Chapter 12

Question 1

Needs["DiscreteMath`RSolve`"]

Needs["Graphics`MultipleListPlot`"]

Needs["Graphics`Legend`"]

Simplify[
      Solve[y == 110 + 0.75 (y1 + 80 - 0.2  y1) + 320 + 0.1  y1 + 330 + 
            440 - 10 - 0.2  y1, y]]

soly0 = 
      RSolve[{y[t] == 1320 + 0.5  y[t - 1], y[0] == 2500}, y[t], 
        t]

oldy0 = Simplify[soly0[[1, 1, 2]]]

The difference equation which results is
	(Subscript[Y, t]) = 1320 + 0.5  (Subscript[Y, t - 2])
In order to solve this second order equation we require to make assumptions about the level of income in period 0 and period 1. We assume that in period 0 income is 2500 and in period 1 it is 2570.

soly1 = 
      RSolve[{y[t] == 1320 + 0.5  y[t - 2], y[0] == 2500, y[1] == 2570}, 
        y[t], t]

newy1 = Simplify[soly1[[1, 1, 2]]]

pathy01 = Table[{t, oldy0, newy1}, {t, 0, 20}];
TableForm[pathy01, 
      TableHeadings -> {{}, {"t", "y0(t)", "y1(t)"}}]

points0 = Table[oldy0, {t, 0, 20}];

points1 = Table[newy1, {t, 0, 20}];

\(\(MultipleListPlot[points0, points1, PlotRange -> All, 
        AxesOrigin -> {0, 2500}, PlotJoined -> True, 
        AxesLabel -> {"t", \*""\!\(Y\_t\)""}, 
        PlotLegend -> {"y0(t)", "y1(t)"}, 
	
        LegendPosition -> {\(-0.35\), \(-1.2\)}, 
	
        LegendOrientation -> Horizontal, 
	
        LegendSize -> {1,  .5}];\)\)

Question 2

(i)

The new resulting difference equation is
	\(TraditionalForm
      \`Y\_t = 1320 + 0.7  Y\_\(t - 1\) - 0.2  Y\_\(t - 2\)\)

soly21 = 
      RSolve[{y[t] == 1320 + 0.7  y[t - 1] - 0.2  y[t - 2], y[0] == 2500, 
          y[1] == 2570}, y[t], t]

newy21 = Simplify[soly21[[1, 1, 2]]]

points21 = Table[newy21, {t, 0, 20}];

\(\(MultipleListPlot[points0, points21, PlotRange -> All, 
        AxesOrigin -> {0, 2500}, PlotJoined -> True, 
        AxesLabel -> {"t", \*""\!\(Y\_t\)""}, 
        PlotLegend -> {"y0(t)", "y21(t)"}, 
	
        LegendPosition -> {\(-0.35\), \(-1.2\)}, 
	
        LegendOrientation -> Horizontal, 
	
        LegendSize -> {1,  .5}];\)\)

(ii)

The resulting difference equation is
	(Subscript[Y, t]) = 1320 + 0.5  (Subscript[Y, t - 1])
which is the same equation as in Question 1, and hence the same time path for income.

Question 3

Since (Subscript[NX, t])=(Subscript[X, t]) - (Subscript[M, t]) and (Subscript[X, t]) = (Subscript[X, 0]) while (Subscript[M, t]) = (Subscript[M, 0]) + m (Subscript[Y, t]), then
	(Subscript[NX, t])=((Subscript[X, 0]) - (Subscript[M, 0])) - m (Subscript[Y, t])
	(Subscript[ΔNX, t])=-m (Subscript[ΔY, t])
But
	(Subscript[k, t])=((Subscript[ΔY, t])/(ΔG))
Therefore
	(Subscript[ΔNX, t])=-m (Subscript[k, t]) ΔG
Also
	\(TraditionalForm
      \`\(Lim\+\(t -> ∞\)\)(m\ k\_t) = 
        \(m\ \(Lim\+\(t -> ∞\)\) k\_t = m\ k\)\)

Question 4

(i) - (ii)

Simplify[
      Solve[y == 110 + 0.75 (y1 + 80 - 0.2  y1) + 320 + 0.1  y1 + 330 + 
            440 - 10 - 0.2  y1, y]]

Solve[ystar1 == 1250 + 0.5  ystar1, ystar1]

Simplify[
      Solve[y == 110 + 0.75 (y1 + 80 - 0.2  y1) + 320 + 0.1  y1 + 330 + 
            440 - 10 - 0.3  y1, y]]

Solve[ystar2 == 1250 + 0.4  ystar2, ystar2]

Simplify[
      Solve[y == 110 + 0.75 (y1 + 80 - 0.2  y1) + 320 + 0.1  y1 + 330 + 
            440 - 10 - 0.4  y1, y]]

Solve[ystar3 == 1250 + 0.3  ystar3, ystar3]

\(\(Plot[{y, 1250 + 0.5  y, 1250 + 0.4  y, 1250 + 0.3  y}, {y, 0, 3000}, 
        AxesLabel -> {"Y", "AE"}, 
	
        PlotStyle -> {{}, {Dashing[{ .01}]}, {Dashing[{ .02}]}}];\)\)

(iii)

sol1 = 
      RSolve[{y[t] == 2500 + 0.5  y[t - 1], y[0] == 2000}, y[t], 
        t]

path1 = Simplify[sol1[[1, 1, 2]]]

list1 = Table[{path1}, {t, 0, 20}];

sol2 = 
      RSolve[{y[t] == 2500 + 0.4  y[t - 1], y[0] == 2000}, y[t], 
        t]

path2 = Simplify[sol2[[1, 1, 2]]]

list2 = Table[path2, {t, 0, 20}];

sol3 = 
      RSolve[{y[t] == 2500 + 0.3  y[t - 1], y[0] == 2000}, y[t], 
        t]

path3 = Simplify[sol3[[1, 1, 2]]]

list3 = Table[path3, {t, 0, 20}];

TableForm[{{list1, list2, list3}}, 
      TableHeadings -> {{}, {"m=0.2", "m=0.3", "m=0.4"}}]

(iv)

Thus, the higher the marginal propensity to import, the lower the equilibrium value of income and the sooner the economy reaches equilibrium for a given level of income.

Question 5

The IS curve is given by equation (12.13)
	\(TraditionalForm\`r = \(a + nx\_0 + \((f + g)\) R\)\/h - \(\([1 - c(1 - t) - j + m]\) y\)\/h\)
while the LM curve is given by equation (12.19)
	r = (m0/u) - ((k/u)) y

Note, however, that in Table 12.2 nx0 is contained in the value of the parameter a. The BP curve is given by
	\(TraditionalForm\`r = \((\(r\^*\) - \(bp0 + \((f + g)\) R\)\/v)\) + \((m\/v)\) y\)

The initial values are given by

intIS = ((a + (f + g) R)/h)

slopeIS = -(((1 - c (1 - t) - j + m))/h)

intIS0 = 
      intIS /. {a -> 43.5, f -> 5, g -> 2, R -> 1.764, h -> 2}

slopeIS0 = 
      slopeIS  /. {c -> 0.75, t -> 0.3, j -> 0, m -> 0.2, h -> 2}

intLM = -m0/u

slopeLM = k/u

intLM0 = -m0/u   /. {m0 -> 3, u -> 0.5}

slopeLM0 = slopeLM  /. {k -> 0.25,  u -> 0.5}

intBP = rstar - ((bp0 + (f + g) R)/v)

slopeBP = (m/v)

intBP0 = 
      intBP  /. {rstar -> 15, bp0 -> -3.5, f -> 5, g -> 2, R -> 1.764, 
          v -> 1}

slopeBP0 = slopeBP  /. {m -> 0.2, v -> 1}

Initial equilibrium values are given by

Solve[{r == intIS0 + slopeIS0*y, r == intLM0 + slopeLM0*y}, {y, 
        r}]

The BP curve also intersects at the same values since

Solve[{r == intIS0 + slopeIS0*y, r == intBP0 + slopeBP0*y}, {y, 
        r}]

Also note that

bp = (x0 - z0) + (f + g) R - m y  +  cf0 + 
        v (r - rstar)

bp0 = 
      bp  /. {x0 -> 0, z0 -> 24,  f -> 5, g -> 2, R -> 1.764, m -> 0.2, 
          cf0 -> 20.5, v -> 1,  rstar -> 15,  y -> 40.506, 
          r -> 14.2532}

(i) Rise in a from 43.5 to 50

The change in the parameter a affects only the IS curve, shifting it to the right. The IS curve will intersect the LM curve at a higher income level and a higher interest rate. With the higher interest rate and the real exchange rate constant, there will be a capital inflow and the balance of payments will go into surplsu. To verify these statements:

intIS1 = 
      intIS  /. {a -> 50, f -> 5, g -> 2, R -> 1.764, h -> 2}

Solve[{r == intIS1 + slopeIS0*y, r == intLM0 + slopeLM0*y}, {y, 
        r}]

bp1 = 
      bp  /. {x0 -> 0, z0 -> 24,  f -> 5, g -> 2, R -> 1.764, m -> 0.2, 
          cf0 -> 20.5, v -> 1,  rstar -> 15,  y -> 44.3869, 
          r -> 16.1934}

Plot[{intIS0 + slopeIS0*y, intLM0 + slopeLM0*y, intBP0 + slopeBP0*y, 
          intIS1 + slopeIS0*y}, {y, 0, 50}, PlotRange -> {0, 35}, 
        AxesLabel -> {"y", "r"}, 
        PlotStyle -> {{}, {}, {}, {Dashing[{ .02}]}}];

(ii) A fall in m0 from 3 to 2

A fall in the money supply shifts the LM curve left, leaving the IS curve and the BP curve unaffected. There is a resulting rise in the rate of interest and a fall in the level of income. The new LM curve intersects the IS curve above the BP curve. The increased interest rate leads to a capital inflow and an improvement in the balance of payments.

intLM2 = intLM  /. {m0 -> 2, u -> 0.5}

Solve[{r == intIS0 + slopeIS0*y, r == intLM2 + slopeLM0*y}, {y, 
        r}]

bp2 = 
      bp  /. {x0 -> 0, z0 -> 24,  f -> 5, g -> 2, R -> 1.764, m -> 0.2, 
          cf0 -> 20.5, v -> 1,  rstar -> 15,  y -> 38.1182, 
          r -> 15.0591}

Plot[{intIS0 + slopeIS0*y, intLM0 + slopeLM0*y, intBP0 + slopeBP0*y, 
          intLM2 + slopeLM0*y}, {y, 0, 50}, PlotRange -> {0, 35}, 
        AxesLabel -> {"y", "r"}, 
        PlotStyle -> {{}, {}, {}, {Dashing[{ .02}]}}];

(iii) Devaluation: rise in R from 1.764 to 2

A devaluation, rise in R, shifts the IS curve right and the BP curve down, leaving the LM curve unaffected. There results a rise in income and a rise in the rate of interest. The rise in the interest rate leads to a capital inflow and an improvement in the balance of payments. In the case of all shifts only the intercepts change.

intIS3 = 
      intIS  /. {a -> 43.5, f -> 5, g -> 2, R -> 2, h -> 2}

intBP3 = 
      intBP  /. {rstar -> 15, bp0 -> -3.5, f -> 5, g -> 2, R -> 2, 
          v -> 1}

Solve[{r == intIS3 + slopeIS0*y, r == intLM0 + slopeLM0*y}, {y, 
        r}]

bp3 = 
      bp  /. {x0 -> 0, z0 -> 24,  f -> 5, g -> 2, R -> 2, m -> 0.2, 
          cf0 -> 20.5, v -> 1,  rstar -> 15,  y -> 41.4925, 
          r -> 14.7463}

\(\(Plot[{intIS0 + slopeIS0*y, intLM0 + slopeLM0*y, intBP0 + slopeBP0*y, 
          intIS3 + slopeIS0*y, intBP3 + slopeBP0*y}, {y, 0, 50}, 
        PlotRange -> {0, 35}, AxesLabel -> {"y", "r"}, 
        PlotStyle -> {{}, {}, {}, {Dashing[{ .02}]}, {Dashing[{ .02}]}}];\)\)

Question 6

(i) Rise in autonomous spending by 20

With a rise in autonomous spending of 10, the new intercept for the IS curve rises by Δa/h = 20/2 = 10, to the value 37.924.

Solve[{r == 37.924 - 0.3375  y, r == -6 + 0.5  y}, {y, 
        r}]

To solve for bp we use the information in Table 12.2 and the solution values just derived.

newbp = 
      bp  /.  {x0 -> 0, z0 -> 24,  f -> 5, g -> 2, R -> 1.764, m -> 0.2, 
          cf0 -> 20.5, v -> 1,  rstar -> 15,  y -> 52.4466, 
          r -> 20.2233}

Under a floating exchange rate, the resulting surplus on the balance of payments leads to an appreciation of the exchange rate (S appreciates, and with P and P* constant, R appreciates by the same amount). The value of R falls to 1.33.

intIS6 = 
      intIS  /. {a -> 63.5, f -> 5, g -> 2, R -> 1.33, h -> 2}

intBP6 = 
      intBP  /. {rstar -> 15, bp0 -> -3.5, f -> 5, g -> 2, R -> 1.33, 
          v -> 1}

Solve[{r == 36.405 - 0.3375  y, r == -6 + 0.5  y}, {y, 
        r}]

bp6 = 
      bp  /. {x0 -> 0, z0 -> 24,  f -> 5, g -> 2, R -> 1.33, m -> 0.2, 
          cf0 -> 20.5, v -> 1,  rstar -> 15,  y -> 50.6328, 
          r -> 19.3164}

Plot[{27.924 - 0.3375  y, -6 + 0.5  y, 6.152 + 0.2  y, 
          37.924 - 0.3375  y, 9.19 + 0.2  y},  {y, 0, 60}, 
        PlotRange -> {0, 40},  
        PlotStyle -> {{}, {}, {}, {Dashing[{ .02}]}, {Dashing[{ .02}]}}];

(ii) A rise in the money supply from 3 to 4

Solve[{r == 27.924 - 0.3375  y, r == -8 + 0.5  y}, {y, 
        r}]

To solve for bp we use the information in Table 12.2 and the solution values just derived.

newbp = 
      bp  /.  {x0 -> 0, z0 -> 24,  f -> 5, g -> 2, R -> 1.764, m -> 0.2, 
          cf0 -> 20.5, v -> 1,  rstar -> 15,  y -> 42.8943, 
          r -> 13.4472}

Under a floating exchange rate, the resulting deficit on the balance of payments leads to a depreciation of the exchange rate (S depreciates, and with P and P* constant, R depreciates by the same amount). The value of R rises to 1.92.

intIS62 = 
      intIS  /. {a -> 43.5, f -> 5, g -> 2, R -> 1.92, h -> 2}

intBP62 = 
      intBP  /. {rstar -> 15, bp0 -> -3.5, f -> 5, g -> 2, R -> 1.92, 
          v -> 1}

Solve[{r == 28.47 - 0.3375  y, r == -8 + 0.5  y}, {y, 
        r}]

bp62 = 
      bp  /. {x0 -> 0, z0 -> 24,  f -> 5, g -> 2, R -> 1.92, m -> 0.2, 
          cf0 -> 20.5, v -> 1,  rstar -> 15,  y -> 43.5463, 
          r -> 13.7731}

Plot[{27.924 - 0.3375  y, -6 + 0.5  y, 
          6.152 + 0.2  y, -8 + 0.5  y, 28.47 - 0.3375  y, 
          5.06 + 0.2  y},  {y, 0, 60}, PlotRange -> {0, 40},  
        PlotStyle -> {{}, {}, {}, {Dashing[{ .02}]}, {Dashing[{ .02}]}, {Dashing[{ .02}]}}];

Question 7

Clear[y, r, s]

Solve[{0 == 2.675 - 0.03375  y - 0.1  r + 0.35  s, 
        0 == -2.4 + 0.2  y - 0.4  r, 
        0 == -0.00185 - 0.00002  y + 0.0001  r + 0.0007  s}, {y, r, 
        s}]

RSolve[{y[t + 1] == 
          2.675 + (1 - 0.03375) y[t] - 0.1  r[t] + 0.35  s[t], 
        r[t + 1] == -2.4 + 0.2  y[t] + 0.6  r[t], 
        s[t + 1] == -0.00185 - 0.00002  y[t] + 0.0001  r[t] + 
            1.0007  s[t], y[0] == 40.506, r[0] == 14.253, 
        s[0] == 1.764}, {y[t], r[t], s[t]}, t]

Therefore, Mathematica provides no output for this problem.

Question 8

Solve[{0 == 2.675 - 0.03375  y - 0.1  r + 0.35  s, 
        0 == -2.4 + 0.2  y - 0.4  r, 
        0 == -0.925 - 0.01  y + 0.05  r + 0.35  s}, {y, r, s}]

RSolve[{y[t + 1] == 
          2.675 + (1 - 0.03375) y[t] - 0.1  r[t] + 0.35  s[t], 
        r[t + 1] == -2.4 + 0.2  y[t] + 0.6  r[t], 
        s[t + 1] == -0.925 - 0.01  y[t] + 0.05  r[t] + 1.35  s[t], 
        y[0] == 40.506, r[0] == 14.253, s[0] == 1.764}, {y[t], r[t], s[t]}, 
      t]

Again Mathematica does not provide an answer because of the numerical inaccuracies.