Unformatted text preview: IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 The Basic Theory of Interest
Feb.3.2010 ISysE Dept., KAIST Woo Chang Kim Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Investment Science Chapter 2.1 BASICS OF INTEREST Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Introduction
• Example
– – – – If you deposit $1 in your bank account for a year, You will get $1.08. Q) Why do you get additional $0.08? A) The bank can make money by lending your money to others such as manufacturing companies, service companies, and individuals. • Interest
– Time value of money Spring 2010 Woo Chang Kim, KAIST 3 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Principal and Interest
• Principal: amount invested • Interest: extra money paid on principal • Let’s recall the previous example
– $1.08 = $1 + $0.08 – $1 is the principal – $0.08 is the interest or the time value of money • Let
– A: principal – r: interest per unit period
A A(1+r) • After one unit period, you get
– V = A(1+r)
Spring 2010 Woo Chang Kim, KAIST 0 1 time
4 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Principal and Interest
• Interest rate
– – – – – The interest of one unit principal per unit time Generally, unit time = 1 year Interest rate = Interest/(principal x time) In the previous example, (annual) interest rate = $0.08/$1 = 8% • Then, what about the interest over multiple periods?
– Option 1: interest will be paid only on the principal => simple interest – Option 2: interest will be paid on the principal and accrued interest ??? => compound interest A 0
Spring 2010 Woo Chang Kim, KAIST 10 time
5 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Simple Interest
• Interest earned is proportional to the total time of investment • Linear growth
A(1+r) A A(1+nr) 0 1 n time • After n years V = A(1 + nr) • After t (potentially fractional) V = A(1 + tr)
Spring 2010 Woo Chang Kim, KAIST 6 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Simple Interest
Q1) If the initial amount of money deposited in your bank account is $10,000 and the annual simple interest rate is 10%, how much will you receive in 5 years? A) V = A(1 + nr) = $10,000(1 + 5(10%)) = $15,000 Q2) The same situation as Q1) but, now the time is 4.5 years. A) V = A(1 + tr) = $10,000(1 + 4.5(10%)) = $14,500 Spring 2010 Woo Chang Kim, KAIST 7 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Compound Interest
• (Yearly) compound interest
– principal produces interest every year – earned interest will produce interest in the future on a yearly basis • Geometric growth
A A(1+r) A(1+r)n 0 1 n time • After n years V = A(1 + r) (1 + r) … (1 + r) = A(1+r)n • After t (potentially fractional) V = A(1 + r)t
Spring 2010 Woo Chang Kim, KAIST 8 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Compound Interest
Q1) If the initial amount of money deposited in your bank account is $10,000 and the annual compound interest rate is 10%, how much will you receive in 5 years? A) V = A(1 + r)n = $10,000(1 + 10%)5 = $16,105.1 Q2) The same situation as Q1) but, now the time is 4.5 years. A) V = A(1 + r)t = $10,000(1 + 10%)4.5 = $15,355.6 Spring 2010 Woo Chang Kim, KAIST 9 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Compound Interest
• Rule of 72
– Money invested at r% compounded annually doubles in about 72/r years • Example
– If $10,000 is invested at 10% with annual compounding, – It becomes $20,000 in 7.27 years (exact solution) – From the rule of 72, it is 7.2 years. Spring 2010 Woo Chang Kim, KAIST 10 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Frequent Compounding
• Compounding period
– The period at the end of which interest is paid – Ex) 1 year, 6 months, 1 month, bi‐weekly, daily, etc.
r/m
0 m periods r/m r/m r/m
1 Spring 2010 Woo Chang Kim, KAIST 11 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Frequent Compounding
• Convention: nominal interest rate
– Rates are quoted yearly – But only the fraction of interest corresponding to the compounding period is applied – r% compounded m periods per year for k periods = (1+r/m)k – Ex) $10,000 invest at 10% compounded bi‐yearly for 2.5 years: $10,000(1 + 10%/2)5 = $12,762.8 • Effective interest rate: – The equivalent yearly rate that would produce the same result after 1 year without com‐pounding Spring 2010 Woo Chang Kim, KAIST 12 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Nominal vs. Effective Interest Rates
• Nominal yearly interest rate: rnom • Compounded m times a year
r/m
0 m periods r/m r/m r/m
1 • After k period, the growth factor is (1+rnom/m)k • After a year, the growth factor is (1+rnom/m)m • Thus, the effective interest rate (reff):
1+reff = (1+rnom/m)m → reff = (1+rnom/m)m ‐1
13 Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Nominal vs. Effective Interest Rates
• Ex) What is the effective yearly rate of 10% compounded bi‐ yearly? reff = (1 + 10%/2)2 – 1 = 1.1025 – 1 = 10.25% Spring 2010 Woo Chang Kim, KAIST 14 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Continuous Compounding
• What if the compounding happens very frequently, say, continuously? r/m
0 r/m r/m r/m
1 m periods Now, m goes to infinity! Spring 2010 Woo Chang Kim, KAIST 15 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Continuous Compounding
• Annual growth factor :
m→∞ lim (1+r/m)m = er • Effective interest rate: reff = er – 1 • Growth factor for t years: lim (1+r/m)mt m→∞ = ert • The money grows exponentially in time!
Spring 2010 Woo Chang Kim, KAIST 16 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Continuous Compounding
• Nominal and effective rates for continuous compounding
Interest rate (%) Nominal Effective 1.00 1.01 5.00 5.13 10.00 10.52 20.00 22.14 30.00 34.99 50.00 64.87 75.00 111.70 100.00 171.83 Spring 2010 Woo Chang Kim, KAIST 17 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Continuous Compounding Growth factor for continuous compounding at 10%
Spring 2010 Woo Chang Kim, KAIST 18 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Money Markets Spring 2010 Woo Chang Kim, KAIST 19 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Money Markets
• Discount Rate – The rate at which member banks may borrow short term funds directly from a federal reserve bank. • Federal Funds Rate – The rate that banks charge each other for the use of federal funds. • Prime Rate – The interest rate that commercial banks charge their most creditworthy borrowers. • LIBOR – London Inter‐Bank Offer Rate: The interest rate that the banks charge each other for loans (usually in Eurodollars). • Eurodollars – Deposits denominated in United States dollars at banks outside the United States
Spring 2010 Woo Chang Kim, KAIST 20 IE471‐FEP311 Introduction to Financial Engineering Investment Science Chapter 2.2 – 2.3 PRESENT AND FUTURE VALUES Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Which Offer Will You Accept?
• Case 1
– Offer 1: Receive $1,000 now – Offer 2: Receive $1,000 in a year • Case 2
– Offer 1: Receive $1,000 now – Offer 2: Receive $1,100 now • Case 3
– Offer 1: Receive $1,000 now – Offer 2: $200 $200
$200 $200 $200 $200 time Spring 2010 Woo Chang Kim, KAIST 22 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Present Value
• How to compare cash flows at different time points? • Present value (PV): The equivalent current value of certain amount to be received at a future time • Discounting
– The process of evaluating a future account as a present value – Discounting factor: the factor by which the future value is discounted • Compounding vs. discounting
– Compounding: every time a cash flow is moved forward one period, it is multiplied by (1+r). – Discounting: every time a cash flow is moved backward one period, it is divided by (1+r). Spring 2010 Woo Chang Kim, KAIST 23 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Present Value and Discounting Factor
• PV(A) = d∙A
Where A: amount of money to be received in the future PV(A): present value of A d: discounting factor • Discounting factor
– Discounting factor after 1 year with simple interest r – Discounting factor after k periods with m compounding periods in a year where the nominal rate is r – Discounting factor after time t with continuous compounding where the nominal rate is r
Spring 2010 Woo Chang Kim, KAIST 24 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Assumption for Present Value: Ideal Bank
• Same interest rate for deposits and loans for any size of principal • No service charges or transaction fees • Separate transactions in an account are additive in their effect of future balances. • Different types of accounts (e.g., 1‐year CD, 2‐year CD, etc) could still have different interest rates. • Constant ideal bank: interest rates are independent of the length of time for which it applies • In short: no fee, no asymmetry between borrowing and lending (e.g. 씨앗은행) Spring 2010 Woo Chang Kim, KAIST 25 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 How to Use Ideal Bank?
• Two operations of ideal bank for PV
– Moving a cash flow forward in time (compounding). ⇒ Deposit money in the ideal bank. – Moving a cash flow backward in time (discounting). ⇒ Get a loan from the bank to be paid back later. • Comparison principle: compare different cash flow streams by moving them to the current time (PV). Ideal Bank Spring 2010 Woo Chang Kim, KAIST 26 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Recall: Which Offer Will You Accept?
• Interest rate is 8% with yearly compounding • Case 1
– Offer 1: Receive $1,000 now → PV = $1,000 – Offer 2: Receive $1,000 in a year → PV = $1,000/(1.08) = $925.9 • Case 2
– Offer 1: Receive $1,000 now → PV = $1,000 – Offer 2: Receive $1,100 in a year → PV = $1,100/(1.08) = $1,018.5 Spring 2010 Woo Chang Kim, KAIST 27 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Recall: Which Offer Will You Accept?
• Interest rate is 8% with continuous compounding • Case 1
– Offer 1: Receive $1,000 now → PV = $1,000 – Offer 2: Receive $1,000 in a year → PV = $1,000/exp(0.08) = $923 • Case 2
– Offer 1: Receive $1,000 now → PV = $1,000 – Offer 2: Receive $1,100 in a year → PV = $1,100/exp(0.08) = $1,015 Spring 2010 Woo Chang Kim, KAIST 28 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Present Value of Cash Flow Stream
• • • • Consider a fixed time period for compounding with an interest rate r per period and a cash flow stream (x0, x1, … , xn), and assume cash flows occur at the end of each period Spring 2010 Woo Chang Kim, KAIST 29 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Recall: Which Offer Will You Accept?
• Interest rate is 8% with yearly compounding • Case 3
– Offer 1: Receive $1,000 now → PV = $1,000 – Offer 2:
Time 0 1 2 3 4 5 6 Total
$200 $200 $200 $200 $200 $200 time → PV?
PV 0 $200/(1.08)1 $200/(1.08)2 $200/(1.08)3 $200/(1.08)4 $200/(1.08)5 $200/(1.08)6 $924.6
30 Cash Flow 0 $200 $200 $200 $200 $200 $200 Discount Factor 1 1/(1.08)1 1/(1.08)2 1/(1.08)3 1/(1.08)4 1/(1.08)5 1/(1.08)6 Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Recall: Which Offer Will You Accept?
• Interest rate is 8% with continuous compounding • Case 3
– Offer 1: Receive $1,000 now → PV = $1,000 – Offer 2:
Time 0 1 2 3 4 5 6 Total
$200 $200 $200 $200 $200 $200 time → PV?
PV 0 $200/exp(0.08∙1) $200/exp(0.08∙2) $200/exp(0.08∙3) $200/exp(0.08∙4) $200/exp(0.08∙5) $200/exp(0.08∙6) $915.4
31 Cash Flow 0 $200 $200 $200 $200 $200 $200 Discount Factor 1 1/exp(0.08∙1) 1/exp(0.08∙2) 1/exp(0.08∙3) 1/exp(0.08∙4) 1/exp(0.08∙5) 1/exp(0.08∙6) Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Future Value
• Future value (FV): the equivalent future value of certain amount to be received in the previous time • Comparison principle: compare different cash flow streams by moving them to the future (FV). Ideal Bank Spring 2010 Woo Chang Kim, KAIST 32 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Recall: Which Offer Will You Accept?
• Interest rate is 8% with yearly compounding • Case 1
– Offer 1: Receive $1,000 now → FV = $1,000*(1.08) = $1,080 – Offer 2: Receive $1,000 in a year → FV = $1,000 • Case 2
– Offer 1: Receive $1,000 now → FV = $1,000*(1.08) = $1,080 – Offer 2: Receive $1,100 in a year → FV = $1,100 Spring 2010 Woo Chang Kim, KAIST 33 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Recall: Which Offer Will You Accept?
• Interest rate is 8% with continuous compounding • Case 1
– Offer 1: Receive $1,000 now → FV = $1,000*exp(0.08) = $1,083 – Offer 2: Receive $1,000 in a year → FV = $1,000 • Case 2
– Offer 1: Receive $1,000 now → FV = $1,000*exp(0.08) = $1,083 – Offer 2: Receive $1,100 in a year → FV = $1,100 Spring 2010 Woo Chang Kim, KAIST 34 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Future Value of Cash Flow Stream
• • • • Consider a fixed time period for compounding with an interest rate r per period and a cash flow stream (x0, x1, … , xn), and assume cash flows occur at the end of each period Spring 2010 Woo Chang Kim, KAIST 35 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Recall: Which Offer Will You Accept?
• Interest rate is 8% with yearly compounding • Case 3
– Offer 1: Receive $1,000 now → FV = $1,000*(1.08)6 = $1,586 – Offer 2:
Time 0 1 2 3 4 5 6 Total
$200 $200 $200 $200 $200 $200 time → FV?
PV 0 $200*(1.08)5 $200*(1.08)4 $200*(1.08)3 $200*(1.08)2 $200*(1.08)1 $200*(1.08)0 $1,467
36 Cash Flow 0 $200 $200 $200 $200 $200 $200 Compound Factor (1.08)6 (1.08)5 (1.08)4 (1.08)3 (1.08)2 (1.08)1 (1.08)0 = 1 Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Recall: Which Offer Will You Accept?
• Interest rate is 8% with continuous compounding • Case 3
– Offer 1: Receive $1,000 now → FV = $1,000*exp(0.08) = $1,616 – Offer 2:
Time 0 1 2 3 4 5 6 Total
$200 $200 $200 $200 $200 $200 time → FV?
PV 0 $200*exp(0.08∙5) $200*exp(0.08∙4) $200*exp(0.08∙3) $200*exp(0.08∙2) $200*exp(0.08∙1) $200*exp(0.08∙0) $1,479
37 Cash Flow 0 $200 $200 $200 $200 $200 $200 Discount Factor exp(0.08∙6) exp(0.08∙5) exp(0.08∙4) exp(0.08∙3) exp(0.08∙2) exp(0.08∙1) exp(0.08∙0) = 1 Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Present Value vs. Future Value
• For the same cash flow stream, if the future value is evaluated at the n‐th period, • Q) What about continuous compounding? Spring 2010 Woo Chang Kim, KAIST 38 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Main Theorem on Present Value
• Two cash flow streams are equivalent for an ideal bank iff the present values are the same. • Implications
– Present value is the only number required to characterize a cash flow stream when an ideal bank is available. Spring 2010 Woo Chang Kim, KAIST 39 IE471‐FEP311 Introduction to Financial Engineering Investment Science Chapter 2.4 INTERNAL RATE OF RETURN Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Another Way to Evaluate an Investment
• Example
– You pay $1,000 now. – You receive $1,100 in a year • People typically say that the investment above has 10% return on investment
– Return on investment = ($1,100 ‐ $1,000)/$1,000 Q) Why is the return on investment 10%? A) If the interest rate is 10%, PV of the investment above is 0 => break‐even interest rate is 10%. • Naturally, if this “break‐even interest rate” is higher, the investment has the higher value.
Spring 2010 Woo Chang Kim, KAIST 41 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Internal Rate of Return (IRR)
Let x=(x0, x1, ... , xn) be a cash flow stream. Then, the internal rate of return (IRR) of this stream is a number r satisfying: PV(x) = 0.
Or, Sometimes, it is more convenient to solve for Spring 2010 Woo Chang Kim, KAIST 42 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Example 1
Q) Find the IRR for (‐2,1,1,1) A) We need to find the solution for 0 = −2 + c + c + c
2 3 – From trial and error (or using solvers in Excel, Mathematica, or Matlab), we find c = 0.81
– Thus, 1 c= 1+ r
Spring 2010 Woo Chang Kim, KAIST r = 0.23 43 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Example 1
• Compare to PV, we don’t need to know the interest rate • What is good about IRR IRR reflects the internal structure of the stream without referring to a prevailing interest rate. Spring 2010 Woo Chang Kim, KAIST 44 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Example 2
Q) Find the IRR for (‐2,‐1,2,1) More than one solution?
Spring 2010 Woo Chang Kim, KAIST 45 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Example 2
• We might have more than one candidate for IRR • But fortunately for us, we have… • Main theorem of IRR Suppose the cash flow stream x=(x0, x1, ..., xn) has x0 < 0, and for all k > 0 where at least one term is strictly positive. Then there is a unique positive root to the equation: Moreover if , then the IRR is positive.
Spring 2010 Woo Chang Kim, KAIST 46 IE471‐FEP311 Introduction to Financial Engineering Investment Science Chapter 2.5 EVALUATION CRITERIA Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Present Value vs. Internal Rate of Return
• Main criteria for investment evaluation
– Present vale – Internal rate of return • Under PV
– An investment with a positive net present value (NPV) is worthy of consideration. – Among others, the one with the highest NPV will be selected. • Under the IRR
– An investment with an IRR higher than the prevailing interest rate is worthy of consideration. – Among others, the one with the largest IRR will be selected. Spring 2010 Woo Chang Kim, KAIST 48 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Net Present Value (NPV)
• NPV
– Compute present value of all cash flows associated with an investment • Tree cutting example
– You have the opportunity to plant trees that later can be sold for lumber. There is an initial cost of purchasing and planting the seedlings. Assume the interest rate is 10%. There are two options as to when to harvest: (a) Cut early: (−1, 2) (b) Cut later: (−1, 0, 3) Spring 2010 Woo Chang Kim, KAIST 49 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Net Present Value (NPV)
• Net present values for two options
– (a) NPV = ‐1 + 2/1.1 = 0.82 – (b) NPV = ‐1 + 3/(1.1)2 = 1.48 • Choose (b) Spring 2010 Woo Chang Kim, KAIST 50 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 IRR
• IRR for two options
– (a) ‐1+2c=0 – (b) ‐1+3c2=0 => => c=0.5 => c=0.577 => r=1.0 r=0.7 • Choose (a) • PV (or NPV) and IRR do not necessarily give the same recommendation! Spring 2010 Woo Chang Kim, KAIST 51 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 NPV vs. IRR
• Suppose the tree cutting example is scaled by 10
– Option (a): (‐10,20) – Option (b): (‐10,0,30) • Under NPV (with 10% interest rate)
– Option (a): NPV = 8.18 – Option (b): NPV = 14.8 • Under IRR
– Option (a): r = 1.0 – Option (b): r = 0.7 • NPV can detect the change of thescale • IRR is scale invariant
Spring 2010 Woo Chang Kim, KAIST 52 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 NPV vs. IRR
• Suppose the tree cutting example is repeated infinitely
– Option (a): (‐1,2,‐1,2,…) – Option (b): (‐1,0,3,‐1,0,‐3,…) • Under NPV (with 10% interest rate)
– Option (a): NPV = 4.7 – Option (b): NPV = 5.9 • Under IRR
– Option (a): r = 1.0 – Option (b): r = 0.7 • NPV can detect the repetition. • IRR cannot
Spring 2010 Woo Chang Kim, KAIST 53 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 NPV vs. IRR
• Suppose the tree cutting example with zero NPV cash flow stream added
– Option (a): (‐1,2) + (‐10,10*1.1) = (‐11,13) – Option (b): (‐1,0,3) + (‐10,0,10*1.12) = (‐11,0,15.1) • Under NPV (with 10% interest rate)
– Option (a): NPV = 0.82 – Option (b): NPV = 1.48 • Under IRR
– Option (a): r = 0.182 – Option (b): r = 0.172 • While IRR can evaluate the added zero NPV cash flow stream correctly, NPV cannot.
Spring 2010 Woo Chang Kim, KAIST 54 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 NPV vs. IRR
• In general, NPV and IRR do not have to agree. • NPV is preferred because it distinguishes between cash flows that IRR does not. • But, this means that modeling of cash flows and assumptions need to be considered carefully before the NPV analysis is done • In conclusion, use NPV as the first evaluation criterion, and check IRR the confirm the results Spring 2010 Woo Chang Kim, KAIST 55 IE471‐FEP311 Introduction to Financial Engineering Investment Science Chapter 2.6 APPLICATIONS AND EXTENSIONS Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Net Flows
• In conducting a cash flow analysis
– The net of income minus expense (net profit) should be used as the cash flow each period • Example) Simplico gold mine
Consider leasing a gold mine for 10 years. Gold can be extracted at a rate of up to 10,000 ounce/year at a cost of $200/ounce. Currently the market price is $400/ounce and the interest rate is 10%. What is the present value of this lease? First, it is obvious that, if the mine is leased, the gold should be extracted at its full capacity => produced gold/year = 10,000 oz/yr Thus, profit = 10,000*($400 ‐ $200) = $2M/yr $2 M PV = = $12.29 M ∑ (1.1) k k =1
Spring 2010 Woo Chang Kim, KAIST 57 10 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Cycle Problems
• Cycle problems relate to activities that can be repeated. • When using NPV, you will arrive at a different answer depending on how many times an activity is repeated. • When the NPV criterion is used to evaluate cycle problems, the activities must be evaluated over the same time horizon.
– Option 1: repeat both cash flow streams until they terminate at the same time – Option 2: repeat both cash flow streams infinitely Spring 2010 Woo Chang Kim, KAIST 58 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Cycle Problems – Automobile Purchase Example
• There are two cars that you are considering to buy
– Car A: low price ($20,000), low maintenance ($1,000/yr), short life time (4 years) – Car B: high price ($30,000), high maintenance ($2,000/yr), long life time (6 years) – Neither car has a salvage value and interest rate is 10% • Main issue: life time spans of two cars are different • Solution: repeat the activities so that both will terminate at the same time! Spring 2010 Woo Chang Kim, KAIST 59 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Cycle Problems – Automobile Purchase Example
• Car A:
– PV for One cycle (PVA)= 20,000 + 1,000∑
1 = $22,487 k k =1 (1.1)
3 – PV for three cycles (PVA3)= PVA(1+1/1.14+1/1.18)=$48,336 • Car B:
– PV for One cycle (PVB)= 30,000 + 2,000∑
1 = $37,582 k k =1 (1.1)
5 – PV for two cycles (PVB3)= PVA(1+1/1.16)=$58,795 • What is your choice?
Spring 2010 Woo Chang Kim, KAIST 60 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Taxes
• Taxes alter the cash flows that a company receives, and hence must be taken into account. • Investment decisions are based on actual cash flows. • But the calculation of taxes often depends on fictitious cash flows, such as depreciation. • Taxes can turn profitable investments into unprofitable ones. Spring 2010 Woo Chang Kim, KAIST 61 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Taxes – Machine Replacement Example
• • • • • Machine cost: $10,000 Life span: 4 years with a salvage value of $2000 at the end. Revenue from operation the machine: $3000/yr Straight line depreciation is used Tax rate is 43%, interest rate is 10%. Before tax Depreciation income of machine ‐$10,000 $3,000 $3,000 $3,000 $5,000 $876 0 ‐$2,000 ‐$2,000 ‐$2,000 ‐$2,000 Taxable income 0 $1,000 $1,000 $1,000 $1,000 After tax income ‐$10,000 $2,570 $2,570 $2,570 $4,570 ‐$487
62 Time 0 1 2 3 4 NPV Tax 0 $430 $430 $430 $430 Spring 2010 Woo Chang Kim, KAIST IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Inflation
• Inflation is characterized by an increase in prices with time. • Characterization of inflation: inflation rate f
– – – – Price of a product at t = 0: A Price of a product at t = 1: A(1+f) Price of a product at t = 2: A(1+f)2 Price of a product at t = k: A(1+f)k • Alternative approach: the purchasing power of a dollar decreases by a factor of 1/(1+f) every year. Spring 2010 Woo Chang Kim, KAIST 63 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Inflation
• Constant (or real) dollar vs. actual (or nominal) dollar
– Constant dollar: dollars that have a constant purchasing power – Nominal dollar: dollars that we use every day which are eroded in value by inflation • Real interest rate vs. nominal interest rate
– Real interest rate (r0): the rate that real dollars would earn – Nominal interest rate (r): the rate that we observe with inflation • Relationship between r and r0
1+r0 = (1+r)/(1+f) or, r0 = (r‐f)/(1+f) Spring 2010 Woo Chang Kim, KAIST 64 IE471‐FEP311 Introduction to Financial Engineering Lecture Note 02 Summary – Things to Remember
• Simple interest, compounding interest (with various compounding frequency)
– Nominal vs. effective interest rates • PV and FV
– How to calculate PV and FV under various compounding frequency? • IRR
– How to calculate IRR for a given cash flow stream • NPV vs. IRR
– Pros and cons of two approaches • Cycle problems • Real vs. nominal interest rates
Spring 2010 Woo Chang Kim, KAIST 65 ...
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This note was uploaded on 02/22/2010 for the course FEP 311 taught by Professor Shin during the Spring '10 term at 카이스트, 한국과학기술원.
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