Answer to Question #117347 in Microeconomics for joud ahdab

Question #117347
what is the saving/investment approach to equilibrium?
1
Expert's answer
2020-05-21T11:57:03-0400

What is the saving/investment approach to equilibrium?

The interaction between the saving functions and the investment function helps ascertain the equilibrium national income. In a frugal economy where households typically exhibit saving habits, the equilibrium national income is ascertained by equating the desired/planned level of saving for firms and the desired/planned level of savings by households (S=I). Any distortion in the circular flow of income brought about by imbalances between expansionary investment forces and contractionary saving forces leads to variation in the national income. Thus, when investment level is lower than the savings level, the aggregate demand falls short of the aggregate supply and this consequently lowers the output level since the investors are not be able to sell all their entire output at the given prices. Consequently, this results to a decline in the national income. In a two-sector economy, withdrawals are represented by savings while injections are represented by investments. At the instance that the injection is greater than the withdrawal, the nation income expands and the vice versa happens when the savings level is greater than the investment level.

                                                 


The equilibrium level is represented by E while the investment is represented by the horizontal line I. The upward sloping line represents the saving equation. Equilibrium national income is ascertained at the point where the investment level is equal to the saving level. From the graph, any move that distorts the equilibrium attracts forces that push the shift back towards the equilibrium point E.

The equilibrium national income is derived as follows;

Using the consumption function to get the savings function

C = a + by

S = -a + (1-b)Y

In equilibrium, S = I; thus,

I = -a + (1-b)Y

(1-b)Y = I + a

Y= (I + a)/(1-b)

 



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