Chapter 11, Problem 6NP

a.

To determine

To describe: The full employment values of real interest rate, consumption, investment and the price level.

Answer to Problem 6NP

The full employment value of real interest rate is $\text{10\%}$ , consumption is $\text{700}$ , investment is $\text{150}$ and the price level is $\text{15}$ .

Explanation of Solution

Given data,

Desired Consumption	$\text{C=325+0}\text{.5}\left(\text{Y-T}\right)\text{-500r}$
Desired Investment	$\text{I=200-500r}$
Government purchases	$\text{G=150}$
Taxes	$\text{T=150}$
Full employment output	$\text{<mover accent="true"><mo>Y</mo><mo>¯</mo></mover> =1000}$
Money supply	$\text{M=6,000}$
Real money demand	$\text{L=0}\text{.5Y-1000r}$

Whereas, the formula used is as under:

${\text{Y=C}}^{\text{d}}{\text{+I}}^{\text{d}}\text{+G}$

Substitute the values from above to the equation, we get:

$\begin{array}{l}\text{1000=325+0}\text{.5(1000-150)-500r+200-500r+150}\\ \text{1000=325+425-1000r+350}\\ \text{1000=1100-1000r}\\ \text{1000r=100}\\ \text{r=0}\text{.10}\end{array}$

And,

Formula used to calculate price level is as under:

$\frac{\text{M}}{\text{P}}\text{=L}$

Substitute M=6000 and $\text{L=0}\text{.5Y-1000r}$ , we get:

$\begin{array}{l}\frac{\text{6000}}{\text{P}}\text{=0}\text{.5Y-1000r}\\ \frac{\text{6000}}{\text{P}}\text{=0}\text{.5X1000-1000X0}\text{.10}\\ \frac{\text{6000}}{\text{P}}\text{=500-100}\\ \frac{\text{6000}}{\text{P}}\text{=400}\\ \text{P=}\frac{\text{6000}}{\text{400}}\\ \text{P=15}\end{array}$

Calculation of desired consumption is as under:

${\text{C}}^{\text{d}}\text{=325+0}\text{.5(Y-T)-500r}$

Substitute, the value of Y=1000, T=150 and r= 0.10, in the above equation:

$\begin{array}{l}\text{=325+0}\text{.5(1000-150)-500X0}\text{.10}\\ \text{=325+425-50}\\ \text{=700}\end{array}$

And,

$\begin{array}{l}{\text{I}}^{\text{d}}\text{=200-500r}\\ \text{=200-500X0}\text{.10}\\ \text{=200-50}\\ \text{=150}\end{array}$

b.

To determine

To compute: The values of the ${\text{α}}_{\text{IS}}{\text{,β}}_{\text{IS}}{\text{,α}}_{\text{LM}}{\text{,β}}_{\text{LM}}{\text{ and l}}_{\text{r}}$ for the proposed economy.

Answer to Problem 6NP

The values of the ${\text{α}}_{\text{IS}}\text{ is 0}{\text{.6,β}}_{\text{IS}}\text{ is 0}{\text{.0005, α}}_{\text{LM}}{\text{ is zero, β}}_{\text{LM}}\text{ is 0}{\text{.0005 and l}}_{\text{r}}\text{ is 1000}\text{.}$ for the proposed economy

Explanation of Solution

Given data,

$\begin{array}{l}{\text{c}}_{\text{o}}=325\\ {\text{i}}_{\text{o}}=200\\ \text{G =150}\\ {\text{c}}_{\text{Y}}=0.5\\ {\text{t}}_{\text{o}}=150\\ {\text{c}}_{\text{r}}=500\\ {\text{i}}_{\text{r}}=500\end{array}$

Using these equations,

$\begin{array}{l}{\text{α}}_{\text{IS}}\text{=}\frac{{\text{c}}_{\text{o}}{\text{+i}}_{\text{o}}{\text{+G-c}}_{\text{Y}}{\text{t}}_{\text{o}}}{{\text{c}}_{\text{r}}{\text{+i}}_{\text{r}}}\\ {\text{β}}_{\text{IS}}\text{=}\frac{\text{[1-(1-t)cY]}}{{\text{c}}_{\text{r}}{\text{+i}}_{\text{r}}}\\ {\text{α}}_{\text{LM}}\text{=}\frac{{\text{l}}_{\text{0}}}{{\text{l}}_{\text{r}}}{\text{-p}}_{\text{e}}\\ {\text{β}}_{\text{LM}}\text{=}\frac{{\text{l}}_{\text{Y}}}{{\text{l}}_{\text{y}}}\\ \end{array}$

$\begin{array}{l}{\text{α}}_{\text{IS}}\text{=}\frac{{\text{c}}_{\text{o}}{\text{+i}}_{\text{o}}{\text{+G-c}}_{\text{Y}}{\text{t}}_{\text{o}}}{{\text{c}}_{\text{r}}{\text{+i}}_{\text{r}}}\\ {\text{α}}_{\text{IS}}\text{=}\frac{\text{325+200+150-(0}\text{.5×150)}}{\text{500+500}}\\ {\text{α}}_{\text{IS}}\text{=0}\text{.6}\end{array}$

And,

$\begin{array}{l}{\text{β}}_{\text{IS}}\text{=}\frac{\text{[1-(1-t)cY]}}{{\text{c}}_{\text{r}}{\text{+i}}_{\text{r}}}\\ {\text{β}}_{\text{IS}}\text{=}\frac{\text{[1-(1-0)0}\text{.5]}}{\text{500+500}}\\ {\text{β}}_{\text{IS}}=\frac{0.5}{1000}\\ {\text{β}}_{\text{IS}}\text{=0}\text{.0005}\end{array}$

Then,

$\begin{array}{l}{\text{α}}_{\text{LM}}\text{=}\frac{{\text{l}}_{\text{0}}}{{\text{l}}_{\text{r}}}{\text{-p}}_{\text{e}}\\ {\text{α}}_{\text{LM}}\text{=}\frac{\text{0}}{\text{1000}}\text{-0}\\ {\text{α}}_{\text{LM}}\text{=0}\end{array}$

And,

$\begin{array}{l}{\text{β}}_{\text{LM}}\text{=}\frac{{\text{l}}_{\text{Y}}}{{\text{l}}_{\text{r}}}\\ {\text{β}}_{\text{LM}}\text{=}\frac{\text{0}\text{.5}}{\text{1000}}\\ {\text{β}}_{\text{LM}}\text{=0}\text{.0005}\end{array}$

c.

To determine

To describe: The values under short run equilibrium for output and real interest if the price level is fixed at $ $15$ .

Answer to Problem 6NP

The value under short run equilibrium for output is $1,000$ and real interest rate is $10\%$ .

Explanation of Solution

Given data,

Desired Consumption	$\text{C=325+0}\text{.5}\left(\text{Y-T}\right)\text{-500r}$
Desired Investment	$\text{I=200-500r}$
Government purchases	$\text{G=150}$
Taxes	$\text{T=150}$
Full employment output	$\text{<mover accent="true"><mo>Y</mo><mo>¯</mo></mover> =1000}$
Money supply	$\text{M=6,000}$
Real money demand	$\text{L=0}\text{.5Y-1000r}$

Whereas, the formula used is as under:

${\text{Y=C}}^{\text{d}}{\text{+I}}^{\text{d}}\text{+G}$

Substitute the values from above to the equation, we get:

$\begin{array}{l}\text{1000=325+0}\text{.5(1000-150)-500r+200-500r+150}\\ \text{1000=325+425-1000r+350}\\ \text{1000=1100-1000r}\\ \text{1000r=100}\\ \text{r=0}\text{.10}\end{array}$

And,

Formula used to calculate price level is as under:

$\frac{\text{M}}{\text{P}}\text{=L}$

Substitute M=6000 and $\text{L=0}\text{.5Y-1000r}$ , we get:

$\begin{array}{l}\frac{\text{6000}}{\text{P}}\text{=0}\text{.5Y-1000r}\\ \frac{\text{6000}}{\text{P}}\text{=0}\text{.5X1000-1000X0}\text{.10}\\ \frac{\text{6000}}{\text{P}}\text{=500-100}\\ \frac{\text{6000}}{\text{P}}\text{=400}\\ \text{P=}\frac{\text{6000}}{\text{400}}\\ \text{P=15}\end{array}$

d.

To determine

To describe:

The short-run equilibrium level of output and ${\text{α}}_{\text{IS}}$ if price level unchanged at $15$ and G will be $250$ instead of $150$ .

Answer to Problem 6NP

The short-run equilibrium level of output is $1,100$ and ${\alpha }_{IS}=0.7$ .

Explanation of Solution

Given data,

$\begin{array}{l}{\text{c}}_{\text{o}}=325\\ {\text{i}}_{\text{o}}=200\\ \text{G = 250}\\ {\text{c}}_{\text{Y}}=0.5\\ {\text{t}}_{\text{o}}=150\\ {\text{c}}_{\text{r}}=500\\ {\text{i}}_{\text{r}}=500\\ {\text{β}}_{\text{LM}}\text{=0}\text{.0005}\\ {\text{β}}_{\text{IS}}\text{=0}\text{.0005}\\ \text{P = 15}\\ \text{M = 6000}\end{array}$

The formula used here is,

$\begin{array}{l}{\text{α}}_{\text{IS}}\text{=}\frac{{\text{c}}_{\text{o}}{\text{+i}}_{\text{o}}{\text{+G-c}}_{\text{Y}}{\text{t}}_{\text{o}}}{{\text{c}}_{\text{r}}{\text{+i}}_{\text{r}}}\\ {\text{α}}_{\text{IS}}\text{=}\frac{\text{325+200+250-(0}\text{.5×150)}}{\text{500+500}}\\ {\text{α}}_{\text{IS}}\text{=0}\text{.7}\end{array}$

Therefore,

$\frac{\text{M}}{\text{P}}\text{=}\frac{\text{6000}}{\text{15}}\text{=400}$

And,

To calculate equilibrium output the following formula is used:

${\text{Y=[α}}_{\text{IS}}{\text{-α}}_{\text{LM}}{\text{+(1/l}}_{\text{r}}{\text{)(M/P)]/[β}}_{\text{IS}}{\text{+β}}_{\text{LM}}\text{]}$

Substitute the values from above:

$\begin{array}{l}\text{Y=[0}\text{.7-0+(400/1000)]/(0}\text{.0005+0}\text{.0005)}\\ \text{Y=1}\text{.1/0}\text{.001}\\ \text{Y=1100}\end{array}$

e.

To determine

To describe:

The government purchases multiplier, given that $\text{G=150}$ to $\text{G=250}$ .

Answer to Problem 6NP

The government purchases multiplier is $\text{1}$ .

Explanation of Solution

Formula for purchases multiplier:

$\frac{\text{dY}}{\text{dG}}{\text{=1/[(c}}_{\text{r}}{\text{+i}}_{\text{r}}{\text{)(β}}_{\text{IS}}{\text{+β}}_{\text{LM}}\text{)]}$

Given information as calculated above,

$\begin{array}{l}{\text{c}}_{\text{r}}=500\\ {\text{i}}_{\text{r}}=500\\ {\text{β}}_{\text{LM}}\text{=0}\text{.0005}\\ {\text{β}}_{\text{IS}}\text{=0}\text{.0005}\end{array}$

Substitute the above information in the equation:

$\begin{array}{l}\frac{\text{dY}}{\text{dG}}{\text{=1/[(c}}_{\text{r}}{\text{+i}}_{\text{r}}{\text{)(β}}_{\text{IS}}{\text{+β}}_{\text{LM}}\text{)]}\\ \text{=1/[(500+500)(0}\text{.0005+0}\text{.0005)]}\\ \text{=1}\end{array}$