ECON 6022A
Fall 2013
Solutions to Problem Set 1
September 30, 2013
1
National Income Accounting
ABC Computer Company has a $20,000,000 factory in Silicon Valley. During the current year ABC builds
$2,000,000 worth of computer components. ABC’s costs are labor, $1,000,000; interest on debt, $100,000; and taxes, $200,000.
ABC sells all its output output to XYZ Supercomputer. Using ABC’s components, XYZ builds four supercomputers at a cost of $800,000 each ($500,000 worth of components, $200,000 in labor costs, and
$100,000 in taxes per computer). XYZ has a $30,000,000 factory.
XYZ sells three of the supercomputers for $1,000,000 each. At year’s end, it had not sold the fourth.
The unsold computer …show more content…
Nominal GDP = 8000 + 42000 + 128000 = 178000 (Base year Real GDP = Base year nominal GDP =
56000)
Percentage increase since the base year =
178000−56000
56000
× 100% = 217.9%
c. GDP deflator of current year = 100× Current year nominal GDP = 100× 200000 = 12.4% Price level changes
Current year real GDP
178000
from the base year to the current year = 12.4%
d. No. It is because the increase in price level is only 12.4%, which is only small compared to the 257.1% increase in nominal output. In other words, the increase in output is mainly due to the physical increase
(in terms of Quantity) in outputs rather then inflation effect.
4
Ratio scale (optional)
Suppose that a country’s money demand at time t, Md (t), is a function of its national income Y (t) and interest rate i = i(t), i.e., Md (t) = G(Y (t), i(t)). Show that the growth rate of the money demand, denoted as, gMd , can be expressed as the weighted sum of the growth rates of the national income, gY , and interest rate, gi : gMd = εMd ,Y gY + εMd ,i gi where εMd ,Y and εMd ,i are the elasticity of Md with respect to Y (t) and i(t) respectively.
Hint: Growth rate of a function can be expressed as the time derivative of the natural log of that function.
Solution: Differentiate ln(Md (t) w.r.t. t to get the growth rate of Md (t), i.e.,
3
dln(Md (t)
1 dMd (t)
=
dt
Md (t) dt dG(Y (t), i(t))
1
=
G(Y (t), i(t)) dt 1
∂G(Y (t), i(t)) dY (t) ∂G(Y (t), i(t)) di(t)
=
+
G(Y (t),