5.1 Compound Amounts. 5: Uniform Series. Uniform Series Compound Amount Factor. Observations. Example 5.1 Uniform Series CA. Example 5.

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1 5: niform Series Cash flows of uniform series Equal Occur each compounding period Also known as annuities, even if not yearly se one series factor instead of several single payment factors Two situations 5.1 Compound Amounts Given the cash flows, determine the compound amount Given the compound amount, determine the cash flows niform Series Compound Amount Factor E s = (F A, i, s-r) r r+1 r+2 s E s = s - (r +1) + s - (r +2) + + E s = [ s - (r +1) + s - (r +2) + + 1] E s = [ ( s - r 1) / i ] E s = (F A, i, m = s-r) r r +1 r +2 E S E s = (F A, i, s-r) Observations s ***Position*** Same time as last cash flow Number flows = one before niform series CA factor or F given A Find a future CA given an annuity, interest rate, and number of series flows Example 5.1 niform Series CA Purchase a car in 3 years Save $2 per month E end-of-month deposits i = 1/2 % per month Amount in savings E 36 = $7, = 2 (F A, 1/2%, 36-) Example 5.2 Delayed CA 36 deposits of $2 at 1/2% / mo Beginning-ofmonth deposits Savings at t = 36 and t = 48? E 36 E CA must be at time of last flow to use F A, so multi-step problem

2 First Determine E 35 E At time of last flow E 35 = $7, = 2 (F A, 1/2%, 35 (-1) ) Have $7, in savings at t = 35 No longer need original cash flows One before Then Other Equivalents E E 36 E E 36 = $7,96.56 = E 35 (F P, 1/2%, ) To From E 48 = $8, = E 35 (F P, 1/2%, ) Sinking Funds Equal deposits to accumulate a given CA Given CA, find E = c s (A F, i, s-r) prior flows Relationship of CA to its prior deposits known from F A r r +1 r +2 s CA = S [ ( s - r - 1 ) / i ] S = CA [ i / ( s - r - 1 ) ] E = c s [ i / ( s - r - 1 )] E = c s (A F, i, s-r ) c s Observations Sinking fund factor also Known as A F Find prior annuity E = c s (A F, i, s-r) given future CA, interest rate, and number of payments r r +1 r +2 s E = c s (A F, i, s-r ) At time of CA minus One Before c s Example 5.3 End-of-Month Sinking Fund Want $1, for car in 3 years 36 end-of-month deposits i = 1/2 % / mo Amount of each deposit? E = $ = 1, (A F, 1/2%, 36-) 1 E 2 1, 36 Different Diagrams Same Meaning Preceeding diagram shows equivalents Dashed, same direction Could show deposit flows Solid lines Depositor s viewpoint Deposits flow from Withdrawal flows to 1 1 E 2 2 1, 1, 36 36

3 Example 5.4 Beginning-of-Month Sinking Fund Still make 36 deposits at 1/2 % per month Beginning-of-month Need CA at t = 35 to use A F E 1, So ask, How much is needed at time 35 to have $1, at time 36? If diagram does not exactly match diagram used to derive factor, then multi-step problem! First Step Amount necessary at t = 35 to have $1, at t = 36 is its discounted amount E 35 1, E 35 = $9,95.25 = 1, (P F, 1/2%, 36-35) Second Step What deposits at, 1, 2,, 35 compound to E 35 = $9,95.25? -1 1 E 35 E E = $ = E 35 (A F, 1/2%, 35 (-1) ) Challenges Position of equivalents Fourth parameter Careful! Two situations 5.2 Discounted Amounts Given the cash flows, determine the discounted amount Given the discounted amount, determine the cash flows niform Series DA Factor E r = (P A, i, s-r) r r+1 r+2 s E r = -(r +1 - r ) + -(r +2 - r ) + + -(s - r ) E r = [ (s - r ) ] E r = [ 1 (1+i) -(s - r ) ] / i E r = (P A, i, s-r)

4 Before first flow Observations E r r r +1 r+2 Er = (P A, i, s-r) Number of flows = One Before niform series discounted amount factor P A since looking for prior given annuity s Example 5.5 niform Series DA Borrow for car Afford $2 / mo 36 end-of-month payments i = 3/4 % / mo E How much car? Amount loaned = Discounted payments E = $6, = 2(P A, 3/4%, 36-) Have $1, (7, ,289.36) more for car by saving $2 / mo instead Example 5.6 DA with Delayed Series Still 36 notes of $2 at 3/4% / mo Delay for 12 months E 2 Determine amount borrowed P / A requires equivalent to be before first cash flow, so multi-step problem What must the debt be at t = 12 to require payments of $2? First Step Time 12 chosen so that P A can be used One period before first flow! E E 12 = $6, = 2(P A, 3/4%, 48-12) Second Step If the debt is E 12 at time 12, then what was it at time? E E The Moral to These Examples Is... Borrowing instead of saving and delaying payments decreases the amount available for a purchaser E = $5, = E 12 (P F, 3/4%, 12-) Now can borrow only $5, The incredible shrinking car

5 Capital Recovery Factor At what rate must c r invested capital be recovered to earn a specified rate of return? r Know from deriving P / A r +1 r +2 DA = S [ 1 (1+i) -(s-r) ] / i S = DA i / [ 1 (1+i) -(s-r) ] E = c r i / [ 1 (1+i) -(s-r) ] E = c r (A P i, s-r) E = c r (A P, i, s-r) s Before first equivalent Observations r +1 r +2 E = c r (A P i, s-r) (niform series) Capital recovery factor A given P since finding an equivalent annuity given a prior cash flow c r Number of flows = One Before r E = c r (A P, i, s-r) s Example 5.7 Capital Recovery Borrow $1, to buy a car 36 EOM notes i = 3/4 % / mo Payments that provide a return of 3/4% / mo? 1, E = $318. = 1,(A P, 3/4%, 36-) Saved $1, with 36 deposits of $254.22, so $63.78 ( ) more each mo E Example 5.8 CR with Delayed Payments 1, E Same loan but delay payments by 12 months Determine payments A P requires equivalent to be one period before first cash flow, so multi-step problem 1, First Step E If $1, owed at time, then how much owed at time 12? Time 12 chosen one period before notes so A P can be used in step 2 E 12 = $1,938.7 = 1,(F P, 3/4%, 12-) Second Step E 12 E Know debt at time ($1,938.7), 14 so 48no longer need original cash flows Note position of E 12, one before payments E = $ = E 12 (A P, 3/4%, 48-12) Delay costs $94.88 ( ) more per month than saving for the car

6 Borrowing is BAD All Together Now Multiple Series Series of deposits followed by series of withdrawals fairly common Strategy se F A or P A on known series se F P or P F on single equivalent Move to either last flow or one period before first flow of unknown series se A F or A P for unknown series Number flows = First - Ex 5.9 Known Deposits, nknown Withdrawals 3, Start deposits in 2 years 8% until t = 3 College from t = 31 to 38. Drawn from account s perspective Retire on from t = 39 to 63? First question? First Step a) How much in savings at t = 3? 3, E First Step b) How much in savings at t = 3? E 3 3, E 3 = $311, = 3,(F A, 8%, 3-1) First Step c) How much in savings at t = 3? E 3 3, E 3 = $311, = 3,(F A, 8%, 3-1) No longer need original flows

7 First Step d) Original problem is now E E 3 = $311, = 3,(F A, 8%, 3-1) Next question? Second Step a) How much in savings at t = 38? E 3 E Second Step b) E 3 E E 38 = $577, 31.8 = E3 (F P, 8%, 38-3) Future Prior Second Step c) E E 38 = $577, 31.8 = E3 (F P, 8%, 38-3) Next question? Third Step Withdrawals? E = $54,8.86 = E38 (A P, 8%, 63-38) Ex 5.1 Known Withdrawals, nknown Deposits As before, but want withdrawals of $6, , Deposits? What is the first question?

8 First Step a) What must be in savings at t = 38? First Step b) What must be in savings at t = 38? E 38 6, E 38 6, E 38 = $64, = 6, (P A, 8%, 63-38) Do not need original flows First Step c) What must be in savings at t = 38? E Second Step a) What must be in savings at t = 3? E 3 E E 38 = $64, = 6, (P A, 8%, 63-38) Next question? Second Step b) What must be in savings at t = 3? Second Step c) What must be in savings at t = 3? E 3 E E E 3 = $346,34.97 = E38 (P F, 8%, 38-3) Future Prior E 3 = $346,34.97 = E38 (P F, 8%, 38-3) Next question?

9 Third Step Deposits? E 3 = $3, = E3 (A F, 8%, 3-1) 5.4 Bonds Government and private industry borrow directly from the public Bypass banks Contract to make future payments Sold on open market Max price = Discounted payments Terminology Terminology Maturity date: time of E final payment (1) 8 1, Coupon: series paid to bondholder ($8) Redemption, face, or par value: paid at maturity date in addition to coupon ($1,) Coupon rate: rate used to compute the series payments (8%) Coupon = Face Value Coupon Rate 8 = 1, 8% Max to pay (E ) Example 5.11 Purchase of a New Bond E Max to pay to earn 9%? 1, E = $ = 8(P A, 9%, 1-) + 1, (P F, 9%, 1-) Offer ($935.82) is below par ($1,) when IRR > Coupon Rate 8% Max to pay to earn 8%? E , E = $1, = 8(P A, 8%, 1-) + 1, (P F, 8%, 1-) Offer ($1,) is at par ($1,) when IRR = Coupon Rate 7% Max to pay to earn 7%? E , E = $1,7.24 = 8(P A, 7%, 1-) + 1, (P F, 7%, 1-) Offer ($1,7.24) is above par ($1,) when IRR < Coupon Rate

10 Example 5.12 Purchase of an Existing Bond Max to pay at time 3 for same bond if third coupon just paid to earn an IRR of 9% 3 E , E3 = $ = 8 (P A, 9%, 1-3) + 1, (P F, 9%, 1-3) Offer ($949.67) is below par ($1,) when IRR > Coupon Rate 5.5 Loans Balloon notes Final payment larger or balloons Principal and interest Components of each payment Early repayment Effect on duration of loan Example 5.13 Balloon Note Borrow $8, for house 3/4% / mo for 5 years M as for 3 yr loan Balance B t = 6 M B , First, determine monthly note M M = 8, (A P, ¾%, 36-) = $ , Step 2 Second, determine balloon payment At time 6, with payment 6 just made Payments 61,, 36 remain Easiest to discount remaining payments B M B = $76,74.11 = (P A, ¾%, 36-6) Alternative Step 2 Could use balance equation once M known M B , = 8, (F P, ¾%, 6-) (F A, ¾%, 6-) B B = $76,74.11, as before Discounting easier Principal and Interest Two parts of each note or payment Loan Payment = Interest + Principal Principal component reduces the debt Interest reduces individual s income tax Home loan Business loan Interest on several payments Interest = Loan Payments Debt Reduction Easier than period-by-period balances

11 Example 5.14 Principal and Interest Payment House loan for $1, 3/4% / mo 3 years Notes 3, 4,, 14 in next tax year. Interest? M , First, determine the monthly payment M = $84.62 = 1, (A P, ¾%, 36-) Interest = Loan Payments Debt Reduction Loan Payments = $9, = Debt Reduction = Debt before Debt after Before = E2 = $99,89.35 Step 2 = (P A, ¾%, 36-2) After = E14 = $99, = (P A, ¾%, 36-14) Reduction = Before After = $ Interest = Loan Payments Debt Reduction = 9, Interest = $8, E E Dollars Principle & Interest Components over Time 8 Interest = $189, Principal Month Early Repayment Notes on a loan Always pay interest first Remainder reduces principal Extra payment reduces the debt Amount of legally required note not changed Number of payments is reduced Some lenders allow, others charge penalty Example 5.15 Early Loan Repayment Step 2 89, = 84.62(P A, ¾%, s-12) Same loan E 89, /4% (P A, ¾%, s-12) = notes of $84.62 (P A, ¾%, ) = (P A, ¾%, ) = Pay $1, extra at t = 12 (P A, i, m) = [1 (1+i) Before = (P A, ¾%, 36-12) -m s ]/ i = 99, $84.62 until 252 too much, until 251 too little After = $89, = 99, , 251 notes of $84.62, then pay off balance $84.62 / mo continues until time s when B = $ = 89,316.48(F P, ¾%, ) 251 $89, = DA of Remaining Payments 84.62(F A, ¾%, ) Note 252 = $ = (F P, ¾%, )

12 Step 3 Eliminate 18 (36-252) notes of $84.62 Invest savings at 1/2% / mo until t = 36 $116,15 = ( )(F P, ½%, ) (F A, ½%, ) $116,15 36 than $1, on car If invest 12 at 1/2% instead of paying off 3/4% loan $56,727= $1,(F P, ½%, 36-12) Eliminate 3/4% 1/2% $59,288 = 116,15-56,727 Yeah, the course is a pain, but Multiple Interest Rates Same concepts as single rates Might not be able to use easy formulas Simplifications for constant regions r Compound Amounts r+1 r+2 s i r+1 i r+2 i s CA CA = r+2 r+3 s + r+3 s + + CA = ( r+2 r+3 s + r+3 s + +1) = CA / ( r+2 r+3 s + r+3 s + +1) Example 5.16 Multiple Rate Series and CAs = $1,, CA? $4, = 1, [ (1.6)(1.65)(1.7) + (1.65)(1.7) ] CA = 2,,? % 6.5% 7% $ = 2, / [ (1.6)(1.65)(1.7) + (1.65)(1.7) ] CA DA Discounted Amounts r r+1 r+2 s i r+1 i r+2 i s DA = ( r+1 ) ( r+1 r+2 s ) -1 DA = [ ( r+1 ) ( r+1 r+2 s ) -1 ] = DA / [( r+1 ) ( r+1 r+2 s ) -1 ]

13 Example 5.17 Multiple Rate Series and DA DA DA Step 2 9 6% % 7% 7.5% = $1,, DA? $3, = 1,[(1.6) -1 + ( ) -1 + ( ) -1 + ( ) -1 ] 9 6% % 7% 7.5% DA = $2,,? $ = 2, / [ (1.6) -1 + ( ) -1 + ( ) -1 + ( ) -1 ] Regions with Constant Rates Example 5.18 Series CA with Regions E CA 7 Discount and compound up to boundaries Equation relating series and DA or CA Solve equation for unknown % 6% E 7 = (F A, 5%, 7-4) CA = E 7 (F P, 6%, 12-7) + (F A, 6%, 12-7) CA = Step 2 CA = If equals $1,, then CA = $9,855.8 = , If CA equals $2,, then = $2,29.26 = 2, / Example 5.19 Series DA with Regions E 7 DA % 6% E 7 = (P A, 6%, 12-7) DA = E 7 (P F, 5%, 7-4) + (P A, 5%, 7-4) DA = 6.362

14 Step 2 DA = If equals $1,, then DA = $6,362. = , If DA equals $2,, then = $3, = 2, / Au revoir Making money is rough work!

4: Single Cash Flows and Equivalence

4.1 Single Cash Flows and Equivalence Basic Concepts 28 4: Single Cash Flows and Equivalence This chapter explains basic concepts of project economics by examining single cash flows. This means that each