Discounting Intro/Refresher H. Scott Matthews 12-706/73-359 Lecture 12 - Oct. 11, 2004.

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Presentation transcript:

Discounting Intro/Refresher H. Scott Matthews / Lecture 12 - Oct. 11, 2004

Admin Issues zPS 2 Handed Back yAverage ~40/50 zMidterm Handed Out, due next Wed.

IRA example zWhile thinking about careers.. zGovernment allows you to invest $2k per year in a retirement account and deduct from your income tax yInvestment values will rise to $5k soon zStart doing this ASAP after you get a job. zSee ‘IRA worksheet’ in RealNominal

Notes on Notation zPV = $FV / (1+i) n = $FV * [1 / (1+i) n ] yBut [1 / (1+i) n ] is only function of i,n y$1, i=5%, n=5, [1/(1.05) 5 ]= = (P|F,i,n) zAs shorthand: yFuture value of Present: (P|F,i,n) xSo PV of $500, 5%,5 yrs = $500*0.784 = $392 yPresent value of Future: (F|P,i,n) yAnd similar notations for other types

Timing of Future Values zNoted last time that we assume ‘end of period’ values zWhat is relative difference? zConsider comparative case: y$1000/yr Benefit for 5 5% yAssume case 1: received beginning yAssume case 2: received end

Timing of Benefits zDraw 2 cash flow diagrams zNPV 1 = / / / / yNPV 1 = = $4,545 zNPV 2 = 1000/ / / / / yNPV 2 = = $4,329 zNPV 1 - NPV 2 ~ $216 zNotation: (P|U,i,n)

Relative NPV Analysis zIf comparing, can just find ‘relative’ NPV compared to a single option yE.g. beginning/end timing problem yNet difference was $216 zAlternatively consider ‘net amounts’ yNPV 1 = = $4,545 yNPV 2 = = $4,329 y‘Cancel out’ intermediates, just find ends yNPV 1 is $216 greater than NPV 2

Uniform Values - Theory zAssume end of period values zStream = F/(1+i) +F/(1+i) F/(1+i) n z(P|U,i,n) = $1[(1+i) -1 +(1+i) (1+i) -n ] z[(1+i) -1 +(1+i) (1+i) -n ] = “A” z[1+(1+i) -1 +(1+i) (1+i) 1-n ] = A(1+i) zA(1+i) - A = [1 - (1+i) -n ] zA = [1 - (1+i) -n ] / i zStream = F*(P|U,i,n) = F*[1 - (1+i) -n ] / i

Uniform Values - Application zRecall $1000 / year for 5 years example zStream = F*(P|U,i,n) = F*[1 - (1+i) -n ] / I z(P|U,5%,5) = zStream = 1000*4.329 = $4,329 = NPV 2

Why Finance? zTime shift revenues and expenses - construction expenses paid up front, nuclear power plant decommissioning at end. z“Finance” is also used to refer to plans to obtain sufficient revenue for a project.

Borrowing zNumerous arrangements possible: ybonds and notes ybank loans and line of credit ymunicipal bonds (with tax exempt interest) zLenders require a real return - borrowing interest rate exceeds inflation rate.

Issues zSecurity of loan - piece of equipment, construction, company, government. More security implies lower interest rate. zProject, program or organization funding possible. (Note: role of “junk bonds” and rating agencies. zVariable versus fixed interest rates: uncertainty in inflation rates encourages variable rates.

Issues (cont.) zFlexibility of loan - can loan be repaid early (makes re-finance attractive when interest rates drop). Issue of contingencies. zUp-front expenses: lawyer fees, taxes, marketing bonds, etc.- 10% common zTerm of loan zSource of funds

Sinking Funds zAct as reverse borrowing - save revenues to cover end-of-life costs to restore mined lands or decommission nuclear plants. zLow risk investments are used, so return rate is lower.

Borrowing zSometimes we don’t have the money to undertake - need to get loan zi=specified interest rate zA t =cash flow at end of period t (+ for loan receipt, - for payments) zR t =loan balance at end of period t zI t =interest accrued during t for R t-1 zQ t =amount added to unpaid balance zAt t=n, loan balance must be zero

Equations zi=specified interest rate zA t =cash flow at end of period t (+ for loan receipt, - for payments) zI t =i * R t-1 zQ t = A t + I t zR t = R t-1 + Q t R t = R t-1 + A t + I t z R t = R t-1 + A t + (i * R t-1 )

Uniform payments zAssume a payment of U each year for n years on a principal of P zR n =-U[1+(1+i)+…+(1+i) n-1 ]+P(1+i) n zR n =-U[( (1+i) n -1)/i] + P(1+i) n zUniform payment functions in Excel zSame basic idea as earlier slide

Example zBorrow $200 at 10%, pay $ at end of each of first 2 years zR 0 =A 0 =$200 zA 1 = - $115.24, I 1 =R 0 *i = (200)(.10)=20 zQ 1 =A 1 + I 1 = zR 1 =R 0 +Q t = zI 2 =10.48; Q 2 = ; R 2 =0

Repayment Options zSingle Loan, Single payment at end of loan zSingle Loan, Yearly Payments zMultiple Loans, One repayment