Fix extra notebooks from chapter 3 + LP relaxation notation convention

notebooks/02/bim-rawmaterialplanning.ipynb

@@ -659,7 +659,7 @@
     "\n",
     "where $(\\Omega_p)_{p \\in P}$ is the vector with the desired end inventories.\n",
     "\n",
-    "Here is the Pyomo implementation of this LP."
+    "Here is the Pyomo implementation of this linear problem."
    ]
   },
   {

notebooks/03/bim-production-revisited.ipynb

@@ -57,7 +57,7 @@
     "    \n",
     "else:\n",
     "    from pyomo.environ import SolverFactory\n",
-    "    SOLVER = SolverFactory('appsi_highs')\n",
+    "    SOLVER = SolverFactory('cbc')\n",
     "\n",
     "assert SOLVER.available(), f\"Solver {SOLVER} is not available.\""
    ]
@@ -574,7 +574,7 @@
     "    unitary_products,\n",
     "    unitary_holding_costs,\n",
     "):\n",
-    "    m = pyo.ConcreteModel(\"Product acquisition and inventory with sophisticated prices\")\n",
+    "    m = pyo.ConcreteModel(\"BIM product acquisition and inventory problem\")\n",
     "\n",
     "    periods = demand.columns\n",
     "    products = demand.index\n",
@@ -2829,7 +2829,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.11.2"
+   "version": "3.10.10"
   }
  },
  "nbformat": 4,

notebooks/03/cryptarithms.ipynb

@@ -72,9 +72,9 @@
     "\n",
     "There are several possible approaches to modeling this puzzle in Pyomo. \n",
     "\n",
-    "[One approach](https://stackoverflow.com/questions/67456379/pyomo-model-constraint-programming-for-sendmore-money-task)  would be to using a matrix of binary variables $x_{a,d}$ indexed by letter $a$ and digit $d$ such that $x_{a,d} = 1$ designates the corresponding assignment. The problem constraints can then be implemented by summing the binary variables along the two axes. The arithmetic constraint becomes a more challenging.\n",
+    "[One approach](https://stackoverflow.com/questions/67456379/pyomo-model-constraint-programming-for-sendmore-money-task) would be to using a matrix of binary variables $x_{a,d}$ indexed by letter $a$ and digit $d$ such that $x_{a,d} = 1$ designates the corresponding assignment. The problem constraints can then be implemented by summing the binary variables along the two axes. The arithmetic constraint becomes a more challenging.\n",
     "\n",
-    "[Another approach](https://www.gecode.org/doc/6.0.1/MPG.pdf) is to use Pyomo integer variables indexed by letters, then setup an linear expression to represent the puzzle. If we use the notation $n_a$ to represent the digit assigned to letter $a$, the algebraic constraint becomes\n",
+    "[Another approach](https://www.gecode.org/doc/6.0.1/MPG.pdf) is to use Pyomo integer variables indexed by letters, then set up a linear expression to represent the puzzle. If we use the notation $n_a$ to represent the digit assigned to letter $a$, the algebraic constraint becomes\n",
     "\n",
     "$$\n",
     "\\begin{align*}\n",
@@ -97,7 +97,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": 2,
+   "execution_count": 11,
    "metadata": {
     "colab": {
      "base_uri": "https://localhost:8080/"
@@ -121,6 +121,17 @@
      "name": "stdout",
      "output_type": "stream",
      "text": [
+      "The solution of the puzzle is:\n",
+      "S = 9\n",
+      "E = 5\n",
+      "N = 6\n",
+      "D = 7\n",
+      "M = 1\n",
+      "O = 0\n",
+      "R = 8\n",
+      "Y = 2\n",
+      "\n",
+      "\n",
       "     9 5 6 7\n",
       "  +  1 0 8 5\n",
       "  ----------\n",
@@ -132,7 +143,7 @@
     "import pyomo.environ as pyo\n",
     "import pyomo.gdp as gdp\n",
     "\n",
-    "m = pyo.ConcreteModel()\n",
+    "m = pyo.ConcreteModel(\"Cryptarithms Problem\")\n",
     "\n",
     "m.LETTERS = pyo.Set(initialize=[\"S\", \"E\", \"N\", \"D\", \"M\", \"O\", \"R\", \"Y\"])\n",
     "m.PAIRS = pyo.Set(initialize=m.LETTERS * m.LETTERS, filter=lambda m, a, b: a < b)\n",
@@ -170,21 +181,25 @@
     "    return [m.n[a] >= m.n[b] + 1, m.n[b] >= m.n[a] + 1]\n",
     "\n",
     "\n",
-    "# assign a \"dummy\" objective to avoid solver errors\n",
+    "# assign a \"dummy\" constant objective to avoid solver errors\n",
     "@m.Objective()\n",
     "def dummy_objective(m):\n",
-    "    return m.n[\"M\"]\n",
+    "    return 0\n",
     "\n",
     "\n",
     "pyo.TransformationFactory(\"gdp.bigm\").apply_to(m)\n",
     "SOLVER.solve(m)\n",
     "\n",
+    "print(\"The solution of the puzzle is:\")\n",
+    "for l in m.LETTERS:\n",
+    "    print(f\"{l} = {int(m.n[l]())}\")\n",
+    "\n",
     "\n",
     "def letters2num(s):\n",
     "    return \" \".join(map(lambda s: f\"{int(m.n[s]())}\", list(s)))\n",
     "\n",
     "\n",
-    "print(\"    \", letters2num(\"SEND\"))\n",
+    "print(\"\\n\\n    \", letters2num(\"SEND\"))\n",
     "print(\"  + \", letters2num(\"MORE\"))\n",
     "print(\"  ----------\")\n",
     "print(\"= \", letters2num(\"MONEY\"))"
@@ -202,13 +217,6 @@
     "\n",
     "2. There are [many more examples](http://cryptarithms.awardspace.us/puzzles.html) of cryptarithm puzzles. Refactor this code and create a function that can be used to solve generic puzzles of this type."
    ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
   }
  ],
  "metadata": {
@@ -233,7 +241,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.9.16"
+   "version": "3.10.10"
   }
  },
  "nbformat": 4,