predavanja-FisherjevModelTrga.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Fisherjev model trga"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "%display latex"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Primer ($m = 3$, $n = 2$)\n",
    "\n",
    "Podajmo podatke za Fisherjev model trga."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [],
   "source": [
    "a = [20, 60, 100]\n",
    "b = [2, 4]\n",
    "U0 = Matrix([[8, 10], [5, 30], [5, 20]])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Problem bomo rešili z Eisenberg-Galeovim konveksnim programom, v katerem predpostavimo, da imamo vsake dobrine eno enoto. Izračunajmo torej novo matriko zadovoljstev."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
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       "\\end{array}\\right)</script></html>"
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       "\\begin{math}\n",
       "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left(\\begin{array}{rr}\n",
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       "10 & 80\n",
       "\\end{array}\\right)\n",
       "\\end{math}"
      ],
      "text/plain": [
       "[ 16  40]\n",
       "[ 10 120]\n",
       "[ 10  80]"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "U = Matrix([[k*v for k, v in zip(b, u)] for u in U0])\n",
    "U"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Pripravimo si spremenljivke in enačbe ter poiščimo rešitve slednjih."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "m, n = U.dimensions()\n",
    "X = Matrix([[var(f\"x{i}{j}\") for j in range(n)] for i in range(m)])\n",
    "lm = Matrix([[var(f\"lambda{i}{j}\") for j in range(n)] for i in range(m)])\n",
    "p = var(' '.join(f\"p{j}\" for j in range(n)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
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       "<html><script type=\"math/tex; mode=display\">\\newcommand{\\Bold}[1]{\\mathbf{#1}}17</script></html>"
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       "\\begin{math}\n",
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     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
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   ],
   "source": [
    "spremenljivke = [*X.coefficients(), *lm.coefficients(), *p]\n",
    "enacbe = [*(sum(X[i, j] for i in range(m)) == 1 for j in range(n)),\n",
    "          *(lm[i, j] * X[i, j] == 0 for i in range(m) for j in range(n)),\n",
    "          *(p[j] == lm[i, j] + a[i] * U[i, j] / sum(U[i, j] * X[i, j] for j in range(n))\n",
    "            for i in range(m) for j in range(n))]\n",
    "resitve = solve(enacbe, spremenljivke)\n",
    "len(resitve)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Upoštevajmo še, da morajo biti vse spremenljivke nenegativne."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "data": {
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       "<html><script type=\"math/tex; mode=display\">\\newcommand{\\Bold}[1]{\\mathbf{#1}}1</script></html>"
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       "\\begin{math}\n",
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      "text/plain": [
       "1"
      ]
     },
     "execution_count": 6,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "nenegativne = [res for res in resitve if all(r.rhs() >= 0 for r in res)]\n",
    "len(nenegativne)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Imamo torej samo eno rešitev. Izpišimo dobljene vrednosti."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
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       "\\end{array}\\right)</script></html>"
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     "metadata": {},
     "output_type": "display_data"
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   ],
   "source": [
    "res, = nenegativne\n",
    "display(X.subs(res))\n",
    "display(lm.subs(res))\n",
    "display([c.subs(res) for c in p])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Preverimo, da vse enačbe držijo."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<html><script type=\"math/tex; mode=display\">\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}</script></html>"
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      "text/latex": [
       "\\begin{math}\n",
       "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\n",
       "\\end{math}"
      ],
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 8,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "all(e.subs(res) for e in enacbe)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Izpišimo še količine in cene glede na prvotno podani vektor $b$."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
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       "\\begin{math}\n",
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       "0 & \\frac{5}{2}\n",
       "\\end{array}\\right)\n",
       "\\end{math}"
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      "text/plain": [
       "[  2   0]\n",
       "[  0 3/2]\n",
       "[  0 5/2]"
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     },
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     "output_type": "display_data"
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    {
     "data": {
      "text/html": [
       "<html><script type=\"math/tex; mode=display\">\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[10, 40\\right]</script></html>"
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      "text/latex": [
       "\\begin{math}\n",
       "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[10, 40\\right]\n",
       "\\end{math}"
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       "[10, 40]"
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     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "X0 = Matrix([[k*y for k, y in zip(b, x)] for x in X])\n",
    "display(X0.subs(res))\n",
    "display([c.subs(res)/k for k, c in zip(b, p)])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Primer ($m = 2$, $n = 2$)\n",
    "\n",
    "Podajmo podatke za Fisherjev model trga. Tokrat bomo že v začetku predpostavili, da imamo na voljo eno enoto vsake dobrine."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [],
   "source": [
    "a = [2, 1]\n",
    "U = Matrix([[2, 2], [1, 2]])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Pripravimo si spremenljivke in enačbe ter poiščimo rešitve slednjih."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [],
   "source": [
    "m, n = U.dimensions()\n",
    "X = Matrix([[var(f\"x{i}{j}\") for j in range(n)] for i in range(m)])\n",
    "lm = Matrix([[var(f\"lambda{i}{j}\") for j in range(n)] for i in range(m)])\n",
    "p = var(' '.join(f\"p{j}\" for j in range(n)))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<html><script type=\"math/tex; mode=display\">\\newcommand{\\Bold}[1]{\\mathbf{#1}}5</script></html>"
      ],
      "text/latex": [
       "\\begin{math}\n",
       "\\newcommand{\\Bold}[1]{\\mathbf{#1}}5\n",
       "\\end{math}"
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      "text/plain": [
       "5"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "spremenljivke = [*X.coefficients(), *lm.coefficients(), *p]\n",
    "enacbe = [*(sum(X[i, j] for i in range(m)) == 1 for j in range(n)),\n",
    "          *(lm[i, j] * X[i, j] == 0 for i in range(m) for j in range(n)),\n",
    "          *(p[j] == lm[i, j] + a[i] * U[i, j] / sum(U[i, j] * X[i, j] for j in range(n))\n",
    "            for i in range(m) for j in range(n))]\n",
    "resitve = solve(enacbe, spremenljivke)\n",
    "len(resitve)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Upoštevajmo še, da morajo biti vse spremenljivke nenegativne."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<html><script type=\"math/tex; mode=display\">\\newcommand{\\Bold}[1]{\\mathbf{#1}}1</script></html>"
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       "\\newcommand{\\Bold}[1]{\\mathbf{#1}}1\n",
       "\\end{math}"
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      "text/plain": [
       "1"
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "nenegativne = [res for res in resitve if all(r.rhs() >= 0 for r in res)]\n",
    "len(nenegativne)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Spet imamo samo eno rešitev. Izpišimo dobljene vrednosti."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "data": {
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       "[  1 1/3]\n",
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      "text/plain": [
       "[3/2, 3/2]"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "res, = nenegativne\n",
    "display(X.subs(res))\n",
    "display(lm.subs(res))\n",
    "display([c.subs(res) for c in p])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Preverimo, da vse enačbe držijo."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<html><script type=\"math/tex; mode=display\">\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}</script></html>"
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       "\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\mathrm{True}\n",
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      "text/plain": [
       "True"
      ]
     },
     "execution_count": 15,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "all(e.subs(res) for e in enacbe)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Prepričajmo se še, da vsak kupec kupuje samo iz svojega optimalnega svežnja. Izračunajmo torej zadovoljstva na denarno enoto."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
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      "text/plain": [
       "[4/3 4/3]\n",
       "[2/3 4/3]"
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     },
     "execution_count": 16,
     "metadata": {},
     "output_type": "execute_result"
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   ],
   "source": [
    "Matrix([[v/c.subs(res) for c, v in zip(p, u)] for u in U])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Izračunajmo še zadovoljstva obeh kupcev."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<html><script type=\"math/tex; mode=display\">\\newcommand{\\Bold}[1]{\\mathbf{#1}}\\left[\\frac{8}{3}, \\frac{4}{3}\\right]</script></html>"
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      "text/latex": [
       "\\begin{math}\n",
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       "\\end{math}"
      ],
      "text/plain": [
       "[8/3, 4/3]"
      ]
     },
     "execution_count": 17,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "[sum(v*y.subs(res) for y, v in zip(x, u)) for x, u in zip(X, U)]"
   ]
  }
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