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        "```{index} single: solver; cbc\n",
        "```\n",
        "```{index} single: solver; highs\n",
        "```\n",
        "\n",
        "# BIM production for worst case"
      ]
    },
    {
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      "execution_count": 1,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
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      "source": [
        "# install dependencies and select solver\n",
        "%pip install -q amplpy\n",
        "\n",
        "SOLVER = \"highs\"\n",
        "\n",
        "from amplpy import AMPL, ampl_notebook\n",
        "\n",
        "ampl = ampl_notebook(\n",
        "    modules=[\"highs\"],  # modules to install\n",
        "    license_uuid=\"default\",  # license to use\n",
        ")  # instantiate AMPL object and register magics"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "## Minmax objective function\n",
        "\n",
        "Another class of seemingly complicated objective functions that can be easily rewritten as an LP are those stated as maxima over several linear functions. Given a finite set of indices $K$ and a collection of vectors $\\{c_k\\}_{k \\in K}$, the minimax problem given by\n",
        "\n",
        "$$\n",
        "\\begin{align}\n",
        "        \\min \\; \\max_{k \\in K} \\; c^\\top_{k} x\n",
        "\\end{align}\n",
        "$$\n",
        "\n",
        "General expressions like the latter can be linearized by introducing an auxiliary variable $z$ and setting\n",
        "\n",
        "$$\n",
        "\\begin{align*}\n",
        "    \\min \\quad & z  \\\\\n",
        "    \\text{s.t.} \\quad & c^\\top_{k} x \\leq z \\qquad \\forall\\, k \\in K.\n",
        "\\end{align*}\n",
        "$$\n",
        "\n",
        "This trick works because if *all* the quantities corresponding to different indices $ k \\in K$ are below the auxiliary variable $z$, then we are guaranteed that also their maximum is also below $z$ and vice versa. Note that the absolute value function can be rewritten $|x_i|= \\max\\{x_i,-x_i\\}$, hence the linearization of the optimization problem involving absolute values in the objective functions is a special case of this. \n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "## BIM problem variant: Maximizing the lowest possible profit\n",
        "\n",
        "In the same way we can minimize a maximum like above, we can also maximize the minimum. Let us consider the [BIM microchip production problem](bim.ipynb), but suppose that there is uncertainty regarding the selling prices of the microchips. Instead of just the nominal prices 12 € and 9 €, BIM estimates that the prices may more generally take the values $P=\\{ (12,9), (11,10), (8, 11) \\}$. The optimization problem for a production plan that achieves the maximum among the lowest possible profits can be formulated using the trick mentioned above and can be implemented in AMPL as follows."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 2,
      "metadata": {},
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "Writing bim_maxmin.mod\n"
          ]
        }
      ],
      "source": [
        "%%writefile bim_maxmin.mod\n",
        "\n",
        "var x1 >= 0;\n",
        "var x2 >= 0;\n",
        "var z;\n",
        "\n",
        "set costs dimen 2;\n",
        "\n",
        "maximize profit: z;\n",
        "    \n",
        "s.t. maxmin {(c1,c2) in costs}:\n",
        "    z <= c1 * x1 + c2 * x2;\n",
        "\n",
        "s.t. silicon: x1 <= 1000;\n",
        "s.t. germanium: x2 <= 1500;\n",
        "s.t. plastic: x1 + x2 <= 1750;\n",
        "s.t. copper: 4 * x1 + 2 * x2 <= 4800;"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 3,
      "metadata": {
        "id": "m33AGCU_PSJw"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "HiGHS 1.5.1: HiGHS 1.5.1: optimal solution; objective 17500\n",
            "4 simplex iterations\n",
            "0 barrier iterations\n",
            "x=(583.3,1166.7) revenue=17500.000\n"
          ]
        }
      ],
      "source": [
        "def BIM_maxmin(costs):\n",
        "    ampl = AMPL()\n",
        "    ampl.read(\"bim_maxmin.mod\")\n",
        "\n",
        "    ampl.set[\"costs\"] = costs\n",
        "\n",
        "    return ampl\n",
        "\n",
        "\n",
        "m = BIM_maxmin([[12, 9], [11, 10], [8, 11]])\n",
        "m.solve(solver=SOLVER)\n",
        "assert m.solve_result == \"solved\", m.solve_result\n",
        "print(\n",
        "    \"x=({:.1f},{:.1f}) revenue={:.3f}\".format(\n",
        "        m.var[\"x1\"].value(), m.var[\"x2\"].value(), m.obj[\"profit\"].value()\n",
        "    )\n",
        ")"
      ]
    }
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