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Answer to Selected Problems in Section 3.1 Problem 3.1.5 Betty’s monthly rate is i(12)/12 = .12/12 = .01. In each payment Kt, PRt = 1980036 = 550. So after the tth payment, OBt = 19800−550t. The interest payment in Kt is It = OBt−1 ·(.01) = (19800−550(t−1))(.01) = 198−5.5(t−1) = 198−5.5t + 5.5. So It = 203.5−5.5t. Adding PRt and It (both in blue), we have Kt = 550 + 203.5−5.5t = 753.5−5.5t. After the 16th payment, Bank Y have the right to collect all the future 20 payments (i.e. K17, K18, ..., K36). The rate Bank Y wants is i(2) = .14 which translate to monthly rate j = .0113402601. Using the formula of Kt above, we have (for example, K17 = 753−5.5·17 = 660): K17 K18 K19 ... ... K35 K36 660 654.50 649 ... ... 561 555.50 The price Bank Y pays to Bank X is the present value of all these payments at time 16 with monthly rate j. We write 660 = 665.50 − 5.5, 654.5 = 665.50 − 2 · 5.5, 649 = 665.50 − 3 · 5.5 ... 555.50 = 665.50−20·5.5. ... ...16 17 18 19 35 36 665.5 665.5 665.5 665.5 665.5 −5.5 −5.5·2 −5.5·3 −5.5·19 −5.5·20 The present value of all these payments at time 16 is (v = 1/(1 + j)) 665.5a20|j + (−5.5v −5.5·2v2 −5.5·3v3 −···−5.5·19v19 −5.5·20v20) Factoring out −5.5, we have1 665.5a20|j −5.5(v + 2v2 + 3v3 +···+ 19v19 + 20v20) = 665.5a20|j −5.5(Ia)20|j = = 10857.27764 Answer $10,857.28 1For formula lovers, here I used the notation (Ia)n|j = v + 2v2 + 3v3 + ··· + nvn = 1j parenleftBig1 − vn 1 − v − nv n parenrightBig = 1j (¨an|j − nvn). Problem 3.1.6 There will be 5×12 = 60 payments. The monthly rate is j = i (12) 12 = .09 12 = .0075. The paymentsare K1 K2 K3 ... ... Kt ... ... K59 K60 1000 (.98)1000 (.98)21000 ... ... (.98)t−11000 ... ... (.98)581000 (.98)591000 Using prospective form of OB40, we have OB40 = K41v + K42v2 + K43v3 +···+ K59v19 + K60v20 = = (.98)401000v + (.98)411000v2 + (.98)421000v3 +···+ (.98)581000v19 + (.98)591000v20. Factoring out (.98)401000v, we get OB40 = (.98)401000v[1 + (.98)v + (.98)2v2 +···+ (.98)18v18 + (.98)19v19] = = (.98)401000v[1 + (.98v) + (.98v)2 +···+ (.98v)18 + (.98v)19] = (.98)401000v · 1−(.98v) 20 1−(.98v) Substituting v = 11 + j = 11.0075 into the above equation, we get OB40 = 6889.114798. Answer $6,889.11 Problem 3.1.7 In the first ten years, each payment only pays the interest, so the outstanding balance does not change: OBt = 1000 for 1 ≤ t ≤ 10. In the second ten years, each payment is 150% of interest due, so Kt = 1.5·iOBt−1 = iOBt−1 + (.5)·iOBt−1 = It + PRt. So PRt = .5iOBt−1 = .05OBt−1 (keep in mind that i = .1) and OBt = OBt−1 − PRt = OBt−1 − (.05)OBt−1 = .95OBt−1. So, OBt = .95OBt−1 for 11 ≤ t ≤ 20. The payment schedule looks like this: (only the blue entries are useful.) t Kt It PRt OBt 0 – – – 1000 1 1000i = 100 1000i = 100 0 1000 ... ... ... ... ... 10 1000i = 100 1000i = 100 0 1000 11 1.5·1000i = 150 1000i = 100 150−100 = 50 (.95)1000 12 1.5·950i = 142.5 950i = 95 142.5−95 = 47.5 (.95)21000 ... ... ... ... ... 20 ... ... ... (.95)101000 21 X ... ... ... 22 X ... ... ... ... ... ... ... ... 30 X ... ... ... OB20 is to be paid by the ten payments in the last ten years. By prospective form of OB20, we have OB20 = Xa10|.1, and so X = OB20a 10|.1 = 97.44167961. Answer X = $97.44 Problem 3.1.9 Sketch of Solution In both cases, let X be the monthly payment. (i) Consider OB12 in both retrospective and prospective forms: OB12 = 20000−12X = Xa36|j. The last equation above gives (note j = .06/12 = .005). X = 200012 + a 36|j = 445.7220216. So, OB12 = 20000−12X = 14651.33574. Answer X = $445.72, and OB12 = $14,651.34 (ii) Similarly OB12 = 20000 parenleftbigg 1 + .0312 parenrightbigg12 −Xs12|j = Xa36|jprime where j = .03/12 = .0025 and jprime = .05/12 = .004166666.... Solving the last equation, we get X = 20000 parenleftBig 1 + .0312 parenrightBig12 s12|j + a36|jprime = 452.6109345 So, OB12 = Xa36|jprime = 15101.68125. Answer X = $452.61, and OB12 = $15,101.68

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