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{
"metadata": {
"name": "",
"signature": "sha256:e388f34dd67bfe5948d20d52a601f7e8e53b92aa882e9a754ed91176569417c0"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Future Worth Method"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.1,Page 54"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variable\n",
"#alternate A\n",
"P=5000000.0;#in Rs\n",
"A=2000000.0;#in Rs\n",
"i=18.0;#in % per annum\n",
"n=4.0;#in years\n",
"\n",
"#calculation\n",
"FW_A=(-P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100));#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of alternative A in RS. \",round(FW_A,3);\n",
"\n",
"#alternate B\n",
"P=4500000.0;#in Rs\n",
"A=1800000.0;#in Rs\n",
"i=18.0;#in % per annum\n",
"n=4.0;#in years\n",
"\n",
"#calculation\n",
"FW_B=(-P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100));#in RS\n",
"print \"The future worth amount of alternative B in RS. \", round(FW_B,3);\n",
"print \"The future worth of alternative A is greater than that of alternative B. Thus, alternative A should be selected.\";\n",
"print \" Calculation in the book is not accurate.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The future worth amount of alternative A in RS. 736975.2\n",
"The future worth amount of alternative B in RS. 663277.68\n",
"The future worth of alternative A is greater than that of alternative B. Thus, alternative A should be selected.\n",
" Calculation in the book is not accurate.\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.2,Page 58"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variable\n",
"#alternate 1\n",
"FC=2000000.0;#in Rs\n",
"AI=800000.0;#in Rs\n",
"ATax=80000.0;#in Rs\n",
"NetAI=AI-ATax;#in Rs\n",
"i=12.0;#in % per annum\n",
"n=20.0;#in years\n",
"\n",
"#calculation\n",
"FW_1=(-FC*(1+i/100)**n)+(NetAI*(((1+i/100)**n)-1)/(i/100));#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of alternative 1 in RS. : \", round(FW_1,3)\n",
"\n",
"#alternative2\n",
"FC=3600000.0;#in Rs\n",
"AI=980000.0;#in Rs\n",
"ATax=150000.0;#in Rs\n",
"NetAI=AI-ATax;#in Rs\n",
"i=12.0;#in % per annum\n",
"n=20.0;#in years\n",
"\n",
"#calculation\n",
"FW_2=(-FC*(1+i/100)**n)+(NetAI*(((1+i/100)**n)-1)/(i/100));#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of alternative 2 in RS.\",round(FW_2,3)\n",
"print\"The future worth of alternative 1 is greater than that of alternative 2. Thus, building the gas station is the best alternative.\";\n",
"print \" Calculation in the book is not accurate.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The future worth amount of alternative 1 in RS. : 32585172.373\n",
"The future worth amount of alternative 2 in RS. 25076872.093\n",
"The future worth of alternative 1 is greater than that of alternative 2. Thus, building the gas station is the best alternative.\n",
" Calculation in the book is not accurate.\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.3,Page 59"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variable\n",
"#part a\n",
"#alternative 1\n",
"P=500000.0;#in Rs\n",
"A1=50000.0;#in Rs\n",
"G=50000.0;#in Rs\n",
"i=8.0;#in % per annum\n",
"n=6.0;#in years\n",
"\n",
"#calculation\n",
"FW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of alternative 1 in RS. \", round(FW_1,3);\n",
"\n",
"#alternative 2\n",
"P=700000.0;#in Rs\n",
"A1=70000.0;#in Rs\n",
"G=70000.0;#in Rs\n",
"i=8.0;#in % per annum\n",
"n=6.0;#in years\n",
"\n",
"#calculation\n",
"FW_2=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of alternative 2 in RS. \", round(FW_2,3);\n",
"print \"The future worth of alternative 2 is greater than that of alternative 1. Thus, alternative 2 must be selected.\";\n",
"\n",
"\n",
"#part b\n",
"#alternative a\n",
"P=500000.0;#in Rs\n",
"A1=50000.0;#in Rs\n",
"G=50000.0;#in Rs\n",
"i=9.0;#in % per annum\n",
"n=6.0;#in years\n",
"\n",
"#calculation\n",
"FW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of alternative 1 in RS. \", round(FW_1,3);\n",
"\n",
"#altenative 2\n",
"P=700000.0;#in Rs\n",
"A1=70000.0;#in Rs\n",
"G=70000.0;#in Rs\n",
"i=9.0;#in % per annum\n",
"n=6.0;#in years\n",
"\n",
"#calculation\n",
"FW_1=(-P*(1+i/100)**n)+(A1+G*(((1+i/100)**n)-i*n/100-1)/(((i/100)*(1+i/100)**n)-i/100))*(((1+i/100)**n)-1)/(i/100);#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of alternative 2 in RS.\", round(FW_1,3);\n",
"print \"The negative sign of alternatives future worth indicates that alternative 2 incurs loss. Thus, none of the two alternatives should be selected. \";\n",
"print \" Calculation in the book is not accurate.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The future worth amount of alternative 1 in RS. 408314.938\n",
"The future worth amount of alternative 2 in RS. 571640.914\n",
"The future worth of alternative 2 is greater than that of alternative 1. Thus, alternative 2 must be selected.\n",
"The future worth amount of alternative 1 in RS. 383913.653\n",
"The future worth amount of alternative 2 in RS. 537479.115\n",
"The negative sign of alternatives future worth indicates that alternative 2 incurs loss. Thus, none of the two alternatives should be selected. \n",
" Calculation in the book is not accurate.\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.4 Page 62"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variable\n",
"#alternative 1\n",
"P=8000000.0;#in Rs\n",
"A=800000.0;#in Rs\n",
"i=20.0;#in % per annum\n",
"n=12.0;#in years\n",
"Salvage=500000;#in Rs\n",
"\n",
"#calcualtion\n",
"FW1=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n",
"\n",
"#result\n",
"print \"The future worth for this alternative in RS.\",round(FW1,3)\n",
"\n",
"#alternative 2\n",
"P=7000000.0;#in Rs\n",
"A=900000.0;#in Rs\n",
"i=20.0;#in % per annum\n",
"n=12.0;#in years\n",
"Salvage=400000;#in Rs\n",
"\n",
"#calculation\n",
"FW2=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n",
"\n",
"#result\n",
"print \"The future worth for this alternative in RS.\",round(FW2,3);\n",
"\n",
"#alternative 3\n",
"P=9000000.0;#in Rs\n",
"A=850000.0;#in Rs\n",
"i=20.0;#in % per annum\n",
"n=12.0;#in years\n",
"Salvage=700000;#in Rs\n",
"\n",
"#calculation\n",
"FW3=P*(1+i/100)**n+A*(((1+i/100)**n)-1)/(i/100)-Salvage;#in RS\n",
"\n",
"print \"The future worth for this alternative in RS. \",round(FW3,3)\n",
"print \"The future worth of alternative 2 is less than that of other two alternatives. Thus, Ms. Krishna castings should buy the annealing furnace from manufacturer 2.\";\n",
"print \" Calculation in the book is not accurate.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The future worth for this alternative in RS. 102493205.379\n",
"The future worth for this alternative in RS. 97635155.155\n",
"The future worth for this alternative in RS. 113188330.939\n",
"The future worth of alternative 2 is less than that of other two alternatives. Thus, Ms. Krishna castings should buy the annealing furnace from manufacturer 2.\n",
" Calculation in the book is not accurate.\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 5.5 Page 64"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initialisation of variable\n",
"#Machine A\n",
"P=400000.0;#in Rs\n",
"A=40000.0;#in Rs\n",
"S=200000.0;#in Rs\n",
"i=12.0;#in % per annum\n",
"n=4.0#in years\n",
"\n",
"#calculation\n",
"FW_A=(P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100))-S;#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of Machine A in RS. \",round(FW_A,3);\n",
"\n",
"#Machine B\n",
"P=800000.0;#in Rs\n",
"A=0.0;#in Rs\n",
"S=550000.0;#in Rs\n",
"i=12.0;#in % per annum\n",
"n=4.0#in years\n",
"\n",
"#calculation\n",
"FW_B=(P*(1+i/100)**n)+(A*(((1+i/100)**n)-1)/(i/100))-S;#in RS\n",
"\n",
"#result\n",
"print \"The future worth amount of Machine B in RS. \",round(FW_B,3);\n",
"print \"The future worth of Machine A is less than that of Machine B. Thus, Machine A should be selected.\";\n",
"print \" Calculation in the book is not accurate\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The future worth amount of Machine A in RS. 620580.864\n",
"The future worth amount of Machine B in RS. 708815.488\n",
"The future worth of Machine A is less than that of Machine B. Thus, Machine A should be selected.\n",
" Calculation in the book is not accurate\n"
]
}
],
"prompt_number": 13
}
],
"metadata": {}
}
]
}
|
