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   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Time optimal control problem"
     ]
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "In this example we will set-up a simple 4-state system,\n",
      "and find a time-optimal trajectory that avoids some obstacles in the world."
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Problem setup"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "%matplotlib inline"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 1
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from casadi import *\n",
      "from pylab import *\n",
      "from optoy import *\n",
      "\n",
      "# Time\n",
      "t = time()\n",
      "\n",
      "# An optimization variable for the end time\n",
      "T = var(lb=0,init=4)\n",
      "\n",
      "# States: position and velocity\n",
      "p = state(2,init=vertcat([3*sin(2*pi/T.init*t),3*cos(2*pi/T.init*t)]))\n",
      "v = state(2)\n",
      "\n",
      "# Control\n",
      "u = control(2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 2
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Specify the system dynamics as ODE"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "p.dot = v\n",
      "v.dot = -10*(p-u)-v*sqrt(sum_square(v)+1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 3
    },
    {
     "cell_type": "markdown",
     "metadata": {},
     "source": [
      "Set up the path constraints"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Specify some parameters for circular obstacles\n",
      "#            ( position,      radius)\n",
      "#\n",
      "circles = [  (vertcat([2,2]),      1),\n",
      "             (vertcat([0.5,-2]), 1.5),\n",
      "          ]\n",
      "\n",
      "# List of path constraints\n",
      "h = []\n",
      "for center, radius in circles:\n",
      "  h.append(\n",
      "     norm_2(p-center) >= radius  # Don't hit the obstacles\n",
      "  )"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Solve the OCP problem"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "ocp(T,h+[p.start[0]==0],regularize=[0.1*u/sqrt(2)],N=30,T=T,verbose=True,periodic=True,integration_intervals=2)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "(1, 1)\n",
        "\n",
        "******************************************************************************\n",
        "This program contains Ipopt, a library for large-scale nonlinear optimization.\n",
        " Ipopt is released as open source code under the Eclipse Public License (EPL).\n",
        "         For more information visit http://projects.coin-or.org/Ipopt\n",
        "******************************************************************************\n",
        "\n"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "This is Ipopt version 3.11.9, running with linear solver mumps.\n",
        "NOTE: Other linear solvers might be more efficient (see Ipopt documentation).\n",
        "\n",
        "Number of nonzeros in equality constraint Jacobian...:      969\n",
        "Number of nonzeros in inequality constraint Jacobian.:      120\n",
        "Number of nonzeros in Lagrangian Hessian.............:      811\n",
        "\n",
        "Total number of variables............................:      185\n",
        "                     variables with only lower bounds:        1\n",
        "                variables with lower and upper bounds:        0\n",
        "                     variables with only upper bounds:        0\n",
        "Total number of equality constraints.................:      125\n",
        "Total number of inequality constraints...............:       60\n",
        "        inequality constraints with only lower bounds:        0\n",
        "   inequality constraints with lower and upper bounds:        0\n",
        "        inequality constraints with only upper bounds:       60\n",
        "\n",
        "iter    objective    inf_pr   inf_du lg(mu)  ||d||  lg(rg) alpha_du alpha_pr  ls\n",
        "   0  4.0000000e+00 3.28e+00 6.13e-01  -1.0 0.00e+00    -  0.00e+00 0.00e+00   0\n",
        "   1  3.9721767e+00 3.22e+00 5.57e+01  -1.0 5.29e+00   0.0 9.30e-01 1.91e-02h  1"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "   2  4.5743296e+00 1.71e+00 4.11e+01  -1.0 6.38e+00  -0.5 3.17e-01 1.00e+00h  1\n",
        "   3  4.5160620e+00 4.78e-02 2.89e+01  -1.0 1.29e+00  -0.1 5.57e-01 1.00e+00h  1\n",
        "   4  4.5387073e+00 1.18e-02 6.84e+00  -1.0 5.09e-01  -0.5 8.05e-01 1.00e+00f  1"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "   5  4.5008564e+00 2.37e-02 2.60e+00  -1.0 6.10e-01  -1.0 4.87e-01 1.00e+00f  1\n",
        "   6  4.3904478e+00 4.20e-02 8.12e-01  -1.0 8.20e-01  -1.5 6.32e-01 1.00e+00f  1\n",
        "   7  3.8063326e+00 7.62e-02 8.91e-02  -1.7 1.02e+00  -2.0 1.00e+00 1.00e+00h  1"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "   8  3.1482676e+00 1.44e-01 2.28e-02  -2.5 1.53e+00  -2.4 1.00e+00 8.52e-01h  1\n",
        "   9  2.7625842e+00 2.69e-01 1.38e-02  -2.5 3.64e+00  -2.9 9.47e-01 8.03e-01h  1\n",
        "iter    objective    inf_pr   inf_du lg(mu)  ||d||  lg(rg) alpha_du alpha_pr  ls\n"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "  10  2.6035215e+00 4.97e-01 2.95e-02  -2.5 5.41e+00  -3.4 5.34e-01 6.97e-01h  1\n",
        "  11  2.5788066e+00 5.84e-01 4.58e-02  -2.5 7.11e+01    -  4.13e-02 3.11e-02h  1\n",
        "  12  2.5527467e+00 2.96e-01 3.33e-02  -2.5 3.74e+00  -3.9 7.11e-01 1.00e+00h  1\n",
        "  13  2.5597431e+00 9.62e-02 1.67e-02  -2.5 2.31e+00    -  1.00e+00 1.00e+00h  1"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "  14  2.5598411e+00 1.97e-03 7.88e-04  -2.5 4.93e-01    -  1.00e+00 1.00e+00h  1\n",
        "  15  2.5334465e+00 4.00e-02 1.75e-02  -3.8 2.38e+00    -  9.22e-01 5.69e-01h  1\n",
        "  16  2.5176035e+00 3.65e-02 1.16e-02  -3.8 1.31e+00    -  1.00e+00 7.97e-01h  1"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "  17  2.5117268e+00 9.60e-03 1.70e-03  -3.8 6.50e-01    -  1.00e+00 9.45e-01h  1\n",
        "  18  2.5107429e+00 3.37e-04 7.40e-05  -3.8 1.79e-01    -  1.00e+00 1.00e+00h  1\n",
        "  19  2.5090323e+00 6.61e-04 1.69e-03  -5.7 2.13e-01    -  9.95e-01 8.63e-01h  1"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "iter    objective    inf_pr   inf_du lg(mu)  ||d||  lg(rg) alpha_du alpha_pr  ls\n",
        "  20  2.5087117e+00 5.08e-05 1.15e-04  -5.7 5.52e-02    -  1.00e+00 9.83e-01h  1\n",
        "  21  2.5086992e+00 5.20e-07 5.68e-07  -5.7 1.21e-02    -  1.00e+00 1.00e+00f  1\n",
        "  22  2.5086749e+00 1.34e-07 3.10e-06  -8.6 4.57e-03    -  1.00e+00 9.94e-01h  1"
       ]
      },
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "\n",
        "  23  2.5086747e+00 7.71e-11 6.55e-11  -8.6 1.65e-04    -  1.00e+00 1.00e+00h  1\n",
        "\n",
        "Number of Iterations....: 23\n",
        "\n",
        "                                   (scaled)                 (unscaled)\n",
        "Objective...............:   2.5086746746469544e+00    2.5086746746469544e+00\n",
        "Dual infeasibility......:   6.5457194679976140e-11    6.5457194679976140e-11\n",
        "Constraint violation....:   7.7104544971007272e-11    7.7104544971007272e-11\n",
        "Complementarity.........:   2.5613845271550734e-09    2.5613845271550734e-09\n",
        "Overall NLP error.......:   2.5613845271550734e-09    2.5613845271550734e-09\n",
        "\n",
        "\n",
        "Number of objective function evaluations             = 24\n",
        "Number of objective gradient evaluations             = 24\n",
        "Number of equality constraint evaluations            = 24\n",
        "Number of inequality constraint evaluations          = 24\n",
        "Number of equality constraint Jacobian evaluations   = 24\n",
        "Number of inequality constraint Jacobian evaluations = 24\n",
        "Number of Lagrangian Hessian evaluations             = 23\n",
        "Total CPU secs in IPOPT (w/o function evaluations)   =      0.064\n",
        "Total CPU secs in NLP function evaluations           =      0.472\n",
        "\n",
        "EXIT: Optimal Solution Found.\n",
        "                   proc           wall      num           mean             mean\n",
        "                   time           time     evals       proc time        wall time\n",
        "       eval_f     0.013 [s]      0.013 [s]    24       0.53 [ms]        0.53 [ms]\n",
        "  eval_grad_f     0.016 [s]      0.016 [s]    25       0.65 [ms]        0.65 [ms]\n",
        "       eval_g     0.015 [s]      0.015 [s]    24       0.61 [ms]        0.61 [ms]\n",
        "   eval_jac_g     0.133 [s]      0.133 [s]    26       5.10 [ms]        5.10 [ms]\n",
        "       eval_h     0.324 [s]      0.324 [s]    24      13.50 [ms]       13.49 [ms]\n",
        "    main loop     0.545 [s]      0.543 [s]\n"
       ]
      },
      {
       "metadata": {},
       "output_type": "pyout",
       "prompt_number": 5,
       "text": [
        "2.5086746746469544"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Plotting of results"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "# Plot the nominal trajectory\n",
      "plot(value(p[0]),value(p[1]),'o-')\n",
      "xlabel('x position')\n",
      "ylabel('y position')\n",
      "title('time-optimal trajectory')\n",
      "axis('equal')\n",
      "\n",
      "# Plot the obstacles\n",
      "theta = linspace(0,2*pi,1000)\n",
      "for center, radius in circles:\n",
      "  fill(radius*cos(theta) + center[0],radius*sin(theta) + center[1],'r') "
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "metadata": {},
       "output_type": "display_data",
       "png": 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       "text": [
        "<matplotlib.figure.Figure at 0x7fcd97017810>"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [],
     "language": "python",
     "metadata": {},
     "outputs": []
    }
   ],
   "metadata": {}
  }
 ]
}