UNIVERSITY OF TORONTO Joseph L. Rotman School of Management SOLUTIONS

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1 UNIVERSITY OF TORONTO Joseph L. Rotman School of Management Oct., 08 Corhay/Kan RSM MID-TERM EXAMINATION Yang/Wang SOLUTIONS. a) The optimal consumption plan is C 0 = Y 0 = 0 and C = Y = 0. Therefore, the utility would be U0, 0)0) 0) =.60. b) If we have access to the capital market but not the production opportunity, the consumption allocation problem can be set up as follows: max C 0,C C 0 C s.t. C 0 + C. With the budget constraint we have = = C 0 = 9.09 C., which can be substituted into the utility function so that we maximize the utility function with respect to C, i.e., max 9.09 C ) C. C. Taking the first order derivative of UC 0, C ) with respect to C and setting it to zero, we obtain du = dc C 9.09 C ) C ) C = 0.. Multiplying C ) 9.09 C on both sides of the above equation, we have C ) C.. = 0 and C =.. Therefore, C 0 = 9.09 C = Today you need to lend. Y 0 C 0 = 0. units of goods to the capital market.

2 c) When we have access to the production opportunity but not the capital market, the consumption allocation problem becomes max C 0 C C 0,C s.t. C 0 = 0 I 0 C = I 0 Substituting the budget constraints into the utility function and we can maximize the utility with respect to I 0, i.e., max0 I 0 ) 0 + 0I 0 ). I 0 Taking the first order derivative of UC 0, C ) with respect to I 0 and setting it to zero, we obtain du = di 0 0 I 0) ) I 0 0 I 0) I0 ) 0 = 0. I 0 Multiplying 0 I 0 ) I0 ) on both sides of the above equation, we have I 0 = 0 I 0 ) 0 I0 4I 0 + I 0 40 = 0. Solving the quadratic equation, we have I 0 =.04 and hence I 0 = 9.4. Therefore, C 0 = Y 0 I 0 = 0.76 and C = I 0 = d) Finally, when we have access to both the production opportunity and the capital market, we know that the investment problem and the consumption can be solved separately according to Fisher s Separation Theorem. First, we solve the investment problem. Since the production displays a decreasing marginal return to investment, the optimal investment amount is achieved when the marginal return on investment is equal to the interest rate in the capital market, i.e., dfi 0 ) di 0 = 0 I0 = Therefore, I 0 = 0/. and I 0 = Once we determine the optimal investment amount, we can set up the consumption allocation problem by taking the optimal investment amount as given, i.e., max C 0,C C 0 C s.t. C 0 + C + r = Y 0 + Y + r + fi 0) + r I 0,

3 where the last two terms on the right hand side of the budget constraint represents the net present value of the investment in the production function. Therefore, the budget constraint becomes C 0 + C. 0 = =.74.. Substituting the above constraint into the utility function, we have UC 0, C ) =.74 C ) C.. Taking the first order derivative of UC 0, C ) with respect to C and setting it to zero, we obtain du = dc.74 C ) C +.74 C ) C = 0. = Multiplying C.74 C ) on both sides of the above equation, we have..74 C ) C.. = 0 and C = Therefore, C 0 =.74 C. = 7.5. We need to borrow C 0+I 0 Y 0 = units of goods today to finance our consumption and investment.. a) If Larry does not invest in any project, his utility level would be min[000, 700] = 700. If he invests in project A, his utility level would be min[000 00, ] = min[900, 900] = 900. If he invests in project B, then his utility level would be min[000 00, ] = min[900, 800] = 800. If he invests in both project, then his utility level would change to min[ , ] = min[800, 000] = 800. So, he would only invest in project A. b) Following logic of part a), we can compute: under no investment, the utility level is = 788.7; under investment in project A, the utility level is = 900; under investment in project B, the utility level is = 8.0; and under investment in both projects, the utility level is = 98.. So, Marry would invest in both projects. 00 c) The NPV of project A is: 00 = 8.88 > 0. The NPV of project B is: = 9.09 < 0. So, Marry will only invest in project A. For the optimal +0. consumption, we need to solve the following system of equations: C C 0 =., C 0 + C. = $000 + $ $8.88.

4 Solving the above system, we have: C 0 = $57.7, C = $ Since Marry is endowed with $000 today, and has spent $00 to invest in project A, she will save $000 $00 $ = $7.7. d) By Fisher separation theorem, Bob continues to invest in project A only.. a) Since the mortgage interest rate in Canada is a semi-annually compounded rate, the monthly interest rate is given by r m = ) 6 = We plan to repay the mortgage using 0 = 40 monthly payments. Therefore, the monthly mortgage payment is C = = r m A 40 = r m +r m) 40 b) After the rd payment, you still owe the bank 7 payments, so the outstanding balance at the beginning of the 4th month is CA 7 r m = C ] [ = r m + r m ) 7 Therefore, the interest payment for the 4th month is r m = 74.4 and the principal repayment for the 4th month is c) The weekly interest rate is r w = = ) 6 = We plan to repay the mortgage using 0 5 = 040 weekly payments. Therefore, the weekly mortgage payment is d) The interest rate for a -year period is C = = r w A 040 =.. r w +r w) 040 r = e 0.05 = e 0.5 =

5 Applying the annuity formula, we figure out the present value of the 00 payments at year 7 is P V 7 = 50A 00 r = 50 [ ] = r + r ) 00 It follows that the present value of these cashflows is P V = P V 7 e = a) To find the first two missing information, we solve: y = 975 y =.564%, $08.86 = c % c + 4%) c = 5%. For Bond, we have c = y, so that the bond trades at par, i.e., P = $000. Alternatively, P = %) %) %) = $000. b) The spot rates satisfy the following system of equations: 975 = r r =.564%, = r + r ) r = 4.06%, = + r ) r ) r ) r = 5.076%. c) Let x and x be the number of units of Bonds and in a bond portfolio that replicates the payoff of Bond 4. The replicating portfolio satisfies the following system of equations: 60 = 000x + 50x 060 = 050x Solving the system, we find x =.0095 and x = The cost of the replicating portfolio is therefore P R = 975x x = The price of Bond 4 is higher than the cost of the replicating portfolio. So, there is an arbitrage opportunity. To take advantage of it, one needs to sell Bond 4 and buy the replicating portfolio, i.e., buy unit of Bond and.0095 units of Bond. The resulting profit from the arbitrage is = $.5. Note that this 5

6 is an arbitrage because the trading strategy pays a positive cash-flows today without any obligations in the future. d) The price of Bond in a year is: P = The holding period return is: ) ) = $ HP R = =.57%. e) True, in the absence of arbitrage, spot rates are always positive and thus zero coupon bonds always trade at a discount. 5. a) The spot rates satisfy the following set of equations: + r ) = + f ) r = %, + r ) = + f ) + f ) r =.50%, + r ) = + f ) + f ) + f ) r =.499%, + r 4 ) 4 = + f ) + f ) + f ) + f 4 ) r 4 =.87%. b) The price of the bond is given by: P = 80 + r ) r ) r ) = $ c) The no-arbitrage forward rate, f 4, satisfies: + f 4 ) = + f ) + f 4 ) f 4 =.499%. d) The market value of the forward contract is given by the present value of future cash-flows. The forward loan consists of two cash-flows: ) +$0000, two years from now ) $ f 4 ) = 07, four years from now By construction, the value of the forward contract at signature is zero: The value of the contract in a year is: P f 0 = r ) 07 + r 4 ) 4 = 0. P f = r + r ) = = $47.. Therefore you will make a profit of $47.. 6

7 6. a) Let E 0 = 0, r = 0.5, g = 0.0, where g is the steady-state earnings growth rate starting from year 4 and forward). For a), we just need the first three years dividends: E = E 0.5 = 5 D = 0.5 E =.50, E = E.0 = 0 D = 0.5 E = 5.00, E = E.5 = 4.5 D = 0.5 E = 7.5. Applying the Gordon formula, one obtains the price of year : Including the first two dividends, P = D r g = P 0 = D + r + D + r) + P + r) = That is, the estimate of the share price today is $8.08. b) Solution : Note that D 4 = D. = Calculate the present value of the four penalties: P penalty =.50 + r r) r) r) 4 = So the new estimate of the share price is P 0 = = Solution : In terms of calculation, this case is equivalent to assuming that D = D = D = D 4 = 0. To use DDM, we need to get D 5 : D 5 = 0.5 E 5 = 0.5 E + g) = Applying the Gordon formula, one obtains the price at year 4: P 4 = D 5 r g = 47.45, 7

8 and so P 0 = P 4 + r) 4 = That is, the estimate of the share price today is $8.68. c) First, repeat the steps of Solution in b) with Bob s assumed value of r bob = 0.6. Now re-estimate P 4 and P 0. and so P 4 = D 5 r bob g = , P 0 = P 4 + r bob ) 4 = 9.. That is, Bob s estimate of the share price today is $9.. marks for this part) Let g cindy stand for the assumed steady-state earnings growth rate of Cindy. Now D 5 is different due to the new growth rate: D 5 = 0.5 E + g cindy ). Applying the Gordon formula, we now have a new expression for the price of year 4: P 4 = D 5 r g cindy = 0.5 E + g cindy ) r g cindy. Discount P 4 and set the estimated current share prices of Cindy and Bob to be equal: 0.5 E + g cindy ) r g cindy + r) = Let z = +g cindy. Then the equation above reduces to the following quadratic equation: 7.5z + 6.0z = 0. Solving it gives g cindy = That is, the assumed steady-state growth rate of Cindy is 8.9%. 8