Numerical chapter 2
5.
Given data: I  40, G  30, GNP  200, CA  –20  NX  NFP, T  60, TR  25, INT  15, NFP 
7 –9  –2. Since GDP  GNP – NFP, GDP  200 – (–2)  202  Y. Since NX  NFP  CA, NX  CA –
NFP  –20 – (–2)  –18. Since Y  C  I  G  NX, C  Y – (I  G  NX)  202 – (40  30  (–18)) 
150.
Spvt  (Y  NFP – T  TR  INT) – C  (202  (–2) – 60  25  15) –150  30. Sgovt  (T – TR – INT)
– G  (60 – 25 – 15) – 30  –10. S  Spvt  Sgovt  30  (–10)  20.
(a) Consumption  150
(b) Net exports  –18
(c) GDP  202
(d) Net factor payments from abroad  –2
(e) Private saving  30
(f) Government saving  –10
(g) National saving  20
6.
Base-year quantities at current-year prices
at base-year prices
3000  $3  $ 9,000
6000  $2  $12,000
8000  $5  $40,000
$61,000
3000  $2  $ 6,000
6000  $3  $18,000
8000  $4  $32,000
$56,000
Apples
Bananas
Oranges
Total
Current-year quantities at current-year prices
Apples
Bananas
Oranges
Total
4,000  $3  $ 12,000
14,000  $2  $ 28,000
32,000  $5  $160,000
$200,000
at base-year prices
4,000  $2  $ 8,000
14,000  $3  $ 42,000
32,000  $4  $128,000
$178,000
(a) Nominal GDP is just the dollar value of production in a year at prices in that year. Nominal GDP
is $56 thousand in the base year and $200 thousand in the current year. Nominal GDP grew
257% between the base year and the current year: [($200,000/$56,000) – 1]  100%  257%.
(b) Real GDP is calculated by finding the value of production in each year at base-year prices.
Thus, from the table above, real GDP is $56,000 in the base year and $178,000 in the current year.
In percentage terms, real GDP increases from the base year to the current year by
[($178,000/$56,000) – 1]  100%  218%.
(c) The GDP deflator is the ratio of nominal GDP to real GDP. In the base year, nominal GDP
equals real GDP, so the GDP deflator is 1. In the current year, the GDP deflator is
$200,000/$178,000  1.124. Thus the GDP deflator changes by [(1.124/1) – 1]  100%  12.4%
from the base year to the current year.
(d) Nominal GDP rose 257%, prices rose 12.4%, and real GDP rose 218%, so most of the increase
in nominal GDP is because of the increase in real output, not prices. Notice that the quantity of
oranges quadrupled and the quantity of bananas more than doubled.
Answers to problems assigned on chapter 3
1)
(a) To find the growth of total factor productivity, you must first calculate the value of A in the
production function. This is given by A  Y/(K.3N.7). The growth rate of A can then be
calculated as
[(Ayear 2 – Ayear 1)/Ayear 1]  100%. The result is:
1960
1970
1980
1990
2000
A
12.484
14.701
15.319
17.057
19.565
% increase in A
—
17.8%
4.2%
11.3%
14.7%
(b) Calculate the marginal product of labor by seeing what happens to output when you add 1.0 to
N; call this Y2, and the original level of output Y1. [A more precise method is to take the
derivative of output with respect to N; dY/dN 
The result is the same (rounded).]
1960
1970
1980
1990
2000
Y1
2502
3772
5162
7113
9817
Y2
2529
3805
5198
7155
9867
.
MPN
27
33
36
42
50
2)
(a) The MPK is 0.2, because for each additional unit of capital, output increases by 0.2 units. The
slope of the production function line is 0.2. There is no diminishing marginal productivity of
capital in this case, because the MPK is the same regardless of the level of K. This can be seen
below because the production function is a straight line.
(b)
When N is 100, output is Y  0.2(100  100.5)  22. When N is 110, Y is 22.0976. So the MPN
for raising N from 100 to 110 is (22.0976 – 22)/10  0.00976. When N is 120, Y is 22.1909. So
the MPN for raising N from 110 to 120 is (22.1909 – 22.0976)/10  0.00933. This shows
diminishing marginal productivity of labor because the MPN is falling as N increases. Below this
is shown as a decline in the slope of the production function as N increases.
4)
MPN  A(100 – N)
(a) A  1. MPN  100 – N.
(1) W  $10. w  W/P  $10/$2  5. Setting w  MPN, 5  100 – N, so N  95.
(2) W  $20. w  W/P  $20/$2  10. Setting w  MPN, 10  100 – N, so N  90.
These two points are plotted as line NDa below. If labor supply  95, then the equilibrium real
wage is 5.
(b) A  2. MPN  2(100 – N).
(1) W  $10. w  W/P  $10/$2  5. Setting w  MPN, 5  2(100 – N), so 2N  195, so N 
97.5.
(2) W  $20. w  W/P  $20/$2  10. Setting w  MPN, 10  2(100 – N), so 2N  190, so N 
95.
These two points are plotted as line NDb in Figure above. If labor supply  95, then the
equilibrium real wage is 10.
5.
(a) If the lump-sum tax is increased, there’s an income effect on labor supply, not a substitution
effect (since the real wage isn’t changed). An increase in the lump-sum tax reduces a worker’s
wealth, so labor supply increases.
(b) If T  35, then NS  22  12w  (2  35)  92  12 w. Labor demand is given by w  MPN 
309 – 2N, or 2N  309 – w, so N  154.5 – w/2. Setting labor supply equal to labor demand gives
154.5 – w/2  92  12w, so 62.5  12.5w, thus w  62.5/12.5  5. With w  5, N  92  (12  5) 
152.
(c) Since the equilibrium real wage is below the minimum wage, the minimum wage is binding.
With w  7, N  154.5 – 7/2  151.0. Note that NS  92  (12  7)  176, so NS > N and there is
unemployment.
7)
(a) At any time, 25 people are unemployed: 5 who have lost their jobs at the start of the month and 20
who have lost their jobs either on January 1 or July 1. The unemployment rate is 25/500  5%.
(b) Each month, 5 people have one-month spells. Every six months, 20 people have six-month
spells. The total number of spells during the year is (5  12)  (20  2)  100. Sixty of the spells
(60% of all spells) last one month, while 40 of the spells (40% of all spells) last six months.
(c) The average duration of a spell is (0.60  1 month)  (0.40  6 months)  3 months.
(d) On any given date, there are 25 people unemployed. Twenty of them (80%) have long spells of
unemployment, while 5 of them (20%) have short spells.
9)
Since ( Y – Y) /Y  2(u – u ), this can be rewritten as Y – Y  2(u – u ) Y or Y  [1 – 2(u – u )] Y , or
Y  Y/[1 – 2(u – u )].
(a) Using the formula above, this table shows the value of Y , given values for u and Y.
Year
1
2
3
4
u
0.08
0.06
0.07
0.05
Y
950
1030
1033.5
1127.5
Y
989.6
1030.0
1054.6
1105.4
b. The first calculation of  Y/Y comes from calculating the percent change in Y from part a.
The second calculation of Δ Y/Y comes from using Eq. (3.6): Y/Y   Y/Y – 2 u, so  Y/Y  Y/Y
2 u.
Year
 Y/Y 
Y
1
989.6
—
2
1030.0 0.041
3
1054.6 0.024
4
1105.4 0.048
YY u
—
—
0.084 –0.02
0.003 0.01
0.091 –0.02
The two methods give close answers.
 Y/Y
—
0.044
0.023
0.051