Question One:
Consider the following model of growth. Output is given by
𝑌(𝑡) = 𝐾(𝑡)𝛼 [𝐴(𝑡)𝐿(𝑡)]1−𝛼
Capital and Knowledge accumulate according to the equations
𝐾̇ (𝑡) = 𝑠𝑌(𝑡)
𝑠 ∈ (0,1),
𝐴̇(𝑡) = 𝑌(𝑡)𝜑
𝜑 ∈ (0,1)
The population grows at the rate n
i)
Give an economic interpretation of the equation for the accumulation of knowledge.
Solutions:
The specification 𝐴̇(𝑡) = 𝑌(𝑡)𝜑 , suggests a learning by doing model. In this model, knowledge
accumulation occurs as a side effect of goods production. The intuition here is that as individuals
produce goods, they inevitably think of ways of improving the production process. For example,
A study by economist Zvi Griliches (1979) found that productivity in the manufacturing sector
increases as workers gain experience on the job. Griliches observed that workers who have been
in the same job for a longer period of time are more productive than those who are new to the job.
This suggests that knowledge accumulation through learning-by-doing is an important driver of
productivity in manufacturing. Another Example is A study by economists Alain de Janvry and
Elisabeth Sadoulet examined the impact of training programs on small-scale farmers in developing
countries. The study found that farmers who received training on new technologies and farming
practices were able to improve their productivity and increase their incomes. This suggests that
knowledge accumulation through education and training can be an important driver of productivity
in agriculture. A study by economist Paul Romer (1990) examined the relationship between
research and development (R&D) spending and economic growth. Romer found that countries that
invest more in R&D tend to experience faster rates of economic growth over time. This suggests
that knowledge accumulation through R&D can be an important driver of technological progress
and economic growth. Hence from these we can conclude that the specification can be interpreted
as a Learning-by-doing model. The type of knowledge being accumulated here is as a result of
deliberate effort as opposed to one driven by research and development.
For what value of φ does this model reduce to the Solow model?
ii)
Solution:
If 𝜑 = 0 then 𝐴̇(𝑡) = 1. What this means is 𝐴(𝑡) = 𝐴(0) + 𝑡. Given that also that the growth rate in
knowledge is given as 𝑔𝐴 (𝑡) =
𝐴̇(𝑡)
𝐴(𝑡)
1
this will entail that 𝑔𝐴 (𝑡) = 𝐴(0)+𝑡. This means that in the long
run, the technological progress will become negligible and hence the economy will be a Solow
economy without technological advancements.
Let 𝑔𝐴 (𝑡) =
iii)
𝐴̇(𝑡)
𝐴(𝑡)
and 𝑔𝑘 (𝑡) =
𝐾̇(𝑡)
𝐾(𝑡)
be the growth rate of Knowledge and capital,
respectively. Find the expression of 𝑔𝐴 (𝑡) and 𝑔𝑘 (𝑡) in terms of 𝐴(𝑡), K(𝑡) and
Parameters.
Solution:
Growth rate of knowledge:
Recall that 𝐴̇(𝑡) = 𝑌(𝑡)𝜑 this means
𝑔𝐴 (𝑡) =
𝐴̇(𝑡)
𝐴(𝑡)
resolves to 𝑔𝐴 (𝑡) =
𝑌(𝑡)𝜑
𝐴(𝑡)
Also recall that 𝑌(𝑡) = 𝐾(𝑡)𝛼 [𝐴(𝑡)𝐿(𝑡)]1−𝛼 , hence
𝑌(𝑡)𝜑 = [𝐾(𝑡)𝛼 [𝐴(𝑡)𝐿(𝑡)]1−𝛼 ]𝜑
𝜑
hence 𝑔𝐴 (𝑡) =
𝑔𝐴 (𝑡) =
[𝐾(𝑡)𝛼 [𝐴(𝑡)𝐿(𝑡)]1−𝛼 ]
𝐴(𝑡)
𝐾(𝑡)𝛼𝜑 𝐴(𝑡)(1−𝛼)𝜑 𝐿(𝑡)(1−𝛼)𝜑
𝐴(𝑡)
. Applying simple algebra this will become,
which then finally becomes
𝑔𝐴 (𝑡) = 𝐾(𝑡)𝛼𝜑 𝐴(𝑡)(1−𝛼)𝜑−1 𝐿(𝑡)(1−𝛼)𝜑
Growth rate of Capital:
̇
𝐾(𝑡)
Recall again that 𝐾̇ (𝑡) = 𝑠𝑌(𝑡) and 𝑔𝑘 (𝑡) = 𝐾(𝑡)
This means 𝑔𝑘 (𝑡) =
𝑠𝑌(𝑡)
𝐾(𝑡)
Given also that 𝑌(𝑡) = 𝐾(𝑡)𝛼 [𝐴(𝑡)𝐿(𝑡)]1−𝛼
The expression now becomes 𝑔𝑘 (𝑡) =
𝑠{𝐾(𝑡)𝛼 [𝐴(𝑡)𝐿(𝑡)]1−𝛼 }
resolving to
𝐾(𝑡)
𝑔𝑘 (𝑡) = 𝑠{𝐾(𝑡)𝛼−1 [𝐴(𝑡)𝐿(𝑡)]1−𝛼 } which with algebraic manipulation essentially becomes
𝑔𝑘 (𝑡) = 𝑠
[𝐴(𝑡)𝐿(𝑡)]1−𝛼
𝐾(𝑡)1−𝛼
and finally can be written as
1−𝛼
𝐴(𝑡)𝐿(𝑡)
𝑔𝑘 (𝑡) = 𝑠 [
]
𝐾(𝑡)
Differentiate to obtain expressions for 𝑔̇𝐴(𝑡) and 𝑔̇𝐴(𝑡) in terms of 𝑔𝐴 (𝑡), 𝑔𝑘 (𝑡) and
iv)
the parameters. Draw the phase diagram in (𝑔𝐴 , 𝑔𝑘 ) space.
Solutions:
For this segment, we solve the question following the rules that the growth rate of a variable
equals the time derivative of its log such that:
a)
𝒁̇(𝒕)
𝒁(𝒕)
𝑿̇(𝒕)
𝒀̇(𝒕)
= 𝑿(𝒕) + 𝒀(𝒕) if 𝑋(𝑡), Y(𝑡), 𝑎𝑛𝑑 𝑍(𝑡) are functions of time and 𝑍(𝑡) = 𝑋(𝑡). 𝑌(𝑡)
and
b) The growth rate of the ratio of two variables equals the difference of their
growth rates. That is;
if 𝑍(𝑡) =
c)
𝒀̇(𝒕)
𝒀(𝒕)
𝑋(𝑡)
𝒁̇(𝒕)
𝑿̇(𝒕)
𝒀̇(𝒕)
, then 𝒁(𝒕) = 𝑿(𝒕) + 𝒀(𝒕)
𝑌(𝑡)
𝑿̇(𝒕)
= 𝜶 𝑿(𝒕) if 𝑋(𝑡), 𝑎𝑛𝑑 𝑌(𝑡) are functions of time and 𝑌(𝑡) = 𝑋(𝑡)𝛼
We have that 𝑔𝐴 (𝑡) = 𝐾(𝑡)𝛼𝜑 𝐴(𝑡)(1−𝛼)𝜑−1 𝐿(𝑡)(1−𝛼)𝜑 .
ln[ 𝑔𝐴 (𝑡)] = 𝛼𝜑 𝑙𝑛 𝐾(𝑡) + [(1 − 𝛼)𝜑 − 1]𝑙𝑛[𝐴(𝑡)] + (1 − 𝛼)𝜑 ln[𝑙(𝑡)]
Differentiating this and following the above rules we get
𝑔̇ 𝐴
𝐾̇ (𝑡)
𝐴̇(𝑡)
𝐿̇(𝑡)
= 𝛼𝜑
+ [(1 − 𝛼)𝜑 − 1] (
) + (1 − 𝛼)𝜑 (
)
𝑔𝐴
𝐾(𝑡)
𝐴(𝑡)
𝐿(𝑡)
Given that labor and knowledge grow at a constant rate, 𝐿̇(𝑡) = 𝑛𝐿(𝑡) and 𝐴̇(𝑡) = 𝐴(𝑡)𝑔𝐴 (𝑡).
𝐾̇(𝑡)
Also, and 𝑔𝑘 (𝑡) = 𝐾(𝑡).
𝑔̇ 𝐴
𝑔𝐴 𝐴(𝑡)
𝑛𝐿(𝑡)
= 𝛼𝜑 𝑔𝑘 (𝑡) + [(1 − 𝛼)𝜑 − 1] (
) + (1 − 𝛼)𝜑 (
)
𝑔𝐴
𝐴(𝑡)
𝐿(𝑡)
𝑨𝒏𝒔𝒘𝒆𝒓 𝟏:
𝑔̇ 𝐴
𝑔𝐴
= 𝛼𝜑 𝑔𝑘 (𝑡) + [(1 − 𝛼)𝜑 − 1](𝑔𝐴 (𝑡)) + (1 − 𝛼)𝜑𝑛 (derived)
Also given that 𝑔𝑘 (𝑡) = 𝑠 [
𝑙𝑛[𝑔𝑘 (𝑡)] = ln(𝑠) + 𝑙𝑛 [
𝐴(𝑡)𝐿(𝑡) 1−𝛼
𝐾(𝑡)
]
𝐴(𝑡)𝐿(𝑡) 1−𝛼
]
𝐾(𝑡)
since 𝑠 ∈ (0,1), letting 𝑠 = 1
We obtain ln(𝑠) = ln(1) = 0 hence the above becomes 𝑙𝑛[𝑔𝑘 (𝑡)] = 𝑙𝑛 [
1−𝛼
𝐴(𝑡)𝐿(𝑡)
𝑙𝑛[𝑔𝑘 (𝑡)] = 𝑙𝑛 [
]
𝐾(𝑡)
𝐴(𝑡)𝐿(𝑡) 1−𝛼
𝐾(𝑡)
]
𝐴(𝑡)𝐿(𝑡)
= (1 − 𝛼) 𝑙𝑛 [
]
𝐾(𝑡)
This then becomes 𝑙𝑛[𝑔𝑘 (𝑡)] = (1 − 𝛼){ln[𝐴(𝑡)] + ln[𝐿(𝑡)] − ln[𝐾(𝑡)]}
Differentiating this then becomes:
𝑔̇ 𝑘
𝐴̇(𝑡) 𝐿̇(𝑡) 𝐾̇ (𝑡)
= (1 − 𝛼) {
+
−
}
𝑔𝑘
𝐴(𝑡) 𝐿(𝑡) 𝐾(𝑡)
Given that
𝐴̇(𝑡)
= 𝑔𝐴 (𝑡),
𝐴(𝑡)
𝑨𝒏𝒔𝒘𝒆𝒓 𝟐:
𝑔̇ 𝑘
𝑔𝑘
𝐿̇ (𝑡)
𝐾̇(𝑡)
= 𝑛 and 𝑔𝑘 (𝑡) = 𝐾(𝑡) the above now translates to
𝐿(𝑡)
= (1 − 𝛼)[𝑔𝐴 (𝑡) + 𝑛 − 𝑔𝑘 (𝑡)] (derived)
To draw the phase diagrams, refer to Answer 1 and Answer 2 derived above.
𝑨𝒏𝒔𝒘𝒆𝒓 𝟏:
𝑔̇ 𝐴
𝑔𝐴
= 𝛼𝜑 𝑔𝑘 (𝑡) + [(1 − 𝛼)𝜑 − 1](𝑔𝐴 (𝑡)) + (1 − 𝛼)𝜑𝑛 let 𝑔̇ 𝐴 = 0 so that
𝛼𝜑 𝑔𝑘 (𝑡) + [(1 − 𝛼)𝜑 − 1](𝑔𝐴 (𝑡)) + (1 − 𝛼)𝜑𝑛 = 0 solving for 𝑔𝑘 gives
−𝛼𝜑 𝑔𝑘 (𝑡) = [(1 − 𝛼)𝜑 − 1](𝑔𝐴 (𝑡)) + (1 − 𝛼)𝜑𝑛 dividing throughout by −𝛼𝜑 this becomes
𝑔𝑘 (𝑡) =
[(1−𝛼)𝜑−1]
−𝛼𝜑
(𝑔𝐴 (𝑡)) +
(1−𝛼)𝜑
−𝛼𝜑
𝑛 which becomes
[1−(1−𝛼)𝜑]
𝑔𝑘 =
𝛼𝜑
(𝑔𝐴 ) −
(1−𝛼)𝜑
𝛼𝜑
𝑛 which gives our first schedule***
Also:
𝑨𝒏𝒔𝒘𝒆𝒓 𝟐:
𝑔̇ 𝑘
𝑔𝑘
= (1 − 𝛼)[𝑔𝐴 (𝑡) + 𝑛 − 𝑔𝑘 (𝑡)]
let 𝑔̇ 𝐾 = 0 so that : 0 = (1 − 𝛼)[𝑔𝐴 (𝑡) + 𝑛 − 𝑔𝑘 (𝑡)] expanding this
0 = 𝑔𝐴 (𝑡) + 𝑛 − 𝑔𝑘 (𝑡) − 𝛼𝑔𝐴 (𝑡) − 𝛼𝑛 + 𝛼𝑔𝑘 (𝑡)
[𝑔𝐴 (𝑡) − 𝛼𝑔𝐴 (𝑡)] + [𝑛 − 𝛼𝑛] + [−𝑔𝑘 (𝑡) + 𝛼𝑔𝑘 (𝑡)] = 0
[𝑔𝐴 (𝑡) − 𝛼𝑔𝐴 (𝑡)] + [𝑛 − 𝛼𝑛] = [𝑔𝑘 (𝑡) − 𝛼𝑔𝑘 (𝑡)]
𝑔𝐴 [1 − 𝛼] + 𝑛[1 − 𝛼] = 𝑔𝑘 [1 − 𝛼] solving for 𝑔𝑘 we divide through out by [1 − 𝛼] which gives
𝑔𝑘 = 𝑔𝐴 + 𝑛 which gives our second schedule***
The point of equilibrium in the schedules will be where the two schedules intersect. That is,
The point where:
𝑔𝐴 + 𝑛 =
𝑔𝐴 − {
[1−(1−𝛼)𝜑]
𝛼𝜑
(𝑔𝐴 ) −
(1−𝛼)𝜑
𝛼𝜑
𝑛 . from here we solve our first equilibrium point (𝑔∗𝐴 ).
[1 − (1 − 𝛼)𝜑]
(1 − 𝛼)𝜑
(𝑔𝐴 )} = −𝑛 −
𝑛
𝛼𝜑
𝛼𝜑
(𝑔𝐴 ) (1 −
[1 − (1 − 𝛼)𝜑]
(1 − 𝛼)𝜑
) = (−𝑛) (1 +
)
𝛼𝜑
𝛼𝜑
(𝑔𝐴 ) (1 −
[1 − (𝜑 − 𝛼𝜑]
(𝜑 − 𝛼𝜑)
) = (−𝑛) (1 +
)
𝛼𝜑
𝛼𝜑
[𝛼𝜑−{1−(𝜑−𝛼𝜑]}
𝛼𝜑+(𝜑−𝛼𝜑)
𝛼𝜑
𝛼𝜑
(𝑔𝐴 ) (
) = (−𝑛) (
) simplifying this we have
[𝛼𝜑 − 1 + 𝜑 − 𝛼𝜑]
𝛼𝜑 + 𝜑 − 𝛼𝜑)
(𝑔𝐴 ) (
) = (−𝑛) (
)
𝛼𝜑
𝛼𝜑
(𝑔𝐴 ) (
𝜑−1
𝜑
𝜑−1
1
) = (−𝑛) ( ) 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 (𝑔𝐴 ) (
) = (−𝑛) ( )
𝛼𝜑
𝛼𝜑
𝛼𝜑
𝛼
We then solve for (𝑔𝐴 ) ℎ𝑒𝑟𝑒 𝑜𝑛 𝑤ℎ𝑖𝑐ℎ 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑜𝑢𝑟 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚 (𝑔𝐴 ∗ )by multiplying both LHS
and RHS of the equation by
(𝑔𝐴 ) (
(𝑔𝐴 ∗ ) = (−𝑛)
𝜑−1
𝛼𝜑
1
𝛼𝜑
)(
) = (−𝑛) ( ) (
)
𝛼𝜑
𝜑−1
𝛼 𝜑−1
𝜑
𝜑𝑛
which can be written as (𝑔𝐴 ∗ ) = 1−𝛼 {point 1}
𝜑−1
Given this we can solve for the equilibrium (𝑔𝐾 ∗ ) By replacing (𝑔𝐴 ∗ ) into the equation
𝑔𝑘 = 𝑔𝐴 + 𝑛 which was derived above
𝑔𝐾 ∗ =
𝜑
𝑛 + 𝑛,
1−𝛼
𝑤ℎ𝑖𝑐ℎ 𝑔𝑖𝑣𝑒𝑠 𝑔𝐾 ∗ =
𝜑𝑛 + 𝑛 − 𝜑𝑛
1−𝛼
𝑛
Which then resolves to 𝑔𝐾 ∗ = 1−𝛼 {𝑝𝑜𝑖𝑛𝑡 2} Thus, the schedules will intersect at the point
(𝑔𝐴 ∗ , 𝑔𝐾 ∗ ) =
v)
𝜑𝑛
,
𝑛
1−𝛼 1−𝛼
Does the economy converge to a balanced growth path? If so, what are the growth
rates of K, A and Y on the balanced growth path?
Solutions:
From the phase diagram, we can see that the economy converges to a balanced growth path. There,
A grows at the rate of growth 𝒈𝑨 ∗ , and Y and K will grow at the 𝒈𝑲 ∗ .
Question Two:
a) Assume that households have the following lifetime utility function
∞
𝑈=∫
𝑡=0
𝑒 −𝜌𝑡 [𝐶(𝑡)𝛼 + ∆]
𝑁(𝑡)
𝑑𝑡
𝐻
where 𝛼 < 1, ∆> 0, 𝜃 ≠ 1, 𝜌 is the discount rate, C(t) is consumption per worker, N(t)
is the number of workers and H is the number of households.
i.
Express the household’s lifetime utility in terms of consumption per unit of effective
labor c(t), using the fact that 𝑁(𝑡) = 𝑁(0)𝑒 𝑛𝑡 , 𝐴(𝑡) = 𝐴(0)𝑒 𝑔𝑡
Solutions: