Algorithm 1 Answer to Question 1 (pseudo-code)
α ← 0.35
δ ← 0.06
σ ← 0.20
k∗
← σ
δ
1
1−α
k1 ← 0.5 · k∗
for t ← 1 to 50 do
yt ← kα
t
it ← σyt
kt+1 ← (1 − δ) kt + it
end for
return k51/k∗
Algorithm 2 Answer to Question 1 (Matlab implementation)
c l e a r a l l ;
alpha = 0 . 3 5 ;
delta = 0 . 0 6 ;
sigma = 0 . 2 0 ;
kstar = ( sigma/ delta )ˆ(1/(1 − alpha ) ) ;
k (1) = 0.5∗ kstar ;
f o r t = 1 : 50
y ( t ) = k ( t )ˆ alpha ;
i ( t ) = sigma∗y ( t ) ;
k ( t+1) = (1− delta )∗k ( t ) + i ( t ) ;
end
disp ( k (51)/ kstar )
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Algorithm 3 Answer to Question 2 (pseudo-code)
α ← 0.35
δ ← 0.06
σ ← 0.20
k∗
← σ
δ
1
1−α
k1 ← 0.5 · k∗
for t ← 1 to 50 do
yt ← kα
t
it ← σyt
ct ← yt − it
kt+1 ← (1 − δ) kt + it
if t ≥ 2 then
gt−1 = yt
yt−1
− 1
end if
end for
plot {t, ct}
50
t=1 and {t, it}
50
t=1 on the same vertical axis
plot {t, yt}
50
t=1 and {t, gt}
49
t=1 on separate vertical axes
Algorithm 4 Answer to Question 2 (Matlab implementation)
c l e a r a l l ;
alpha = 0 . 3 5 ;
delta = 0 . 0 6 ;
sigma = 0 . 2 0 ;
kstar = ( sigma/ delta )ˆ(1/(1 − alpha ) ) ;
k (1) = 0.5∗ kstar ;
f o r t = 1 : 50
y ( t ) = k ( t )ˆ alpha ;
i ( t ) = sigma∗y ( t ) ;
c ( t ) = y ( t ) − i ( t ) ;
k ( t+1) = (1− delta )∗k ( t ) + i ( t ) ;
i f t >= 2
g ( t −1) = y ( t )/y ( t −1) − 1;
end
end
plot ( 1 : 5 0 , c , 1 : 5 0 , i ) ;
legend ( ’ c ’ , ’ i ’ ) ;
f i g u r e ;
plotyy ( 1 : 5 0 , y , 1 : 4 9 , g ) ;
legend ( ’ y ’ , ’ g ’ ) ;
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