Class Notes (810,969)
Economics (470)
ECON 371 (1)
Huang Hui (1)
Lecture

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School
University of British Columbia
Department
Economics
Course
ECON 371
Professor
Huang Hui
Semester
Winter

Description
ECON 371 in Fall 2009 Suggested Answers to Assignment 1 Q1. You are an investment advisor who has been approached by a client for help on his ▯nancial strategy. He has \$250,000 in savings in the bank. He is 55 years old and expects to work for 10 more years, making \$100,000 a year. He expects to make a return of 5% on his investment for the foreseeable future. You can ignore taxes. a). Once he retires 10 years from now, he would like to be able to withdraw \$80,000 a year for the following 25 years. His doctor tells him he will probably live to be 90 years old. How much would he need in the bank 10 years from now when he retires to be able to do this? (the withdrawals will start when he would be 66 years old, i.e. one year after this retirement) b). How much of his income would he need to save each year for the next 10 years to be able to a▯ord these planned withdrawals after the 10th year? c). Assume that interest rates decline to 4% 10 years from now. How much, if any, would your client have to lower his annual withdrawal, so that PV of withdrawals stays the same as in part (a)? An: a). The present value of annual withdrawals at retirement is given by: ▯ ▯ PV = PMT ▯ 1 ▯ 1 r r(1 + r)n ▯ ▯ = \$80;000 ▯ 1 ▯ 1 5% 5%(1 + 5%) 25 = \$1;127;515:57 b). Future value of existing savings at retirement: \$250;000 ▯ (1 + 5%)0= \$407;223:66 (1) Shortfall in savings = 1;127;515:57 ▯ 407;223:66 = \$720;291:91. The annual savings needed to cover the shortfall is given by: ▯ n ▯ FV = PMT ▯ (1 + r) ▯ 1 r ▯ 10 ▯ \$720;291:91 = PMT ▯ (1 + 5%) ▯ 1 5% PMT = \$57;266:50 c). The annual withdrawal is given by: ▯ ▯ 1 1 \$1;127;515:57 = PMT ▯ 4% ▯ 4%(1 + 4%)25 1 PMT = \$72;174:48 Annual withdrawals have to be decreased by 80;000 ▯ 72;174:48 = \$7;825:52 Q2. Your buddy in mechanical engineering has invented a money machine. The main drawback of the machine is that it is slow. It takes six months to manu- facture \$60. Once built, the machine will last forever, although it does require maintenance (that is, manufactures no money) in years 2 and 3 (this is the only maintenance required). After the maintenance, it will need six months to be able to manufacture money again. The machine can be built immediately, but it will cost \$1;500 to build. Your buddy wants to know if he should invest the money to construct it. If the interest rate is 9% per year, what should you buddy do? An: The six-month interest rate is (1 + 9%) 1=2▯ 1 = 4:4031% 60 The present value of a perpetuity of 60 at the above rate is 4:4031% = 1362:6763. Up to now we can already decide that your buddy should not do it because the PV of the cash ows of this money machine is less than 1362:6763 which is lower than the cost 1;500. Q3. You are planning buying a house. The house costs \$350;000; you will put 10% down and obtain a 30-year ▯xed rate mortgage at 5:5% (APR semi-annually compounded) for the rest. a). Assuming that monthly payments begin in one month, what will each pay- ment be? b). How much interest will you pay (in dollars) over the life of the loan? c). How much is still owed to the bank at the end of year 20? 2=12 An: The monthly rate is rm = (1 + 0:055=2) ▯ 1 = 0:004532 a). ▯ ▯ 1 1 PV = PMT ▯ t rm rm(1+r m PV 350;000 ▯ 90% PMT = = = 1776:38 1 ▯ 1 t 1 ▯ 1 30▯12 rm rm(1+m) 0:004532 0:004532▯(1+0:004532) b). Interest paid = Total payments - Principal = 1776:38▯30▯12▯350;000▯ 0:9 = 324496:80. 2 c). The outstanding balance equals the present value of the remaining pay- ments. So ▯ ▯ 1 1 PV = PMT ▯ ▯ t ▯m rm(1+r m ▯ 1 1 PV = 1776:38 ▯ ▯ 0:004532 0:004532 ▯ (1 + 0:004532)▯12 PV = 164143:29
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