Get started today!

Good to have you back!
If you've signed in to StudyBlue with Facebook in the past, please do that again.

clifford y.

Advertisement

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

Advertisement

"StudyBlue is great for studying. I love the study guides, flashcards and quizzes. So extremely helpful for all of my classes!"

Alice , Arizona State University"I'm a student using StudyBlue, and I can 100% say that it helps me so much. Study materials for almost every subject in school are available in StudyBlue. It is so helpful for my education!"

Tim , University of Florida"StudyBlue provides way more features than other studying apps, and thus allows me to learn very quickly!Â I actually feel much more comfortable taking my exams after I study with this app. It's amazing!"

Jennifer , Rutgers University"I love flashcards but carrying around physical flashcards is cumbersome and simply outdated. StudyBlue is exactly what I was looking for!"

Justin , LSU
StudyBlue is not sponsored or endorsed by any college, university, or instructor.

© 2015 StudyBlue Inc. All rights reserved.

© 2015 StudyBlue Inc. All rights reserved.