EXERCISE:
Consider the following model of the economy where
the expressions for the AD and SRAS curves are
given by:
AD Æ Y = 710 – 30P + 5G
SRAS Æ Y = 10 + 5P – 2POIL
(a) Explain the various terms in the AD curve. What
is the value of the multiplier?
AD Æ Y = 710 – 30P + 5G
30P = 710 + 5G – Y
P = [(710 + 5G) / 30] – (1/30) Y
Vertical Intercept: (710 + 5G) / 30
Slope: – 1 /30
βAE = ∆Y / ∆G = 5
P
710 + 5G
30
Slope = –1/30
AD
710 + 5G
Y
P
710 + 5G
30
Horizontal shift:
∆Y = βAE∆G = 5∆G
AD
710 + 5G
AD’
Y
(b) Explain the various terms in the SRAS curve.
Explain why the price of oil enters negatively.
SRAS Æ Y = 10 + 5P – 2POIL
5P = –10 + 2POIL – Y
P = [(–10 + 2POIL) / 5] + (1/5)Y
Vertical Intercept: (–10 + 2POIL) / 5
Slope: 1/5
P
SRAS
Slope = 1/5
–10 + 2POIL
5
Y
(c) Solve for the equilibrium value of real GDP and
the price level.
AD Æ Y = 710 – 30P + 5G
SRAS Æ Y = 10 + 5P – 2POIL
AD = SRAS
710 – 30P + 5G = 10 + 5P – 2POIL
35P = 700 + 5G + 2POIL
P* = 20 + (1/7)G + (2/35)POIL
SRAS Æ Y = 10 + 5P – 2POIL
Y* = 10 + 5[20 + (1/7)G + (2/35)POIL] – 2POIL
= 10 + 100 + (5/7)G + (2/7)POIL – 2POIL
= 110 + (5/7)G – (12/7)POIL
P
SRAS
710 + 5G
30
P*
AD
–10 + 2POIL
5
Y*
Y
(d) Using your solution to part c), what is the effect
of a change in G on equilibrium Y and P?
P* = 20 + (1/7)G + (2/35)POIL
∆P* = (1/7)∆G
Y* = 110 + (5/7)G – (12/7)POIL
∆Y* = (5/7)∆G
P
SRAS
710 + 5G
30
P**
P*
AD’
AD
–10 + 2POIL
5
Y*
Y**
Y
(e) Using your solution to part c), what is the effect
of a change in POIL on equilibrium Y and P?
P* = 20 + (1/7)G + (2/35)POIL
∆P* = (2/35)∆ POIL
Y* = 110 + (5/7)G – (12/7)POIL
∆Y* = –(12/7)∆POIL
P
SRAS’
SRAS
710 + 5G
30
P**
P*
AD
–10 + 2POIL
5
Y** Y*
Y