Domar Growth Model

# Cord idea

- In equilibrium condition, aggregate demand equals potential output.
- Investment I, appears in demand function and output function.
- So, there's a function of I. (It appears a First-Order Differential Equation)

# Mathematical Model

Change in $I(t)$ effects change in $Y(t)$ via a multiplier $k=1/s$. $s$: saving rate, $0<s<1$

(1)\begin{align} y'=I'\frac{1}{s} \end{align}

Output per capital is a constant

(2)\begin{align} \frac{\kappa}{K}\equiv \rho\;\;\; (\kappa\, is\, potential\, output) \end{align}

then

(3)\begin{align} \kappa'=\rho K'=\rho I \end{align}

Demand equals output

(4)\begin{align} I'\frac{1}{s}=\rho I\;\;\;(reason \;of \;dynamic) \end{align}

Note: $I=I(t)$

## Solve the Differential Equation

(5)\begin{align} I(t)=I(0){e}^{s \rho t} \end{align}

Refer: Differential Equations

# Razor's Edge

$\rho s$ is the equilibrium growth rate, r is the actual growth rate.

If $r>\rho s$, actual $I(t)$ curve will be higher than the optimal one. (same original point)

.

Now, that $I(t)=I(0){e}^{rt}$, then $I'=rI(0){e}^{rt}$

so,

\begin{align} Y'=\frac{1}{s}I'=\frac{r}{s}{e}^{rt} \end{align}

(7)
\begin{align} \kappa '=\rho I(t)=\rho I(0){e}^{rt} \end{align}

We get

(8)\begin{align} \frac{Y'}{\kappa '}=\frac{r}{\rho s} \end{align}

If $r>\rho s$, then $Y'>\kappa '$, demand growth exceeds output growth, causing shortage of capacity.

This section learned from *Fundamental Methods of Mathematical Economics (4th edition)*, Chiang & Wainwright

page revision: 6, last edited: 17 Apr 2009 20:30