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Tom added a lecture on input-output analysis; not in toc yet
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lectures/input_output.md

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---
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jupytext:
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text_representation:
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extension: .md
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format_name: myst
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format_version: 0.13
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jupytext_version: 1.14.4
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kernelspec:
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display_name: Python 3 (ipykernel)
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language: python
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name: python3
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---
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(input_output)=
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In this lecture, we will need the following library.
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# Input-Output Models
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## Overview
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We adopt notation in chapters 8 and 9 of the classic book {cite}`DoSSo`.
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We let
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* $X_0$ be the amount of a single exogenous input to production. We'll call this input labor
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* $X_j, j = 1,\ldots n$ be the gross output of final good $j$
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* $C_j, j = 1,\ldots n$ be the net output of final good $j$ that is available for final consumption
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* $x_{ij} $ be the quantity of good $i$ allocated to be an input to producing good $j$ for $i=1, \ldots n$, $j = 1, \ldots n$
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* $x_{0j}$ be the quantity of labor allocated to produce one unit of good $j$.
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* $a_{ij}$ be the number of units of good $i$ required to produce one unit of good $j$, $i=0, \ldots, n, j= 1, \ldots n$.
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* $w >0$ be the exogenous wage of labor, denominated in dollars per unit of labor
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* $p$ be an $n \times 1$ vector of prices of produced goods $i = 1, \ldots , n$.
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The production function for goods $j \in \{1, \ldots , n\}$ is the **Leontief** function
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$$
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X_j = \min_{i \in \{0, \ldots , n \}} \left( \frac{x_{ij}}{a_{ij}}\right)
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$$
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To illustrate ideas, we'll begin by setting $n =2$.
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Feasible allocations must satisfy
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$$
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\begin{aligned}
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(1 - a_{11}) X_1 - a_{12} X_2 & \geq C_1 \cr
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-a_{21} X_1 + (1 - a_{22}) X_2 & \geq C_2 \cr
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a_{01} X_1 + a_{02} X_2 & \leq X_0
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\end{aligned}
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$$
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or more generally
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$$
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\begin{aligned}
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(I - a) X & \geq C \cr
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a_0^\top X & \leq X_0
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\end{aligned}
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$$ (eq:inout_1)
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where $a$ is the $n \times n$ matrix with typical element $a_{ij}$ and $a_0^\top = \begin{bmatrix} a_{01} & \cdots & a_{0n} \end{bmatrix}$.
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If we solve the first block of equations of {eq}`eq:inout_1` for gross output $X$ we get
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$$
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X = (I -a )^{-1} C \equiv A C
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$$ (eq:inout_2)
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where $A = (I-a)^{-1}$.
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The coefficient $A_{ij} $ is the amount of good $i$ that is required as an intermediate input to produce one unit of final output $j$.
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We assume the **Hawkins-Simon condition**
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$$
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\det (I - a) > 0
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$$
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to assure that the solution $X$ of {eq}`eq:inout_2` is a positive vector.
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## Production Possibility Frontier
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The second equation of {eq}`eq:inout_1` can be written
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$$
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a_0^\top X = X_0
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$$
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or
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$$
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A_0^\top C = X_0
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$$ (eq:inout_frontier)
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where
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$$
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A_0^\top = a_0^\top (I - a)^{-1}
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$$
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The $i$th Component $A_0$ is the amount of labor that is required to produce one unit of final output of good $i$ for $i \in \{1, \ldots , n\}$.
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Equation {eq}`eq:inout_frontier` sweeps out a **production possibility frontier** of final consumption bundles $C$ that can be produced with exogenous labor input $X_0$.
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## Prices
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{cite}`DoSSo` argue that relative prices of the $n$ produced goods must satisfy
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$$
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p = a^\top p + a_0 w
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$$
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which states that the price of each final good equals the total cost
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of production, which consists of costs of intermediate inputs $a^\top p$
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plus costs of labor $a_0 w$.
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This equation can be written as
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$$
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(I - a^\top) p = a_0 w
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$$ (eq:inout_price)
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which implies
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$$
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p = (I - a^\top)^{-1} a_0 w
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$$
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Notice how {eq}`eq:inout_price` with {eq}`eq:inout_1` form a
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**conjugate pair** through the appearance of operators
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that are transposes of one another.
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This connection surfaces again in a classic linear program and its dual.
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## Linear Programs
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A **primal** problem is
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$$
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\min_{X} w a_0 ^\top X
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$$
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subject to
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$$
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(I -a ) X \geq C
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$$
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The associated **dual** problem is
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$$
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\max_{p} p^\top C
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$$
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subject to
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$$
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(I -a)^\top p \leq a_0 w
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$$
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The primal problem chooses a feasible production plan to minimize costs for delivering a pre-assigned vector of final goods consumption $C$.
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The dual problem chooses prices to maxmize the value of a pre-assigned vector of final goods $C$ subject to prices covering costs of production.
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Under sufficient conditions discussed XXXX, optimal value of the primal and dual problems coincide:
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$$
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w a_0^\top X^* = p^* C
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$$
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where $^*$'s denote optimal choices for the primal and dual problems.
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+++
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## Exercise
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{cite}`DoSSo`, chapter 9, carries along an example with the following
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parameter settings:
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$$
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a = \begin{bmatrix} .1 & 1.46 \cr
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.16 & .17 \end{bmatrix}
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$$
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$$
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a_0 = \begin{bmatrix} .04 & .33 \end{bmatrix}
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$$
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$$
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C = \begin{bmatrix} 50 \cr 60 \end{bmatrix}
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$$
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$$
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X_0 = \begin{bmatrix} 250 \cr 120 \end{bmatrix}
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$$
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$$
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X = 50
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$$
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```{code-cell} ipython3
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```
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```{code-cell} ipython3
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```

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