---
# System prepended metadata

title: 'DP: Wine and Bread Problem'
tags: [math, ' macro', ' memo']

---

---
tags: math, memo, macro
---


# DP: Wine and Bread Problem

## The Question

A vintner has one unit of labor to use each day. He can allocate that labor between the making of bread and the pressing of grapes for grape juice. The bread he makes today he can consume today. The grape juice he makes today will become tomorrow’s wine (he doesn’t care for grape juice). The production technology is linear: it produces one unit of bread per unit of labor allocated to baking, and one unit of juice per unit of labor allocated to grape pressing, and one unit of wine per unit of grape juice left to ferment. The transformation of juice into wine requires no labor, only time. The vintner allocates his labor so as to maximize the utility of his own consumption. His utility function has the form:
$$
\sum_{t=0}^\infty \beta^t \sqrt{b_t w_t},
$$

where $b_t$ and $w_t$ are the bread and wine consumption, respectively, in period $t$. The initial wine consumption $w_0$ is given. The discount factor is $\beta \in (0, 1)$.

## The Sequential Problem

The technoly/budget constraint is $b_t+ w_{t+1}=1$, so we can write down the following sequential problem,

\begin{align*}
\max_{\{w_t\}_{t=1}^\infty}\sum_{t=0}^\infty \beta^t \sqrt{(1-w_{t+1}) w_t}\\
\text{s.t } w_t \ge 0
\end{align*}

## The Recursive Problme

The Bellman equation corresponding to the sequential problem is

$$
v(w) = \max_{w' \in [0,1]} \{ \sqrt{(1-w') w} + \beta v(w')\}
$$

## Guess and Verify

Guess that the value function has the form $v(w) = a \sqrt{g + w}$, where $a$ and $g$ are unknown parameters. Plugging the guess into the Bellman equation,

$$
a \sqrt{g + w} = \max_{w' \in [0,1]} \{ \sqrt{(1-w') w} + \beta a \sqrt{g + w'}\}.
$$

The first-order condition of RHS is

$$
 \left ((1-w')w \right )^{-\frac{1}{2}}w = a \beta (g + w')^{-\frac{1}{2}}.
$$

Take square on both sides,

\begin{align*}
\left ((1-w')w \right )^{-1}w^2 &= (a \beta)^2 (g + w')^{-1}\\
(g + w')w^2 &= (a \beta)^2 \left ((1-w')w \right )\\
(w^2 + a^2 \beta^2 w)w' &= (a \beta)^2 w - w^2 g.\\
\end{align*}

Hence,

$$
w' = \frac{(a \beta)^2 w - w^2 g}{w^2 + (a \beta)^2 w}.
$$


Plugging the result into the Bellman equation,

\begin{align*}
a \sqrt{g + w} &= \sqrt{(1-w') w} + \beta a \sqrt{g + w'}\\
&=\sqrt{(1-\frac{(a \beta)^2 w - w^2 g}{w^2 + (a \beta)^2 w}) w} + \beta a \sqrt{g + \frac{(a \beta)^2 w - w^2 g}{w^2 + (a \beta)^2 w}}\\
&= \sqrt{\frac{w^2 (1+g)}{w^2 + (a \beta)^2 w} w} +  a \beta \sqrt{ \frac{ (1+g) (a \beta)^2 w }{w^2 + (a \beta)^2 w}}\\
&= w \sqrt{\frac{ (1+g)}{w + (a \beta)^2 } } +  (a \beta)^2 \sqrt{ \frac{ (1+g)   }{w + (a \beta)^2 }}\\
&= \left(w + (a \beta)^2 \right) \sqrt{\frac{ (1+g)}{w + (a \beta)^2 } }\\
&= \sqrt{\left(w + (a \beta)^2 \right) (1+g) }\\
&= \sqrt{1+g} \sqrt{(a \beta)^2 +w}
\end{align*}

### Matching Coefficients

Let $a^2 = 1+g$, and $g = (a \beta)^2$, we can solve,

\begin{align*}
a&=\frac{1}{1-\beta^2},\\
g&=\frac{\beta^2}{1-\beta^2},\\
w'(w)&= \frac{\beta^2 (1-w)}{w(1-\beta^2)+\beta^2}. 
\end{align*}


## Reference

https://jasandford.com/590_f11/590_f11_hw5a.pdf