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+{
+ "metadata": {
+  "name": "",
+  "signature": "sha256:12638b4b6feef6ce42affa16d2293ad388bb21932c76952667badfebd6be692b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter 11 : Distribution System Reliability"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.1  Page No : 598"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "Rsys = 0.99         #Minimum Acceptable System Reliabilty\n",
+      "n = 15.;         #Number of identical Components\n",
+      "\n",
+      "# Calculations\n",
+      "q = (1-Rsys)/n;         #Probability of component failure\n",
+      "Ri = 1-q;         #Approximate value of the component reliability\n",
+      "\n",
+      "# Results\n",
+      "print 'The Approximate Value of The component reliability is %g'%(Ri)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Approximate Value of The component reliability is 0.999333\n"
+       ]
+      }
+     ],
+     "prompt_number": 1
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.2  Page No : 606"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Variables\n",
+      "L = 4.;         #Total Length of the cable\n",
+      "Lov = 3.;        #Length of Overhead Cable\n",
+      "Lu = L-Lov;         #Length of Underground Cable\n",
+      "Nct = 2.;         #Number of circuit terminations\n",
+      "T = 10.;         #No of years for which the record is shown\n",
+      "\n",
+      "Fov = 2.;         # Faults Per Mile of the Over Head Cable\n",
+      "Fu = 1.;         #Faults Per Mile of The Underground cable\n",
+      "\n",
+      "Ct = 0.3/100        # Cable Termination Fault Rate\n",
+      "\n",
+      "#Repair Time\n",
+      "Tov = 3.;         #Over Head\n",
+      "Tu = 28.;         #Underground\n",
+      "Tct = 3.;         #Cable Termination\n",
+      "\n",
+      "# Calculations\n",
+      "lamdaFDR =  (Lov*Fov/T)+(Lu*Fu/T)+(2*Ct);         #Total Annual Fault Rate\n",
+      "rFDR = ((Tov*Lov*Fov/T)+(Tu*Lu*Fu/T)+(2*Ct*Tct))/lamdaFDR;         #Annual Fault Restoration Time\n",
+      "mFDR = 8760-rFDR;         #Annual Mean Time of Failure\n",
+      "UFDR = rFDR*100/(rFDR+mFDR);         #Unavailability of Feeder\n",
+      "AFDR = 100-UFDR;         #Availability of Feeder\n",
+      "\n",
+      "# Results\n",
+      "print 'a) The Total Annual Fault Rate is %g faults per year'%(lamdaFDR)\n",
+      "print 'b) The Annual Fault Restoration Time is %g hours per fault per year'%(rFDR)\n",
+      "print 'c) Unavailability of the feeder is %g percent'%(UFDR)\n",
+      "print 'd) Availability of the feeder is %g percent'%(AFDR)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Total Annual Fault Rate is 0.706 faults per year\n",
+        "b) The Annual Fault Restoration Time is 6.54108 hours per fault per year\n",
+        "c) Unavailability of the feeder is 0.0746698 percent\n",
+        "d) Availability of the feeder is 99.9253 percent\n"
+       ]
+      }
+     ],
+     "prompt_number": 2
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.3  Page No : 608"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "#Annual average Fault rates\n",
+      "Fm = 0.08;\n",
+      "Fl = 0.2;\n",
+      "\n",
+      "\n",
+      "#Average Repair Times\n",
+      "Rm = 3.5;         #Main\n",
+      "Rl = 1.5;         #Lateral\n",
+      "Rs = 0.75;        #Manual Sections\n",
+      "\n",
+      "# Dismath.tances of the Lateral Feeders of A,B, and C respectively\n",
+      "Lla = 2.;\n",
+      "Llb = 1.5;\n",
+      "Llc = 1.5;\n",
+      "\n",
+      "# Dismath.tances of the Main Feeders of A,B, and C respectively\n",
+      "Lma = 1.;\n",
+      "Lmb = 1.;\n",
+      "Lmc = 1.;\n",
+      "\n",
+      "# Calculations\n",
+      "TFm = (Lma*Fm)+(Lmc*Fm)+(Lmb*Fm);         #Annual Fault of the Main Sections\n",
+      "\n",
+      "def SusInt(y): \n",
+      "    return TFm+(Fl*y)\n",
+      "\n",
+      "#Sustained Interruption Rates for A,B and C\n",
+      "IrA = SusInt(Lla);\n",
+      "IrB = SusInt(Llb);\n",
+      "IrC = SusInt(Llc);\n",
+      "\n",
+      "#Annual Repair time for A,B and C\n",
+      "rA = ((Lma*Fm*Rm)+(Lmb*Fm*Rs)+(Lmc*Fm*Rs)+(Lla*Fl*Rl))/IrA;\n",
+      "rB = ((Lma*Fm*Rm)+(Lmb*Fm*Rm)+(Lmc*Fm*Rs)+(Llb*Fl*Rl))/IrB;\n",
+      "rC = ((Lma*Fm*Rm)+(Lmb*Fm*Rm)+(Lmc*Fm*Rm)+(Llc*Fl*Rl))/IrC;\n",
+      "\n",
+      "# Results\n",
+      "print 'i The Annual Sustained Interruption Rates for:'\n",
+      "print 'Customer A : %g faults per year'%(IrA)\n",
+      "print 'Customer B : %g faults per year'%(IrB)\n",
+      "print 'Customer C : %g faults per year'%(IrC)\n",
+      "print 'ii The Average Annual Repair Time Restoration Time for:'\n",
+      "print 'Customer A : %g hours per fault per year'%(rA)\n",
+      "print 'Customer A : %g hours per fault per year'%(rB)\n",
+      "print 'Customer A : %g hours per fault per year'%(rC)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "i The Annual Sustained Interruption Rates for:\n",
+        "Customer A : 0.64 faults per year\n",
+        "Customer B : 0.54 faults per year\n",
+        "Customer C : 0.54 faults per year\n",
+        "ii The Average Annual Repair Time Restoration Time for:\n",
+        "Customer A : 1.5625 hours per fault per year\n",
+        "Customer A : 1.98148 hours per fault per year\n",
+        "Customer A : 2.38889 hours per fault per year\n"
+       ]
+      }
+     ],
+     "prompt_number": 3
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.4  Page No : 612"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "\n",
+      "# Variables\n",
+      "Ri = 0.85;\n",
+      "\n",
+      "def relp(y,z):\n",
+      "    return 1-((1-(Ri**y))**z)        #Equal Parallel Combination\n",
+      "\n",
+      "def rels(y,z):\n",
+      "    return (1-((1-Ri)**y))**z         #Equal Series Combination\n",
+      "\n",
+      "# Calculations\n",
+      "#Case 1: 4 elements in series\n",
+      "\n",
+      "Req1 =  rels(1,4);\n",
+      "\n",
+      "#Case 2: Two Comination of 4 elements in series, parallel to each other\n",
+      "\n",
+      "Req2 = relp(4,2);\n",
+      "\n",
+      "#Case 3 : ((two elements in series)        #(two elements in series))in series with ((two elements in series)        #(two elements in series))\n",
+      "\n",
+      "#Two Segments\n",
+      "R1 = relp(2,2);\n",
+      "R2 = relp(2,2);\n",
+      "Req3 = R1*R2;\n",
+      "\n",
+      "#Case 4 : (two elements in parallel)in series with ((three elements in series)        #(three elements in series))\n",
+      "\n",
+      "#Two Segments\n",
+      "R1 = relp(1,2);\n",
+      "R2 = relp(3,2);\n",
+      "Req4 = R1*R2;\n",
+      "\n",
+      "#Case 5, 4 groups of (2 elements in parallel) connected in series to each other\n",
+      "Req5 = rels(2,4);\n",
+      "\n",
+      "# Results\n",
+      "print 'The Equivalent System reliability for:'\n",
+      "print 'a) Configuration A : %g'%(Req1)\n",
+      "print 'b) Configuration B : %g'%(Req2)\n",
+      "print 'c) Configuration C : %g'%(Req3)\n",
+      "print 'd) Configuration D : %g'%(Req4)\n",
+      "print 'e) Configuration E : %g'%(Req5)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The Equivalent System reliability for:\n",
+        "a) Configuration A : 0.522006\n",
+        "b) Configuration B : 0.771522\n",
+        "c) Configuration C : 0.851917\n",
+        "d) Configuration D : 0.831951\n",
+        "e) Configuration E : 0.912992\n"
+       ]
+      }
+     ],
+     "prompt_number": 5
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.5  Page No : 614"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Design the system to meet the given Equivalent System Reliability\n",
+      "#Page 614\n",
+      "\n",
+      "# Variables\n",
+      "#Individual System Reliabilities\n",
+      "Ra = 0.8;\n",
+      "Rb = 0.95;\n",
+      "Rc = 0.99;\n",
+      "Rd = 0.90;\n",
+      "Re = 0.65;\n",
+      "\n",
+      "# Calculations\n",
+      "#When All Are Connected in Series\n",
+      "\n",
+      "Req = Ra*Rb*Rc*Rd*Re;         #Equivalent System Reliability\n",
+      "\n",
+      "Rr = 0.8;         #Required\n",
+      "\n",
+      "Rae = Rr/(Rb*Rc*Rd);\n",
+      "\n",
+      "#Since Connecting the elements in parallel will increase their reliability\n",
+      "def rel(Ri,y,):\n",
+      "    return  (1-((1-Ri)**y))         #Equal Only Parallel Combination\n",
+      "\n",
+      "#Since Connecting the elements in parallel will increase their reliability\n",
+      "#Conditions to Find The Number of Elements to be used\n",
+      "for i in range(1,11):\n",
+      "    L = i;         #Number of Time Element A is used\n",
+      "    R1 = rel(Ra,i);\n",
+      "    X = R1-Rae;\n",
+      "    if(abs(X)+X == 0):\n",
+      "        continue;\n",
+      "    else:\n",
+      "        break;\n",
+      "\n",
+      "for i in range(1,11):\n",
+      "    M = i;        #Number of Time Element E is used\n",
+      "    R2 = rel(Re,i);\n",
+      "    X = R2-Rae;\n",
+      "    if(abs(X)+X == 0):\n",
+      "        continue;\n",
+      "    else:\n",
+      "        break;\n",
+      "\n",
+      "print 'a) The Equivalent system Reliability is %g'%(Req)\n",
+      "print 'b) One Each of B,C and D all connected in series are connected in serieswith the series combination of XComination of\\\n",
+      " %g elements of A, \\nAll Connected in Parallel)and YComination of %g elements of E, \\nAll Connected in Parallel) to\\\n",
+      "  achieve %g Equivalent System Realibility'%(L,M,Rr)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) The Equivalent system Reliability is 0.440154\n",
+        "b) One Each of B,C and D all connected in series are connected in serieswith the series combination of XComination of 2 elements of A, \n",
+        "All Connected in Parallel)and YComination of 3 elements of E, \n",
+        "All Connected in Parallel) to  achieve 0.8 Equivalent System Realibility\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.6  Page No : 614"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Find The Probability on the reliability of transformers\n",
+      "#Page 614\n",
+      "\n",
+      "# Variables\n",
+      "#Reliabilities of The Three Transformers\n",
+      "Pa = 0.9;\n",
+      "Pb = 0.95;\n",
+      "Pc = 0.99;\n",
+      "\n",
+      "#Faliures of Three Transformers\n",
+      "Qa = 1-Pa;\n",
+      "Qb = 1-Pb;\n",
+      "Qc = 1-Pc;\n",
+      "\n",
+      "# Calculations\n",
+      "#Probability of NO Transformer Failing\n",
+      "Pnf = Pa*Pb*Pc;\n",
+      "\n",
+      "PfA = Qa*Pb*Pc        #Probability of Transformer A Failing\n",
+      "PfB = Pa*Qb*Pc        #Probability of Transformer B Failing\n",
+      "PfC = Pa*Pb*Qc        #Probability of Transformer C Failing\n",
+      "\n",
+      "PfAB = Qa*Qb*Pc        #Probability of Transformer A and B Failing\n",
+      "PfBC = Pa*Qb*Qc        #Probability of Transformer B and C Failing\n",
+      "PfCA = Qa*Pb*Qc        #Probability of Transformer C and A Failing\n",
+      "\n",
+      "Pf = Qa*Qb*Qc;         #Probability of All Transformers failing\n",
+      "\n",
+      "# Results\n",
+      "print 'a) Probability of No Transformer Failing is %g'%(Pnf)\n",
+      "print 'b'\n",
+      "print 'Probability of Transformer A Failing is %g'%(PfA)\n",
+      "print 'Probability of Transformer B Failing is %g'%(PfB)\n",
+      "print 'Probability of Transformer C Failing is %g'%(PfC)\n",
+      "print 'c'\n",
+      "print 'Probability of Transformers A and B Failing is %g'%(PfAB)\n",
+      "print 'Probability of Transformers B and C Failing is %g'%(PfBC)\n",
+      "print 'Probability of Transformers C and A Failing is %g'%(PfCA)\n",
+      "print 'd) Probability of All Three Transformers Failing is %g'%(Pf)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a) Probability of No Transformer Failing is 0.84645\n",
+        "b\n",
+        "Probability of Transformer A Failing is 0.09405\n",
+        "Probability of Transformer B Failing is 0.04455\n",
+        "Probability of Transformer C Failing is 0.00855\n",
+        "c\n",
+        "Probability of Transformers A and B Failing is 0.00495\n",
+        "Probability of Transformers B and C Failing is 0.00045\n",
+        "Probability of Transformers C and A Failing is 0.00095\n",
+        "d) Probability of All Three Transformers Failing is 5e-05\n"
+       ]
+      }
+     ],
+     "prompt_number": 8
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.7  Page No : 619"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Determine Probabilities Using Markovian Principle\n",
+      "#Page 619\n",
+      "\n",
+      "# Variables\n",
+      "#Conditional Probabilites Present Future\n",
+      "Pdd = 2./100;         #Down Down\n",
+      "Pud = 5./100;         #Up Down\n",
+      "Pdu = 1-Pdd;         #Down up\n",
+      "Puu = 1-Pud;         #Up Up\n",
+      "\n",
+      "# Calculations\n",
+      "P = [[Pdd,Pdu],[Pud,Puu]];         #Transition Matrix\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Conditional Probabilites for'\n",
+      "print 'Transformers Down in Present and Down in Future is %g'%(Pdd)\n",
+      "print 'Transformers Down in Present and Up in Future is %g'%(Pdd)\n",
+      "print 'Transformers Up in Present and Down in Future is %g'%(Pdd)\n",
+      "print 'Transformers Up in Present and Up in Future is %g'%(Pdd)\n",
+      "print 'b The Transition Matrix is',\n",
+      "print (P)\n",
+      "print 'c The Transition Diagram can be viewed with the result file attached to this code'\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Conditional Probabilites for\n",
+        "Transformers Down in Present and Down in Future is 0.02\n",
+        "Transformers Down in Present and Up in Future is 0.02\n",
+        "Transformers Up in Present and Down in Future is 0.02\n",
+        "Transformers Up in Present and Up in Future is 0.02\n",
+        "b The Transition Matrix is [[0.02, 0.98], [0.05, 0.95]]\n",
+        "c The Transition Diagram can be viewed with the result file attached to this code\n"
+       ]
+      }
+     ],
+     "prompt_number": 9
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.8  Page No : 620"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "#To Determine the Conditional Outage Probabilites\n",
+      "#Page 620\n",
+      "\n",
+      "# Calculations\n",
+      "#Conditional Outage Probabilites From The Table Given\n",
+      "P11 = 40./100;\n",
+      "P12 = 30./100;\n",
+      "P13 = 30./100;\n",
+      "P21 = 20./100;\n",
+      "P22 = 50./100;\n",
+      "P23 = 30./100;\n",
+      "P31 = 25./100;\n",
+      "P32 = 25./100;\n",
+      "P33 = 50./100;\n",
+      "\n",
+      "#Transition Matrix\n",
+      "P = [[P11,P12,P13],[P21,P22,P23],[P31,P32,P33]];\n",
+      "\n",
+      "print \"a The Conditional Outage Probabilites for:\"\n",
+      "print \"Presently Outaged Feeder is 1, Next Outaged Feeder is 1 is %g\"%(P11)\n",
+      "print \"Presently Outaged Feeder is 1, Next Outaged Feeder is 2 is %g\"%(P12)\n",
+      "print \"Presently Outaged Feeder is 1, Next Outaged Feeder is 3 is %g\"%(P13)\n",
+      "print \"Presently Outaged Feeder is 2, Next Outaged Feeder is 1 is %g\"%(P21)\n",
+      "print \"Presently Outaged Feeder is 2, Next Outaged Feeder is 2 is %g\"%(P22)\n",
+      "print \"Presently Outaged Feeder is 2, Next Outaged Feeder is 3 is %g\"%(P23)\n",
+      "print \"Presently Outaged Feeder is 3, Next Outaged Feeder is 1 is %g\"%(P31)\n",
+      "print \"Presently Outaged Feeder is 3, Next Outaged Feeder is 2 is %g\"%(P32)\n",
+      "print \"Presently Outaged Feeder is 3, Next Outaged Feeder is 3 is %g\"%(P33)\n",
+      "print \"b Transition Matrix is\"\n",
+      "print (P)\n",
+      "print \"c The Transition figure is print layed in the result file attached to this code\"\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Conditional Outage Probabilites for:\n",
+        "Presently Outaged Feeder is 1, Next Outaged Feeder is 1 is 0.4\n",
+        "Presently Outaged Feeder is 1, Next Outaged Feeder is 2 is 0.3\n",
+        "Presently Outaged Feeder is 1, Next Outaged Feeder is 3 is 0.3\n",
+        "Presently Outaged Feeder is 2, Next Outaged Feeder is 1 is 0.2\n",
+        "Presently Outaged Feeder is 2, Next Outaged Feeder is 2 is 0.5\n",
+        "Presently Outaged Feeder is 2, Next Outaged Feeder is 3 is 0.3\n",
+        "Presently Outaged Feeder is 3, Next Outaged Feeder is 1 is 0.25\n",
+        "Presently Outaged Feeder is 3, Next Outaged Feeder is 2 is 0.25\n",
+        "Presently Outaged Feeder is 3, Next Outaged Feeder is 3 is 0.5\n",
+        "b Transition Matrix is\n",
+        "[[0.4, 0.3, 0.3], [0.2, 0.5, 0.3], [0.25, 0.25, 0.5]]\n",
+        "c The Transition figure is print layed in the result file attached to this code\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Example 11.9  Page No : 624"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "import math \n",
+      "from numpy import array\n",
+      "\n",
+      "# Variables\n",
+      "P = array([[0.6,0.4],[0.3,0.7]]);         #One Step Transition Matrix\n",
+      "\n",
+      "Po = array([0.8,0.2]);         #Initial State Probability Vector\n",
+      "\n",
+      "# Calculations\n",
+      "#Funtion to determine the Vector of State Probability\n",
+      "def VSP(y): \n",
+      "    return (Po*(P**y))\n",
+      "\n",
+      "P1 = VSP(1);         #Vector of State Probability at Time t1\n",
+      "P4 = VSP(4);         #Vector of State Probability at Time t4\n",
+      "P8 = VSP(8);         #Vector of State Probability at Time t8\n",
+      "\n",
+      "# Results\n",
+      "print 'a The Vector of State Probability at time t1 is',\n",
+      "print (P1)\n",
+      "print 'a The Vector of State Probability at time t4 is',\n",
+      "print (P4)\n",
+      "print 'a The Vector of State Probability at time t8 is',\n",
+      "print (P8)\n"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "a The Vector of State Probability at time t1 is [[ 0.48  0.08]\n",
+        " [ 0.24  0.14]]\n",
+        "a The Vector of State Probability at time t4 is [[ 0.10368  0.00512]\n",
+        " [ 0.00648  0.04802]]\n",
+        "a The Vector of State Probability at time t8 is [[  1.34369280e-02   1.31072000e-04]\n",
+        " [  5.24880000e-05   1.15296020e-02]]\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
+\ No newline at end of file