Prof Dmitry Livdan 12 October 2011 University of California Walter A Haas School of Business UGBA 103 Introduction to Finace Solutions to the MIDTERM 1 a 5 points The monthly interest rate on the Savings Account is r 0 005 0 06 12 The present value of Anna s contributions to the Savings Account is PV 100 1 005 0 005 1 cid 0 1 1 005 60 5 198 42 The future value of her contributions on her 18th birthday will therefore be On the other hand the price of the car will be FV PV 1 005 60 7 011 89 P 16 000 1 02 5 17 665 29 so that the amount that will need to be cid 12 nanced is 17 665 29 cid 0 7 011 89 10 653 40 b 4 points The monthly interest rate charged by the dealer is r 1 08 1 12 cid 0 1 0 00643 Therefore the amount C of Anna s monthly payment will solve 10 653 40 C 1 cid 0 1 0 00643 1 00643 48 C 258 69 2 a 5 points Since the price of every bond must be the sum of its discounted cash cid 13 ows the discount factors must solve Using 1 we have DF1 0 92 Using this value for DF1 in 2 we get 100DF1 92 00 8DF1 108DF2 101 32 1 2 101 32 cid 0 8 0 92 DF2 108 0 87 UGBA 103 MIDTERM Solutions 2 b 3 points The discount factors can be written as Therefore DFt 1 1 rt t 1 DF1 1 cid 0 1 8 69565 and cid 0 1 7 21125 r1 r2 DF 1 2 2 c 3 points he bond that you would like to purchase will pay 5 200 10 at the end of the cid 12 rst year and 210 at the end of the second year Let us form a portfolio containing a quantity nA of bond A and nB of bond B We would like this portfolio to pay 10 at the end of the cid 12 rst year and 210 at the end of the second year Mathematically we would like nA and nB to satisfy Using 2 we have nB 210 104 2 02 Using this value for nB in 1 we get 100 nA 4 nB 10 104 nB 210 1 2 10 cid 0 4 2 02 100 nA 0 0192 Therefore the portfolio that would replicate the 5 coupon bond would consist in buying 0 0192 units of bond A and 2 02 units of bond B 3 The present value of the costs associated with operating the new machine in thousands of dollars is NPV cid 0 3 000 1 cid 0 1 1 12 5 1 1 12 5 Initital investment PV maintenance costs PV depreciation tax shield PV salvage value cid 0 1 cid 0 0 34 500 0 12 0 34 cid 1 600 1 cid 0 1 cid 0 0 34 500 0 12 1 12 5 cid 0 3 266 95 The equivalent annual cost for the new machine then solves 3 266 95 EAC 0 12 1 cid 0 1 1 12 5 EAC 906 284 Since the machine will have to be replaced every cid 12 ve years in all three scenarios we know that the equivalent annual cash cid 13 ows after year 2 will be exactly the same cid 0 906 284 UGBA 103 MIDTERM Solutions 3 every year in every scenario So we only need to compare the present value of the costs for the cid 12 rst 2 years for each of the three scenarios Replacing immediately generates a cash cid 13 ow C0 1 400 cid 0 0 34 1 400 cid 0 1 000 1 264 now and cash cid 13 ows of C1 C2 cid 0 906 284 in years 1 and 2 The present value of these cash cid 13 ows is cid 0 906 284 1 12 2 cid 0 267 666 PV0 1 264 cid 0 906 284 1 12 Similarly replacing after 1 year we have C0 0 C1 cid 0 1 cid 0 0 34 1 500 0 34 200 1 200 cid 0 0 34 1 200 cid 0 800 142 C2 cid 0 906 284 PV1 142 1 12 cid 0 906 284 1 12 2 cid 0 595 698 Finally replacing after 2 years we have C0 0 C1 cid 0 1 cid 0 0 34 1 500 0 34 200 cid 0 922 C2 cid 0 1 cid 0 0 34 1 500 0 34 200 800 cid 0 0 34 800 cid 0 600 cid 0 190 so that and P V2 cid 0 922 1 12 cid 0 190 1 12 2 cid 0 974 681 Hence the best alternative is to replace immediately