Given
V = K e 2 t V=Ke^{2\sqrt t} V = K e 2 t
Rice can be stored in a way it maximized the present value of rice, if the interest rate is r.
Then, the present value is given as:
P = V . e − r t P=V.e^{-rt} P = V . e − r t
P = K e 2 t × e − r t P=Ke^{2\sqrt {t}} \times e^{-rt} P = K e 2 t × e − r t
Apply log on both sides as follows:
l n P = l n k + 2 t − r t lnP=lnk+2\sqrt t-rt l n P = l nk + 2 t − r t
δ i n P δ t = 2 2 t − r = 0 \frac{\delta in P}{\delta t}=\frac{2}{2\sqrt t}-r=0 δ t δ in P = 2 t 2 − r = 0
1 t = r \frac{1}{\sqrt t}=r t 1 = r
1 r = t \frac{1}{r}=\sqrt t r 1 = t
Optimal time t ∗ = 1 r 2 t^*=\frac{1}{r^2} t ∗ = r 2 1
In the second order derivative
δ 2 v δ 2 t = v δ δ t ( t − 1 2 − r ) δ v δ t \frac{\delta ^2 v}{\delta ^2 t}=v\frac{\delta}{\delta t}(t^{\frac{-1}{2}}-r)\frac{\delta v}{\delta t} δ 2 t δ 2 v = v δ t δ ( t 2 − 1 − r ) δ t δ v
Here, the product rule is used;
Then, the second term vanishes when: δ v δ t = 0 \frac{\delta v}{\delta t}=0 δ t δ v = 0
Thus,
δ v δ t = − v 2 t 3 < 0 \frac{\delta v}{\delta t}=\frac{-v}{2}\sqrt{t^3}<0 δ t δ v = 2 − v t 3 < 0
This satisfies the second order condition for maximum.
Thus the dealer should store the rice up to t = 1 r 2 t=\frac{1}{r^2} t = r 2 1
Comments