ANSWER KEY THIRD TAKE-HOME EXERCISE SET [13 Points Total] L. Tesfatsion DUE DATE: Thursday, Feb 29, 9:30 A.M. Econ 302/Spring 96 ----------------------------------------------------------------- ----------------------------------------------------------------- EXERCISE 3.1 [3 Points]: Hall and Taylor, Chapter 4, NUMERICAL Exercise Number 1, page 118. Be sure to justify your assertions and to label graphs carefully. ---------- #Part a)#: The labor supply function is given by (1) N = 1000 + 12[W/P] and labor demand is (2) N = 2000 - 8[W/P] . Draw a diagram showing these schedules. Find the equilibrium level of employment and the real wage. ---------- #Answer Outline for Ex 3.1 (a) [One Point]#: W/P | | | 250 * | 200 | | 150 | | 100 | | (W/P)*=50 |-------------------------------. E* = (1600,50) | . 0 |-------------------*-------------------*----- N | 500 1000 1600 2000 -50 | | -100 | | For the labor supply curve, the N-intercept is 1000, the W/P-intercept is -83.33, and the slope dN/d(W/P) is 12. For the labor demand curve, the N-intercept is 2000, the (W/P)-intercept is 250, and the slope dN/d(W/P) is -8. When these linear functions are graphed in the N-(W/P) plane (see above), they intersect at the point E* = (N*,[W/P]*) = (1600,50). Economically, E* = (1600,50) represents the unique equilibrium for the labor market, in the sense that the supply of labor supply equals the demand for labor at E* and only at E*. Algebraically, the intersection point E* can be found by solving the two equations (1) and (2) for the two unknowns N and (W/P). First, use (1) to substitute out for N in (2), obtaining (3) 1000 + 12[W/P] = 2000 - 8[W/P] . Equation (3) represents one equation in the one unknown, W/P. Collecting terms in [W/P], one obtains (4) (12 + 8)[W/P] = 2000 - 1000 = 1000. Solving for W/P, one obtains the solution value (5) [W/P]* = 1000/20 = 50. Replacing [W/P] in (1) by 50, and solving for N, gives the solution value for N: (6) N* = 1000 + (12 x 50) = 1,600. --------- #Part b)#: Given existing technology and the capital stock, output is given by the function Y = 100 x [square root of N], Graph the production function. Does the production function exhibit diminishing marginal product of labor? ---------- #Answer Outline for Ex 3.1 (b) [One Point]#: By assumption, the production function is given by (7) Y = F(N) = 100 x [square root of N], where, for notational simplicity, dependence on the current capital stock (captured in the coefficient 100) has been omitted. To graph the production function as a function of N, one needs to calculate a table of values for N and Y = F(N). The production function exhibits diminishing marginal product of labor if and only if the marginal product of labor, i.e., the derivative F_N(N) = dF(N)/dN of F(N) with respect to N---decreases with increases in N for all positive N. Also listed, then, are values for F_N(N), which are seen to decrease as N increases. Production Function ------------------------------------- N Y F_N(N) ------------------------------------- 0 0 - 1 100.0 50.0 2 141.2 35.4 3 173.2 28.9 4 200.0 25.0 9 300.0 16.7 16 400.0 12.5 Y | 400| | | | 300| | | Production Function | Y = F(N) 200| | | | 100| | | | ---------------------------------------------- N 0 5 10 15 20 For any positive N, the derivative F_N(N) of F(N) with respect to N---that is, the marginal product of labor evaluated at N---is given by (8) F_N(N) = (1/2) x 100 (1/2) x 100 x 100 5000 ---------------- = ------------------------ = ------ . square root of N 100 x [square root of N] F(N) It follows from (8) that F_N(N) is #positive# for all positive N. The fact that F(N) exhibits diminishing marginal product of labor---that is, that F_N(N) decreases with increases in N for all positive N---follows directly from (8). For (8) shows that F_N(N) is positive for all positive N, which implies F(N) #increases# with increases in N and hence F_N(N) = 5000/F(N) #decreases# with increases in N. The fact that F_N(N) decreases with increases in N can also be determined more formally by taking the derivative of F_N(N) with respect to N, which yields the second derivative of F(N): - 5000 x F_N (9) F_N_N(N) = ------------- . F(N)^2 Clearly (9) is #negative# for all positive N because F_N and F(N)^2 are both positive for all positive N. **IMPORTANT:*** Note that a graph of the production function can at best #illustrate# diminishing marginal product of labor for a limited range of positive N values but cannot #establish# diminishing product of labor for #all# positive N. For the latter, one needs to use an analytical argument such as given in the previous two paragraphs. ------------------------ #Part c)#: Using the labor market from Part a) and the production function from Part b), determine the equilibrium level of output for this economy. ------------------------- #Answer Outline for Ex 3.1 (c) [One Point]#: From Part a), the equilibrium level of employment is N* = 1600. It follows from Part b) that the equilibrium level of output is given by Y* = F(N*) = 100 x [square root of 1600] = 100 x 40 = 4000 . ----------------------------------------------------------------- ----------------------------------------------------------------- EXERCISE 3.2 [2 Points]: Hall and Taylor, Chapter 4, NUMERICAL Exercise Number 2, page 118. Be sure to justify your assertions. ---------------------------- #First Part#: Assume that over a 10 year period the growth rate of capital is 4 percent, the growth rate of employment is 2 percent, and the growth rate of real output is 5 percent. Calculate the growth rate of total factor productivity. ---------------------------- #Answer Outline for Exercise 3.2 First Part [One Point]#: Following HT4, assume the production function takes the form Y = AF(N,K), where A denotes total factor productivity. Using the economic growth formula, the growth rate of total factor productivity is then given by DA DY DN DK ---- = ---- - .70 ---- - .30 ---- A Y N K = .05 - .70[.02] - .30[.04] = .05 - .0140 - .0120 = .0240 . That is, the growth rate of total factor productivity is 2.4 percent. [Note that round off error can significantly affect your answer here. The calculations above are rounded off to four decimal places.] ---------------------------- #Second Part#: Suppose that a permanent cut in the budget deficit increases investment, and the growth rate of capital rises by 1 percent. How much does the growth rate of output increase? Suppose that a tax reduction increases the supply of labor by 1 percent in one year. What happens to the growth rate of real output? ---------------------------- #Answer Outline for Exercise 3.2 Second Part [One Point]#: Using the econonomic growth formula, if DK/K rises by 0.01 over some period of time, then DY/Y rises by 0.30[0.01] = 0.003 (i.e., by 0.3 percent) over that period of time. And if DN/N rises by 0.01 over some period of time, then DY/Y increases by 0.70[0.01] = 0.007 (i.e., by 0.7 percent) over that same period of time. ----------------------------------------------------------------- ----------------------------------------------------------------- EXERCISE 3.3 [3 Points]: Hall and Taylor, Chapter 6, NUMERICAL Exercise Number 1, page 173. Be sure to justify your assertions and to label graphs carefully if you make use of them. Suppose the model of the economy is given by (1) Y = C + I + G + X (NOTE: X = NE) (2) C = a + bY_d (3) Y_d = [1-t]Y (4) X = g - mY where I = $900 billion, G = $1,200 billion, m = 0.1, b = 0.9, t = 0.3, a = $220 billion, and g = $500 billion. ------------------------- #Part a)#: Show that the value of GDP at the point of spending balance is $6,000 billion. Compared to the example on page 165 with exogenous net exports, is the (investment) multiplier larger or smaller? ------------------------- #Answer Outline for Exercise 3.3 (a) [One Point}#: Use equations (2), (3), and (4) to substitute out for C, Y_d, and X in equation (1). The result is: Y[1 - b[1-t] + m] = a + I + G + g , which yields the spending balance solution for Y: a + I + G + g Y^o = ---------------------- . 1 - b[1-t] + m Substituting in the given numerical values for the terms on the right hand side, one obtains Y^o = $6,000 billion. It follows from this formula for Y^o that the (investment) multiplier for the economy at hand is given by dY^o 1 1 ------ = --------------- = ------ = 2.13. dI 1 - b[1-t] + m 0.47 This is #smaller# than the (investment) multiplier 2.70 obtained by HT on page 165 with exogenously given net exports X (or NE). The reason for this finding is that the change in net exports in response to a change in income Y works to partially offset the change in Y. For example, as discussed in HT6, if investment I decreases by $100, then the "first round" effect is a decrease in Y by $100, which in turn leads to a decrease in consumption C by b[1-t]$100 and hence a further decrease in Y by b[1-t]$100in the "second round." However, given relation (4) for net exports, these (and further round) decreases in Y are partially offset by #increases# in net exports in response to decreases in Y. ----------------------- #Part b)#: What proportion of investment is private saving? Government saving? Saving by the rest of the world? ----------------------- #Answer Outline for Exercise 3.3 (b) [One Point]#: We know that I = S_p + S_g + S_r $900 billion where S_p = Y_d - C = [1-t]Y - (a + b[1-t]Y) = Y( [1-t] - b[1-t] ) - a = Y[1-b][1-t] - a = Y[1-0.9][1-0.3] - $220 = Y[0.1][0.7] - $220 = Y[0.07] - $220 = $6000[0.07] - $220 = $200 . Therefore S_p/I = 200/900 = 0.222 (or 22.2 percent). Also, S_g = tY - G = 0.3[$6000] - $1,200 = $600. Therefore S_g/I = 600/900 = 0.667 (or 66.7 percent). Finally, S_r = I - S_p - S_g = $900 - $200 - $600 = $100 . Therefore S_r/I = 100/900 = 0.111 (or 11.1 percent) . Note that 22.2 + 66.7 + 11.1 = 100. --------------------------- #Part c)#: Now suppose that I increases by $100 billion. By what proportion of the increase in investment do each of the three categories of savings increase? --------------------------- #Answer Outline for Exercise 3.3 (c) [One Point]#: By Part a), dY^o/dI = 2.13. Thus, if I increases by $100, then the change in Y is approximately given by Delta(Y) = 2.13 x $100 = $213. As seen in Part b), S_p = Y[0.07] - $220, implying that dS_p/dY = 0.07. Thus, the change in S_p due to the increase in I by $100 is approximately given by Delta(S_p) = 0.07 x Delta(Y) = 0.07 x $213 = $15 (rounded off to the nearest dollar) . Similarly, S_g = tY - G = [0.30]Y - $1200 implies that dS_g/dY = 0.30 Thus, the change in S_g due to the increase in I by $100 is approximately given by Delta(S_g) = 0.30 x Delta(Y) = 0.30 x $213 = $64 (rounded off to the nearest dollar) . Finally, S_r = I - S_p - S_g = I - (Y[0.07] - $220) - ([0.30]Y - $1200) = I + $1200 + $220 - Y[0.07 + 0.30] = I + $1420 - Y[0.37] . Thus, the change in S_r due to an increase in I by $100 is approximately given by Delta(S_r) = $100 - Delta(Y)[0.37] = $100 - $213[0.37] = $100 - $78.81 = $21 (rounded off to the nearest dollar). Another way to calculate Delta(S_r) is by noting that, for the economy at hand, V = 0 (because household income prior to income taxes is assumed in equation (3) to just be given by Y, not by Y+V+F+N---see Hall and Taylor (2-1), page 51.) Thus, using equation (4), S_r = - X = - [g - mY] = - [500 - .1Y] = -500 + .1Y, implying that the change in S_r due to the change in I is approximately given by Delta(S_r) = .1 x Delta(Y) = .1 x $213 = $21 (rounded off to the nearest dollar). As a simple calculational check, note that 15 + 64 + 21 = 100. The desired proportions of savings to investement can now be determined as follows: Delta(S_p)/Delta(I) = 15/100 = .15 (15 percent) Delta(S_g)/Delta(I) = 64/100 = .64 (64 percent) Delta(S_r)/Delta(I) = 21/100 = .21 (21 percent) ----------------------------------------------------------------- ----------------------------------------------------------------- EXERCISE 3.4 [5 Points Total---1 point for Part (a) and 2 points each for Parts (b) and (c)]: [Compare Hall and Taylor, Chapter 6, ANALYTICAL Exercise Number 6, page 175.] Consider an economy described by the following equations: (1) Y = C + I + G + NE; (2) C = a + b[1-t]Y ; (3) NE = g - mY ; (4) G = tY (balanced budget restriction on government) Exogenous Variables: I, a, b, t, where all terms are strictly positive and b and t are also strictly less than 1. Endogenous Variables: Y, C, NE, G Please answer each of the following questions, being sure to justify your assertions and to label graphs carefully. ------------------------- #Part (a)#: Explain carefully why G is #endogenous# in model (1)-(4). ------------------------- #Answer Outline for Exercise 3.4 (a) [One Point]#: G is endogenous in model (1)-(4) because there is an equation in the model, namely equation (4), that determines the value of G once Y is determined by equations (1) through (3). That is, the value of G is determined #within the model#, and this is the definition of an endogenous variable for a model. -------------------------- #Part (b)#: Determine an analytical expression for the #investment# multiplier for model (1)-(4). Is this investment multiplier larger or smaller than in the case (covered in HT6) where G is exogenous? Provide a careful explanation for your finding. -------------------------- #Answer Outline for Exercise 3.4 (b) [Two Points]#: The model given by equations (1), (2), and (3) alone, with G classified as an exogenous variable, is the same model as developed in the class lecture notes HT6. As determined in those notes, the investment multiplier for model (1)-(3) is given by: #Investment Multiplier with Exogenous G#: dY^o 1 (*) ------ = ---------------- . dI 1 - b[1-t] + m #With# equation (4) included in the model, we need to determine the solution value for Y as a function of I. Using equations (2), (3), and (4) to substitute out for C, NE, and G in equation (1), one obtains one equation in Y: Y = ( a + b[1-t]Y ) + I + tY + (g - mY). Collecting terms in Y, one obtains Y x (1 - b[1-t] - t + m) = a + I + g or Y x ( [1-t][1-b] + m) = a + I + g . Dividing through by the term in parentheses on the left, one obtains: #The Solution Value for Y as Determined by Model (1)-(4)#: a + I + g (**) Y+ = ---------------- . [1-t][1-b] + m It follows from (**) that the investment multiplier for model (1)-(4) is given by dY+ 1 (+) ----- = -------------- . dI [1-t][1-b] + m Is the multiplier (+) for endogenous G larger or smaller than the multiplier (*) with exogenous G? The multiplier (+) is #larger# than the multiplier (*) if and only if the denominator of (+) is #smaller# than the denominator of (*), that is, if and only if [1-t][1-b] + m is smaller than 1 - b[1-t] + m . or equivalently, writing out terms, if and only if (++) 1 - b - t + bt + m is smaller than 1 - b + bt + m Cancelling identical terms on each side, on sees that (++) holds as long as -t is smaller than 0; but this is true because t is assumed to be positive as part of the given specifications for model (1)-(4). Therefore, it has been established mathematicaly that the multiplier (+) with endogenous G is larger than the multiplier (*) with exogenous G. But what is the reason for this difference in more intuitive economic terms? The reason for the difference is that tax revenues in model (1)-(4) do not "leak out" of the economy as they do for model (1)-(3) with exogenous G. Rather, they enter back in by way of changes in G. Consequently, any "first round" increase in Y due to an increase in I is #amplified# in all future rounds because it leads to increased tax revenues and hence increases in G which further increase Y. And similarly for any first round decrease in Y due to a decrease in I. ------------------------------- #Part (c)#: Suppose t increases. Determine carefully whether the solution value for Y determined by model (1)-(4) increases, decreases, or stays the same in response to this increase in t. Provide a careful explanation for your finding. ---------------------- #Answer Outline for Exercise 3.4 (c) [Two Points]#: The solution value for Y as determined by model (1)-(4) is given in (**). Recall that the income tax rate t is restricted to lie between 0 and 1. As t increases towards 1, the denumerator in (**) becomes #smaller#, hence the solution value of Y given by (**) becomes #larger#. More formally, differentiating (**) with respect to t, one obtains: #Tax Multiplier for Model (1)-(4)#: dY+ - [a + I + g] ----- = ----------------- x ( - [1-b] ) dt ([1-t][1-b] + m)^2 [1-b]Y+ = ----------------- , [1-t][1-b] + m which is positively valued given the assumed sign restrictions on b,t, and m. Thus Y+ #increases# with increases in t. What is the reason for this in more intuitive economic terms? Recall, for example, that Y #decreases# with increases in t in the HT6 model (1)-(3) with exogenously given G, the reverse of the result we have just obtained for model (1)-(4). The reason why Y+ increases when t increases, with government expenditures G determined endogenously by equation (4), is because the subsequent increase in G more than offsets the reduction in household spending caused by the higher income tax rate t. This is because part of the increase in the taxes assessed on household income is paid out of what would have been household savings. In other words, part of the income that households would previously have saved (a leakage out of the circular flow of income) is now retained in the circular flow because government spends this income on goods and services. -------------------------- End of Answer Key for Exercise Set 3