Question

Assume that in a certain economy, the LM curve is given by Y = 2,000r – 2,000 + 2(M/P) + u, where u is a shock that is equal to +200 half the time and –200 half the time, and the IS curve is given by Y = 8,000 – 2,000r. The price level (P) is fixed at 1.0. The natural rate of output is 4,000. The government wants to keep output as close as possible to 4,000 and does not care about anything else. Consider the following two policy rules: i. Set the money supply M equal to 1,000 and keep it there. ii. Manipulate M from day to day to keep the interest rate constant at 2 percent. a. Under rule i, what will Y be when u = +200? Under rule i, what will Y be when u = –200? b. Under rule ii, what will Y be when u = +200? Under rule ii, what will Y be when u = –200? c. Which rule will keep output closer to 4,000?

Answer 1

a-1. Y = 4,100

a-2. Y = 3,900

b-1. Y = 7,960

b-2. Y = 7,960

c. The rule that will keep output closer to 4,000 is rule i. Set the money supply M equal to 1,000 and keep it there.

Step-by-step explanation:

M curve: Y = 2,000r – 2,000 + 2(M/P) + u ………………… (1)

IS curve: Y = 8,000 – 2,000r ………………………………………. (2)

At equilibrium, LM = IS. To obtain this, we equate equations (1) and (2) and solve as follows:

2,000r – 2,000 + 2(M/P) + u = 8,000 – 2,000r

2,000r – 2,000 + 2(M/P) + u - 8,000 + 2,000r = 0

4,000r – 10,000 + 2(M/P) + u = 0

Since P = 1.0 and fixed, we have:

4,000r – 10,000 + 2(M/1.0) + u = 0 …………………………. (3)

a-1. Under rule i, what will Y be when u = +200?

With this condition, we have:

M = 1,000

Substituting the relevant values into equation (3) and solve for r, we have:

4,000r – 10,000 + (2*(1,000/1.0)) + 200 = 0

4,000r – 7,800 = 0

4,000r = 7,800

r = 7,800 / 4000

r = 1.95

Substituting r = 1.95 into either equation (1) or (2), in case we use equation (2), we have:

Y = 8,000 – (2,000 * 1.95)

Y = 4,100

a-2. Under rule i, what will Y be when u = –200?

With this condition, we have:

M = 1,000

Substituting the relevant values into equation (3) and solve for r, we have:

4,000r – 10,000 + (2*(1,000/1.0)) - 200 = 0

4,000r – 8,200 = 0

r = 8,200 / 4,000

r = 2.05

Substituting r = 2.05 into equation (2), we have:

Y = 8,000 – (2,000 * 2.05)

Y = 3,900

b-1. Under rule ii, what will Y be when u = +200?

With this condition, we have:

r = 2%, or 0.02

Substituting the relevant values into equation (3) and solve for M, we have:

(4,000 * 0.02) - 10,000 + (2*(M/1.0)) + 200 = 0

80 - 10,000 + 200 + (2*(M/1.0)) = 0

-9720 + (2*(M/1.0)) = 0

2*(M/1.0) = 9720

M / 1 = 9720 / 2

M = 4,860

Substituting r = 4,860 and other given values into equation (1), we have:

Y = (2000 * 0.02) - 2,000 + (2 * (4860/1)) + 200

Y = 7,960

b-2. Under rule ii, what will Y be when u = –200?

With this condition, we still have:

r = 2%, or 0.02

Substituting the relevant values into equation (3) and solve for M, we have:

(4,000 * 0.02) - 10,000 + (2*(M/1.0)) - 200 = 0

80 - 10,000 - 200 + (2*(M/1.0)) = 0

-10120 + (2*(M/1.0)) = 0

2*(M/1.0) = 10120

M / 1 = 10120 / 2

M = 5,060

Substituting r = 5,060 and other given values into equation (1), we have:

Y = (2000 * 0.02) - 2,000 + (2 * (5060/1)) - 200

Y = 7,960

c. Which rule will keep output closer to 4,000?

Based on the above asnwers to a-1 where Y = 4,100 and a-2 where Y = 3,900; the rule that will keep output closer to 4,000 is rule i. Set the money supply M equal to 1,000 and keep it there.