The Transportation Problem (simple example)

The Transportation Problem.ipynb

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 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# The Transportation Problem \n",
    "\n",
    "Data Table Representation \n",
    "\n",
    "|          | Warehouse 1 | Warehouse 2 | Warehouse 3 | Supply |\n",
    "| -------- | ----------: : ----------: | ----------: | -----: |\n",
    "|Factory 1 |          30 |          20 |          10 |   100  |\n",
    "|Factory 2 |          25 |          15 |           5 |   200  |\n",
    "| Demand   |         100 |         120 |          80 |        |\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2×3 Array{Int64,2}:\n",
       " 30  20  10\n",
       " 25  15   5"
      ]
     },
     "execution_count": 1,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "cost = [ 30 20 10 ; \n",
    "         25 15  5 ]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2-element Array{Int64,1}:\n",
       " 100\n",
       " 200"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "supply = [100, 200]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3-element Array{Int64,1}:\n",
       " 100\n",
       " 120\n",
       "  80"
      ]
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "demand = [100, 120, 80]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Min 30 x[1,1] + 20 x[1,2] + 10 x[1,3] + 25 x[2,1] + 15 x[2,2] + 5 x[2,3]\n",
      "Subject to\n",
      " factory[1] : x[1,1] + x[1,2] + x[1,3] = 100.0\n",
      " factory[2] : x[2,1] + x[2,2] + x[2,3] = 200.0\n",
      " warehouse[1] : x[1,1] + x[2,1] = 100.0\n",
      " warehouse[2] : x[1,2] + x[2,2] = 120.0\n",
      " warehouse[3] : x[1,3] + x[2,3] = 80.0\n",
      " x[1,1] ≥ 0.0\n",
      " x[2,1] ≥ 0.0\n",
      " x[1,2] ≥ 0.0\n",
      " x[2,2] ≥ 0.0\n",
      " x[1,3] ≥ 0.0\n",
      " x[2,3] ≥ 0.0\n"
     ]
    }
   ],
   "source": [
    "m = 2      # number of factories (suppliers)\n",
    "n = 3      # number of warehouses (demanders)\n",
    "\n",
    "using JuMP, GLPK\n",
    "tp = Model(GLPK.Optimizer)\n",
    "@variable(tp, x[1:m, 1:n] >= 0)\n",
    "@objective(tp, Min, sum(cost[i,j] * x[i,j] for i in 1:m, j in 1:n))\n",
    "@constraint(tp, factory[i in 1:2], sum(x[i,j] for j in 1:n) == supply[i])\n",
    "@constraint(tp, warehouse[j in 1:3], sum(x[i,j] for i in 1:m) == demand[j])\n",
    "print(tp)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "raw_status(tp) = \"Solution is optimal\"\n",
      "objective_value(tp) = 5200.0\n",
      "value.(x) = [100.0 0.0 0.0; 0.0 120.0 80.0]\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "2×3 Array{Float64,2}:\n",
       " 100.0    0.0   0.0\n",
       "   0.0  120.0  80.0"
      ]
     },
     "execution_count": 17,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "optimize!(tp)\n",
    "@show raw_status(tp)\n",
    "@show objective_value(tp)\n",
    "@show value.(x)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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