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{
"metadata": {
"name": "EE-11"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": "Inflation Adjusted Decisions"
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 11.1 Page 148"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nprint \"Few steps are as follows : \";\nprint \"Step 1 : The estimated future requirementt per year in terms of todays rupees from his age 61 through 70 is Rs. 24000.\";\nprint \"Step 2 : The formula which is given below is used to get future equivalent of Rs. 24000 with the inflation of 9% per year.\";\nprint \"Formula : P*(1+IR)^n\";\nP=24000.0;#in Rs.\nIR=9.0;#in % per year\nAge=61.0;#in years\nfor n in range(21,31) :\n IV=P*(1+IR/100)**n;#in Rs\n print \"Age in years : \",Age,\". \"\"Inflated value of Rs. 24000 at end of year : \",n,\" in Rs. \",round(IV);\n Age=Age+1;#in Years\n\nPW=0;#For initialising\nn=21.0;#in years\nfor m in range(10) :\n A=P*(1+IR/100)**n;#in Rs\n n=n+1;\n i=15.0;#in % per year\n PW=PW+A/(1+i/100)**(m+1);\n\n#result\nprint \"Present equivalent of withdrawls in Rs. : \",round(PW,3);\n\n#The annual equivalent amount :\ni=15.0;#in % per year\nn=20.0;#in years\n#Formula : (A/F,i,n) : (i/100)/(((1+i/100)^n)-1) \nA=PW*(i/100)/(((1+i/100)**n)-1);#in Rs.\nprint \"The annual equivalent amount in Rs. : \",round(A,3);\nprint\"RECOMMENDATION : The person has to invest an amount of Rs. \",round(A,3),\" at the end of every year starting from his age 41 through 60 which will enable him to make a withdrawl at the end of every year starting from his age 61 through 70.\"; ",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "Few steps are as follows : \nStep 1 : The estimated future requirementt per year in terms of todays rupees from his age 61 through 70 is Rs. 24000.\nStep 2 : The formula which is given below is used to get future equivalent of Rs. 24000 with the inflation of 9% per year.\nFormula : P*(1+IR)^n\nAge in years : 61.0 . Inflated value of Rs. 24000 at end of year : 21 in Rs. 146611.0\nAge in years : 62.0 . Inflated value of Rs. 24000 at end of year : 22 in Rs. 159806.0\nAge in years : 63.0 . Inflated value of Rs. 24000 at end of year : 23 in Rs. 174189.0\nAge in years : 64.0 . Inflated value of Rs. 24000 at end of year : 24 in Rs. 189866.0\nAge in years : 65.0 . Inflated value of Rs. 24000 at end of year : 25 in Rs. 206954.0\nAge in years : 66.0 . Inflated value of Rs. 24000 at end of year : 26 in Rs. 225580.0\nAge in years : 67.0 . Inflated value of Rs. 24000 at end of year : 27 in Rs. 245882.0\nAge in years : 68.0 . Inflated value of Rs. 24000 at end of year : 28 in Rs. 268011.0\nAge in years : 69.0 . Inflated value of Rs. 24000 at end of year : 29 in Rs. 292132.0\nAge in years : 70.0 . Inflated value of Rs. 24000 at end of year : 30 in Rs. 318424.0\nPresent equivalent of withdrawls in Rs. : 1013631.794\nThe annual equivalent amount in Rs. : 9894.537\nRECOMMENDATION : The person has to invest an amount of Rs. 9894.537 at the end of every year starting from his age 41 through 60 which will enable him to make a withdrawl at the end of every year starting from his age 61 through 70.\n"
}
],
"prompt_number": 7
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 11.2 Page 150"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nprint\"The method of finding the economic life of the machine witha discounting factor of 20% at zero inflation rate is summarized in table below. From the table it is clear that total annual cost is minimum if the machine is used for 14 years. Hence the economic life of the machine is 14 years.\";\nprint\"End of year Op_cost Main_cost Op+Main P/F,i,n PW Cummulative Salvage PW_S TPW A/P,i,n AEM\";\ni=20.0;#in per year\nCum=0.0;#initialising\nOp_cost=40000.0;#in RS.\nMain_cost=60000.0;#in Rs.\nOpMain=Op_cost+Main_cost;#in Rs.\nS=400000.0;#in Rs.\nfor n in range (1,16) :\n PF=1/((1+i/100)**n);\n PW=OpMain*PF;#in Rs.\n Cum=Cum+PW\n PW_S=PF*S;#in RS.\n TPW=500000.0+Cum-PW_S;#in Rs.\n AP=((i/100)*(1+i/100)**n)/(((1+i/100)**n)-1);\n AEM=TPW*AP;#in RS\n print \" \",n,\" \",Op_cost,\" \",Main_cost,\" \",OpMain,\" \",PF,\" \",PW,\" \",Cum,\" \",S,\" \",PW_S,\" \",TPW,\" \",AP,\" \",AEM;\n Op_cost=Op_cost+5000.0;#in Rs.\n Main_cost=Main_cost+6000.0;#in Rs.\n S=S-50000.0;#in Rs.\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "The method of finding the economic life of the machine witha discounting factor of 20% at zero inflation rate is summarized in table below. From the table it is clear that total annual cost is minimum if the machine is used for 14 years. Hence the economic life of the machine is 14 years.\nEnd of year Op_cost Main_cost Op+Main P/F,i,n PW Cummulative Salvage PW_S TPW A/P,i,n AEM\n 1 40000.0 60000.0 100000.0 0.833333333333 83333.3333333 83333.3333333 400000.0 333333.333333 250000.0 1.2 300000.0\n 2 45000.0 66000.0 100000.0 0.694444444444 69444.4444444 152777.777778 350000.0 243055.555556 409722.222222 0.654545454545 268181.818182\n 3 50000.0 72000.0 100000.0 0.578703703704 57870.3703704 210648.148148 300000.0 173611.111111 537037.037037 0.474725274725 254945.054945\n 4 55000.0 78000.0 100000.0 0.48225308642 48225.308642 258873.45679 250000.0 120563.271605 638310.185185 0.386289120715 246572.280179\n 5 60000.0 84000.0 100000.0 0.401877572016 40187.7572016 299061.213992 200000.0 80375.5144033 718685.699588 0.33437970329 240313.910987\n 6 65000.0 90000.0 100000.0 0.33489797668 33489.797668 332551.01166 150000.0 50234.6965021 782316.315158 0.300705745867 235247.011053\n 7 70000.0 96000.0 100000.0 0.279081647234 27908.1647234 360459.176383 100000.0 27908.1647234 832551.01166 0.277423926347 230969.570539\n 8 75000.0 102000.0 100000.0 0.232568039361 23256.8039361 383715.980319 50000.0 11628.4019681 872087.578351 0.260609422409 227274.240084\n 9 80000.0 108000.0 100000.0 0.193806699468 19380.6699468 403096.650266 0.0 0.0 903096.650266 0.248079461672 224039.730836\n 10 85000.0 114000.0 100000.0 0.16150558289 16150.558289 419247.208555 -50000.0 -8075.27914449 927322.4877 0.238522756883 221187.516286\n 11 90000.0 120000.0 100000.0 0.134587985742 13458.7985742 432706.007129 -100000.0 -13458.7985742 946164.805703 0.231103794152 218662.276491\n 12 95000.0 126000.0 100000.0 0.112156654785 11215.6654785 443921.672608 -150000.0 -16823.4982177 960745.170825 0.225264964904 216422.227187\n 13 100000.0 132000.0 100000.0 0.0934638789872 9346.38789872 453268.060506 -200000.0 -18692.7757974 971960.836304 0.220620001083 214434.000758\n 14 105000.0 138000.0 100000.0 0.0778865658226 7788.65658226 461056.717089 -250000.0 -19471.6414557 980528.358544 0.216893055222 212669.791417\n 15 110000.0 144000.0 100000.0 0.0649054715189 6490.54715189 467547.264241 -300000.0 -19471.6414557 987018.905696 0.213882119838 211105.69587\n"
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 11.3 Page 152"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nIR=5.0;#in % per year\ni=15.0;#in % per year\n#Machine X :\nPpx=1500000.0;#in Rs.\nn=7.0;#in years\nS=200000.0;#in Rs.\nAMC=300000.0;#in Rs.\nprint \"End of year AMC InflationFactor InflatedAmount P/F PW\";\nPw=0;#For initialising\nfor n in range (1,8) :\n FP=(1+IR/100)**n\n IA=AMC*FP;#in Rs.\n PF=1/((1+i/100)**n);\n PW=IA*PF;#in Rs.\n Pw=Pw+PW;#in Rs.\n print\" \",n,\" \",AMC,\" \",FP,\" \",IA,\" \",PF,\" \",PW;\n\nprint \"Present worth of inflated annual operating and maintenance cost in Rs. : \",round(Pw,3);\nPWX=Ppx+Pw-S*1/((1+i/100)**n);\nprint \"Present worth of machine X in Rs. : \",round(PWX,3);\n\n#Machine Y\n\nPpy=2000000.0;#in Rs.\nn=7.0;#in years\nS=300000.0;#in Rs.\nAMC=250000.0;#in Rs.\nprint \"End of year AMC InflationFactor InflatedAmount P/F PW\";\nPw=0;#For initialising\nfor n in range (1,8) :\n FP=(1+IR/100)**n\n IA=AMC*FP;#in Rs.\n PF=1/((1+i/100)**n);\n PW=IA*PF;#in Rs.\n Pw=Pw+PW;#in Rs.\n print \" \",n,\" \",AMC,\" \",FP,\" \",IA,\" \",PF,\" \",PW;\n\nPWY=Ppy+Pw-S*1/((1+i/100)**n);\n\n#result\nprint \"Present worth of inflated annual operating and maintenance cost in Rs. : \",round(Pw,3);\nprint \"Present worth of machine Y in Rs. : \",round(PWY,3);\nprint \"Since the present worth of Machine X is less than Machine Y, select Machine X\"\nprint \" Calculations are not accurate in the book\"",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "End of year AMC InflationFactor InflatedAmount P/F PW\n 1 300000.0 1.05 315000.0 0.869565217391 273913.043478\n 2 300000.0 1.1025 330750.0 0.756143667297 250094.517958\n 3 300000.0 1.157625 347287.5 0.657516232432 228347.168571\n 4 300000.0 1.21550625 364651.875 0.571753245593 208490.893043\n 5 300000.0 1.2762815625 382884.46875 0.497176735298 190361.25017\n 6 300000.0 1.34009564063 402028.692188 0.432327595912 173808.097981\n 7 300000.0 1.40710042266 422130.126797 0.375937039923 158694.35033\nPresent worth of inflated annual operating and maintenance cost in Rs. : 1483709.322\nPresent worth of machine X in Rs. : 2908521.914\nEnd of year AMC InflationFactor InflatedAmount P/F PW\n 1 250000.0 1.05 262500.0 0.869565217391 228260.869565\n 2 250000.0 1.1025 275625.0 0.756143667297 208412.098299\n 3 250000.0 1.157625 289406.25 0.657516232432 190289.307142\n 4 250000.0 1.21550625 303876.5625 0.571753245593 173742.410869\n 5 250000.0 1.2762815625 319070.390625 0.497176735298 158634.375141\n 6 250000.0 1.34009564063 335023.910156 0.432327595912 144840.081651\n 7 250000.0 1.40710042266 351775.105664 0.375937039923 132245.291942\nPresent worth of inflated annual operating and maintenance cost in Rs. : 1236424.435\nPresent worth of machine Y in Rs. : 3123643.323\nSince the present worth of Machine X is less than Machine Y, select Machine X\n Calculations are not accurate in the book\n"
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": "Example 11.4 Page 156"
},
{
"cell_type": "code",
"collapsed": false,
"input": "#initiation of variable\nIR=6.0;#in % per year\ni=18.0;#in % per year\nAFR=5000000.0;#in Rs.\nn=7.0;#in years\nAI=500000.0;#in Rs.\nprint\"End of year AFR InflationFactor InflatedAmount P/F PW\";\nTPW=0;#Initialising\n\nfor n in range (1,6) :\n IF=(1+IR/100)**n;\n IA=IF*AFR;#in Rs.\n PF=1/((1+i/100)**n);\n PW=PF*IA;#in Rs.\n TPW=PW+TPW;#in Rs.\n print\" \",round(n),\" \",round(AFR,3),\" \",round(IF,3),\" \",round(IA,3),\" \",round(PF,3),\" \",round(PW,3);\n AFR=AFR+AI;#in Rs.\n\n \n#result\nprint \"The value of the single deposit to be made now to recieve the specified series for the next five years is Rs. : \",round(TPW,3)\n",
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": "End of year AFR InflationFactor InflatedAmount P/F PW\n 1.0 5000000.0 1.06 5300000.0 0.847 4491525.424\n 2.0 5500000.0 1.124 6179800.0 0.718 4438236.139\n 3.0 6000000.0 1.191 7146096.0 0.609 4349334.645\n 4.0 6500000.0 1.262 8206100.24 0.516 4232615.212\n 5.0 7000000.0 1.338 9367579.043 0.437 4094655.134\nThe value of the single deposit to be made now to recieve the specified series for the next five years is Rs. : 21606366.553\n"
}
],
"prompt_number": 19
}
],
"metadata": {}
}
]
}
|
