Asked by May
demand function for kerosene: Kd=a0+a1Pk
supply function for kerosene: Ks=b0+b1Pk+b2Pg
demand function for gasoline: Gd=c0+c1Pg+u1 (ul here shows a exogenous random shock to the demand of gasoline)
supply function for gasoline: Gs=d0+d1Pk+d2Pg
a) if gasoline and kerosene are joint products,what are the signs of b2 and d2 ?
ok,for this question I assumes the signs of b2 and d2 are the same because they are generated simultaneously using a common input, so Gs=Ks
but I am not sure how to find out the signs of b2 and d2 as I am not sure of the signs of b0,b1,d0,d1??
b)if there is an exogenous increase in the demand for gasoline, (delta ul >0) what is the effect on the price of kerosene and quantity of kerosene produced?
for this question I had quantity of kerosene increases, and price of kerosene decreases, but not sure about the exact workingout.
c)assume Kd=Ks
and Gd=Gs
we can write:
(a1-b1)Pk-b2Pg=(b0-a0) and
-d1Pk+(c1-d2)Pg=(d0-c0)-ul
use cramer's rule to solve this system of equations for Pk and Pg.
this is the question that I am most confused about.by cramer's rule, using matrix and determinant I got Apk=(b0-a0)(c1-d2)-(-b2((d0-c0)-ul)
but this answer seemed so complicated and strange that I think I have done something wrong...
can somebody please help? thanks.
supply function for kerosene: Ks=b0+b1Pk+b2Pg
demand function for gasoline: Gd=c0+c1Pg+u1 (ul here shows a exogenous random shock to the demand of gasoline)
supply function for gasoline: Gs=d0+d1Pk+d2Pg
a) if gasoline and kerosene are joint products,what are the signs of b2 and d2 ?
ok,for this question I assumes the signs of b2 and d2 are the same because they are generated simultaneously using a common input, so Gs=Ks
but I am not sure how to find out the signs of b2 and d2 as I am not sure of the signs of b0,b1,d0,d1??
b)if there is an exogenous increase in the demand for gasoline, (delta ul >0) what is the effect on the price of kerosene and quantity of kerosene produced?
for this question I had quantity of kerosene increases, and price of kerosene decreases, but not sure about the exact workingout.
c)assume Kd=Ks
and Gd=Gs
we can write:
(a1-b1)Pk-b2Pg=(b0-a0) and
-d1Pk+(c1-d2)Pg=(d0-c0)-ul
use cramer's rule to solve this system of equations for Pk and Pg.
this is the question that I am most confused about.by cramer's rule, using matrix and determinant I got Apk=(b0-a0)(c1-d2)-(-b2((d0-c0)-ul)
but this answer seemed so complicated and strange that I think I have done something wrong...
can somebody please help? thanks.
Answers
Answered by
MathMate
You're on the right track, but you need to complete the process.
Your answer for x is partially correct, namely it consists of the denominator (ce-bf, see text below), but you have not divided by the denominator.
Cramer's rule for a 2x2 system of equations works as follows:
ax+by=c
dx+ey=f
then δ=ae-bd (≠0)
and
x=(ce-bf)/δ
y=(af-cd)/δ
Using the given equations,
x=Pk
y=Pg
a=(a1-b1)
b=-b2
c=(b0-a0)
d=-d1
e=(c1-d2)
f=(d0-c0)-ul
For your information, the denominator δ would be as follows:
δ
= ae-bd
= (a1-b1)*(c1-d2)-b2*d1
and it does not look strange to me.
Continue the calculations for x and y, post your answer for checking if you wish.
Your answer for x is partially correct, namely it consists of the denominator (ce-bf, see text below), but you have not divided by the denominator.
Cramer's rule for a 2x2 system of equations works as follows:
ax+by=c
dx+ey=f
then δ=ae-bd (≠0)
and
x=(ce-bf)/δ
y=(af-cd)/δ
Using the given equations,
x=Pk
y=Pg
a=(a1-b1)
b=-b2
c=(b0-a0)
d=-d1
e=(c1-d2)
f=(d0-c0)-ul
For your information, the denominator δ would be as follows:
δ
= ae-bd
= (a1-b1)*(c1-d2)-b2*d1
and it does not look strange to me.
Continue the calculations for x and y, post your answer for checking if you wish.
Answered by
May
my answer:
x=(b0-a0)*(c1-d2)-(-b2((d0-c0)-ul) divided by(a1-b1)*(c1-d2)-b2*d1
y=[(d0-c0)-ul]*(a1-b1)+[d1(b0-a0)]
divided by(a1-b1)*(c1-d2)-b2*d1
but I can't simplify it further...
Also for question (a) I somehow got a negative sign for b2, is it correct? how do I work out the sign for d2?
And can you please give me some hints on how to do the working out for question (c)? thanks!
x=(b0-a0)*(c1-d2)-(-b2((d0-c0)-ul) divided by(a1-b1)*(c1-d2)-b2*d1
y=[(d0-c0)-ul]*(a1-b1)+[d1(b0-a0)]
divided by(a1-b1)*(c1-d2)-b2*d1
but I can't simplify it further...
Also for question (a) I somehow got a negative sign for b2, is it correct? how do I work out the sign for d2?
And can you please give me some hints on how to do the working out for question (c)? thanks!
Answered by
MathMate
The answer for x is correct.
There is a mistake of the sign in the answer for y:
y=[(d0-c0)-ul]*(a1-b1) <b>-</b> [d1(b0-a0)]
divided by(a1-b1)*(c1-d2)-b2*d1
I'm not sure if I understand your question "how to do the working out for question (c)".
Do you mean where those formulae came from, or how to derive those formulae? These are the standard equations for Cramer's rule for a system of two equations, which you can find in your textbook, or if you don;t have one, you can find it on the Internet.
For questions related to (a) and (b), I suggest you repost with a subject of macroeconomics.
There is a mistake of the sign in the answer for y:
y=[(d0-c0)-ul]*(a1-b1) <b>-</b> [d1(b0-a0)]
divided by(a1-b1)*(c1-d2)-b2*d1
I'm not sure if I understand your question "how to do the working out for question (c)".
Do you mean where those formulae came from, or how to derive those formulae? These are the standard equations for Cramer's rule for a system of two equations, which you can find in your textbook, or if you don;t have one, you can find it on the Internet.
For questions related to (a) and (b), I suggest you repost with a subject of macroeconomics.
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