Section 9

A Worked Example, End to End

Sections 2–8 built the machinery in symbols. This section runs the whole engine once with numbers, so you can see every quantity land. We start from two Cobb-Douglas agents and their endowments, derive the market-clearing price by hand, read off the equilibrium allocation, and confirm both agents are strictly better off than where they started. Nothing here is new theory — it is §5 with the symbols replaced by numbers.

The primitives

Two agents, $A$ and $B$ , two goods. Both have Cobb-Douglas preferences but different tastes: $A$ leans toward good 1, $B$ leans toward good 2.

u^A(x_1, x_2) = x_1^{0.6}\, x_2^{0.4}, \qquad u^B(x_1, x_2) = x_1^{0.3}\, x_2^{0.7}

\alpha_A = 0.6

and

\alpha_B = 0.3

in the notation of §5.

Each agent arrives at the market holding exactly one good. $A$ owns all of good 1; $B$ owns all of good 2 — the classic "two traders, one good each" setup that makes the gains from trade vivid.

\omega^A = (\omega^A_1, \omega^A_2) = (10,\ 0), \qquad \omega^B = (\omega^B_1, \omega^B_2) = (0,\ 10)

Tastes: $\alpha_A = 0.6$ , $\alpha_B = 0.3$ .
Total endowments: $\Omega_1 = \omega^A_1 + \omega^B_1 = 10$ , $\Omega_2 = \omega^A_2 + \omega^B_2 = 10$ — a square box.
Numéraire: Good 2, so $p_2 = 1$ and we solve only for $p_1$ .

Step 1 — Wealth at price $p_1$

Wealth is the market value of the endowment, $m^i = p_1 \omega^i_1 + \omega^i_2$ . Because each agent holds only one good, their wealth is especially simple:

m^A = p_1 \cdot 10 + 0 = 10\,p_1, \qquad m^B = p_1 \cdot 0 + 10 = 10

$A$ 's wealth scales with the price of the good they own; $B$ 's wealth is fixed at 10 because good 2 is the numéraire.

Step 2 — Cobb-Douglas demands

Each agent spends an $\alpha_i$ fraction of wealth on good 1 and the rest on good 2: $x^i_1 = \alpha_i m^i / p_1$ and $x^i_2 = (1-\alpha_i) m^i$ . Substituting the wealths from Step 1:

Step 1
$x^A_1 = \frac{0.6 \cdot 10\,p_1}{p_1} = 6, \qquad x^A_2 = 0.4 \cdot 10\,p_1 = 4\,p_1$
$A$ 's demand for good 1 is a constant 6 — the price cancels because $A$ 's wealth is itself proportional to $p_1$ .
Step 2
$x^B_1 = \frac{0.3 \cdot 10}{p_1} = \frac{3}{p_1}, \qquad x^B_2 = 0.7 \cdot 10 = 7$
$B$ 's demand for good 1 falls as good 1 gets more expensive.

Step 3 — Clear the market for good 1

By Walras's Law we only need one market to clear; pick good 1. Total demand must equal the total endowment $\Omega_1 = 10$ :

Step 1
$x^A_1 + x^B_1 = \Omega_1$
Market-clearing condition for good 1.
Step 2
$6 + \frac{3}{p_1} = 10$
Substitute the demands from Step 2.
Step 3
$\frac{3}{p_1} = 4 \quad\Longrightarrow\quad p_1^* = \frac{3}{4} = 0.75$
Good 1 is cheaper than good 2 ( $p_1^* < 1$ ): $A$ brings a lot of good 1 to sell, so its relative price is bid down.

Step 4 — The Walrasian allocation

Now evaluate wealth at $p_1^* = 0.75$ and read the demands off: $m^A = 10 \cdot 0.75 = 7.5$ and $m^B = 10$ .

Step 1
$x^{A*}_1 = \frac{0.6 \cdot 7.5}{0.75} = 6, \qquad x^{A*}_2 = 0.4 \cdot 7.5 = 3$
$A$ sells 4 units of good 1 and buys 3 units of good 2.
Step 2
$x^{B*}_1 = \frac{0.3 \cdot 10}{0.75} = 4, \qquad x^{B*}_2 = 0.7 \cdot 10 = 7$
$B$ buys 4 units of good 1 and sells 3 units of good 2.

x^{A*} = (6,\ 3), \qquad x^{B*} = (4,\ 7)

Walrasian allocation

Step 5 — Gains from trade

At their endowments each agent holds only one good, so Cobb-Douglas utility is zero — you cannot make $x_1^{\alpha} x_2^{1-\alpha}$ positive with $x_2 = 0$ . Trade lifts both strictly above that floor:

Step 1
$u^A(\omega^A) = 10^{0.6}\, 0^{0.4} = 0 \quad\longrightarrow\quad u^A(x^{A*}) = 6^{0.6}\, 3^{0.4} \approx 4.55$
$A$ moves from nothing to a positive bundle of both goods.
Step 2
$u^B(\omega^B) = 0^{0.3}\, 10^{0.7} = 0 \quad\longrightarrow\quad u^B(x^{B*}) = 4^{0.3}\, 7^{0.7} \approx 5.92$
$B$ likewise strictly gains.

Both agents are strictly better off, which is exactly the individual-rationality guarantee from §6: voluntary trade at a common price never leaves anyone worse than not trading.

Seeing it in the box

Measured from $A$ 's origin, the budget line runs through the endowment $\omega^A = (10, 0)$ with slope $-p_1^* = -0.75$ . $A$ 's chosen bundle $x^{A*} = (6, 3)$ sits where $A$ 's indifference curve is tangent to that line — the defining condition $\mathrm{MRS}^A = p_1^*$ . At that point the indifference curve's slope is $-\tfrac{\alpha_A}{1-\alpha_A}\tfrac{x_2}{x_1} = -\tfrac{0.6}{0.4}\cdot\tfrac{3}{6} = -0.75$ , matching the price line exactly.

Agent $A$ : budget line tangent to the indifference curve at $x^{A*}$

The budget line

0.75\,x_1 + x_2 = 7.5

passes through the endowment

\omega^A = (10,0)

A

's utility-maximising bundle

x^{A*} = (6,3)

is the tangency point, where the indifference curve

u^A = 6^{0.6}3^{0.4} \approx 4.55

just touches the line.