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 May 4th, 2018, 07:21 AM #1 Member   Joined: Mar 2017 From: Israel Posts: 85 Thanks: 2 Macro-economics Hello Can you help me please about the next exercise: In a certain economy, the following behavioral functions are presented (in millions dollars): Disposable Income: Yd Private Consumption: C = 160 + 0.6Yd Investment: I = 50 + 0.15Y Public consumption: G = 100 The government doesn't invest, and assume that in the state of origin (outside of the country) the government finances all its expenses by taxes. Find the private consumption as a function of national product. Thanks! Last edited by IlanSherer; May 4th, 2018 at 07:58 AM.
 May 4th, 2018, 08:26 AM #2 Member   Joined: Mar 2017 From: Israel Posts: 85 Thanks: 2 Oh, I forgot to add something: Product: Y All firm profits are divided. The tax is constant/fixed.
May 4th, 2018, 11:23 AM   #3
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Quote:
 Originally Posted by IlanSherer Hello Can you help me please about the next exercise: In a certain economy, the following behavioral functions are presented (in millions dollars): Disposable Income: Yd Private Consumption: C = 160 + 0.6Yd Investment: I = 50 + 0.15Y Public consumption: G = 100 The government doesn't invest, and assume that in the state of origin (outside of the country) the government finances all its expenses by taxes. Find the private consumption as a function of national product. Thanks!
The highlighted language makes no sense in English.

Are you assuming a closed or an open economy.

I am guessing closed, in which case I assume $Y = C + I + G.$

You gave no definition for $Y_d.$

I am guessing that $Y_d = Y - T \text { and } G = T \text { by assumption.}$

I am further guessing that you meant the tax rate is fixed rather than the taxes themselves.

If my guesses are correct, we have

$Y = C + I + G.$

$Y_d = Y - T.$

$C = 160 + 0.6Y_d.$

$I = 50 + 0.15Y.$

$G = T.$

We have six variables (Y, Y_d, C, I, G, and T) and five equations.

$C = 160 + 0.6Y_d \implies Y_d = \dfrac{C - 160}{0.6}$.

$\text {But } Y_d = Y - T \implies T = Y - Y_d = Y - \dfrac{C - 160}{0.6}.$

$T = G \text { and } Y = C + I + G \implies Y = C + I + T.$

$T = Y - \dfrac{C - 160}{0.6} \text { and } Y = C + I + T \implies$

$Y = C + I + Y - \dfrac{C - 160}{0.6} \implies I = \dfrac{C - 160}{0.6} - C = \dfrac{0.4C - 160 }{0.6}.$

$I = 50 + 0.15Y \text { and } I = \dfrac{0.4C - 160 }{0.6} \implies$

$50 + 0.15Y =\dfrac{0.4C - 160 }{0.6} \implies 30 + 0.09Y = 0.4C - 160 \implies 0.4C = 190 + 0.09Y \implies$

$C = 475 + 0.225Y.$

Does that make any sense? Let's see.

$C + I \le Y \implies 475 + 0.225Y + 50 + 0.15Y \le Y \implies$

$525 \le 0.625Y \implies \le Y.$

$Y = 840 \implies C = 475 + 0.225Y = 664 \text {and }$

$I =50 + 0.15Y = 176 \implies 664 + 176 = 840.$

Y = 840 means that government expenditures and therefore taxes are zero.

$Y_d = Y - T = 840 - 0 = 840 \text { and } 160 + 0.6 * 840 = 664.$

So let's try another.

$Y = 1200 \implies I = 50 + 180 = 230.$

$Y = 1200 \implies C = 475 + 270 = 745.$

$\text {But } T = G = Y - C - I = 1200 - 230 - 745 = 225.$

$\therefore Y_d = 1200 - 225 = 975 \implies 745 = 160 + 0.6 * 975 = 745.$

Our algebra was correct.

Now notice the implication. It is G that determines Y. It actually makes more sense to solve for Y in terms of G.

$I = 50 + 0.15Y \text { and } Y = C + I + G \implies 0.85Y = 50 +C + G.$

$C = 160 + 0.6Y_d \text { and } Y_d = Y - T \implies C = 160 + 0.6Y - 0.6T.$

$G = T \text { and } C = 160 + 0.6Y - 0.6T \implies C = 160 + 0.6Y - 0.6G.$

$C = 160 + 0.6Y - 0.6G \text { and } 0.85Y = 50 +C + G \implies$

$0.85Y = 50 + 160 + 0.6Y - 0.6G + G \implies 0.25Y = 210 + 0.4G \implies Y = 4(210 + .4G).$

We can check that with the values we already found.

$G = 0 \implies Y = 4(210 + 0.4 * 0) = 4 * 210 = 840.$

$G = 225 \implies Y = 4(210 + 0.4 * 225) = 4(210 + 90) = 1200.$

May 5th, 2018, 05:12 AM   #4
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I see that part of my answer got garbled.
Quote:
 Originally Posted by JeffM1 But T = G = Y - C - I = 1200 - 230 - 745 = 225. $\therefore Y_d = 1200 - 225 = 975 \implies 745 = 160 + 0.6 * 975 = 745.$ Our algebra was correct. Now notice the implication. It is G that determines Y. It actually makes more sense to solve for Y in terms of G. $I = 50 + 0.15Y \text { and } Y = C + I + G \implies 0.85Y = 50 +C + G.$ $C = 160 + 0.6Y_d \text { and } Y_d = Y - T \implies C = 160 + 0.6Y - 0.6T.$ $G = T \text { and } C = 160 + 0.6Y - 0.6T \implies C = 160 + 0.6Y - 0.6G.$ $C = 160 + 0.6Y - 0.6G \text { and } 0.85Y = 50 +C + G \implies$ $0.85Y = 50 + 160 + 0.6Y - 0.6G + G \implies 0.25Y = 210 + 0.4G \implies Y = 4(210 + .4G).$ We can check that with the values we already found. $G = 0 \implies Y = 4(210 + 0.4 * 0) = 4 * 210 = 840.$ $G = 225 \implies Y = 4(210 + 0.4 * 225) = 4(210 + 90) = 1200.$

May 5th, 2018, 05:14 AM   #5
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I see that part of my answer got garbled.
Quote:
 Originally Posted by JeffM1 $\text {But } T = G = Y - C - I =$ $1200 -$ $230 - 745 = 225.$ $\therefore Y_d = 1200 - 225 = 975 \implies 745 = 160 + 0.6 * 975 = 745.$ Our algebra was correct. Now notice the implication. It is G that determines Y. It actually makes more sense to solve for Y in terms of G. $I = 50 + 0.15Y \text { and } Y = C + I + G \implies 0.85Y = 50 +C + G.$ $C = 160 + 0.6Y_d \text { and } Y_d = Y - T \implies C = 160 + 0.6Y - 0.6T.$ $G = T \text { and } C = 160 + 0.6Y - 0.6T \implies C = 160 + 0.6Y - 0.6G.$ $C = 160 + 0.6Y - 0.6G \text { and } 0.85Y = 50 +C + G \implies$ $0.85Y = 50 + 160 + 0.6Y - 0.6G + G \implies 0.25Y = 210 + 0.4G \implies Y = 4(210 + .4G).$ We can check that with the values we already found. $G = 0 \implies Y = 4(210 + 0.4 * 0) = 4 * 210 = 840.$ $G = 225 \implies Y = 4(210 + 0.4 * 225) = 4(210 + 90) = 1200.$

May 5th, 2018, 08:03 AM   #6
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Quote:
 Originally Posted by JeffM1 I see that part of my answer got garbled.
It's ok, I already understood you
And sorry for bad english, I tried to find the right words, in my language it's one word when in English it's a few words (one meaning against several meanings).
But you understood and guessed correctly, thanks!

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