MOOC: Finance for non-finance professionals

mooc_finance4non-fin-professionals.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Week 1: Financial Valuation Discounting\n",
    "### Compound interest rate \n",
    "* can be calculated annually or on smaller periods\n",
    "* $ FutureValue = PresentValue (1 + r)^N$, where $r$ is the interest rate, $N$ is the number of periods.\n",
    "* the correct way of calculating monthly rate from yearly rate is $r_m = (r_y)^{1/12}$, but a simpler formula of $r_y/12$ is sometimes also used."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1628.894626777441\n"
     ]
    }
   ],
   "source": [
    "# example: Compound interest of bank deposit\n",
    "s = 1000.\n",
    "n_years = 10\n",
    "rate_yearly = 0.05\n",
    "for i in range(n_years):\n",
    "    s += s*(rate_yearly)\n",
    "    \n",
    "print(s)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Discounted cash flow (DCF)\n",
    "How much a future cash flow costs now, at a given interest rate $r$?\n",
    "- one-time payment in future: $PresentValue = FutureValue / (1+r\\%)^N$, N is number of years (months, quarters)\n",
    "- multiple periodic payments: $PresentValue = \\sum_1^N \\frac{CF_i}{(1+r\\%)^i}$, where $CF_i$ are future cash flows.\n",
    "\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 36,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "1038.2693939108303"
      ]
     },
     "execution_count": 36,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "# example: DCF of bond price\n",
    "coupon = 50\n",
    "rate = 0.03\n",
    "principal = 1000\n",
    "n_years = 2\n",
    "\n",
    "s = 0\n",
    "for i in range(1, n_years+1):\n",
    "    if i == n_years:\n",
    "        s += (principal + coupon)/(1 + rate)**i\n",
    "    else:\n",
    "        s += coupon/(1 + rate)**i\n",
    "s"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Week 2: Capital budgeting\n",
    "### Net Present Value (NPV)\n",
    "* NPV = - IniCost + $\\sum \\frac{CF_i}{(1+r\\%)^i}$, where $CF_i$ are future cash flows.\n",
    "* is NPV > 0?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "361.6240146852392"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "ini = 1000\n",
    "cf = np.array([500, 500, 500])\n",
    "rate = 0.05\n",
    "r_i = np.array([(1 + rate)**(i+1) for i in range(len(cf))])\n",
    "npv = -ini + (cf/r_i).sum()\n",
    "npv"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Payback period\n",
    "- its the time period when IniCost becomes =$\\sum \\frac{CF_i}{(1+r\\%)^i}$\n",
    "- a weaker budgeting tool than NVP, few pros and many cons."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 35,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2.352 years\n"
     ]
    }
   ],
   "source": [
    "# example of payback period calculation (fractional 3. year)\n",
    "ini = 2000\n",
    "cf = np.array([1000, 1000, 1000])\n",
    "rate = 0.1\n",
    "r_i = np.array([(1 + rate)**(i+1) for i in range(len(cf))])\n",
    "t_payback = len(cf) - ((cf/r_i).sum() - ini)/(cf[-1]/r_i[-1])\n",
    "print(f\"{t_payback:2.3f} years\")"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "... to be continued"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python [conda env:py36]",
   "language": "python",
   "name": "conda-env-py36-py"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.6.5"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 2
}