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CAIIB -Financial Management. Module A -Quantitative Techniques and Business Mathematics Madhav K Prabhu M.Tech, MIM, PMP, CISA, CAIIB, CeISB, MCTS, DCL. Agenda. Time Value of Money Bond Valuation Theory Sampling Regression and Correlation. Time Value of Money. Objectives.
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CAIIB -Financial Management Module A -Quantitative Techniques and Business Mathematics Madhav K Prabhu M.Tech, MIM, PMP, CISA, CAIIB, CeISB, MCTS, DCL
Agenda • Time Value of Money • Bond Valuation Theory • Sampling • Regression and Correlation
Objectives • What do we mean by Time value of money • Present Value, Discounted Value, Annuity
Time Value of Money • What is Time Value of Money? • Future Value • Present Value • Future Value: Compounding: How would you do Compounding?
Compounding • Compounding Formula • What if compounding is done on monthly basis?
Compounding Exercise • Exercise: • Prepare a table showing compounding as per following conditions: • Rate of Interest - 5%, 12% and 15% • Compounding 2 & 4 times in a year • Principal Rs.100,000/-
Discounting • Present Value • You have an option to receive Rs. 1,000/- either today or after one year. Which option you will select? Why? • Decision will depend upon the present value of money; which can be calculated by a process called Discounting (opposite of Compounding) • Interest Rate and Time of Receipt of money decide Present Value • What is the present value of Rs. 1,000/- today and a year later? To compute Present Value?
Discounting contd… • Formula to find Present Value of Future Cash Receipt • Where PV = Present Value, P = Principal, i = Rate of Interest, n = Number of Years after which money is received • Assuming Rate of Interest is 10%, value of Rs. 1,000/- to be received after 1 year will be, • Whereas the value of money to be received today will be Rs. 1,000/- • What if you were to choose between: • Receive Rs. 1,000/- every year for 3 years, OR • Receive Rs. 2,500/- today? (assume 10% annual interest rate)
Discounting of a Series contd… • How discounting is done for a series of cashflow? e.g. • Receive Rs. 1,000/- at the end of every year for 3 years OR • Receive Rs. 2,500/- today • Assume Rate of Interest @10% If cashflow was to occur every 6 months instead of 1 year, what impact it will have on Present Value?
Periodic Discounting • What if the receipts are over six months’ interval ? Find Present Value of the money receipts • Periodic Discounting Formula • Receive Rs. 1,000/- at the end of every 6 months for 1-1/2 years OR • Receive Rs. 2,600/- today • Assume Rate of interest @10% Where, P = Principal, i = Rate of Interest, t = Times Payments made in a Year, n = nth Period (in this case it is half year)
Periodic Discounting Formula Expressed mathematically, the equation will look like: Generically expressed, the formula is: Here, N = 3
Charting of Cashflow • For any financial proposition prepare a chart of cashflow: e.g. Interest Received + 100 Sold Bond +2,050 Total +2,150 Interest Received +50 01.01.04 31.12.04 Timeline 30.06.04 30.06.05 Invested in Bonds (1,000) Interest Received + 50 New Bond Purchased (1,020) Net ( 970)
Net Present Value • Net Present Value means the difference between the PV of Cash Inflows & Cash Outflows • How do you compute NPV? • Prepare Cashflow Chart • Net off Inflow & Outflow for each period separately • If Inflow > Outflow, positive cash • If Inflow < Outflow, negative cash • Find present values of Inflows & Outflows by applying Discount Factor (or Present Value Factor) • NPV = (PV of Inflows) LESS (PV of Outflows); Result can be +ve OR -ve • Continuing with our example of Bond Investment: Inflow Interest Received + 100 Sold Bond +2,050 Total +2,150 Interest Received +50 01.01.04 31.12.04 Timeline 30.06.04 30.06.05 Invested in Bonds (1,000) Interest Received + 50 New Bond Purchased (1,020) Net ( 970) Outflow
Description Date Amount In / Out PV Outflow PV Inflow Invested in 10% Bonds 01-Jan-04 (1,000) Outflow (1,000.00) Interest received 30-Jun-04 50 Inflow 47.62 Interest received 31-Dec-04 50 Inflow 45.35 New Bond Purchased from 31-Dec-04 (1,020) Outflow (925.17) Open Market Interest received 30-Jun-05 100 Inflow 86.38 Sold Bond in Open Market 30-Jun-05 2,050 Inflow 1,770.87 Sum (1,925.17) 1,950.22 How these values are arrived at? Net Present Value 25.05 NPV contd… • If Cashflows are discounted at say 10%, the sum of PV is 25.05, a positive number & therefore the IRR has be higher than 10% to make Net Present Value to zero What is IRR?
Internal Rate of Return (IRR) • Definition: The Rate at which the NPV is Zero. It can also be termed as “Effective Rate” • If we want to find out IRR of the bond investment cashflow:
IRR Contd… • To prove that at IRR of 11.38% the NPV of Investment Cashflow is zero, see the formula & table:
IRR - Additional Example • You buy a car costing Rs. 600,000/- • Banker is willing to finance upto Rs. 500,000/- • The loan is repayable over 3 years, in Equated Monthly Installments (EMI) of Rs. 15,000/- • Installments are payable In Arrears • What is the IRR? • How do you express this mathematically? What are the values of each component in the formula? • What will be the impact on IRR if the EMIs are payable In Advance? • Can we use IRR for computing Interest & Principal break-up?
IRR - Additional Examplecontd… • Plot the cashflow: • EMI in Arrears Begin 1 2 3 35 36 +500,000 01.02.2006 01.03.2006 01.04.2006 01.11.2008 01.12.2008 … … … … … … -15,000 -15,000 -15,000 -15,000 -15,000 01.01.2006 End Formula Expression Values in Expression Value of ‘i’ to be determined
IRR - Additional Examplecontd… • Plot the cashflow: • EMI in Advance Begin 1 2 3 35 36 +500,000 01.02.2006 01.03.2006 01.04.2006 01.12.2008 01.01.2009 … … … … … … -15,000 -15,000 -15,000 -15,000 -15,000 -15,000 End 01.01.2006 Formula Expression Values in Expression Value of ‘i’ to be determined
Objectives • Distinguish bond’s coupon rate, current yield, yield to maturity • Interest rate risk • Bond ratings and investors demand for appropriate interest rates
Bond characteristics • Bond - evidence of debt issued by a body corporate or Govt. In India, Govt predominantly • A bond represents a loan made by investors to the issuer. In return for his/her money, the investor receives a legaI claim on future cash flows of the borrower. • The issuer promises to: • Make regular coupon payments every period until the bond matures, and • Pay the face/par/maturity value of the bond when it matures
How do bonds work? • If a bond has five years to maturity, an Rs.80 annual coupon, and a Rs.1000 face value, its cash flows would look like this: • Time 0 1 2 3 4 5 • Coupons Rs.80 Rs.80 Rs.80 Rs.80 Rs.80 • Face Value 1000 • Market Price Rs.____ • How much is this bond worth? It depends on the level of current market interest rates. If the going rate on bonds like this one is 10%, then this bond has a market value of Rs.924.18. Why?
Coupon payments Maturity Face value Lump sum component Annuity component
Bond prices and Interest Rates • Interest rate same as coupon rate • Bond sells for face value • Interest rate higher than coupon rate • Bond sells at a discount • Interest rate lower than coupon rate • Bond sells at a premium
Bond terminology • Yield to Maturity • Discount rate that makes present value of bond’s payments equal to its price • Current Yield • Annual coupon divided by the current market price of the bond Current yield = 80 / 924.18 = 8.66%
Rate of return • Rate of return = Coupon income + price change ---------------------------------------- Investment e.g. you buy 6 % bond at 1010.77 and sell next year at 1020 Rate of return = 60+9.33/1010.77 = 6.86%
Risks in Bonds • Interest rate risk • Short term v/s long term • Default risk • Default premium
Bond pricing • The following statements about bond pricing are always true. • Bond prices and market interest rates move in opposite directions. • When a bond’s coupon rate is (greater than / equal to / less than) the market’s required return, the bond’s market value will be (greater than / equal to / less than) its par value. • Given two bonds identical but for maturity, the price of the longer-term bond will change more (in percentage terms) than that of the shorter-term bond, for a given change in market interest rates. • Given two bonds identical but for coupon, the price of the lower-coupon bond will change more (in percentage terms) than that of the higher-coupon bond, for a given change in market interest rates.
Objectives • Distinguish sample and population • Sampling distributions • Sampling procedures • Estimation – data analysis and interpretation • Testing of hypotheses – one sample data • Testing of hypotheses – two sample data
Types of sampling • Non random or judgement • Random or probability
Methods of sampling • Sampling is the fundamental method of inferring information about an entire population without going to the trouble or expense of measuring every member of the population. Developing the proper sampling technique can greatly affect the accuracy of your results.
Random sampling • Members of the population are chosen in such a way that all have an equal chance to be measured. • Other names for random sampling include representative and proportionate sampling because all groups should be proportionately represented.
Types of Random sampling • Simple random sampling • Systematic Sampling: Every kth member of the population is sampled. • Stratified Sampling: The population is divided into two or more strata and each subpopulation is sampled (usually randomly). • Cluster Sampling: A population is divided into clusters and a few of these (often randomly selected) clusters are exhaustively sampled. • Stratified v/s cluster • Stratified when each group has small variation withn itself but if there is wide variation between groups • Cluster when there is considerable variation within each group but groups are similar to each other
Sampling from Normal Populations • Sampling Distribution of the mean • the probability distribution of sample means, with all samples having the same sample size n. • Standard error of mean for infinite populations • sx = s/n1/2 • Standard Normal probability distribution
Definitions • Density Curve(or probability density function) the graph of a continuous probability distribution • The total area under the curve must equal 1. • Every point on the curve must have a vertical height that is 0 or greater.
Because the total area under the density curve is equal to 1, there is a correspondence between area and probability.
Definition Standard Normal Deviation a normal probability distribution that has a mean of 0 and a standard deviation of 1
-3 -2 -1 0 1 2 3 Definition Standard Normal Deviation a normal probability distribution that has a mean of 0 and a standard deviation of 1 Area = 0.3413 Area 0.4429 z = 1.58 0 Score (z )
Table A-2 Standard Normal Distribution µ = 0 = 1 0 x z
Table for Standard Normal (z) Distribution z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 .0000 .0398 .0793 .1179 .1554 .1915 .2257 .2580 .2881 .3159 .3413 .3643 .3849 .4032 .4192 .4332 .4452 .4554 .4641 .4713 .4772 .4821 .4861 .4893 .4918 .4938 .4953 .4965 .4974 .4981 .4987 .0040 .0438 .0832 .1217 .1591 .1950 .2291 .2611 .2910 .3186 .3438 .3665 .3869 .4049 .4207 .4345 .4463 .4564 .4649 .4719 .4778 .4826 .4864 .4896 .4920 .4940 .4955 .4966 .4975 .4982 .4987 .0080 .0478 .0871 .1255 .1628 .1985 .2324 .2642 .2939 .3212 .3461 .3686 .3888 .4066 .4222 .4357 .4474 .4573 .4656 .4726 .4783 .4830 .4868 .4898 .4922 .4941 .4956 .4967 .4976 .4982 .4987 .0120 .0517 .0910 .1293 .1664 .2019 .2357 .2673 .2967 .3238 .3485 .3708 .3907 .4082 .4236 .4370 .4484 .4582 .4664 .4732 .4788 .4834 .4871 .4901 .4925 .4943 .4957 .4968 .4977 .4983 .4988 .0160 .0557 .0948 .1331 .1700 .2054 .2389 .2704 .2995 .3264 .3508 .3729 .3925 .4099 .4251 .4382 .4495 .4591 .4671 .4738 .4793 .4838 .4875 .4904 .4927 .4945 .4959 .4969 .4977 .4984 .4988 .0199 .0596 .0987 .1368 .1736 .2088 .2422 .2734 .3023 .3289 .3531 .3749 .3944 .4115 .4265 .4394 .4505 .4599 .4678 .4744 .4798 .4842 .4878 .4906 .4929 .4946 .4960 .4970 .4978 .4984 .4989 .0239 .0636 .1026 .1406 .1772 .2123 .2454 .2764 .3051 .3315 .3554 .3770 .3962 .4131 .4279 .4406 .4515 .4608 .4686 .4750 .4803 .4846 .4881 .4909 .4931 .4948 .4961 .4971 .4979 .4985 .4989 .0279 .0675 .1064 .1443 .1808 .2157 .2486 .2794 .3078 .3340 .3577 .3790 .3980 .4147 .4292 .4418 .4525 .4616 .4693 .4756 .4808 .4850 .4884 .4911 .4932 .4949 .4962 .4972 .4979 .4985 .4989 .0319 .0714 .1103 .1480 .1844 .2190 .2517 .2823 .3106 .3365 .3599 .3810 .3997 .4162 .4306 .4429 .4535 .4625 .4699 .4761 .4812 .4854 .4887 .4913 .4934 .4951 .4963 .4973 .4980 .4986 .4990 .0359 .0753 .1141 .1517 .1879 .2224 .2549 .2852 .3133 .3389 .3621 .3830 .4015 .4177 .4319 .4441 .4545 .4633 .4706 .4767 .4817 .4857 .4890 .4916 .4936 .4952 .4964 .4974 .4981 .4986 .4990 * *
Example: If a data reader has an average (mean) reading of 0 units and a standard deviation of 1 unit and if one data reader is randomly selected, find the probability that it gives a reading between 0 and 1.58 units. That is 44.29% of the readings between 0 and 1.58 degrees. Area = 0.4429 P ( 0 < x < 1.58 ) = 0.4429 0 1.58
Central Limit Theorem 1. The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation . 2. Samples all of the same size n are randomly selected from the population of xvalues.
Central Limit Theorem 1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means will be the population mean µ. 3. The standard deviation of the sample means will approach n
Practical Rules Commonly Used: 1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample sizen (not just the values of n larger than 30).
Objectives • Relationship between two or more variables • Scatter diagrams • Regression analysis • Method of least squares