is added to the outstanding balance and that total is used to determine the new periodic payment. Part A 1. Original payment = 2333 4 9K 7777|6.66±K = $159.00 At the end of 1 year ] = 5000(1.0075) EB − 159.00D EB 7777 |3.33\2 = $3480.33 New monthly payment = GYP3.GG 4 5U 7777|6.66K =$154.25 Monthly savings in interest = 159.00 − 154.25 = $4.75 2. Original payment = R333 4 V6 7777| 6.6K5L M5 = $114.00 At the 30th payment ] = 6000(1.0044) G3 − 114.00D G3 7777 |3.33YY = 6844.67 − 3647.43 = $3197.24 New monthly payment = GEF\.BY 4 96 7777|6.669K =$112.46 Monthly savings in interest = 114.00 − 112.46 = $1.54 3. Original payment = P333 4 UL 7777|6.6M = $210.67 Balance after 20 payments ] = 8000(1.01) B3 − 210.67D B3 7777 |3.3E = 9761.52 − 4638.74 = $5122.78 (a) Penalty = 3 × 0.01(5122.78) = $153.68 Total to be refinanced = $5276.46 New monthly payment = 2B\R.YR 4 5L 7777|6.66K = $202.42 → Do refinance; can save 8.25 a month. (b) IRD = 0.01 ̶ 0.005 = 0.005; term remaining = 28; penalty = 5122.78 x 0.005 x 28 = $717.19; total to be refinanced = 5839.97 New monthly payment = 2PGF.F\ 4 5L 7777|6.66K = $224.03; do NOT refinance 4. Original monthly payment = EY33 4 9V 7777|6.66±K = $44.52 Balance after 12 payments = 14000(1.0075) EB − 44.52D EB 7777 |3.33\2 = 1531.33 − 556.84 = $974.49 Penalty = 974.49 × 3.3F EB × 3 = $21.93 Total to be refinanced = 974.49 + 21.93 = $996.42 New monthly payment = FFR.YB 4 5U 7777|6.66K = $44.16 → They should refinance. 5. X E = (1.0525) E R ⁄ − 1 = 0.008564515; Original payment = R3 333 4 966 777777|Z M = $557.00 Balance after 5 years = 60 000(1 + X E ) R3 − 557D R3 7777 |c M = 100 085.76 − 43 450.15 = $56 635.61

4. a) X E = (1.02425) E R ⁄ − 1 = 0.004001424 Find i such that 357.00‘ b 7|c M = 28 410.61 E^(Ejc M ) kl c = BP YE3.RE G2\.33 (1 + X E ) ^b = 0.681560515 i = − mno 3.RPE2R32E2 mno(Ejc M ) i = f 96 months The maturity date of the mortgage is 8 years after January 1, 2020, that is January 1, 2028. b) Final payment @ on January 1, 2028: @ = 28 410.61(1 + X E ) FR − 357.00D F2 7777 |c M (1 + X E ) = 41 684.65 − 41 327.64 = $356.24 Date Payment Interest Principal Balance c) January 1, 2020 February 1, 2020 357.00 357.00 114.65 113.68 242.35 243.32 28 410.61 28 167.29 Note: ·¸¹¡¸¢£ E,B3B3 /¢¥¹ƒ¥/¸m §¤'¢¡¢¸¢£ E,B3B3 /¢¥¹ƒ¥/¸m = 1 + X E Or January 1, 2020 principal = (February 1 ,2020 principal)(1 + X E ) ^E = 243.32(1 + X E ) ^E = $242.35 d) X B = (1.00195) E R ⁄ − 1 = 0.0032239045 New monthly payment = BP YE3.RE 4 “V 7777|Z 5 = $344.57 5. Let 1 E be the original monthly payment, 1 B be the new monthly payment. At the end of 24 months: 1 E ‘ GR 7777 |3.332 = 1 B ‘ EP 7777 |3.332 « 5 « M = 4 9V 7777|6.66K 4 ML 7777|6.66K « 5 « M = 1.91413616 6. a) X E = (1.0225) E R ⁄ − 1 = 0.00371532 Monthly payment = EF3 333 4 966 777777|Z M = $1051.60 b) Outstanding balance after 2 years: 190 000(1 + X E ) BY − 1051.60D BY 7777 |c M = $181 339.13 Outstanding balance after 5 years: 190 000(1 + X E ) R3 − 1051.60D R3 7777 |c M = $166 813.04 c) Penalty = (181 339.13)(X E )(3) = $2021.20 Amount to refinance = 181 339.13 + 2021.20 = $183 360.33 X B = (1.015) E R ⁄ − 1 = 0.002484517 Let @ be the new monthly payment. @ = EPG GR3.GG 4 5±V 777777|Z 5 = $918.76 It would pay to refinance.

7. a) Monthly rate X = (1.0325) E R ⁄ − 1 = 0.005344740 Monthly payment B33 333 4 966 777777|Z = $1339.65 Final payment = 200 000(1 + X) G33 − 1339.65D BFF 77777 |c (1 + X) = 989 767.10 − 988 429.24 = $1337.86 b) In general, the v-th interest payment = 1[1 − (1 − X) ^(b^wjE) ] Accumulated value of the v-th interest payment at i = 1‹1 − (1 + X) ^(b^wjE) ›(1 + X) b^w = 1[(1 + X) b^w − (1 + X) ^E ] = 1(1 + X) ^E [(1 + X) b^wjE − 1] Accumulated value of all interest payments at i = 1(1 + X) ^E [∑ (1 + X) b^wjE − i] b wflE = 1(1 + X) ^E {[(1 + X) b + (1 + X) b^E + ⋯ + (1 + X)] − i} = 1(1 + X)^ − 1 [D b 7|c (1 + X) − i] = 1[D b 7|c − i(1 + X) ^E ] In our example, assuming all payments are equal we substitute for 1 = 1339.65, X = (1.0325) E R ⁄ − 1, i = 300 to obtain the accumulated value of all the interest payments at the end of 25 years: 1339.65‹D G33 77777 |c − 300(1 + X) ^E › = $590 010.50 c) Outstanding balance after 5 years: 200 000(1 + X) R3 − 1339.65D R3 7777 |c = 275 378.86 − 94 467.92 = $180 910.94 d) Find i such that 1339.65‘ b 7|c = 175 910.94 i = 227.0085266 months Final payment = 175 910.94(1 + X) BBP − 1339.65D BB\ 77777 |c (1 + X) = 593 082.54 − 593 071.09 = $11.41 Total payout = (227 + 60)1339.65 + 11.41 + 5000 = $389 490.96 Difference in total payouts = (299 × 1339.65 + 1337.86) − 389 490.96 = $12 402.25 8. º – = (1.025) E R ⁄ − 1 = 0.004123915 Monthly payment = BB2 333 4 966 777777|Z M = $1308.61 Balance after 3 years = 225 000(1 + X) GR − 1308.61D GR 7777 |c M = $210 256.75 IRD = 3.32 EB − 3.3GB EB = 0.0015 New monthly interest rate º ² = (1.016) E R ⁄ − 1 = 0.002649061 With no lump sum payment With no lump sum payment With no lump sum payment With no lump sum payment Penalty = (0.0015)(210 256.75)(24) = $7569.24 Balance after three years = $217 825.99 New monthly payment = BE\ PB2.FF 4 5VU| 7777777 Z 5 = $1148.02 With 10% lump sum payment With 10% lump sum payment With 10% lump sum payment With 10% lump sum payment Lump sum payment = (0.10)(210 256.75) = $21 025.68 Balance after lump sum = 210 256.75 ̶ 21 025.68 = $189 231.07

Penalty = (0.0015)(189 231.07)(24) = $6812.32 (Penalty is $756.92 LESS) Balance after three years = $189 231.07 + 6812.32 = 196 043.39 New monthly payment = EFR 3YG.GF 4 5VU| 7777777 Z 5 = $1033.22 (a difference of $114.80 per month or a savings of 114.80 x 24 = $2755.20 over the 24 months before the mortgage is renewed again) 9. Mortgage A: X E = (1.035) E R ⁄ − 1 = 0.005750039 Monthly payment = G33 333 4 966 777777|Z M = $2102.35 Balance after 3 years = 300 000(1 + X E ) GR − 2101.25D GR 7777 |c 5 = 368 776.60 − 83 777.29 = $284 999.31 Penalty = (284 999.81)(X E )(3) = $4916.27 Mortgage B: X B = (1.0375) E R ⁄ − 1 = 0.006154524 Monthly payment = G33 333 4 966 777777|Z 5 = $2194.66 Balance after 3 years = 300 000(1 + X B ) GR − 2194.66D GR 7777 |c 5 = 374 153.56 − 88 142.13 = $286 011.43 Savings under mortgage A at X G = (1.03) E EB ⁄ − 1 = 0.00246627: (2194.66 − 2101.25)D GR 7777 |c 9 = $3512.04 Mortgage A net amount owing = 284 999.31 + 4916.27 − 3512.04 = $286 403.54 Mortgage B amount owing = $286 011.43 Choose Mortgage B. EXERCISE 5.4 Concept Questions 1. Sum of digits method: Sum of digits method: Sum of digits method: Sum of digits method: Calculates a defined proportion of the total interest paid where the denominator of this proportion is the sum of digits from 1 to n ; Amortization method: Amortization method: Amortization method: Amortization method: Takes the previous period’s outstanding balance and multiplies it by the interest rate per period. 2. There is no difference. In both methods the borrower starts with $ A and finishes with a balance of $0 and pays the same total amount of interest, I , over the life of the loan. 3. The advantage occurs if the borrower wants to pay off the loan in full OR refinance the loan before the end of the original term. This is due to the sum of digits method giving a higher outstanding balance at the point where the refinancing takes place.

Part A 1. Monthly payment = F33 4 V T|6.6M = $155.30 Final payment = 900(1.01) R − 155.30D 2 T |3.3E (1.01) = $155.26 Total interest paid = 5 × 155.30 + 155.26 − 900 = $31.76 Sum of digits from 1 to 6 = 21 Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 1 2 3 4 5 6 155.30 155.30 155.30 155.30 155.30 155.26 9.07 7.56 6.05 4.54 3.02 1.51 146.23 147.74 149.25 150.76 152.28 153.75 900.00 753.77 606.03 456.78 306.02 153.74 -0.01* *1 ¢ error due to round off 2. Monthly payment = E333 4 M5 7777|6.66±K = $87.46 Final payment = 1000(1.0075) EB − 87.46D EE 7777 |3.33\2 (1.0075) = $87.35 Total interest paid = 87.46 × 11 + 97.35 − 1000 = $49.41 Sum of digits from 1 to 12 = 78 Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 1 2 3 4 5 6 7 8 9 10 11 12 87.46 87.46 87.46 87.46 87.46 87.46 87.46 87.46 87.46 87.46 87.46 87.46 7.60 6.97 6.33 5.70 5.07 4.43 3.80 3.17 2.53 1.90 1.27 0.63 79.86 80.49 81.13 81.76 82.39 83.03 83.66 84.29 84.93 85.56 86.19 86.72 1000.00 920.14 839.65 758.52 676.76 594.37 511.34 427.68 343.39 258.46 172.90 86.71 -0.01* *1 ¢ error due to round off 3. Original monthly payment = E2 333 4 9V 7777|6.669±K = $446.21 Final payment = 15 000(1.00375) GR − 446.21D G2 7777 |3.33G\2 (1.00375) = $445.97 Total debt = (446.21 × 35 + 445.97) = 16 063.32 Less interest rebate = \P RRR × 1 063.32 = 124.53 Less payments to date =24× 446.21 = 10 709.04 Outstanding balance after 2 years = $5 229.75 (Amortization balance = 15 000(1.00375) BY − 446.21D BY 7777 |3.33G\2 = $5 226.03)

4. Original monthly payment = B3 333 4 M5 7777|6.65K = $1949.75 Final payment = 20 000(1.025) EB − 1949.75D EE 7777 |3.3B2 (1.025) = $1949.65 Total debt = (1949.75 × 11 + 1949.65) = 23 396.90 Less interest rebate = BE \P × 3396.90 = 914.55 Less payments to date = 6× 1949.75 = 11 698.50 Outstanding balance after 18 months = $10 783.85 (Amortization balance = 20 000(1.025) R − 1949.75D R T |3.3B2 = $10 739.38) 5. Original monthly payment = 2 333 4 9V 7777|6.66±K = $159.00 Final payment = 5 000(1.0075) GR − 159.00D G2 7777 |3.33\2 (1.0075) = $158.94 Total debt = (159 × 35 + 158.94) = 5723.94 Less interest rebate = G33 RRR × 723.94 = 326.10 Less payments to date = 12× 159 = 1908.00 Outstanding balance after 12 months = $3489.84 New monthly payment = GYPF.PY 4 5U 7777|6.66K = $154.67 Monthly savings in interest 159 − 154.67 = $4.33 (vs. Monthly savings in interest = $4.75 Xi 5.3.1) 6. Original monthly payment = R333 4 V6 7777| 6.6K5L M5 = $114.00 Final payment = 6000(1.0044) R3 − 114.00D 2F 7777 |3.33YY (1.0044) = $113.91 Total interest paid on loan = nR − A = (59)(114.00) + 113.91 − 6000 = 839.91 Total debt = (6000 + 839.91) = 6839.91 Less interest rebate = YR2 EPG3 × 839.91 = 213.42 Less payments to date = 30× 114.00 = 3420.00 Outstanding balance after 30 payments = $3206.49 New monthly payment = GB3R.YF 4 96 7777| 6.6U5 M5 = $112.78 Monthly savings in interest 114.00 − 112.78 = $1.22 (vs. Monthly savings in interest = $1.54 in 5.3.2) 7. Monthly payment = E3 333 4 ML6 777777|6.6M5K = $139.96 Final payment = 10 000(1.0125) EP3 − 139.96D E\F 77777 |3.3EB2 (1.0125) = $139.10 a) Total debt = (139.96 × 179 + 139.10) = $25 191.94 Less interest rebate = EB BYR ER BF3 × 15 191.94 = 11 420.53 Less 24 payments = 24 × 139.96 = 3 359.04 Outstanding principal at the end of 2 years = $10 412.37 Outstanding principal at the end of 2 years by the amortization method: 10 000(1.0125) BY − 139.96D BY 7777 |3.3EB2 = 13 473.51 − 3889.22 = $9584.29

b) Total debt = (139.96 × 179 + 139.10) = $25 191.94 Less interest rebate = \BR3 ER BF3 × 15 191.94 = 6 770.63 Less 60 payments = 60 × 139.96 = 8 397.60 Outstanding principal at the end of 5 years = $10 023.71 Outstanding principal at the end of 5 years by the amortization method: 10 000(1.0125) R3 − 139.96D R3 7777 |3.3EB2 = 21 071.81 − 12 396.89 = $8674.92 8. Original monthly payment = 2333 4 UL 7777|6.66±K = $124.43 Final payment = 5000(1.0075) YP − 124.43D Y\ 7777 |3.33\2 (1.0075) = $124.15 Total debt = (124.43 × 47 + 927.36) = 7648.97 Less interest rebate = RRR EE\R × 972.36 = 550.67 Less payments to date = 12× 124.43 = 1493.16 Outstanding balance after 1 year = $3928.53 New monthly payment = GFBP.2G 4 9V 7777|6.6K = $119.52 Monthly savings in interest 124.43 − 119.52 = $4.91 The borrower will save money be refinancing the loan. EXERCISE 5.4 Part B 1. Option 1 (sum of digits): X E = (1.025) E G ⁄ − 1 = 0.00826484 1 E = E2 333 4 M56 777777|Z M = $197.55 Final payment 15 000(1 + X E ) EB3 − 197.55D EEF 77777 |c M (1 + X E ) = $196.33 Total debt = (197.55 × 119 + 196.33) = $23 704.78 Less interest rebate = BRBP \BR3 × 8704.78 = 3 150.99 Less payments to date = 48 × 197.55 = 9 482.40 Outstanding balance after 4 years = $11 071.39 Option 2 (amortization): X B = (1.03) E G ⁄ − 1 = 0.009901634 1 B = E2 333 4 M56 777777|Z 5 = $214.19 (higher) Outstanding principal after 4 years = 15 000(1 + X B ) YP − 214.19D YP 7777 |c 5 = 24 070.60 − 13 080.88 = $10 989.72 (lower) Accumulated savings = (214.19 − 197.55)D YP 7777 |3.332 = $900. 19 New balance of option1: 11 071.39 − 900. 19 = $10 171.20 Balance of option 2 = $10 989.72 Therefore take option 1.

2. a) X E = (1.0425) E R ⁄ − 1 = 0.006961062 1 E = EP 333 4 M56 777777|Z M = $221.77 Final payment 18 000(1 + X E ) EB3 − 221.77D EEF 77777 |c M (1 + X E ) = $220.69 Total debt = (221.77 × 119 + 220.69) = $26 611.32 Total interest = $8 611.32 Sum of digit from 1 to 120 = 7260 Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 1 2 221.77 221.77 142.34 141.15 79.34 80.62 18 000.00 17 920.57 17 839.95 Determine the outstanding principal after 118 payments: Total debt = $26 611.32 Less interest rebate = G \BR3 × 8 611.32 = 3.56 Less payments to date = 118 × 221.77 = 26 168.86 Outstanding balance after 118 payments = $438.90 Payment Number Periodic Payment Interest Payment Principal Repaid Outstanding Principal 118 119 120 221.77 220.69 2.37 1.19 219.40 219.50 438.90 219.50 0 b) Interest portion of the tenth payment = EEE \BR3 × 8 611.32 = $131.66 Principal portion of the tenth payment = 221.77 − 131.66 = $90.11 c) Total debt = 26 611.32 Less interest rebate = YR2R \BR3 × 8 611.32 = 5 522.63 Less payments to date = 24× 221.77 = 5 322.48 Outstanding balance after 2 years = $15 766.21 Amortization balance = 18 000(1 + X E ) BY − 221.77D BY 7777 |c M = 21 260.66 − 5771.12 = $15 489.54 d) X B = (1.04) E R ⁄ − 1 = 0.006558197 1 B = E2 \RR.BE 4 “V 7777|Z 5 = $221.85 → Do not not not not refinance. 3. Monthly rate X = (1.012) E G ⁄ − 1 = 0.003984206 Monthly payment = B3 333 4 M56 777777|Z = $210 (rounded up to the next dollar) Final payment = 20 000(1 + X) EB3 − 210D EEF 77777 |c (1 + X) = $209.37 Total debt (210 × 119 + 209.37) = 25 199.37 Less interest rebate = BRBP \BR3 × 5 199.37 = 1 882.09 Less payment to date = 48 × 210 = 10 080.00 Outstanding balance after 4 years = $13 237.28

EXERCISE 5.5 Part A 1. Yearly deposit = 23 333 ² U T|6.69 = $11 951.35 Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 1 2 3 4 - 358.54 727.84 1108.21 11 951.35 11 951.35 11 951.35 11 951.36 11 951.35 12 309.89 12 679.19 13 059.57 11 951.35 24 261.24 36 940.43 15 000.00 2. Yearly deposit = E33 333 ² K T|6.6UK = $18 279.16 Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 1 2 3 4 5 - 822.56 1682.14 2580.40 3519.08 18 279.16 18 279.16 18 279.16 18 279.16 18 279.18 18 279.16 19 101.72 19 961.30 20 859.56 21 798.26 18 279.16 37 380.88 57 342.18 78 201.74 100 000.00 3. Quarterly deposit = B 333 ² L T|6.6M = $241.38 Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 1 2 3 4 5 6 7 8 0 2.41 4.85 7.31 9.80 12.31 14.85 17.41 241.38 241.38 241.38 241.38 241.38 241.38 241.38 241.40 241.38 243.79 246.23 248.69 251.18 253.69 256.23 258.81 241.38 485.17 731.40 980.09 1231.27 1484.96 1741.19 2000.00 4. At the end of 5 years: 1000(1.015) B3 + 1D B3 7777 |3.3E2 = 10 000 Solving 1 = $374.21 Interest earned by fund = (0.015)1000(1.015) EE + (0.015)(374.21)D EE 7777 |3.3E2 = 17.67 + 66.59 = $84.26 Increase in fund 84.26 + 374.21 = $458.47 Amount in fund after 3 years = 1000(1.015) EB + 374.21D EB 7777 |3.3E2 = 1195.62 + 4880.15 = $6075.77

5. Total annual deposit = \23 333 ² K T|6.6KK = $134 382.33 Deposit per cottager = EGY GPB.GG G3 = $4479.41 Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 1 2 3 4 5 0 7 391.03 15 188.56 23 414.96 32 093.80 134 382.33 134 382.33 134 382.33 134 382.33 134 382.33 134 382.33 141 773.36 149 570.89 157 797.29 166 476.13 134 382.33 276 155.69 425 726.58 583 523.87 750 000.00 6. Quarterly deposit = E3 333 ² U6 7777|6.6MK = $184.27 Amount in fund after 9 years = 184.27D GR 7777 |3.3E2 = $8711.54 Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 36 37 38 39 40 130.67 135.40 140.19 145.06 184.27 184.27 184.27 184.27 314.94 319.67 324.46 329.39 8 711.54 9 026.48 9 346.15 9 670.61 10 000.00 7. Annual deposit = B 333 333 ² MK 7777|6.6U = $99 882.20 Interest earned in 4th year = 99 882.20[(1.04) G − 1] = $12 471.69 Increase in fund in 9th year = 99 882.20(1.04) P = $136 695.69 Amount in fund at end of 11 years = 99 882.20D EE 7777 |3.3Y = $1 347 046.45 8. X = (1.0125) E G ⁄ − 1 = 0.004149425 Monthly deposit = G333 ² 5U 7777|Z = $119.14 9. Find i such that 4500D b 7|3.33P = 200 000 D b 7|3.33P = 44.4444444 (1.008) b = 1.35555555 i = 38.1783 → 38 full deposits Final deposit @ = 200 000 − 4500D GP 7777 |3.33P (1.008) = $200 508.65 No final deposit is needed, after 39 quarters (9.75 years) the account will have $200 508.65 10. Annual deposit = \3 333 ² MK 7777|6.6U = $3495.88 Amount in fund at the end of 10 years = 3495.88D E3 7777 |3.3Y = $41 971.91

EXERCISE 5.5 Part B 1. Monthly rate X = O1 + 3.32 GR2 Q GR2 EB ⁄ − 1 = 0.004175073 Monthly deposit = 23 333 ² 9V 7777|Z = $1290.02 Amount in fund after 34 deposits= 1290.02D GY 7777 |c = $47 021.21 Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 1 2 3 - 5.39 10.79 1290.02 1290.02 1290.02 1290.02 1295.41 1300.81 1 290.02 2 585.43 3 886.24 34 35 36 196.32 202.52 1290.02 1289.93 1486.34 1492.45 47 021.21 48 507.55 50 000.00 2. Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 1 2 3 4 5 0 = 1[1 − 1] X ∙ 1D E T |c = 1[(1 + X) − 1] X ∙ 1D B T |c = 1[(1 + X) B − 1] X ∙ 1D Y T |c = 1[(1 + X) G − 1] X ∙ 1D Y T |c = 1[(1 + X) Y − 1] 1 1 1 1 1 1 1(1 + X) 1(1 + X) B 1(1 + X) G 1(1 + X) Y 1 or 1D E T |c 1D B T |c 1D G T |c 1D Y T |c 1D 2 T |c Total 1(D 2 T |c − 5) 51 1D 2 T |c 3. Let @ be the amount in the fund (v − 1)st deposit. Then @(1.005) + 200 = 5394.69 @ = 5194.69(1.005) ^E @ = $5168.85 EXERCISE 5.6 Concept Questions 1. The outstanding balance of a sinking-fund loan is always equal to the loan amount since there are no payments of principal until the very last payment, at which time the outstanding balance drops to $0. 2. A sinking fund is used by the borrower to accumulate to the amount of the loan by the end of the term, at which time the original principal/loan is paid back in one lump sum. 3. The interest payments on a sinking fund loan are the SAME each period; they do not go down (or up) and thus have no mathematical relationship.

Part A 1. Semi-annual interest = 5000(0.05) = $250 Semi-annual S.F. deposit = 2333 ² M6 7777|6.65 = $456.63 Total semi-annual expense = 250 + 456.63 = $706.63 Amount in the S.F. at the end of 4 years = 456.63D P T |3.3B = $3919.24 2. Annual interest = 1 250 000(0.0625) = $78 125.00 Annual S.F. deposit = E B23 333 ² MK 7777|6.69K = $64 781.34 Total annual expense = 78 125 + 64 781.34 = $142 906.34 3. a) Semi-annual interest = 500 000(0.04) = $20 000 Semi-annual S.F. deposit = 233 333 ² U6 7777|6.65 = $8277.87 Total semi-annual expense = $28 277.87 b) Amount in the S.F. after 15 years = 8277.87D G3 7777 |3.3B = $335 817.29 Company’s indebtedness = 500 000 − 335 817.29 = $164 182.71 4. a) Semi-annual interest = 2 000 000(0.025) = $50 000 Semi-annual S.F. deposit = B 333 333 ² K6 7777|6.6M5K = $29 035.25 Total semi-annual expense = $79 035.25 b) Amount in the S.F. after 15 years = 29 035.25D G3 7777 |3.3EB2 = $1 049 016.55 City’s indebtedness = 2 000 000 − 1 049 016.55 = $950 983.458 5. X = (1.0065) E G ⁄ − 1 = 0.002161989 Monthly interest = 5000(0.0045) = $22.50 Monthly S.F. deposit = 2333 ² V6 7777|Z = $78.14 Total monthly expense = 22.50 + 78.14 = $100.64 6. X = (1.00225) R − 1 = 0.013576166 Semi-annual interest = 200 000(0.03) = $6000 Semi-annual S.F. deposit = B33 333 ² M6 7777|Z = $18 808.34 Total semi-annual expense = 6000 + 18 808.34 = $24 808.34 7. a) Semi-annual rate X = (1.005) B − 1 = 0.010025 Semi-annual S.F. deposit = G3 333 ² L T|Z = $3620.39 Amount in the S.F. after 3 years 3620.39D R T |c = $22 274.09 Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 6 7 8 223.30 261.83 3620.39 3620.39 3843.69 3882.22 22 274.09 26 117.78 30 000.00

b) Semi-annual expense of the loan= (0.052/2)(30 000) + 3620.39 = $4400.39 c) Amount in the S.F. at the end of 2 years 3620.39D Y T |c = $14 700.79 Book value of the loan at the end of 2 years = 30 000 − 14 700.79 = $15 299.21 8. Annual S.F. deposit = E33 333 ² M6 7777|6.65K = $8925.88 Amount in the S.F. after the 5th deposit 8925.88D 2 T |3.3B2 = $46 917.36 Lump sum payment to pay off the loan = 100 000 − 46 917.36 = $53 082.64 EXERCISE 5.6 Part B 1. a) Under the amortization method: 1 E = µ 4 l T|Z M Under the sinking fund method: 1 B = eX B + µ ² l T|Z 9 b) 1 B = eh + µ ² l T|» = e ¶h + E ² l T|» ™ = e ¶ E 4 l T|» ™ (see exercise 3.3.B1b)= 1 E 2. a) X = (1.005) E G ⁄ − 1 = 0.001663897 Monthly interest = 2 000 000 O 3.3R EB Q = $10 000 Monthly S.F. deposit B 333 333 ² ML6 777777|Z = $9 539.32 Total monthly expense 10 000 + 9 539.32 = $19 539.32 b) Amount in the S.F. at the end of the 5th year = 9 539.32D R3 7777 |c = $601 379.01 Company’s indebtedness = 2 000 000 − 601 379.01 = $ 1 398 620.99 3. Monthly interest = 10 000(0.005) = $50 Monthly S.F. deposit = 300 − 50 = $250 Find i such that 250D b 7|3.33B\2 = 10 000 D b 7|3.33B\2 = 40 (1.00275) b = 1.11 i = mno E.EE mno(E.33B\2) = 38.00125261 Need 37 deposits of $250 and a 38 th deposit (at the end of 38 months) of: Final deposit, Y = 10 000 − 250D G\ 7777 |3.33B\2 (1.00275) = 250.35 4. Total debt at year 15 = 100 000(1.03) G3 = $242 726.25 Semi-annual S.F. deposit: = BYB \BR.B2 ² 96 7777|6.6M5K = $6 718.31

5. Annual interest = 10 000(0.10) = 1000.00 Annual S.F deposit = E3 333 ² M6 7777|6.6U = 832.91 Annual amortization payment = E3 333 4 M6 7777|6.M6 = 1627.46 Total annual payment = $3460.37 6. Annual S.F deposit= E3 333 ² M6 7777|6.6K = $795.05 Annual interest payment = 1445.05 − 795.05 = $650 Annual effective rate = R23 E3 333 = 0.065 = 6.5% 7. Semi-annual S.F rate = (1.01) B − 1 = 0.0201 Semi-annual S.F deposit = E3 333 ² M6 7777|6.656M = $912.85 Semi-annual expense = 340 + 912.85 = $1252.85 Find X, then h B = 2X, such that 1252.85‘ E3 7777 |c = 10 000 ‘ E3 7777 |c = 7.9818 Starting value to solve ‘ E3 7777 |c = 7.9818 is X = E^O ±.“LML M6 Q 5 \.FPEP = 0.045467023 or h B = 2X = 9.09% ‘ E3 |c h B { 8.1109 8% } ¼ E% = 3.EBFE 3.EFPB 0.1982 0.1291 7.9818 h B ½ 1% ½ = f 0.65136% 7.9127 9% h B = 9.65136% Annual effective rate h = O1 + 3.3FR2EGR B Q B − 1 = f 9.88% 8. a) Let @ be the semi-annual S.F. deposit, X E = (1.07) E B ⁄ − 1 = 0.034408043 and X B = (1.06) E B ⁄ − 1 = 0.029563014 At the end of 10 years: @D E3 |c M (1.07) 2 EB ⁄ (1.06) 22 EB ⁄ + @D E3 |c 5 = 20 000 @(15.71774428 + 11.44083538) = 20 000 @ = $736.42 Semi-annual cost of loan = 736.42 + 1000 = $1836.42 b) Amount in the S.F. after the August 1, 2029 deposit = 736.42D E3 |c M (1.07) 2 EB ⁄ (1.06) YG EB ⁄ + 736.42D P |c 5 = 10 919.68 + 6538.35 = $17 458.03

c) Date Interest On Fund Deposit Increase In Fund Amount In Fund Aug. 1, 2029 Feb. 1, 2030 Aug. 1, 2030 516.11 553.14 736.42 736.30 252.53 1289.44 17 458.03 18 710.56 20 00.00 9. a) For the first 5 years: Annual interest payment = 15 000(0.07) = $1050 Annual S.F. deposit = 2000 − 1050 = $950 For the second 5 years: Annual interest payment = 15 000(0.05) = $750 Annual S.F. deposit = 2000 − 750 = $1250 Amount in the S.F. deposit during the first 5 years. 950D 2 T |3.3G A1.03) 2 + 1250D 2 T |3.3G = 5847.01 + 6636.42 = $12 483.43 Shortage = 15 000 − 12 483.43 = $2516.57 b) Let @ be the S.F. deposit during the first 5 years. At the end of 10 years: @D 2 T |3.3G A1.03) 2 + 1250D 2 T |3.3G = 15 000 6.154743501@ = 8363.58 @ = $1358.88 Increase in the S.F. needed during the first 5 years is $1358.88 − 950 = $408.88 EXERCISE 5.7 Concept Questions 1. The interest rate on the sinking fund would have to be greater than i 1 . 2. Not necessarily. It will depend on the interest rate earned by the sinking fund. 3. 1. The interest rate on the sinking-fund loan is often lower; 2. The borrower can miss making a sinking-fund deposit every once in a while (as long as they make it up later); 3. The borrower may be able to take some risks in their investments in a sinking fund in order to achieve a higher rate of return. 4. The lending institution may require the borrower to make their sinking-fund deposits with them (at the same time the interest payments are due and by offering a competitive interest rate on the sinking fund).

Part A 1. a) Annual payment = 23 333 4 M6 7777 |6.6V = $6793.40 b) Annual interest = 50 000A0.06) = $3000 Annual S.F. deposit = 23 333 · M6 7777|6.6V = $3 793.40 Total annual expense = $6793.40 Asame as a) c) Annual interest = 50 000A0.06) = $3000 Annual S.F. deposit = 23 333 · M6 7777|6.69 = $4361.53 Total annual expense 3000 + 4361.53 = $7361.53 2. Amortization: ¶ • = EP3 333 4 MK 7777|6.6V = $ 18 533.30 Sinking Fund: Interest 180 000A0.04) = $7200 Deposit = EP3 333 · MK 7777|6.69 = $9677.98 Total annual expense ¶ · = 7200 + 9677.98 = $16 877.98 Sinking fund is cheaper by $1655.32 per year. 3. Amortization: ¶ • = R3 333 4 M6 7777|6.6K = $7770.27 Sinking Fund: Interest 60 000A0.0475) = $2850.00 Deposit = R3 333 · M6 7777|6.65 = $5479.59 Total semi-annual expense ¶ · = 2850 + 5479.59 = $8329.59 Amortization is cheaper by $559.32 semi-annually. 4. Sinking fund: Interest = 100 000A0.03) = $3000.00 Deposit= E33 333 · 56 7777|6.65 = $4115.67 Total semi-annual expense = $7115.67 Amortization: 7115.67‘ B3 7777 |c = 100 000 ‘ B3 7777 |c = 14.0535 Starting value to solve ‘ B3 7777 |c = 14.0535 is X = E^O MU.6K9K 56 Q 5 EY.32G2 = 0.036023 or h B = 2X = 7.20% ‘ B3 |c h B { 14.2124 7% } µ E% = 3.E2PF 3.RBBE 0.6221 0.1589 14.0535 h B ¶ 1% ¶ = f 0.26% 13.5903 8% h B = 7.26% 5. Sinking fund: Interest = 500 000A0.03) = $15 000 Deposit= 233 333 · U6 7777|6.65 = $8 277.87 Total semi-annual expense = $23 277.87 Amortization: 23 277.87‘ Y3 |c = 500 000 ‘ Y3 |c = 21.4796

Starting value to solve ‘ Y3 |c = E^O 5M.U¸“V U6 Q 5 BE.Y\FR = 0.03313 or h B = 2X = 6.63% ‘ Y3 |c h B { { 23.1148 6% } } µ E% = E.RG2B E.\2F\ 1.7597 1.6352 21.4796 h B ¶ 1% ¶ = f 0.93% 21.3551 7% h B = 6.93% 6. Amortization: Annual payment = B33 333 4 M6 |6.6K = $25 900.92; Equivalent j 1 = A1.02) 2 - 1 = 0.0404 Sinking fund: Interest = 200 000A0.0404) = $8080 Deposit= 25 900.92 − 8080 = $17 820.92 17 820.92D E3 |c = 200 000 D E3 |c = 11.2228 Starting value to solve D E3 |c = 11.2228 is X = O MM.555L M6 Q 5 ^E EE.BBBP = 2.31% D E3 |c X = h E { { 10.9497 2% } } µ E% = 3.B\GE 3.2EYB 0.5142 0.2731 11.2228 h E ¶ 1% ¶ = f 0.53% 11.4639 3% h E = 2.53% 7. a) Sinking fund: Quarterly interest = 500 000A0.0125) = 6 250.00 Quarterly deposit = 233 333 · V6 |6.6MM = 5 927.79 Total quarterly expense = $12 177.79 b) Amortization: 12 177.79‘ R3 |c = 500 000 ‘ R3 |c = 41.0583 Starting value to solve ‘ R3 |c = 41.0583 is X = E^O UM.6KL9 V6 Q 5 YE.32PG = 0.01295 or h Y = 4X = 5.18% ‘ R3 |c h Y { { 42.0346 5% } } µ E% = 3.F\RG B.R2YG 2.6543 0.9763 41.0583 h Y ¶ 1% ¶ = f 0.37% 39.3803 6% h Y = 5.37% It would be less expensive to amortize the debt at h Y < 5.37%. 8. a) Amortization: Monthly expense = P3 333 4 V6 |Z b) Sinking fund: Monthly interest= 80 000 O 3.3R2 EB Q = $433.33 Monthly deposit= P3 333 · M5 7777|6.69 = $6545.33 Total monthly expense= $6978.66 Amortization method is cheaper by $56.52 a month.

EXERCISE 5.7 Part B 1. a) Amortization: X E = A1.03) E R ⁄ − 1 = 0.0049386 Monthly payment = B33 333 4 ¸5 7777|Z M = $3307.63 b) Sinking fund: X B = A1.01125) E G ⁄ − 1 = 0.003736025 Monthly interest = 200 000AX B ) = $747.20 X G = O1 + 3.3G GR2 Q 9VK M5 − 1 = 0.002503 Monthly deposit = B33 333 · ¸5 7777|Z 9 = $2538.46 Total monthly expense 747.20 + 2538.46 = $3285.66 Use sinking fund method to save $21.97 a month. 2. a) Sinking fund: Semi-annual interest = 10 000A0.04) = $400 X = A1.005) R − 1 = 0.0303775 Semi-annual deposit = E3 333 · 56 7777|Z = $370.73 Total semi-annual expense = $770.73 b) Amortization: 770.73‘ B3 7777 |c = 10 000 ‘ B3 7777 |c = 12.9747 Starting value to solve ‘ B3 7777 |c = 12.9747 is X = E^O M5.“¸U¸ 56 Q 5 EB.F\Y\ = 0.044636 or h B = 2X = 8.93% ‘ B3 |c h B { 13.0079 9% } µ E% = 3.3GGB 3.2Y2\ 0.5457 0.0332 12.9747 h B ¶ 1% ¶ = f 0.060839% 12.4622 10% h B = 9.060839% Find h Y such that O1 + ´ U Y Q Y = O1 + 3.3F3R3PGF B Q B h Y = 4 „O1 + 3.3F3R3PGF B Q M 5 − 1” = 8.96% 3. Annual cost to amortize $20 000 is B3 333 4 M6 |6.6L = $2980.59 Annual interest on 80 000 is 80 000A0.08) = $6400 S.F. X = A1.0125) Y − 1 = 0.050945337 Annual S.F. deposit to accumulate $80 000 is P3 333 · M6 |Z = $6332.36 Total annual expense = $15 712.95 Annual cost to amortize $100 000 is E33 33 4 M6 |6.6L = $14 902.95 Extra annual cost = 15 712.95 − 14 902.95 = $810.00.

4. Let @ be the amount of the loan, X E = A1.028) E R ⁄ − 1 = 0.00461314, X B = A1.02) E R ⁄ − 1 = 0.00330589 Amortization: Monthly cost = » 4 ML6 |Z M Sinking fund: Monthly interest = @X B Monthly deposit = » · ML6 |Z Find X and h EB = 12X such that the monthly expenses are equal, that is: » 4 ML6 |Z M = @X B + » · ML6 |Z E · ML6 |Z = E 4 ML6 |Z M − X B E · ML6 |Z = 0.004883934 D EP3 |c = f 204.7530 Starting value to solve D EP3 |c = 204.7530 is X = O 56U.¸K96 ML6 Q 5 ^E B3Y.\2BP = 0.0014356 or h EB = 12X = 1.72% D EP3 |c h EB { 194.1140 1% } µ E% = E3.RGF3 E2.2FFE 15.5991 10.639 204.7530 h EB ¶ 1% ¶ = f 0.68% 209.7131 2% h EB = 1.68% 5. Total annual outlay = $1400 Annual interest= 10 000A0.04) = $400 Annual S.F. deposit= 1400 − 400 = $1 000 Find i such that 1 000D b 7|3.3B = 10 000 D b 7|3.3B = 10 A1.02) b = 1.2 i = mno E.B mno E.3B i = 9.20693775 → 10 deposits Last S.F. deposit 10 000 − 1000D F 7 |3.3B A1.02) = $50.28 At the end of 10 years the irregular payment is 400 + 50.28 = $450.28 6. Change: amortization rate j 12 =9%, loan interest rate j 12 =8% Amortization: Quarterly rate X E = O1 + 3.3\2 EB Q G − 1 = 0.018867432 Quarterly payment = 2333 4 56 7777 7777 |Z M = $302.45

Sinking fund: Needs 5000 O1 + 3.3R EB Q R3 = $6744.25 Quarterly deposit = R\YY.B2 · 56 |Z For equal quarterly costs = R\YY.B2 · 56 |Z = 302.45 D B3 |c = 22.2987 Starting value to solve D B3 |c = 22.2987 is X = O 55.5“L¸ 56 Q 5 ^E BB.BFPR = 0.0109 or h Y = 4X = 4.36% D B3 |c h Y { 22.0190 4% } µ E% = 3.B\FR 3.2YY3 0.5440 0.2796 22.2986 h Y ¶ 1% ¶ = f 0.51% 22.5630 5% h Y = 4.51% Annual effective rate h = O1 + 3.3Y2EGF\ Y Q Y − 1 = 4.59% REVIEW EXERCISES 5.8 1. a) 1 = E2 333 4 M56 777777|6.669¸K = $155.46 b) i) Amortization method: Outstanding balance after 36 payments: 15 000A1.00375) GR − 155.46D GR 7777 |3.33G\2 = 17 163.72 − 5979.94 = $11 183.78 ‰ G\ = A11 183.78)A0.00375) = $41.94 ] G\ = 155.46 − 41.94 = $113.52 ii) Sum of digits: Final payment = 15 000A1.00375) EB3 − 155.46D EEF 77777 |3.33G\2 A1.00375) = $155.10 Total debt = 119 × 155.46 + 155.10 = $18 654.84 Interest rebate = G2\3 \BR3 × 3 654.84 = 1 797.21 Payments to date = 36 × 155.46 = 5 596.56 Outstanding balance = $11 261.07 ‰ G\ = PY \BR3 × 3 654.84 = $42.29 ] G\ = 155.46 − 42.29 = $113.17 2. X = O1 + 3.3\2 EB Q G − 1 = 0.018867432 Monthly payment = B3 333 4 56 7777|Z = $1209.81 Outstanding balance after 8 payments = 20 000A1 + X) P − 1209.81D P T |c = 23 225.84 − 10 342.30 = $12 883.54 ‰ F = 12 883.54 × X = $243.08 ] F = 1209.81 − 243.08 = $966.73

3. X = A1.07) B − 1 = 0.1449 Annual payment = E3 333 4 K|¹ 7777 = $2947.22 Final payment = 10 000A1 + X) 2 − 2947.22D Y T |c A1 + X) = $2947.18 Total interest = A4 × 2947.22 + 2947.18) − 10 000 = $4736.06 4. Let @ be the amount of the loan, X E = 3.32R B , X B = O1 + 3.3G EB Q R − 1 Amortization: Semi-annual payment = » 4 56 7777|Z Sinking fund: Semi-annual interest = @X E Semi-annual deposit = » · 56 7777|Z 5 For equal semi-annual costs: » 4 56 7777|Z = @X E + » · 56 7777|Z 5 E 4 56 7777|Z = X E + E · 56 7777|Z 5 E 4 56 7777|Z = 0.071205696 → ‘ B3 7777 |c = f 14.04382 Starting value to solve ‘ B3 7777 |c = 14.04382 is X = E^O MU.6U9L5 56 Q 5 EY.3YGPB = 0.036096 or h B = 2X = 7.22%. ‘ B3 |c h B { 14.2124 7% } µ E% = 3.ERPR 3.RBBE 0.6221 0.1686 14.0438 h B ¶ 1% ¶ = f 0.2710175% 13.5903 8% h B = 7.2710175% h EB = 12 º{1 + 0.072710175 2 | AB EB ⁄ ) − 1¼ = 7.16% 5. Semi-annual rate X = A1.01) B − 1 = 0.0201 Semi-annual S.F. deposit = G333 · L T|Z = $349.41 Balance in the S.F. at the end of 3 years is 349.41D R T |c = $2204.67 Deposit Number Interest On Fund Deposit Increase In Fund Amount In Fund 6 7 8 - 44.31 52.23 - 349.41 349.41 - 393.72 401.64 2204.67 2598.39 3000.03 6. a) X = A1.05) E R ⁄ − 1 = 0.0088164846 Monthly payment = R3 333 4 966|¹ 77777777 = $536.70 b) Outstanding balance after the 48th payment = 60 000A1 + X) YP − 536.70D YP 7777 |c = 88 647.33 − 31 384.59 = $57 262.74 c) ‰ YF = A57 262.74)AX) = $467.54 ] YF = 536.70 − 467.54 = $69.16 d) Principal paid in the first 48 payments = 60 000 − 57 262.74 = $2737.26 Payments made = 48 × 536.70 = $25 761.60 Interest paid in the first 48 payments = 25 761.60 − 2732.26 = $23 024.34

7. X = A1.025) E R ⁄ − 1 = 0.004123915 Monthly payment = E33 333 4 MUU 777777|Z = $922.32 Final payment = 100 000A1 + X) EYY − 922.32D EYG 77777 |c A1 + X) = 922.12 Total debt = 922.32 × 143 + 922.12 = $132 813.88 Interest rebate = G2\3 E3 YY3 × 32 813.88 = 11 220.84 Payment to date = 922.32 × 60 = 55 339.20 Outstanding balance after 5 years = $ 66 253.84 For the sinking fund: Amount due in 7 years: 66 253.84A1.02) EY = $87 420.53 Monthly S.F. deposit = P\ YB3.2G · LU 7777|6.665K = $936.56 Do NOT NOT NOT NOT refinance. 8. a) X E = A1.016) E R ⁄ − 1 = 0.002649061 Initial monthly payment = GG3 333 4 966 777777|Z M = $1595.77 b) Balance after 5 years = 330 000A1 + X E ) R3 − 1595.77D R3 7777 |c = 283 141.81 c) X B = A1.024) E R ⁄ − 1 = 0.003960577 New monthly payment = BPG EYE.PE 4 5U6 777777|Z 5 = $1830.15 d) Balance August 1, 2026= 283 141.81A1 + X B ) \ − 1830.15D \ T |c 5 = $278 121.26 ‰ ¿½`´. = 278 121.26 × X B = $1101.52 ] ¿½`´. = 1830.15 − 1101.52 = $728.63 9. a) X E = A1.025) E R ⁄ − 1 = 0.004123915 Original monthly payment = E23 333 4 5U6 777777|Z M = $985.69 Outstanding balance after 3 years = 150 000A1 + X E ) GR − 985.69D GR 7777 |c M = 135 784.41 Penalty = 135 784.41AX E )A3) = $1 679.88 Amount to be refinanced = 135 784.41 + 1 679.88 = $137 464.29 X B = A1.017) E R ⁄ − 1 = 0.00281347 New monthly payment = EG\ YRY.BF 4 56U 777777|Z 5 = $886.54; worth it to refinance a) IRD Penalty Method IRD= 3.32 EB − 3.3GY EB = 0.001333333 Penalty = 135 784.41A‰1ˆ)A24) = $4345.10 Amount to be refinanced = 135 784.41 + 4345.10 = $140 129.51 X B = A1.017) E R ⁄ − 1 = 0.00281347 New monthly payment = EY3 EBF.2E 4 56U 777777|Z 5 = $903.73; still worth it to refinance, but impact of the IRD penalty is much higher.

10. Amortization: Annual payment = E3 333 4 M6 7777|6.6V = $1358.68 Sinking fund: Annual interest = 10 000A0.045) = $450 Annual S.F. rate X = A1.0075) Y − 1 = 0.030339 Annual S.F. deposit = E3 333 · M6 7777|Z = $870.94 Total annual expense = 450 + 870.94 = $1320.94 Use Sinking fund and save $37.74 a year. 11. Given ] E = 10.00 and ] Y = 13.31, we calculate X from: 10A1 + X) G = 13.31 A1 + X) G = 1.331 X = 0.10 ] B = 10.00A1 + X) = 11.00 ‰ B = 389.00 Hence the periodic payment is $400.00 Hence, ‰ E = $390.00 If e is the original loan, Xe = 390 and e = GF3 3.E3 = $3900. 12. Total interest = $5681.17 Number of payment = 19 × 12 = 228 D b = ABBP)ABBF) B = 26 106 ‰ ERG = RR BR E3R A5681.17) = $14.36 13. a) Monthly rate X = A1.025) E R ⁄ − 1 = 0.004123915 Monthly payment = E33 333 4 5U6 777777|Z = $657.13 Final payment = 100 000A1 + X) BY3 − 657.13D BGF 77777 |c A1 + X) = 655.10 b) Outstanding balance after 5 years = 100 000A1 + X) R3 − 657.13D R3 7777 |c = 83 378.06 Principal paid in the first 5 years = 100 000 − 83 378.06 = $16 621.94 Interest paid in the first 5 years = 60 × 657.13 − 16 621.94 = $22 805.86 c) Amortization: ‰ E = 100 000X = $412.39 ] E = 657.13 − 412.39 = $244.74 Sum of digits: Total interest= 239 × 657.13 + 655.10 − 100 000 = $57 709.17 Sum of digits from 1 to 240 = BY3 B A241) = 28 920 ‰ E = BY3 BP FB3 A57 709.17) = $478.91 ] E = 657.13 − 478.91 = $178.22 d) New monthly rate X = A1.04) E R ⁄ − 1 = 0.006558197 New monthly payment = PG G\P.3R 4 ML6 777777|Z = $790.55

14. a) Principal paid in the 18th payment = 40A1.00625) EY = $43.65 b) Principal paid in the 1st payment = 40A1.00625) ^G = $39.26 Interest paid in the 1st payment = 2000A0.00625) = $12.50 Regular monthly payment = 39.26 + 12.50 = $51.76 15. a) Principal paid in the 14th payment = 62A1.01) R = $65.81 b) Amount of the loan = sum of all principal payments = 62²A1.01) ^\ + A1.01) ^R + ⋯ + A1.01) Y3 ³ = 62A1.01) ^\ AE.3E) UL ^E 3.3E = $3540.41 16. Annual payment = 50 000 + 180 975 = $230 975 Loan amount = 819 025 + 180 975 = $1 000 000 Annual rate of interest = 23 333 E 333 333 = 0.05 @ = 1 000 000A1.05) G − 230 975D G T |3.32 = $429 476.31 17. Monthly interest X = A1.03625) E R ⁄ − 1 = 0.005952383 Monthly payment = GR3 333 4 966 777777|Z = $2577.31 Balance after 12 payments = 360 000A1 + X) EB − 2577.31D EB 7777 |c = 386 573.06 − 31 960.60 = $354 612.46 Principal repaid in the first year = 360 000 − 354 612.46 = $5387.54 18. a) Monthly interest X = A1.0225) E R ⁄ − 1 = 0.00371532 Monthly payment = EP3 333 4 966 777777|Z = $996.25 = $996.30 Arounded up to the next dime) b) Final payment = 180 000A1 + X) G33 − 996.30D BFF 77777 |c A1 + X) = 969.54 c) Balance after 36 payments = 180 000A1 + X) GR − 996.30D GR 7777 |c = $167 408.51 Balance after 48 payments = 180 000A1 + X) YP − 996.30D YP 7777 |c = $162 823.69 Principal paid during the 4th year = 167 408.51 − 162 823.69 = $4584.82 Interest paid during the 4th year = 12 × 996.30 − 4584.82 = $7370.78 d) New balance = 162 823.69 − 5000 = $157 823.69 Find i such that 996.30‘ b 7|c = 157 823.69 A1 + X) ^b = 0.411456946 i = − mno 3.YEEY2RFYR mnoAEjc) i = 239.47 → 240 months The mortgage will be paid off in 48 + 240 = 288 months, that is 12 months sooner. Concluding payment = 157 823.69A1 + X) BY3 − 996.30D BGF 77777 |c A1 + X) = $466.57 Difference total payments = A299 × 996.30 + 969.54) − A287 × 996.30 + 466.57 + 5000) = $7458.57

19. Monthly payment = E3 333 4 K6 7777|6.66V5K = $234 Arounded up to the next dollar) Final payment = 10 000A1.00625) 23 − 234D YF 7777 |3.33RB2 A1.00625) = $204.42 Total interest = A49 × 234 + 204.42) − 10 000 = $1670.42 Total debt = $11 670.42 Less interest rebate = RG3 EB\2 × 1670.42 = 825.39 Less payments to date = 15 × 234 = 3510.00 Outstanding balance after 15 payments = $7335.03 20. a) Annual S.F. deposit = E33 333 · 56 7777|6.65K = $3914.71 b) Annual interest on the loan = 100 000A0.08) = $8 000.00 Total annual cost of the loan = 8 000 + 3914.71 = $11 914.71 c) Find X = h such that 11 914.71‘ B3 7777 |c = 100 000 ‘ B3 7777 |c = 8.3930 Starting value to solve ‘ B3 7777 |c = 8.3930 is X = E^O L.9“96 56 Q 5 P.GFG3 = 0.09816467 ‘ B3 |c X = h { 8.5136 10% } µ E% = 3.EB3R 3.223G 0.5503 0.1206 8.3930 h ¶ 1% ¶ = f 0.22% 7.9633 11% h EB = 10.22% 21. a) Annual payment = 23 333 4 MK 7777|6.6U9 = $4591.90 Final payment = 50 000A1.043) E2 − 4591.90D EY 7777 |3.3YG A1.043) = $4591.88 b) Monthly S.F. deposit = Y2FE.F3 · M5 7777|6.669 = $376.39 Arounded up) c) Total debt = 14 × 4591.90 + 4591.88 = 68 878.48 Less interest rebate = Y2 EB3 × 18 878.48 = 7 079.43 Less payments to date = 6 × 4591.90 = 27 551.40 Outstanding principal after 6 years = $34 247.65 22. a) Monthly payment = E2 333 4 MUU 777777|6.66UMV¸ = $138.74 Final payment = 15 000A1.004167) EYY − 138.74D EYG 77777 |3.33YER\ A1.004167) = $137.47 Total interest = A143 × 138.74 + 137.47) − 15 000 = $4977.29 Amortization: ˜ BY = 15 000A1.004167) BY − 138.74D BY 7777 |3.33YER\ = $13 079.83 Sum of digits: Total debt = 143 × 138.74 + 137.47 = 19 977.29 Less interest rebate= \BR3 E3 YY3 × 4977.29 = 3461.22 Less payments to date = 24 × 138.74 = 3 329.76 ˜ BY = $13 186.31

b) New monthly payment Oat X = 3.3Y EB Q = EG EPR.GE 4 M56 777777|6.6U M5 ⁄ = $133.51 Since, the new monthly payment is less than the original monthly payment, should should should should refinance the loan refinance the loan refinance the loan refinance the loan 23. Original Mortgage: Monthly rate X = A1.03) E R ⁄ − 1 = 0.004938622 Monthly payment = E33 333 4 ML6 777777|Z = $839.89 Balance at the end of 3 years = 100 000A1 + X) GR − 839.89D GR 7777 |c = 86 403.60 IRD = 3.3R EB − 3.32 EB = 0.0008333 Term remaining in guaranteed period=A2)A12)=24 IRD penalty=A86 403.60)A 0.0008333)A24) =1728.07 New rate i = A1.025) 1/6 ̶ 1 = 0.04123915 New monthly payment = = PR Y3G.R3jE\BP.3\ 4 MUU 777777|Z = $812.86 Should refinance Case Study I: Case Study I: Case Study I: Case Study I: Comparison of amortization, sum-of-digits and sinking-fund methods i) Amortization: Monthly payment = E3 333 4 V6 7777|Z = $222.45 Sum of digits: Monthly payment = $222.45 Sinking fund: Monthly interest on the loan = 10 000A0.01) = $100 Monthly rate on S.F. X = O1 + 3.3G2 GR2 Q GR2 EB ⁄ − 1 = 0.002920784 Monthly S.F deposit= E3 333 · V6 7777|Z = $152.73 Total monthly expense= $252.73 ii) Amortization: Balance at the end of 2 years = 10 000A1.01) BY − 222.45D BY 7777 |3.3E = $6697.10 Sum of digits Final payment = 10 000A1.01) R3 − 222.45D 2F 7777 |3.3E A1.01) = $222.00 Total debt = 59 × 222.45 + 222.00 = 13 346.55 Less interest rebate = RRR EPG3 × 3346.55 = 1 217.92 Less payments to date = 24 × 222.45 = 5 338.80 Balance at the end of 2 years = $6 789.83 Sinking fund: Amount in the S.F. at the end of 2 years = 152.73D BY 7777 |c = $3791.32 Balance at the end of 2 years = 10 000 − 3791.32 = $6208.68

iii) Amortization ‰ E = 10 000A0.01) = $100 ] E = 222.45 − 100 = $122.45 Balance after 23 payments = 10 000A1.01) BG − 222.45D BG 7777 |3.3E = $6851.04 ‰ BY = 6851.04A0.01) = $68.51 ] BY = 222.45 − 68.51 = $153.94 Sum of digits: ‰ E = R3 EPG3 × 3346.55 = $109.72 ] E = 222.45 − 109.72 = $112.73 ‰ BY = G\ EPG3 × 3346.55 = $67.66 ] BY = 222.45 − 67.66 = $154.79 Sinking fund: ‰ E = ‰ BY = $100 ] E = ] BY = $152.73 Case Study II: Case Study II: Case Study II: Case Study II: Increasing extra annual payments a) Monthly rate X ¯ = A1.04) E R ⁄ − 1 = 0.006558197 Monthly payment = E33 333 4 966 777777|Z ˘ = $763.30 Arounded up to the next dime) Final payment = 100 000A1 + X ¯ ) G33 − 763.30D BFF 77777 |c ˘ A 1 + X ˙ ) = $682.71 b) Equivalent annual rate X 4 = A1.04) B − 1 = 0.0816 Equivalent annual payment = 763.30D 12 7777 |c ˘ = 9497.32 Outstanding balance ˜ b after i years ˜ b = 100 000A1.0816) b − 9497.32D i T|3.3PER − 300 D i T|3.3PER − 50 · l T|6.6LMV −i 0.0816 By trial and error we want to find i such that ˜ b = 0 . For i = 20 ˜ B3 = 7444.31 Final payment at 20 years and 11 months: 7444.31A1 + X ¯ ) EE − 763.30D 10 7777 |c ˘ A 1 + X ˙ ) = $85.50

Case Study III Case Study III Case Study III Case Study III: Mortgage amortization a) Monthly rate X = A1.021) E R ⁄ − 1 = 0.003469762 Monthly payment = Y23 333 4 966 777777|Z = $2416.14 Final payment = 450 000A1 + X) G33 − $2416.14D BFF 77777 |c A 1 + X ) = $2412.00 b) Total interest = A299 × 2416.14 + 2412.00) − 450 000 = 274 837.86 c) Payment Number Monthly Payment Interest Payment Principal Payment Outstanding Balance 1 2 3 2416.14 2416.14 2416.14 1561.39 1558.43 1555.45 854.75 857.71 860.69 450 000.00 449 145.25 448 287.54 447 426.85 d) Retrospective method: ˜ BY = 450 000A1 + X) BY − 2416.14D BY 7777 |c = $428 646.31 Prospective method: ˜ BY = 2416.14 ‘ 275 77777 |X + 2412A1 + X) ^B\R = $428 646.31 e) Principal paid = ˜ 3 − ˜ BY = 450 000 − 428 646.31 = $21 353.69 Interest paid = 24 × 2416.14 − 21 353.69 = $36 633.67 f) Outstanding balance after 2 years = 428 646.31 − 10 000 = $418 646.31 Find i such that 2416.14 ‘ i 7 |X = $418 646.31 i = f 265.41 months Final payment = 418 646.31A1 + X) BRR − 2416.14D BR2 77777 |c A 1 + X ) = $991.75 Total interest = ²10 000 + A24 + 265)2416.14 + 991.75} − 450 000 = $259 256.21 Savings = 274 837.86 − 259 256.21 = $15 581.65 g) Monthly payment = Y23 333 4 5U6 777777|Z = $2765.89 Final payment= 450 000A1 + X) BY3 − 2765.89D BGF 77777 |c A 1 + X ) = $2764.98 Total interest = A239 × 2765.89 + 2764.98) − 450 000 = $213 812.69 Paying an extra A2765.89 − 2416.14) = $349.75 per month reduces the total interest paid by A274 837.86 − 213 812.69) = $61 025.17 h) Penalty= 3XA428 646.31) = $4461.90 New monthly rate X = O1 + 3.3GG B Q E R ⁄ − 1 = 0.002731282 New monthly payment = YBP RYR.GEjYYRE.F3 ‘ 5¸V 777777|Z = $2236.37 < $2416.14 It does pay to refinance

Case Study IV Case Study IV Case Study IV Case Study IV: Accelerated mortgage payments a) 20 year amortization 20 year amortization 20 year amortization 20 year amortization at monthly rate X ¯ = A1.019) E R ⁄ − 1 = 0.00314188444 Monthly payment = B33 333 ‘ 240 77777 |X ˘ = $1187.89 Interest paid over life of mortgage=A20)A12)A 1187.89) − 200 000 = $85 094 Interest saved = 109 141 − 85 094 = $24 047 b) Doubling Doubling Doubling Doubling-up the 4 up the 4 up the 4 up the 4¨¾ monthly payment each year monthly payment each year monthly payment each year monthly payment each year: Equivalent annual payment = 1187.89‹D 12 7777 |c ˘ + A 1 + X ˙ ) 8 › = $15 721.68 Equivalent annual rate X 4 = A1.019) B − 1 = 0.038361 Find i such that 15 721.68‘ b 7|3.3GPGRE = 200 000 i = f 17.8349536 years = f 17 years and 10 month A214 months) Time paid off sooner = 25 years − 17 years 10 months = 7 years 2 months Final Payment = 200 000A1 + X ¯ ) BEY − ²15 695.23D E\ 7777 |3.3GPGRE A 1 + X ˙ ) 10 + 1187.89D 9| T X ˙ A 1 + X ˙ ) + 1187.89 A 1 + X ˙ ) 6 ³ = $857.43 Total interest = ²A213 + 18) × 1187.89 + 857.43³ − 200 000 = $75 260.02 Interest saved = 109 141 − 75 260.02 = $33 880.98 = $33 881 Doubling Doubling Doubling Doubling-up the 4 up the 4 up the 4 up the 4 th th th th and 10 and 10 and 10 and 10 th th th th monthly payment each year monthly payment each year monthly payment each year monthly payment each year Equivalent annual payment = 1187.89 ˚D 12 7777 |c ˘ + A1 + X ˙ ) 8 + A 1 + X ˙ ) 2 ¸ = $16 917.05 Equivalent annual rate X 4 = A1.019) B − 1 = 0.038361 Find i such that 16 917.05 ‘ i 7 |0.038361 = 200 000 i = f 16.052 years = f 16 years and 1 month A193 months) Time paid off sooner = 25 years − 16 years 1 month = 8 years 11 months Final Payment = 200 000A1 + X ¯ ) EFG − 16 917.05 D ER 7777 |3.3GPGRE A 1 + X ˙ ) = $865.87 Total interest = ²A192 + 32) × 1187.89 + 865.87³ − 200 000 = $66 953.23 = $66 953 Interest saved = 109 141 − 66 953 = $42 188 Doubling Doubling Doubling Doubling-up each 3 up each 3 up each 3 up each 3À˝ payment Amonths 3, 6, 9 and 12): payment Amonths 3, 6, 9 and 12): payment Amonths 3, 6, 9 and 12): payment Amonths 3, 6, 9 and 12): Equivalent quarterly payment = 1187.89˛D G T |c ˘ + 1 ˇ = $4762.77 Aat end of each 3- month period) Equivalent quarterly rate X — = A1.019) E B ⁄ − 1 = 0.009455299 Find i such that 4762.77‘ b 7|c Á = 200 000 i = f 53.759247 quarters = f 13 years 6 months A162 months) Time paid off sooner = 25 years − 13 years 6 months = 11 years 6 months Final payment = 200 000A1 + X ¯ ) ERB − 4762.77D 2G 7777 |c Á A 1 + X ˙ ) 3 − 1187.89 D B T |c ˘ A 1 + X ˙ ) = $1233.22 Total interest = ²A161 + 53) × 1187.89 + 1233.22} − 200 000 = $55 441.68 = $55 442 Interest saved = 109 141 − 55 442 = $53 699

Case S Case S Case S Case Study V udy V udy V udy V: Sub-Prime Mortgages a) Monthly rate X E = A1.02) E R ⁄ − 1 = 0.00330589 Monthly payment = Y33 333 ‘ UL6 777777|Z M = $1663.57 Total interest paid = 480 ∗ 1663.57) − 400 000 = $398 513.60 b) Outstanding balance ˜ GR after 3 years ˜ GR = 400 000A1 + X E ) GR − 1663.57D GR 7777 |c M = $386 978.28 New monthly rate X B = A1.035) E R ⁄ − 1 = 0.005750039 New monthly payment = GPR F\P.BP 4 UUU 777777|Z 5 = $2414.48 Total interest paid = 36A1663.57) + 444A2414.48) − 400 000 = $731 917.64 which is a whopping $344 939.36 more than the situation in part a and it is due to the interest rate rising to a rate that was more in line with what the rate should have been at the start of the mortgage c) Had house prices dropped after 3-years, the Chens, seeing that their outstand- ing balance of $386,978,28 was more than the value of the house, may have been tempted to default on their mortgage payments, which could have led to them simply packing up and leaving the house behind.