Financial Instruments: Bonds, Shares, and Commercial Paper - Interest Rates & Metrics - Pr, Apuntes de Administración de Empresas

¡Descarga Financial Instruments: Bonds, Shares, and Commercial Paper - Interest Rates & Metrics - Pr y más Apuntes en PDF de Administración de Empresas solo en Docsity! Chapter 2 Financial transactions: price and value creation. Question Is 1 euro worth the same today than within 100 years? No! recall VK ≥ 0 Financial Transactions An example A UN PRECIO INCREÍBLE. inside” HP Paválion 1360 11-k103ns HP Pavilion 360 11-410215 Con Windows 10 incluido, 1 3485: 12 cuela RE 4,11% TINO%6 4 09,1 DE | impacta tota cl crécio adeuciacdo M1BA5E Food for thought… NPV, IRR, APR annual percentage rate (fees contained) AER annual effective rate (VAN, TIR, TAE) APR Sections 1. The price of a financial transaction 2. Discounting and capitalizing cash flows 3. Decisions on investment and financing 4. Creating value in financial transactions 5. Investment selection using NPV criteria 6. Transactions with more than two periods 7. An example of calculating the price of a financial transaction 8. Simple vs. compounded interest rate 9. Some mathematical simplifications 1. The price of financial transactions: the interest rate Interest rate: The price p of money today (t = 0) in terms of future money (t) The additional percentage of future money that is being exchanged for each euro today (t=0) It is usually expressed as a percentage rate increase, where i is the interest rate 100 1 i p  1. The price of financial transactions: the interest rate Examples of financial transactions and interest rates: 1. Company issues bonds so that receives € 1 today and must return € 1.04 a year after. • p = 1.04 • Annual interest rate i = 4% 2. The firm receives a bank loan of € 6,000 to be returned in a year with annual interest of € 300 • p = 6,300/6,000 = 1.05 • Annual interest rate i = 5% 3. Discount trade receivables (bills from clients) for an amount of € 1,000 in a year. Fee is 2% discount today. • p = 1,000 / (1,000 * 0.98) = 1,000/980 = 1.0204 • Annual interest rate i = 2.04% 2. Discounting and capitalizing cash flows Capitalization Given: Interest of a transaction Amount of money exchanged today Calculate: The amount of money that is exchanged in the future. 2. Discounting and capitalizing cash flows Discounting Given: Interest of a transaction Future amount of money exchanged Calculate: The current amount of money that is exchanged today 2. Discounting and capitalizing cash flows Discounting : Price of X euros in the future (t) expressed in euros today: i X 1 2. Discounting and capitalizing cash flows Examples financial transactions and discounting : 1. Issuing bonds • Discounting 1.04 euros at an annual interest rate of 4% is 1 euro 2. Bank loan • Discounting 6,300 euros at an annual interest rate of 5% is 6,000 euro 3. Discount trade receivables • Discounting 1,000 euros at an annual interest rate of 2.0408% is 980 euros 3. Decisions on investment and financing Investor (Seller of money today): Price: Interest rate (i) Return on investment (IRR: Internal rate of return) Value: Assumption: For every person j there is an interest rate ij such that this person is indifferent between 1 € today (t = 0) or 1 + ij € in the future (t). (j = Inv) Minimum required return (iInv) Investors make an investment (for a sum X): Price> Value  In unit terms (1 € now): (1 + i)> (1 + iInv )  Return on investment (i)> minimum required return, that is: i>iInv  Total amount (X is euros today): X(1+i) > X(1+iInv)  Xi = I > X iInv = VInv (X) Interest to be received: Xi = I Net value of X euros: X iInv = VInv (X) 3. Decisions on investment and financing The investor chooses the transaction that generates the bigger return (IRR), provided that it exceeds the minimum required return (iInv) Examples: 1. Bank loan: annual interest rate of 5% 2. Discount: annual interest rate of 2.04% • What are the cash flows for the bank in the examples? • Which investment project does the bank choose?  For the investor (bank), the annual yield of the loan is greater than the discount.  If the bank must choose between a loan or a discount for the same amount, the bank prefers the bank loan (higher yield) 3. Decisions on investment and financing The company will choose the transaction that bears a lower cost of capital, provided that it is less than the maximum cost capital (iFin) Examples: 1. Bank loan: annual interest rate of 5% 2. Discount: annual interest rate of 2.04% • What are the cash flows for the firm in the examples? • Which investment project does the firm choose? • For the company, the cost of capital of bank loan is greater than the commercial discount • If the company needs to choose between borrow from bank or use a trade receivables discount, by the same amount, prefers the discount. 3. Decisions on investment and financing For a financial transaction to take place it is necessary that the maximum cost of capital that the person financed (borrower) is willing to pay is greater than or equal to the minimum return required by investors, that is, it is necessary that the buyer (financed) of the good (money today) values it more than the seller (investor): iFin > = iInv Value generated for every euro transacted = iFin - iInv 3. Decisions on investment and financing Value created in euros of time t:  In unit terms (per euro): iFin - iInv  Total amount (X € exchanged): XiFin - XiInv = VFin (X) - VInv(X) = X (iFin -iInv ) The amount paid / received as interest determines the distribution of the value generated between the investor and borrower:  Value appropriated by borrower: VFin(X) - I = X (iFin - i)  Value appropriated by investor: I - VInv(X) =X (i - iInv ) 4. Creating value in financial transactions 4. Creating value in financial transactions  Example: Discount iFin = 6% iInv = 4% X = 980 i = 2.04% Appropriated value by the bank = X (i-iInv ) = -19.2 Appropriated value by the firm = X (iFin -i) = 38.8 Total value generated = 19.6  Why is the value for the investor negative?  Do you think that this transaction will take place? 4. Creating value in financial transactions Example: Tie company (Chapter 1)  The entrepreneur contributes 45,000 euros at t=0 and obtains a profit (residual rent) of B=9,000 plus the initial contribution, i.e. cash flow at t=1 year is 54,000  Question: What is the anual return of this project? 54,000/45,000=1.20  return is 20% anual  Assume now that iInv = 10% (Vk=4,500)  What is the economic profit of the entrepreneur?  B-Vk=9,000-4,500=4,500  X (1 + i) - X (1 + iInv) = 54,000-45,000*1.10=54,000-49,500=4,500  So far, we have measured cash flows in t=1. However, it is more common to value transactions at time t = 0 4. Today: (cont) Creating value in financial transactions, NPV Recall the example: Tie company (Chapter 1)  The entrepreneur contributes 45,000 euros at t=0, B=9,000. cash flow at t=1 year is 54,000  Question: What is the anual return of this project? 54,000/45,000=1.20  return is 20% anual  Assume that iInv = 10% (we used that in chapter 1…Vk=4,500)  economic profit ? B-Vk=9,000-4,500=4,500  So far, we measured in t=1. However, it is more common to value transactions at time t = 0….. economic profit at t=0? = 4,090.91 t=0 t=1 -45,000 +54,000 Today: Value in Financial Transactions Money exchanges, interest, move forward or backwards. NPV Criteria for financial transactions Transactions with several periods Compound interest rates Questions Do we create value? Before… recall VCL ≥ p ≥ VK Now… iFin ≥ i ≥ iInv (later) When comparing several options how do we choose? 4. Creating value in financial transactions 69.57 04.1 300,6 000,6 1 )1( NPVInv     invi iX X Example Bank Loan : Calculate NPV iFin = 6% iInv = 4% X = 6,000 i= 5% Total value generated (today) = 114.29 60.56 06.1 300,6 000,6 1 )1( NPV Fin Fin     i iX X 29.114 06.1 1 04.1 1 300,6 1 1 1 1 )1(NPV Total FinInv                   ii iX 4. Creating value in financial transactions Example Bank Loan: Summary of findings -400 -300 -200 -100 0 100 200 300 400 0 0,005 0,01 0,015 0,02 0,025 0,03 0,035 0,04 0,045 0,05 0,055 0,06 0,065 0,07 0,075 0,08 0,085 0,09 0,095 0,1 NPV (investor) NPV (financed) NPVinv 4. Creating value in financial transactions Example Bank Loan: Summary of findings -400 -300 -200 -100 0 100 200 300 400 0 0,005 0,01 0,015 0,02 0,025 0,03 0,035 0,04 0,045 0,05 0,055 0,06 0,065 0,07 0,075 0,08 0,085 0,09 0,095 0,1 NPV (investor) NPV (financed) iFin = 6% iInv = 4% 4. Creating value in financial transactions Example Bank Loan Now assume iFin = iInv = 5% X = 6,000 NPV (bank) = -6,000 +6,300 / 1.05 = 0 NPV (company) = +6,000-6,300/1.05 = 0 Total value generated (today) = 0 Example loan and discount: Both transactions of the same amount = 6,300 iFin = 6% iInv = 4% Loan interest rate = 5% Discount interest rate = 2.04% From the viewpoint of the bank (investor): The bank would prefer the loan (does not depend on iInv) 5. Investment selection using NPV criteria 69.57 05.01 1 1 1 300,6 04.1 300,6 000,6)(             invi loanNPV 307.116 0204.01 1 1 1 300,6 04.1 300,6 174,6)(             invi discountNPV 174 05.01 1 0204.01 1 300,6.)()(           disNPVloanNPV Example loan and discount: (Continued) From the viewpoint of the company (borrower): The company prefers discount (does not depend on iFin) 5. Investment selection using NPV criteria 6037.56 1 1 05.01 1 300,6 06.1 300,6 000,6)(           Fini loanNPV 6037.230 1 1 0204.01 1 300,6 06.1 300,6 174,6)(           Fini discountNPV 174 05.01 1 0204.01 1 300,6)(.)(           loanNPVdisNPV 5. Investment selection using NPV criteria Where does the bank obtain the resources to grant loans or discounts? Deposits Wholesale funding (interbank)  Bonds  Etc..  Should the bank issue bonds to lend?  Evaluation of aggregated projects 6. Transactions with more than two periods  For example: Bank gives loan € 6,000 to be repaid at the end of the second year, and at the end of every year the bank receives an interest payment of € 300.  Cash flows from the point of view of the company: C0= +6,000 C1= -300 C2= -6,300 t = 0 year 1 year 2  What is the interest rate on the loan?  Interest rate = 300/6,000=0.05  5% annually Loan repayment table: Period t Amount loan (balance) Pt-1 Amount paid Ct Interest payment It=i P t-1 Principal payment Ct- It 1 6,000 300 300 0 2 6,000 6,300 300 6,000 6. Transactions with more than two periods  Implicit assumptions: 1. The transaction (n = 2 years) can be divided into n different transactions of 1 year (n = 2 transactions 1 year) 2. The interest rate for each period is the same for all transactions. 3. At the last period (n) the remaining balance on the loan should be zero (Pt = n= 0) 6. Transactions with more than two periods  Equation to be solved: 6,000(1+i)- 3,000 = 3,465-[6,000(1+i)- 3,000]i -3,465 – 3,000 (1 + i) + 6,000 (1 + i)2 = 0 +6,000 – 3,000 - 3,465 = 0 = NPV (i) (1 + i) (1 + i)2 i= 0.05  5% annual interest rate NOTE: The 2 year interest rate is equal to ibian= (1 + i) 2-1 = 0.1025 that is, 10.25% bi-annual interest rate. 6. Transactions with more than two periods Loan repayment table: t Amount loan (balance) Pt-1 Amount paid Ct Interest payment It=i P t-1 Principal payment Ct- It 1 6,000 3,000 300 2,700 2 3,300 3,465 165 3,300 6. Transactions with more than two periods Generalizing: C0 .... Ct ..... Cn 0 t N  The price of this transaction is the interest rate such that NPV equals zero: NPV(i)=0  The interest rate refers to the time period of the units that measure time (month, year etc..) 0 i)(1 C )( 0 t t     n t iNPV 6. Transactions with more than two periods 7. An example A UN PRECIO INCREÍBLE. inside” HP Paválion 1360 11-k103ns HP Pavilion 360 11-410215 Con Windows 10 incluido, 1 3485: 12 cuela RE 4,11% TINO%6 4 09,1 DE | impacta tota cl crécio adeuciacdo M1BA5E 7. An example of calculating the price of a financial transaction 7. An example of calculating the price of a financial transaction  Example PC: Cash flows from the buyer´s viewpoint (cash flows are the same for the bank with reversed sign)  Calculate the monthly interest rate Mes 0 1 4 7 10 13 16 19 22 Dinero 549.900 -72.931 -72.931 -72.931 -72.931 -72.931 -72.931 -72.931 -72.931 0 i)(1 C )( 1 t t 0     n t CiNPV 8. Simple vs. compound interest rate The price of financial transactions is generally reported as Annual Percentage Rate (APR) or Annual Equivalent or Effective Rate (AER) Compound interest: (1 +AER) = (1 + it) t it interest rate of one period (t number of periods in a year) AER equivalent annual interest rate 8. Simple vs. compound interest rate In the example of the computer: i12 = 0.00521869  monthly interest rate  What is the annual equivalent rate?  AER = 6.45% Two ways to solve that are equivalent: 1. Expressing the time periods in annual terms 2. Conversion rules from simple to compound interest rate 8. Simple vs. compound interest rate 1. Expressing the time periods in annual terms Año 0 1/12 4/12 7/12 10/12 13/12 16/12 19/12 22/12 Dinero -549.900 72.931 72.931 72.931 72.931 72.931 72.931 72.931 72.931 0 )1( 1 )1( 1 )1( 1 )1( 1 )1( 1 )1( 1 )1( 1 )1( 1 72931549900 12/2212/1912/1612/1312/1012/712/412/1                       iiiiiiii AER= i= 6.45% 8. Simple vs. compound interest rate Simple interest (nominal): it = i t i interest rate of one period it equivalent interest rate for t periods We will only use the nominal interest rate to compute the cash flows of some financial transactions. 8. Simple vs. compound interest rate In the example of the computer: i12 = 0.521869  monthly interest rate Annual Equivalent Rate (AER) = 6.45% A simple form of approximation to AER is simply by multiplying by 12 the monthly interest rate: 12 * 0.521869 = 6.26% It is a approximation and it is not equal to AER 8. Simple vs. compounded interest rate Example: The monthly interest rate of a financial transaction is i12 =0.5%  Compute the equivalent interest rates for a quarter, a semester and a year.  Make an approximation using the simple interest rate rule period n AER Simple quarter 3 1.5075% 1.50% semester 6 3.0378% 3.00% year 12 6.1678% 6.00% 8. Simple vs. compounded interest rate 2. Loan of €100 for one year with a nominal annual interest rate (simple) of 10% payable at maturity. +100 -110 t = 0 t = 1 (1 year) Interest rate is 10% per year (nominal = APR) We prefer loan 2. The nominal interest rate should not be used to make financial decisions 8. Simple vs. compound interest rate  What would be the borrower´s decision if we compute NPV instead of AER?  NPV of loan 1 using interest rate of loan 2  What is the NPV of loan 1 using 10,25% as interest rate?  NPV of loan 2 using interest rate of loan 1  Borrower prefers Loan 2. Same conclusion with AER!!. 02219.0 )05.01( 5 )05.01( 5 100%)10 ,1 ( 2     loanNPV 02219.0 )1025.01( 110 100%)25.10 ,2 ( 1   loanNPV 02219.0 )10.01( 5 )10.01( 5 100%)10 ,1 ( 12/1     loanNPV 8. Simple vs. compounded interest rate Exercise 4 (b) Exercise 5 Proof: i i)(1 1 -1 i)(1 A )( n 1 t      AiPV n t 9. Some mathematical simplifications 9. Some mathematical simplifications In the example of the computer: AER = 6.45% Constant payments of 72,931 euros Calculate the net present value First calculate the quarterly interest = (1.0645)1 / 4= 1.015738 205,544 015738.0 )015738.01( 1 1 931,72)( 8    iPV 9. Some mathematical simplifications In the example of the computer: If calculate the present value two months after we signed the transaction: 544,205 (1.0645)2 / 12= 549,900