mathematical_programming/production_and_inventory.ipynb

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   "source": [
    "Dependencies:\n",
    "\n",
    "numpy: 1.21.0\\\n",
    "cvxpy: 1.1.14\\\n",
    "json : 2.0.9"
   ]
  },
  {
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   "id": "cb715525",
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   "source": [
    "* Demands: \n",
    "    * Day 1: 75\n",
    "    * Day 2: 150\n",
    "    * Day 3: 100\n",
    "    * Day 4: 60\n",
    "* Unit production cost:\n",
    "    * Day 1: \\$4\n",
    "    * Day 2: \\$5\n",
    "    * Day 3: \\$7\n",
    "    * Day 4: \\$4\n",
    "* Inventory cost: $1 per unit per day\n",
    "\n",
    "How many units should we produce per day to minimize the total cost?"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b8c33244",
   "metadata": {},
   "source": [
    "# Problem statement"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a37c25a9",
   "metadata": {},
   "source": [
    "**Input parameters**\n",
    "\n",
    "$d_t$: demand in day $t$, $t=1,..., 4$\n",
    "\n",
    "$c_t$: unit cost in day $t$, $t=1,..., 4$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 136,
   "id": "505ceda3",
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     "start_time": "2021-07-31T18:16:11.523551Z"
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   },
   "outputs": [],
   "source": [
    "d = [75, 150, 100, 60]\n",
    "c = [4, 5, 7, 4]"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "aa3743ce",
   "metadata": {},
   "source": [
    "**Decision variables**:\n",
    "\n",
    "$q_t$: production quantity of day $t$, $t=1, ..., 4$\n",
    "\n",
    "$e_t$: ending inventory of day $t, t=0,1, ..., 4$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 137,
   "id": "799fd25e",
   "metadata": {
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     "start_time": "2021-07-31T18:16:11.975751Z"
    }
   },
   "outputs": [],
   "source": [
    "import cvxpy as cp"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 138,
   "id": "2b139ffb",
   "metadata": {
    "ExecuteTime": {
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   "outputs": [],
   "source": [
    "t = 4\n",
    "q = cp.Variable(t, integer=True)\n",
    "e = cp.Variable(t+1, integer=True)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "8c880a93",
   "metadata": {},
   "source": [
    "**Objectives:**\n",
    "\n",
    "\\begin{align}\n",
    "\\min \\sum_{t=1}^4 c_t q_t + e_t \\times 1\n",
    "\\end{align}"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2045d720",
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   "source": [
    "$$e_t \\times 1 = e_t$$"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "a3e30410",
   "metadata": {},
   "source": [
    "$$\n",
    "\\sum_{t=1}^4 c_t q_t = 4\\times75 + 8 \\times 150 + 12 \\times 100 + 7 \\times 60 \\\\\n",
    "= \\begin{bmatrix}\n",
    "4& 8& 12& 7\n",
    "\\end{bmatrix} \\begin{bmatrix}\n",
    "75\\\\ \n",
    "150\\\\ \n",
    "100\\\\\n",
    "60\n",
    "\\end{bmatrix} \\\\\n",
    "$$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 139,
   "id": "ded6987e",
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     "start_time": "2021-07-31T18:16:12.969432Z"
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   "outputs": [],
   "source": [
    "obj = cp.Minimize(c @ q + sum(e))"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "261f4290",
   "metadata": {},
   "source": [
    "**Constraints:**\n",
    "\n",
    "The ending inventory is equal to the production quantity of day $t$ subtract the demand in day $t$:\n",
    "\n",
    "$$\n",
    "e_1 = e_0 + q_1 - d_1\\\\\n",
    "e_2 = e_1 + q_2 - d_2\\\\\n",
    "$$\n",
    "In general, we have:\n",
    "$$\n",
    "e_t = e_{t-1} + q_t - d_t \\;\\; \\forall t = 1, 2, 3, 4\\\\\n",
    "$$\n",
    "\n",
    "The number of units in inventory at the beginning of day 1 is 0:\n",
    "$$e_0 = 0$$\n",
    "\n",
    "The number of units in inventory cannot be negative:\n",
    "$$e_t \\geq 0\\;\\; \\forall t = 0, 1, 2, 3, 4$$\n",
    "\n",
    "Production quantitiy cannot be negative:\n",
    "$$q_t \\geq 0\\;\\; \\forall t = 1, 2, 3, 4$$"
   ]
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   "cell_type": "code",
   "execution_count": 155,
   "id": "7d5f27cd",
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   "metadata": {},
   "source": [
    "# Solve"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 141,
   "id": "46127637",
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     "end_time": "2021-07-31T18:16:13.786917Z",
     "start_time": "2021-07-31T18:16:13.783567Z"
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   "outputs": [],
   "source": [
    "prob = cp.Problem(obj, constraints)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 142,
   "id": "6a73a396",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-07-31T18:16:14.045808Z",
     "start_time": "2021-07-31T18:16:14.026728Z"
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       "1890.0"
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     "execution_count": 142,
     "metadata": {},
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   "source": [
    "prob.solve()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 143,
   "id": "cb5f81cc",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-07-31T18:16:14.244944Z",
     "start_time": "2021-07-31T18:16:14.240190Z"
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   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "'optimal'"
      ]
     },
     "execution_count": 143,
     "metadata": {},
     "output_type": "execute_result"
    }
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   "source": [
    "prob.status"
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  {
   "cell_type": "code",
   "execution_count": 144,
   "id": "a845f9d1",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-07-31T18:16:14.557757Z",
     "start_time": "2021-07-31T18:16:14.553650Z"
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   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[ 75. 250.   0.  60.]\n"
     ]
    }
   ],
   "source": [
    "print(q.value)"
   ]
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   "cell_type": "code",
   "execution_count": 145,
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   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[  0.   0. 100.   0.   0.]\n"
     ]
    }
   ],
   "source": [
    "print(e.value)"
   ]
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  {
   "cell_type": "markdown",
   "id": "3e208865",
   "metadata": {},
   "source": [
    "# Experiment with Other Possible Combinations"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 146,
   "id": "ff43ea41",
   "metadata": {
    "ExecuteTime": {
     "end_time": "2021-07-31T18:16:17.857141Z",
     "start_time": "2021-07-31T18:16:17.853793Z"
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   },
   "outputs": [],
   "source": [
    "import numpy as np"
   ]
  },
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   "cell_type": "code",
   "execution_count": 147,
   "id": "ffad7ffc",
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    "ExecuteTime": {
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   },
   "outputs": [],
   "source": [
    "def calculate_inventory(demands: np.array, quantities: np.array):\n",
    "    \"\"\"Calculate the ending inventory based on the demand and the production quantities for each day\"\"\"\n",
    "\n",
    "    inventory = []\n",
    "    new_inventory = 0\n",
    "    for i in range(len(quantities)):\n",
    "        new_inventory = new_inventory + quantities[i] - demands[i]\n",
    "        inventory.append(new_inventory)\n",
    "    return inventory"
   ]
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    "demands = np.array([75, 150, 100, 60])\n",
    "cost = np.array([4, 5, 7, 4])\n",
    "quantities = np.array([385, 0, 0, 0])\n",
    "get_cost(demands, cost, quantities)"
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  {
   "cell_type": "markdown",
   "id": "45bdd1af",
   "metadata": {},
   "source": [
    "Produce everything in the first day and second day"
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    "quantities = np.array([300, 85, 0, 0])\n",
    "get_cost(demands, cost, quantities)"
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  {
   "cell_type": "markdown",
   "id": "4cc2f33a",
   "metadata": {},
   "source": [
    "Produce just enough to fulfill the demand for each day"
   ]
  },
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   "cell_type": "code",
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   "id": "0a230f34",
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     "execution_count": 159,
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    "quantities = np.array([75, 150, 100, 60])\n",
    "get_cost(demands, cost, quantities)"
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  {
   "cell_type": "markdown",
   "id": "97899801",
   "metadata": {},
   "source": [
    "Produce everything in the first day and the last day"
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   "source": [
    "quantities = np.array([325, 0, 0, 60])\n",
    "get_cost(demands, cost, quantities)"
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