ISI ME II 2014

[Ridhika]

Thanks a ton Subhayu!! You are awesome!

[Granpa Simpson]

You are most welcome Ridhika..its my pleasure..

[Shefali]

Subhayu...pls explain 5b

[Ridhika]

I have one doubt from ISI ME II 2013 Q10..

Consider an otherwise identical Solow model of economic growth where
the entire income is consumed.
(a) Analyse how wage and rental rate on capital would change over
time.
(b) Can the economy attain steady state equilibrium?


Now if s=0

The fundamental equation which suggests s.k = (d+n)k = 0
Then if d or n > 0 it must be that at equilibrium k=0.. so we have the steady state..

But what about part (a) .. if k =0 then y=0.. so does that mean even w and r = 0 ? or can we say nothing about them?

[Granpa Simpson]

In this case there is no accumulation of capital over time however labor is growing at a rate n, so what happens is that capital per worker ratio decreases which means that MP(L) for a fixed level of capital falls, as MP(L) falls the wage rate will also fall over time.

[Ridhika]

Makes sense.. and the steady state wage will become 0 or no ?

And as k declines it's rental rate will increase .. as for a given level of labor a single addition to k at low levels of k will increase output so mpk will increase.. and at steady state r will be infinite ?

----- Reply message -----
From: "subhayu [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: ISI ME II 2014
Date: Fri, May 9, 2014 12:25 AM

In this case there is no accumulation of capital over time however labor is growing at a rate n, so what happens is that capital per worker ratio decreases which means that MP(L) for a fixed level of capital falls, as MP(L) falls the wage rate will also fall over time.

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[Ridhika]

Shefali copy pasting Amit sir's answer to q5 which he posted on another thread.. hope it helps..

5 In what follows q(i) denotes firm i's output: 

(a) 

Firm 1 maximizes wrt q(1) 

(10-q(1)-q(2))q(1) -q(1) 

and firm 1's reaction function will be 

q(1) = (9-q(2))/2 

Similarly firm 2's reaction function is 

q(2) = (9-q(1))/2 

Solving the above two, 

we get the equilibrium as q(1) = q(2) = 3. 

(b) 

Firm 1 maximizes wrt q(1) 

(10-q(1)-q(2))q(1) -q(1) s.t. q(1) <= 2 

and firm 1's reaction function will be 

q(1) = min{(9-q(2))/2, 2} 

Similarly firm 2's reaction function is 

q(2) = min{(9-q(1))/2, 2} 

Solving the above two, 

we get the equilibrium as q(1) = q(2) = 2. 

----- Reply message -----
From: "Shefali [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: ISI ME II 2014
Date: Thu, May 8, 2014 9:36 PM

Subhayu...pls explain 5b

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[Arushi]

In part 1 b, If u take quantity exported to be equal to 0, and then max the profit of the domestic mkt function. Then shouldn't u get Q=2.
P=10-q.
TR=10q-q^2
MR= 10-2q
MC= 6
then MR=MC gives q=2 right? not 4
Thus quantity exported is 0 and quantity sold in domestic market is 2

[Dreyfus]

@arushi....u m getting the same answer for partb....and for part a .....domestic output=3 and export=97 as export price is greater than mc

[Ridhika]

Yup .. you are correct arushi .. I forgot to divide by 2 in the end ! 

----- Reply message -----
From: "Arushi [via Discussion forum]" <ml-node+[hidden email]>
To: "Ridhika" <[hidden email]>
Subject: ISI ME II 2014
Date: Sat, May 10, 2014 7:49 PM

In part 1 b, If u take quantity exported to be equal to 0, and then max the profit of the domestic mkt function. Then shouldn't u get Q=2.
P=10-q.
TR=10q-q^2
MR= 10-2q
MC= 6
then MR=MC gives q=2 right? not 4
Thus quantity exported is 0 and quantity sold in domestic market is 2

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