1a. From Proof 3.1, we obtain the formula n = (iZ/2J) .5 . The monthly interest rate is .10/12 = 0.00833. Therefore
n = [(0.0083 x 1000)/(2 x 2.50)] .5 = 1.6667 .5 = 1.29 per month.
Then, the optimal money holdings are calculated as M = Z/2n = 1000/2.58 = $387.30. [10]
1b. If there are integer constraints, the optimal number of transactions will be 1 or 2. First calculate which of these has lower costs. If n = 1, total costs are 2.50 + (0.00833 x 1000)/2 = $6.65. If n = 2, total costs are 5.00 + (0.00833 x 1000)/4 = $7.08. From this comparison, only one conversion is optimal and average money holdings are $500. [10]
1c. If interest is paid on the minimum monthly balance, there will be no income derived from a noncheckable savings account since the balance at the end of each month is zero. For that reason, the individual should put the entire $1000 in a demand deposit at the beginning of each month. [10]
1d. Since the checkable daily-interest account has the same cost per cheque as the demand deposit, but the former pays interest whereas the latter does not, the individual should put the entire $1000 in the daily-interest account and write cheques on it. [10]
4(a) The sum is virtually zero; thus the bank neither gained nor lost reserves. [10]
4(b) The standard deviation is 152.314. [10]
4(c) First calculate the cumulative sum for the series to see if the bank reached a negative reserve position. The series is negative on the following days: 6, 7, 8, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, and 30 (17 days). Average borrowing is $113.46 [10]
4(d) Optimal reserves are calculated as: 152.314(.06 - .04)/(.06 + .04) = $30.46. [10]
4(e) If the bank starts with $30.46 in reserves, the cumulative total will be negative on the following days: 7, 8, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, and 30. (15 days); average borrowing is $97.42. [10]
4(f) Holding optimal reserves reduces borrowing by 2 days; total borrowing when optimal reserves are held is $3020, while it is $3517, when starting with zero reserves. [10]
1. The utility function should be written as U = (24 - H).4C.6, so that the two exponents (i.e., the share parameters) add to unity.
1a. Caculate U1 and U2 by taking derivatives of the utility function and set results to zero. Then,
U1/U2 = .4C/.6(24 - H) = W/P = 10.
Substitute C = 10H + 20 into the above expression and solve for H:
.4(10H + 20)/.6(24 - H) = 10 or H = 13.6. [10]
1b. Now
.4(12H + 20)/.6(24 - H) = 12 or H = 13.733,
so that hours of leisure will fall with a higher wage rate. [10]
1c. C = 10 x 8 + 10 = 100. Total utility is U = (24 - 8).4100.6. At this point it is easier to work in natural logs.
lnU = .4ln16 + .6ln100 = 3.872; U = 48.053. [10]
1d. The level of utility available at complete leisure is 24.420.6. Taking logs, lnU = .4ln24 + .6ln20 = 3.069. This must be equal to the natural log of utility when the person works 8 hours, so that
lnU = .4ln(24 - 8) + .6ln(8W/P + 20) = 3.069.
Therefore, ln(8W/P + 20) = 3.266 and taking antilogs, (8W/P + 20) = 26.206.
Solving for W/P = $0.776 gives us the reservation wage. [10]
3(a) The utility of working is U = (365 - 3 x 30).5 100,000.5 = 5243.05. The utility of taking early retirement must be the same; thus we can solve for the "bribe", X, from U = 5243.05 = 365.5 X.5. Hence X.5 = 274.43 and squaring both sides yields X = $75,313.90. [10]
3(b) If Lacklustre worked harder, his utility would be lower. Thus it takes a smaller "bribe" to get him to retire. U = (365 - 5 x 30).5 100,000.5 = 242.77. Now X = 58,936. [10]