CASH FLOW ANALYSIS HANDOUTS

CASH FLOW ANALYSIS HANDOUTS 1

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RateOfCompoundingAnalysis.hava FV1 r m T FV PV1 r m T PV 0.09 1 0.5 1.04403 0.09 1 0.5 0.957826 0.09 1 1 1.09 0.09 1 1 0.917431 0.09 1 2 1.1881 0.09 1 2 0.84168 0.09 1 5 1.53862 0.09 1 5 0.649931 0.09 1 20 5.60441 0.09 1 20 0.178431 0.09 1 40 31.4094 0.09 1 40 0.031838 0.09 2 0.5 1.045 0.09 2 0.5 0.956938 0.09 2 1 1.09202 0.09 2 1 0.91573 0.09 2 2 1.19252 0.09 2 2 0.838561 0.09 2 5 1.55297 0.09 2 5 0.643928 0.09 2 20 5.81636 0.09 2 20 0.171929 0.09 2 40 33.8301 0.09 2 40 0.02956 0.09 4 0.5 1.04551 0.09 4 0.5 0.956474 0.09 4 1 1.09308 0.09 4 1 0.914843 0.09 4 2 1.19483 0.09 4 2 0.836938 0.09 4 5 1.56051 0.09 4 5 0.640816 0.09 4 20 5.93015 0.09 4 20 0.16863 0.09 4 40 35.1666 0.09 4 40 0.028436 0.09 12 0.5 1.04585 0.09 12 0.5 0.956158 0.09 12 1 1.09381 0.09 12 1 0.914238 0.09 12 2 1.19641 0.09 12 2 0.835831 0.09 12 5 1.56568 0.09 12 5 0.6387 0.09 12 20 6.00915 0.09 12 20 0.166413 0.09 12 40 36.1099 0.09 12 40 0.027693 0.09 52 0.5 1.04599 0.09 52 0.5 0.956035 0.09 52 1 1.09409 0.09 52 1 0.914002 0.09 52 2 1.19703 0.09 52 2 0.8354 0.09 52 5 1.5677 0.09 52 5 0.637876 0.09 52 20 6.04024 0.09 52 20 0.165556 0.09 52 40 36.4845 0.09 52 40 0.027409 0.09 365 0.5 1.04602 0.09 365 0.5 0.956003 0.09 365 1 1.09416 0.09 365 1 0.913941 0.09 365 2 1.19719 0.09 365 2 0.835289 0.09 365 5 1.56823 0.09 365 5 0.637664 0.09 365 20 6.04831 0.09 365 20 0.165336 0.09 365 40 36.582 0.09 365 40 0.027336 0.09 INF 0.5 1.04603 0.09 INF 0.5 0.955997 0.09 INF 1 1.09417 0.09 INF 1 0.913931 0.09 INF 2 1.19722 0.09 INF 2 0.83527 0.09 INF 5 1.56831 0.09 INF 5 0.637628 0.09 INF 20 6.04965 0.09 INF 20 0.165299 0.09 INF 40 36.5982 0.09 INF 40 0.027324 FV2 r m T FV PV2 r m T PV 3

0.03 1 0.5 1.01489 0.03 1 0.5 0.985329 0.03 1 1 1.03 0.03 1 1 0.970874 0.03 1 2 1.0609 0.03 1 2 0.942596 0.03 1 5 1.15927 0.03 1 5 0.862609 0.03 1 20 1.80611 0.03 1 20 0.553676 0.03 1 40 3.26204 0.03 1 40 0.306557 0.03 2 0.5 1.015 0.03 2 0.5 0.985222 0.03 2 1 1.03022 0.03 2 1 0.970662 0.03 2 2 1.06136 0.03 2 2 0.942184 0.03 2 5 1.16054 0.03 2 5 0.861667 0.03 2 20 1.81402 0.03 2 20 0.551262 0.03 2 40 3.29066 0.03 2 40 0.30389 0.03 4 0.5 1.01506 0.03 4 0.5 0.985167 0.03 4 1 1.03034 0.03 4 1 0.970554 0.03 4 2 1.0616 0.03 4 2 0.941975 0.03 4 5 1.16118 0.03 4 5 0.86119 0.03 4 20 1.81804 0.03 4 20 0.550042 0.03 4 40 3.30528 0.03 4 40 0.302546 0.03 12 0.5 1.01509 0.03 12 0.5 0.98513 0.03 12 1 1.03042 0.03 12 1 0.970482 0.03 12 2 1.06176 0.03 12 2 0.941835 0.03 12 5 1.16162 0.03 12 5 0.860869 0.03 12 20 1.82075 0.03 12 20 0.549223 0.03 12 40 3.31515 0.03 12 40 0.301646 0.03 52 0.5 1.01511 0.03 52 0.5 0.985116 0.03 52 1 1.03045 0.03 52 1 0.970454 0.03 52 2 1.06182 0.03 52 2 0.941781 0.03 52 5 1.16178 0.03 52 5 0.860745 0.03 52 20 1.8218 0.03 52 20 0.548907 0.03 52 40 3.31897 0.03 52 40 0.301298 0.03 365 0.5 1.01511 0.03 365 0.5 0.985113 0.03 365 1 1.03045 0.03 365 1 0.970447 0.03 365 2 1.06183 0.03 365 2 0.941767 0.03 365 5 1.16183 0.03 365 5 0.860713 0.03 365 20 1.82207 0.03 365 20 0.548825 0.03 365 40 3.31995 0.03 365 40 0.301209 0.03 INF 0.5 1.01511 0.03 INF 0.5 0.985112 0.03 INF 1 1.03045 0.03 INF 1 0.970446 0.03 INF 2 1.06184 0.03 INF 2 0.941765 0.03 INF 5 1.16183 0.03 INF 5 0.860708 0.03 INF 20 1.82212 0.03 INF 20 0.548812 0.03 INF 40 3.32012 0.03 INF 40 0.301194 4

AnnuityCashFlowAnalysis.hava r 0.09 A 6000 g 0.12 T 10 Yearly r1 0.09 SA1 Date CF FV PV 1 6000 6000.0 5504.59 2 6000 12540.0 10554.7 3 6000 19668.6 15187.8 4 6000 27438.8 19438.3 5 6000 35908.3 23337.9 6 6000 45140.0 26915.5 7 6000 55202.6 30197.7 8 6000 66170.8 33208.9 9 6000 78126.2 35971.5 10 6000 91157.6 38505.9 SA2 Date CF FV PV 1 6000.0 6000.0 5504.59 2 6720.0 13260.0 11160.7 3 7526.4 21979.8 16972.4 4 8429.57 32387.6 22944.2 5 9441.12 44743.5 29080.2 6 10574.1 59344.5 35385.2 7 11842.9 76528.5 41863.7 8 13264.1 96680.1 48520.5 9 14855.8 120237 55360.5 10 16638.5 147697 62388.8 Monthly r2 0.0075 SA3 Date CF FV PV 12 500 6253.79 5717.46 24 500 13094.2 10944.6 36 500 20576.4 15723.4 48 500 28760.4 20092.4 60 500 37712.1 24086.7 72 500 47503.5 27738.4 84 500 58213.5 31077.0 96 500 69928.1 34129.2 108 500 82741.6 36919.7 120 500 96757.1 39470.8 SA4 Date CF FV PV 12 557.834 6603.63 6037.29 24 628.582 14664.2 12256.8 36 708.301 24424.7 18664.1 48 798.132 36164.1 25264.8 60 899.355 50203.1 32064.7 72 1013.42 66909.3 39069.9 84 1141.94 86704.2 46286.6 96 1286.77 110070 53721.1 108 1449.96 137560 61380.1 120 1633.85 169806 69270.2 5

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Fixed-Rate Mortgages 1 Mortgage Basics We adopt the following continuous-time notation: LB(t) = the outstanding loan balance at time t. IP (t) = the interest payment on the loan made at time t. P P (t) = the principal payment on the loan made at time t. i = the annual loan interest rate. T = the (remaining) life of the loan. All continuous-time formulas have discrete-time counterparts, which we include as we go. Let c(t) denote the rate of cash flow paid to the mortgage holder (i.e. the bank). With an outstanding loan balance of LB(t) at time t, in the next t units of time the bank increases the loan balance by ilb(t) t, the interest it expects the mortgagee to pay, but will subtract the payment c(t) t it receives. Therefore, the change in loan balance from t to t + t is which, as t 0, implies that LB(t + t) LB(t) = ilb(t) t c(t) t, (1) d LB(t) = ilb(t) c(t). (2) dt The solution to the differential equation (2) may be readily verified as 1 Note that we can invert (3) to reveal that LB(t) = e it{ t } LB(0) e is c(s)ds. (3) 0 t LB(0) = e it LB(t) + e is c(s)ds, (4) 0 which merely states that present value of all cash the bank receives up to time t including repayment of the outstanding loan balance at time t (which pays off the loan) must equal the original loan balance, i.e., the bank is indifferent to these two cash flow streams. 1 Use the product rule of differentiation and the Fundamental Theorem of Calculus. 7

We now proceed to calculate the principal and interest payments at time t. The total payment M = IP (t) + P P (t) (5) is the sum of the principal and interest payments, and since IP (t) = ilb(t), P P (t) = M ilb(t). (6) Since we now know how to compute M and LB(t) we can compute P P (t), too. However, we can do better. Since ilb(t) = IP (t) and c(t) = IP (t) + P P (t), it follows immediately from (2) that d LB(t) = P P (t). (7) dt Now take the time derivative of both sides of (6) and use the identity (7) to obtain that or that d P P (t) = ip P (t), (8) dt P P (t) = P P (0)e it. (9) We conclude that the principal payments increase exponentially. Perhaps this is not too surprising: we know the loan balance decreases exponentially, and this is solely due to the principal payments. In discrete-time, the period n principal payment is given by P P (n) = P P (1)(1 + i/12) n 1. 2 Conventional Fixed-Rate Mortgage 2.1 Fixed-rate payment A conventional fixed-rate mortgage requires the mortgagee to pay a fixed amount M per month for a duration of T years. In continuous-time, the fixed rate is Y = 12M per year. Setting c(s) = Y and LB(T ) = 0 into (4), we have that or that LB(0) = T 0 e is Y ds = Y i ( 1 e it ) (10) Y = ilb(0). (11) 1 e it The corresponding estimate of the monthly payment is then M = Y/12. Example 1. Consider a 30-year, fixed-rate mortgage for 125,000 at 6.75%. In continuous-time, the yearly payment is 0.0675(125, 000) Y = 1 e 0.0675(30) = 9720.55 (12) 8

and the monthly payment is estimated as The actual monthly payment is computed as M = 2.2 Remaining loan balance M = Y/12 = 810.05. (13) (0.0675/12)(125, 000) = 810.75. (14) 1 (1 + 0.0675/12) 360 After replacing c(s) = Y in (3) with its solution (11), and performing a little algebra, we have that { } 1 e i(t t) LB(t) = LB(0) 1 e it. (15) The expression in braces equals the proportion LB(t)/LB(0) of the original loan balance remaining, and T t represents the time (or number of periods) remaining on the loan. To obtain the corresponding discrete-time formula, one simply replace e ix with (1 + i/12) 12x : { } 1 (1 + i/12) 12(T t) LB(t) = LB(0) 1 (1 + i/12) 12T. (16) This is a very useful fact to remember. Remark 1. Suppose it is time t and we wish to take out a loan in the amount of LB(t) for a period of T t at the same rate of interest i. What would be the new (monthly) mortgage payment M t T? Intuition suggests the process is regenerative, namely, the process starts over and this claim is easily verified. From (11) and (15), M t T = 1 12 Y t T = 1 12 ilb(t) 1 e i(t t) = 1 12 ilb(0) 1 e it = 1 12 Y 0 T = M. (17) Example 2. For the loan of Example 1, what is the loan balance after 10 years and 8 months? In continuous-time, the value for t = 10. 6 with T t = 19. 3. Therefore, LB(10. 6) LB(0) = 0.839657, which implies that the loan balance is 104,957.13. In discrete-time, the value of t = 128 with T t = 232 payment periods remaining. Keep in mind you have to convert to monthly periods! Therefore, LB(128) LB(0) = 1 (1 + 0.0675/12) 232 = 0.8392387, 1 (1 + 0.0675/12) 360 which implies the loan balance is 104,904.84. 9

2.3 Prepayment options We now address two practical questions often asked by a mortgagee. In what follows, we let M denote the current monthly payment computed as in Example 1. (1) If I want to pay off my loan in S years, then how much more do I have to pay per month? To answer this question, use (11); that is, M + A = 1 12 from which one may solve for A. In discrete-time, M + A = 1 12 ilb(0), (18) 1 e is ilb(0) 1 (1 + i/12) 12S (19) (2) If I pay an additional A dollars per month, then how many years S will it take to pay off my loan? To answer this question, use (10); that is, 12(M + A) LB(0) = (1 e is ), (20) i from which one may solve for S. In discrete-time, LB(0) = 12(M + A) (1 (1 + i/12) 12S ). (21) i Example 3. Consider once again the loan of Example 1, and suppose the remaining duration of the loan is 19 years and 4 months. Recall that M = 810.05 in continuous-time and M = 810.75 in discrete-time. If we add 100 to our payment each month, then how quickly will the loan be paid off? Let S denote the remaining life of the loan. In continuous-time, we have 104, 957.13 = (910.05)(12) (1 e 0.0675S ), 0.0675 which gives S = 15.4996 years. In discrete-time, we have 104, 904.84 = 910.75 0.0675/12 [1 (1 + 0.675/12) S ], which gives S = 186.1012 months or 15.5084 years. How much do we have to add to our monthly payment to pay off the loan in 10 years (or 120 payments periods)? Let A denote the additional amount. In continuous-time, we have (M + A) = 1 12 104, 957.13(0.0675) 1 e 0.0675(10) = 1202.79, which means that A = 392.74. In discrete-time, we have (M + A) = 1 12 which means that A = 393.81. 104, 904.84(0.0675) = 1205.56, 1 (1 + 0.0675/12) 120 10

3 Homework Problems Consider a 15-year fixed-rate mortgage for 200,000 at 6.25%. In what follows, the first solution is the discrete-time answer while the solution in brackets is the continuous-time answer. 1. What is the monthly payment? M = 1714.85 [M = 1712.16] 2. What is the loan balance after 4 years, 3 months? LB(51) = 160, 792.27 [LB(4.25) = 160, 833.17] 3. Suppose the remaining duration of the loan is 10 years and 9 months. If we pay 2000 each month how quickly will the loan be paid off? Let S denote the remaining life of the loan. S = 104.44 months or 8.70 years [S = 8.68 years] 4. Suppose the remaining duration of the loan is 10 years, 9 months. How much do we have to add to our monthly payment to pay off the loan in 5 years? Let A denote the additional amount. A = 1412.44 [A = 1409.01] Load files MA Hmk.01.02.hava, MA Hmk.03.hava and MA Hmk.04.hava to see more complete solutions to these homework problems. 11

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MA.01.hava T 5 LB 10000 r 0.1 n 1 taxrate 0.4 FixedRateLCF i IP PP netap TP taxshield cashflow LB netlb 0 10000 10000 10000 1 1000.0 1637.98 0 2637.98 400.0-2237.98 8362.02 8362.02 2 836.2 1801.78 0 2637.98 334.48-2303.5 6560.24 6560.24 3 656.02 1981.96 0 2637.98 262.41-2375.57 4578.28 4578.28 4 457.83 2180.15 0 2637.98 183.13-2454.85 2398.13 2398.13 5 239.81 2398.13 0.0 2637.94 95.92-2542.02 0.0 0.0 FixedRatePV 0.000583653 FixedPrincipalLCF i IP PP netap TP taxshield cashflow LB netlb 0 10000 10000 10000 1 1000.0 2000.0 0 3000.0 400.0-2600.0 8000.0 8000.0 2 800.0 2000.0 0 2800.0 320.0-2480.0 6000.0 6000.0 3 600.0 2000.0 0 2600.0 240.0-2360.0 4000.0 4000.0 4 400.0 2000.0 0 2400.0 160.0-2240.0 2000.0 2000.0 5 200.0 2000.0 0.0 2200.0 80.0-2120.0 0.0 0.0 FixedPrincipalPV 1.59162E-12 13

MA.02.hava T 5 LB 10000 r 0.1 n 1 taxrate 0.4 FixedRateLCF i IP PP netap TP taxshield cashflow LB netlb 0 10000 10000 10000 1 1000.0 1637.98 100 2737.98 400.0-2337.98 8362.02 8262.02 2 826.2 1811.78 100 2737.98 330.48-2407.5 6450.24 6350.24 3 635.02 2002.96 100 2737.98 254.01-2483.97 4347.28 4247.28 4 424.73 2213.25 100 2737.98 169.89-2568.09 2034.03 1934.03 5 193.4 1934.03 0.0 2127.43 77.36-2050.07 0.0 0.0 FixedRatePV 0.00357269 FixedPrincipalLCF i IP PP netap TP taxshield cashflow LB netlb 0 10000 10000 10000 1 1000.0 2000.0 100 3100.0 400.0-2700.0 8000.0 7900.0 2 790.0 2000.0 100 2890.0 316.0-2574.0 5900.0 5800.0 3 580.0 2000.0 100 2680.0 232.0-2448.0 3800.0 3700.0 4 370.0 2000.0 100 2470.0 148.0-2322.0 1700.0 1600.0 5 160.0 1600.0 0.0 1760.0 64.0-1696.0 0.0 0.0 FixedPrincipalPV 9.09495E-13 14

MA.03.hava T 5 LB 10000 r 0.1 n 1 taxrate 0.4 FixedRateLCF i IP PP netap TP taxshield cashflow LB netlb 0 10000 10000 10000 1 1000.0 1637.98 100 2737.98 400.0-2337.98 8362.02 8262.02 2 826.2 1811.78 175 2812.98 330.48-2482.5 6450.24 6275.24 3 627.52 2010.46 250 2887.98 251.01-2636.97 4264.78 4014.78 4 401.48 2236.5 325 2962.98 160.59-2802.39 1778.28 1453.28 5 145.33 1453.28 0.0 1598.61 58.13-1540.48 0.0 0.0 FixedRatePV -0.000163605 FixedPrincipalLCF i IP PP netap TP taxshield cashflow LB netlb 0 10000 10000 10000 1 1000.0 2000.0 100 3100.0 400.0-2700.0 8000.0 7900.0 2 790.0 2000.0 175 2965.0 316.0-2649.0 5900.0 5725.0 3 572.5 2000.0 250 2822.5 229.0-2593.5 3725.0 3475.0 4 347.5 2000.0 325 2672.5 139.0-2533.5 1475.0 1150.0 5 115.0 1150.0 0.0 1265.0 46.0-1219.0 0.0 0.0 FixedPrincipalPV 2.27374E-13 15

MA_Refi.01.hava T 5 LB 10000 r 0.1 n 1 taxrate 0.4 new_r 0.08 LCF_FixedRate i IP PP netap TP taxshield cashflow LB netlb 0 10000 10000 10000 1 1000.0 1637.98 0 2637.98 400.0-2237.98 8362.02 8362.02 2 836.2 1801.78 0 2637.98 334.48-2303.5 6560.24 6560.24 3 656.02 1981.96 0 2637.98 262.41-2375.57 4578.28 4578.28 4 457.83 2180.15 0 2637.98 183.13-2454.85 2398.13 2398.13 5 239.81 2398.13 0.0 2637.94 95.92-2542.02 0.0 0.0 LCF_newFixedRate i IP PP netap TP taxshield cashflow LB netlb 0 10000 10000 10000 1 800.0 1704.56 0 2504.56 320.0-2184.56 8295.44 8295.44 2 663.64 1840.92 0 2504.56 265.46-2239.1 6454.52 6454.52 3 516.36 1988.2 0 2504.56 206.54-2298.02 4466.32 4466.32 4 357.31 2147.25 0 2504.56 142.92-2361.64 2319.07 2319.07 5 185.53 2319.07 0.0 2504.6 74.21-2430.39 0.0 0.0 currentloan_pv 0.000583653 currentloan_newpv -342.569 newloan_pv -0.0107963 refibenefit i value 1 114.504 2 205.867 3 274.254 4 319.796 5 342.558 16

CFA.01.hava revyr1 200000 rev_growth 0 costyr1 160000 cost_growth 0 investment 120000 loanamount 0 loaninterestrate 0.1 loantype FIXED_PRINCIPAL propertylife 5 taxrate 0.4 projecthorizon 4 salvagevalue 50000 costofcapital 0.15 loanlcf i IP PP netap TP taxshield cashflow LB netlb 0 0 0 0 1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 2 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 3 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 4 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 AfterTaxCF Date Rev Cost EBITDA Int Depr EBT Taxes NetInc LPCF Depr Inv FCF PV NPV 0 0-120000 -120000 118569-1431 1 200000 160000 40000 0.0 24000 16000 6400 9600 0 24000 0 33600 102754 136354 2 200000 160000 40000 0.0 24000 16000 6400 9600 0 24000 0 33600 84567 118167 3 200000 160000 40000 0.0 24000 16000 6400 9600 0 24000 0 33600 63652 97252 4 200000 160000 40000 0.0 24000 16000 6400 9600 0 24000 39600 73200 0 73200 NPV -1431 17

CFA.02.hava revyr1 200000 rev_growth 0 costyr1 160000 cost_growth 0 investment 120000 loanamount 80000 loaninterestrate 0.1 loantype FIXED_PRINCIPAL propertylife 5 taxrate 0.4 projecthorizon 4 salvagevalue 50000 costofcapital 0.15 loanlcf i IP PP netap TP taxshield cashflow LB netlb 0 80000 80000 80000 1 8000.0 20000.0 0 28000.0 3200.0-24800.0 60000.0 60000.0 2 6000.0 20000.0 0 26000.0 2400.0-23600.0 40000.0 40000.0 3 4000.0 20000.0 0 24000.0 1600.0-22400.0 20000.0 20000.0 4 2000.0 20000.0 0.0 22000.0 800.0-21200.0 0.0 0.0 AfterTaxCF Date Rev Cost EBITDA Int Depr EBT Taxes NetInc LPCF Depr Inv FCF PV NPV 0 80000-120000 -40000 52309 12309 1 200000 160000 40000 8000.0 24000 8000 3200 4800-20000 24000 0 8800 51355 60155 2 200000 160000 40000 6000.0 24000 10000 4000 6000-20000 24000 0 10000 49058 59058 3 200000 160000 40000 4000.0 24000 12000 4800 7200-20000 24000 0 11200 45217 56417 4 200000 160000 40000 2000.0 24000 14000 5600 8400-20000 24000 39600 52000 0 52000 NPV 12309 18

CFA.03.hava revyr1 200000 rev_growth 0 costyr1 160000 cost_growth 0 investment 120000 loanamount 80000 loaninterestrate 0.1 loantype FIXED_RATE propertylife 5 taxrate 0.4 projecthorizon 4 salvagevalue 50000 costofcapital 0.15 loanlcf i IP PP netap TP taxshield cashflow LB netlb 0 80000 80000 80000 1 8000.0 17237.7 0 25237.7 3200.0-22037.7 62762.3 62762.3 2 6276.23 18961.4 0 25237.7 2510.49-22727.2 43800.9 43800.9 3 4380.09 20857.6 0 25237.7 1752.04-23485.6 22943.3 22943.3 4 2294.33 22943.3 0.0 25237.7 917.73-24319.9 0.0 0.0 AfterTaxCF Date Rev Cost EBITDA Int Depr EBT Taxes NetInc LPCF Depr Inv FCF PV NPV 0 80000-120000 -40000 52874 12874 1 200000 160000 40000 8000.0 24000 8000 3200 4800-17238 24000 0 11562 49243 60805 2 200000 160000 40000 6276.23 24000 9724 3890 5834-18961 24000 0 10873 45756 56629 3 200000 160000 40000 4380.09 24000 11620 4648 6972-20858 24000 0 10114 42505 52619 4 200000 160000 40000 2294.33 24000 13706 5482 8224-22943 24000 39600 48881 0 48881 NPV 12874 19

AfterTaxCF Date 0 1 2 3 4 Rev 200000 200000 200000 200000 Cost 160000 160000 160000 160000 EBITDA 40000 40000 40000 40000 Int 8000 6000 4000 2000 Depr 24000 24000 24000 24000 EBT 8000 10000 12000 14000 Taxes 3200 4000 4800 5600 NetInc 4800 6000 7200 8400 LPCF 80000-20000 -20000-20000 -20000 Depr 24000 24000 24000 24000 Inv -120000 0 0 0 39600 FCF -40000 8800 10000 11200 52000 PV 52309 51355 49058 45217 0 NPV 12309 60155 59058 56417 52000 AfterTaxCF Date 0 1 2 3 4 Rev Cost EBITDA Int Depr EBT Taxes NetInc LPCF Depr Inv FCF PV NPV AfterTaxCF Date 0 1 2 3 4 Rev Cost EBITDA Int Depr EBT Taxes NetInc LPCF Depr Inv FCF PV NPV 20

Depreciation A. Tax Effect of Depreciation 1. For the current year a company has income of R and cost of C for a net income before tax of R-C for the year. The company s marginal tax rate is τ. The company s after-tax cash flow for the year is therefore (1- τ)[r C]. Example 1. Say R = 3,000 and C = 1,000 with τ = 40%. Then the company s after-tax cash flow for the year is (1 0.40)[3,000 1,000] = 1,200. Note that this may be equivalently expressed as (0.60)[3,000] (0.60)[1,000] = 1,800 600 = 1,200. That is, the company receives a net of 60 cents for each dollar of revenue it takes in but it only costs the company 60 cents for dollar of expense. On the expense side note that -(1- τ)[c] = -C + τc = -1000 + (0.40)(1000), and so the net effect of being able to deduct expenses it that the company receives a tax benefit of τc = 400, i.e., it wrote a check for 1,000 during the year and receives back 400 due to the expense write-off. 2. Let s assume the revenue R is sufficiently high to offset the expenses. In what follows we shall ignore the after-tax cash flow due to revenue, since it is a constant as far as the analysis of the after-tax cash flow due to cost. Now let s suppose for the current year a company spends I = 1000 on capital investment for equipment. The equipment is classified as 5-year property and straight-line depreciation is used. If the company were able to write-off this capital investment as an expense, then the after-tax cash flow would be 600, exactly as in Example 1. However, the government s perspective on this is that the equipment is being used over a 5- year period and provides economic value over its entire book life of 5 years. (The company may choose to use this equipment for more than 5 years.) Consequently, the government will only allow the company to expense each year an appropriate amount for the wear and tear or depreciation of the equipment. Since straight-line depreciation is being used, the depreciation expense each year is 1000/5 = 200. There is NO out-of-pocket depreciation expense each year. That is, the 1000 is spent now; the 200 is computed only for tax purposes. As explained in #1 above, from the after-tax cash flow perspective, this 200 depreciation allowance yields a tax benefit of 0.40(200) = 80 each year for the next 5 years. That is, the after-tax cash flow due to the capital investment for the next 5 years (ignoring the salvage value due to possible sale of equipment) is: 0 1 2 3 4 5-1,000 80 80 80 80 80 Note how the undiscounted sum of the depreciation allowances over the 5-year horizon adds up to 400, the total tax benefit. Again, the total tax benefit of 400 would be added to the 1,000 if this were an ordinary expense (for materials, labor, etc.). 21

Suppose the company were to use a 6% cost of capital for purposes of determining the Net Present Value of this cash flow stream, then the NPV = -1,000 + {[1 (1.06) -5 ]/(0.06)} 80 = -1,000 + 4.212(80) = -663.01, which translates to a net cost of 66.3% per dollar spent on this capital equipment. Note that if this equipment had been expensed, then the net cost would be 60% per dollar spent. The loss of 6.3% is due to the DELAY of receiving the benefit of 80 per year for 5 years, as above, as opposed to receiving the total of 400 right NOW. Which cash flow would a company prefer? For the purpose of after-tax cash flow, the company would prefer to be able to write-off as much depreciation as quickly as it can. That is, it would prefer accelerated depreciation. For the purpose of reporting profits, however, the company would prefer to delay the expense! The IRS allows a company to keep two sets of books, one for tax purposes and the other for reporting to shareholders. B. Accounting for Depreciation in the Income Statement Example 2. Suppose our company has a net operating income (R C) for the year of 3,000 and a depreciation expense of 1,000. According to what we have learned above, the taxes owed to the government this year would be 0.40(3,000 1,000) = 800. The Flow In is therefore 3,000 and the Flow Out is 800 for a net of 2,200. Here is the income statement approach to this calculation: Income Statement Approach Flow In Flow Out Approach Net Revenue 3,000 Net Revenue 3,000 Expenses Total 1,000 Taxes -800 Income tax 800 Net Income after tax 1,200 Adjustments Depreciation 1,000 After tax cash flow 2,200 After tax cash flow 2,200 In general notation, the two approaches to after tax cash flow are, respectively: Income Statement Approach: Flow In Flow Out Approach: = (1 - τ)[r C Dep] + Dep = (1- τ)[r C] + τ Dep, = [R C] - τ[r C Dep] = (1- τ)[r C] + τ Dep, which produce the same result, as it should. Note that if the depreciation adjustment below the line (i.e. below the Net Income after tax) is NOT made, then the income statement approach would incorrectly report an after tax cash flow of 1,200. The discrepancy would arise because the depreciation allowance is NOT an out of pocket expense. 22

SIMPLE EQUIPMENT SELECTION ANALYSIS I. A company must acquire a piece of equipment. It is considering the following two options: a. Initial cost = 1,000. Annual cost = 300. Salvage value = 100. 10-year lifetime. b. Initial cost = 1,300. Annual cost = 270. Salvage value = 200. 15-year lifetime. Discount rate = 12%. Ignore taxes. ANALYSIS: I. COMPUTE PRESENT VALUE: The present value cost of option (a) is: 1000 + 300{ [ 1 (1.12) -10 ] 0.12} - 100(1.12) -10 = 2662.87. The present value cost of option (b) is: 1300 + 270{ [ 1 (1.12) -15 ] 0.12} - 200(1.12) -15 = 3024.99. 2. COMPUTE ANNUAL EQUIVALENT: The Annual Equivalent (or Worth) of option (a) is: 2662.87 { 0.12 [ 1 (1.12) -10 ] } = 471.28, and its lifetime cost is: 471.28 0.12 = 3927.33. The Annual Equivalent (or Worth) of option (b) is: 3024.99 { 0.12 [ 1 (1.12) -15 ] } = 444.14, and its lifetime cost is: 444.14 0.12 = 3701.18. 23

II. A manufacturer requires a chemical finishing process for a product produced under contract for 4 years. Three options are available. Initial cost of process device A = 100,000. It has an annual cost = 60,000 and a salvage value = 10,000 after 4 years. Initial cost of process device B = 150,000. It has an annual cost = 50,000 and a salvage value = 30,000 after 6 years. It is also possible to subcontract at 100,000 per year. Appropriate cost of capital is 10%. Marginal tax rate = 40%. ANALYSIS: We shall assume process devices are classified as 5-year property with SL depreciation. For purposes of estimating salvage value we shall assume that the market value of both process devices depreciates in a straight line to its terminal projected market (salvage) values. Thus, the salvage value for process device A is 10,000, and for process device B it is 70,000. Finally, we assume there are other profitable ongoing operations sufficient to cover any losses on sale of equipment. Process Device A 0 1 2 3 4 Operations cost 60,000 60,000 60,000 60,000 Depreciation 20,000 20,000 20,000 20,000 Total cost 80,000 80,000 80,000 80,000 Tax benefit 32,000 32,000 32,000 32,000 After-tax cost 48,000 48,000 48,000 48,000 Depreciation 20,000 20,000 20,000 20,000 Investment -100,000 0 0 0 *14,000 After-tax cash flow -100,000-28,000-28,000-28,000-14,000 Process Device B 0 1 2 3 4 Operations cost 50,000 50,000 50,000 50,000 Depreciation 30,000 30,000 30,000 30,000 Total cost 80,000 80,000 80,000 80,000 Tax benefit 32,000 32,000 32,000 32,000 After-tax cost 48,000 48,000 48,000 48,000 Depreciation 30,000 30,000 30,000 30,000 Investment -150,000 0 0 0 **54,000 After-tax cash flow -150,000-18,000-18,000-18,000 36,000 * 14,000 = 10,000 0.40(10,000 20,000) ** 54,000 = 70,000 0.4(70,000 30,000). NPV @ 10% of after-tax cash flow for A = -179,194. Annual Equivalent (AE) for 4 years = -56,530. NPV @ 10% of after-tax cash flow for B = -170,175. AE for 4 years = -53,685. NPV @ 10% of after-tax cash flow for subcontracting = -190,192. AE for 4 years = -60,000. Process device B is the best option. SENSITIVITY ANALYSIS: What is the salvage value of B after 4 years that would cause the manufacturer to be indifferent in choosing between it and A? Let δ SV denote the change in salvage value of B for which we would be indifferent between it and A. The new salvage value = 70 + δ SV. The change to after-tax cash flows for B will simply be (1-τ)δ SV = 0.60δ SV. The present value of this change @ 10% = 0.60δ SV (1.1) -4. This must equal (-179,194) - (-170,175) = -9,019. Thus, δ SV = -22,008 so that the break-even salvage value = 47,992. 24

Selecting a Bid Price A municipality needs to dispose of 1 million tons of refuse each year for the next 5 years. It is requesting bids from firms for the business. A bid of R means that the firm contractually agrees to receive $R/ton each year for the next 5 years from the municipality in exchange for disposing of the refuse. A new firm is planning to bid for the business, and has asked you to recommend a bid price. The firm will have to acquire a new fleet of trucks to pick up the refuse. The cost of the trucks to handle the proposed volume is $6 million. The projected market value of the trucks is $2 million at the end of 5 years. The trucks are classified as 5-year property and straight-line depreciation is used. A landfill is also required for disposal. The municipality s sole landfill owner is currently charging $5/ton each year. Fuel and labor are projected at $1/ton per year. There is an administrative cost of $0.5 million each year to set up contracts, acquire capital goods, plan operations, etc. To be qualified for a bid a firm must finance at least 50% of the initial investment with equity. The loan interest rate offered by the local bank is 10% compounded annually with principal repaid in five equal annual installments. The marginal tax rate of the company is 40%. The after-tax cost of equity capital is 20%. What bid price do you recommend? SAMPLE WORKSHEET: Revenues Cost Landfill cost Labor/fuel cost Administrative cost EBITDA Interest expense Depreciation EBT Taxes Net Income Adjustments Loan principal cash flow Depreciation Investment Free cash flow 0 1 2 3 4 5 25

26

PBA.01.hava cashflows ((0, -175), (1, 110), (2, 121)) marr 0.1 numcompoundperiods 1 ProjectBalances_marr t cashflow projectbalance 0-175 -175.0 1 110-82.5 2 121 30.25 PresentValueAnalysis_marr date flow rpv 0-175 25.0 1 110 220.0 2 121 121.0 irr 0.20322 ProjectBalances_irr t cashflow projectbalance 0-175 -175.0 1 110-100.564 2 121 0.0 PresentValueAnalysis_irr date flow rpv 0-175 0.0 1 110 210.564 2 121 121.0 27

PBA.02.hava cashflows ((0, -120000), (1, 33600), (2, 33600), (3, 33600), (4, 73200)) marr 0.15 numcompoundperiods 1 ProjectBalances_marr t cashflow projectbalance 0-120000 -120000 1 33600-104400 2 33600-86460.0 3 33600-65829.0 4 73200-2503.35 PVA_marr date flow rpv 0-120000 -1431.3 1 33600 136354 2 33600 118167 3 33600 97252.2 4 73200 73200.0 irr 0.144803 ProjectBalances_irr t cashflow projectbalance 0-120000 -120000 1 33600-103776 2 33600-85203.5 3 33600-63941.2 4 73200-0.1105 PVA_irr date flow rpv 0-120000 -0.0643 1 33600 137376 2 33600 118803 3 33600 97541.1 4 73200 73200.0 PBA.03.hava cashflows ((0, -40000), (1, 8800), (2, 10000), (3, 11200), (4, 52000)) marr 0.15 numcompoundperiods 1 ProjectBalances_marr t cashflow projectbalance 0-40000 -40000.0 1 8800-37200.0 2 10000-32780.0 3 11200-26497.0 4 52000 21528.5 PVA_marr date flow rpv 0-40000 12309.0 1 8800 60155.3 2 10000 59058.6 3 11200 56417.4 4 52000 52000.0 irr 0.254885 ProjectBalances_irr t cashflow projectbalance 0-40000 -40000.0 1 8800-41395.4 2 10000-41946.5 3 11200-41438.0 4 52000 0.0389 PVA_irr date flow rpv 0-40000 0.0157 1 8800 50195.4 2 10000 51946.5 3 11200 52638.1 4 52000 52000.0 28

IRA.01.hava numcompoundperiods 1 SpotRates date rate 1 0.03 2 0.04 3 0.06 4 0.07 ForwardRates date1 date2 rate 0 1 0.03 0 2 0.04 0 3 0.06 0 4 0.07 1 2 0.0500971 1 3 0.0753261 1 4 0.0836756 2 3 0.101161 2 4 0.100865 3 4 0.10057 Discounts date1 date2 discountfactor 0 1 0.970874 0 2 0.924556 0 3 0.839619 0 4 0.762895 1 2 0.952293 1 3 0.864808 1 4 0.785782 2 3 0.908132 2 4 0.825147 3 4 0.90862 IRA.02.hava numcompoundperiods 2 SpotRates date rate 1 0.03 2 0.04 3 0.06 4 0.07 ForwardRates date1 date2 rate 0 1 0.03 0 2 0.04 0 3 0.06 0 4 0.07 1 2 0.0500493 1 3 0.0751658 1 4 0.0835077 2 3 0.10059 2 4 0.100441 3 4 0.100292 Discounts date1 date2 discountfactor 0 1 0.970662 0 2 0.923845 0 3 0.837484 0 4 0.759412 1 2 0.951769 1 3 0.862797 1 4 0.782365 2 3 0.90652 2 4 0.822011 3 4 0.906777 29

IRA.03.hava numcompoundperiods 1 SpotRates date rate 0.5 0.0032 1 0.0048 2 0.0123 3 0.0185 5 0.0281 7 0.0346 10 0.0381 ForwardRates date1 date2 rate 0 0.5 0.0032 0 1 0.0048 0 2 0.0123 0 3 0.0185 0 5 0.0281 0 7 0.0346 0 10 0.0381 0.5 1 0.00640255 0.5 2 0.0153516 0.5 3 0.0215879 0.5 5 0.0309045 0.5 7 0.0370557 0.5 10 0.0399701 1 2 0.019856 1 3 0.0254199 1 5 0.0340089 1 7 0.0396519 1 10 0.0418675 2 3 0.0310142 2 5 0.0387701 2 7 0.0436569 2 10 0.0446521 3 5 0.0426699 3 7 0.0468418 3 10 0.046615 5 7 0.0510304 5 10 0.0481973 7 10 0.0463128 Discounts date1 date2 discountfactor 0 0.5 0.998404 0 1 0.995223 0 2 0.975847 0 3 0.946492 0 5 0.870609 0 7 0.788121 0 10 0.688031 0.5 1 0.996814 0.5 2 0.977407 0.5 3 0.948005 0.5 5 0.872001 0.5 7 0.789381 0.5 10 0.689131 1 2 0.980531 1 3 0.951035 1 5 0.874788 1 7 0.791904 1 10 0.691334 2 3 0.969919 2 5 0.892158 2 7 0.807628 2 10 0.705061 3 5 0.919827 3 7 0.832675 3 10 0.726928 5 7 0.905252 5 10 0.790287 7 10 0.873002 30

PVA.01.hava cashflows ((1, 1100), (2, 1210), (3, 1331)) spotrates ((1, 0.1), (2, 0.1), (3, 0.1)) numcompoundperiods 1 cashflowvalue 3000.0 PV_Analysis date flow npv 0 0 3000.0 1 1100 3300.0 2 1210 2420.0 3 1331 1331.0 Discounts date1 date2 discountfactor 0 1 0.909091 0 2 0.826446 0 3 0.751315 1 2 0.909091 1 3 0.826446 2 3 0.909091 ForwardRates date1 date2 rate 0 1 0.1 0 2 0.1 0 3 0.1 1 2 0.1 1 3 0.1 2 3 0.1 ytm 0.0999999 PVA.02.hava cashflows ((1, 1100), (2, 1210), (3, 1331)) spotrates ((1, 0.1), (2, 0.09), (3, 0.08)) numcompoundperiods 1 cashflowvalue 3075.02 PV_Analysis date flow npv 0 0 3075.02 1 1100 3382.53 2 1210 2465.34 3 1331 1331.0 Discounts date1 date2 discountfactor 0 1 0.909091 0 2 0.84168 0 3 0.793832 1 2 0.925848 1 3 0.873215 2 3 0.943152 ForwardRates date1 date2 rate 0 1 0.1 0 2 0.09 0 3 0.08 1 2 0.0800909 1 3 0.0701368 2 3 0.0602744 ytm 0.0865259 31

32

BA.01.hava coupon 0.06 issemiannual false maturity 4 facevalue 1000 numcompoundperiods 1 spotrates 1 0.06 2 0.06 3 0.06 4 0.06 cashflows 1 60.0 2 60.0 3 60.0 4 1060.0 bondprice 1000.0 presentvalueanalysis date flow rpv 0 0 1000.0 1 60.0 1060.0 2 60.0 1060.0 3 60.0 1060.0 4 1060.0 1060.0 ytm 0.0599999 BA.02.hava coupon 0.06 issemiannual false maturity 4 facevalue 1000 numcompoundperiods 1 spotrates 1 0.07 2 0.07 3 0.07 4 0.07 cashflows 1 60.0 2 60.0 3 60.0 4 1060.0 bondprice 966.128 presentvalueanalysis date flow rpv 0 0 966.128 1 60.0 1033.76 2 60.0 1041.92 3 60.0 1050.65 4 1060.0 1060.0 ytm 0.0700002 33

BA.03.hava coupon 0.06 issemiannual false maturity 4 facevalue 1000 numcompoundperiods 1 spotrates 1 0.05 2 0.05 3 0.05 4 0.05 cashflows 1 60.0 2 60.0 3 60.0 4 1060.0 bondprice 1035.46 presentvalueanalysis date flow rpv 0 0 1035.46 1 60.0 1087.23 2 60.0 1078.59 3 60.0 1069.52 4 1060.0 1060.0 ytm 0.0500002 34

BA.04.hava coupon 0.06 issemiannual false maturity 4 facevalue 1000 numcompoundperiods 1 spotrates 1 0.05 2 0.06 3 0.07 4 0.08 cashflows 1 60.0 2 60.0 3 60.0 4 1060.0 bondprice 938.652 presentvalueanalysis date flow rpv 0 0 938.652 1 60.0 985.585 2 60.0 990.464 3 60.0 1014.47 4 1060.0 1060.0 ForwardRates date1 date2 rate 0 1 0.05 0 2 0.06 0 3 0.07 0 4 0.08 1 2 0.0700952 1 3 0.0801424 1 4 0.0901893 2 3 0.0902839 2 4 0.100377 3 4 0.110564 Discounts date1 date2 discountfactor 0 1 0.952381 0 2 0.889996 0 3 0.816298 0 4 0.73503 1 2 0.934496 1 3 0.857113 1 4 0.771781 2 3 0.917192 2 4 0.82588 3 4 0.900443 ytm 0.0784588 35

BA.05.hava coupon 0.06 issemiannual true maturity 4 facevalue 1000 numcompoundperiods 2 spotrates 0.5 0.06 1 0.06 1.5 0.065 2 0.065 2.5 0.07 3 0.07 3.5 0.075 4 0.075 cashflows 0.5 30.0 1.0 30.0 1.5 30.0 2.0 30.0 2.5 30.0 3.0 30.0 3.5 30.0 4.0 1030.0 bondprice 951.148 presentvalueanalysis date flow rpv 0 0 951.148 0.5 30.0 979.682 1.0 30.0 978.173 1.5 30.0 983.747 2.0 30.0 984.743 2.5 30.0 997.765 3.0 30.0 1001.64 3.5 30.0 1022.77 4.0 1030.0 1030.0 36

ForwardRates date1 date2 rate 0 0.5 0.06 0 1 0.06 0 1.5 0.065 0 2 0.065 0 2.5 0.07 0 3 0.07 0 3.5 0.075 0 4 0.075 0.5 1 0.06 0.5 1.5 0.0675045 0.5 2 0.0666694 0.5 2.5 0.0725076 0.5 3 0.0720058 0.5 3.5 0.0775106 0.5 4 0.0771518 1 1.5 0.0750364 1 2 0.0700121 1 2.5 0.0766936 1 3 0.0750182 1 3.5 0.0810305 1 4 0.0800242 1.5 2 0.065 1.5 2.5 0.0775227 1.5 3 0.0750121 1.5 3.5 0.0825318 1.5 4 0.0810232 2 2.5 0.0901214 2 3 0.0800363 2 3.5 0.0884087 2 4 0.0850484 2.5 3 0.07 2.5 3.5 0.0875529 2.5 4 0.0833602 3 3.5 0.105255 3 4 0.0900726 3.5 4 0.075 37

Discounts date1 date2 discountfactor 0 0.5 0.970874 0 1 0.942596 0 1.5 0.90851 0 2 0.879913 0 2.5 0.841973 0 3 0.813501 0 3.5 0.772829 0 4 0.744895 0.5 1 0.970874 0.5 1.5 0.935766 0.5 2 0.90631 0.5 2.5 0.867232 0.5 3 0.837906 0.5 3.5 0.796014 0.5 4 0.767242 1 1.5 0.963838 1 2 0.9335 1 2.5 0.893249 1 3 0.863043 1 3.5 0.819894 1 4 0.790259 1.5 2 0.968523 1.5 2.5 0.926762 1.5 3 0.895423 1.5 3.5 0.850655 1.5 4 0.819908 2 2.5 0.956882 2 3 0.924524 2 3.5 0.878301 2 4 0.846555 2.5 3 0.966184 2.5 3.5 0.917878 2.5 4 0.884702 3 3.5 0.950004 3 4 0.915666 3.5 4 0.963855 ytm 0.0743427 Sensitivity Analysis delta 0.001 newspotrates 0.5 0.061 1 0.061 1.5 0.066 2 0.066 2.5 0.071 3 0.071 3.5 0.076 4 0.076 newbondprice 947.854 presentvalueanalysis2 date flow rpv 0 0 947.854 0.5 30.0 976.764 1.0 30.0 975.64 1.5 30.0 981.592 2.0 30.0 982.994 2.5 30.0 996.413 3.0 30.0 1000.72 3.5 30.0 1022.29 4.0 1030.0 1030.0 durationestimate 3.46255 38

DSRA.01.hava numbonds 4 couponrates (0, 0.035, 0.04, 0.045) maturities (1, 2, 3, 4) facevalues (1000, 1000, 1000, 1000) issemiannual false prices (970.874, 990.896, 974.208, 951.905) bonddata coupon issemiannual maturity facevalue 0 false 1 1000 0.035 false 2 1000 0.04 false 3 1000 0.045 false 4 1000 bondcashflowmatrix 1000 0 0 0 35.0 1035.0 0 0 40.0 40.0 1040.0 0 45.0 45.0 45.0 1045.0 deriveddiscountfactors 1 0.970874 2 0.924556 3 0.863837 4 0.792094 derivedspotrates 1 0.0299998 2 0.0400001 3 0.0500001 4 0.06 NB: Supplied spot rate curve to derive bond price data above spotrates 1 0.03 2 0.04 3 0.05 4 0.06 discountfactors 1 0.970874 2 0.924556 3 0.863838 4 0.792094 derivedprices 1 970.874 2 990.896 3 974.208 4 951.905 39

DSRA.02.hava numbonds 4 couponrates (0.06, 0.07, 0, 0.09) maturities (0.5, 1.5, 1.5, 2) facevalues (1000, 1000, 1000, 1000) issemiannual true prices (1000.97, 1008.65, 909.831, 1043.01) bonddata coupon issemiannual maturity facevalue 0.06 true 0.5 1000 0.07 true 1.5 1000 0 true 1.5 1000 0.09 true 2 1000 bondcashflowmatrix 1030.0 0 0 0 35.0 35.0 1035.0 0 0.0 0.0 1000.0 0 45.0 45.0 45.0 1045.0 deriveddiscountfactors 0.5 0.971816 1.0 0.941753 1.5 0.909831 2.0 0.876514 derivedspotrates 0.5 0.0580037 1.0 0.0609212 1.5 0.0640003 2.0 0.0669991 NB: Supplied spot rate curve to derive bond price data above spotrates 0.5 0.058 1.0 0.061 1.5 0.064 2.0 0.067 discountfactors 0.5 0.971817 1.0 0.941681 1.5 0.909831 2.0 0.876512 derivedprices 1 1000.97 2 1008.65 3 909.831 4 1043.01 40

CASH FLOW ANALYSIS SAMPLE PROBLEMS 1. You invest $1000 today in a bank. Stated annual interest is 4%. How much money will you have (to the nearest penny) if this investment is a. compounded annually for 4 years? b. compounded semi-annually for 4 years? c. compounded continuously for 4 years? 2. You invest $1000 today in a bank. Stated annual interest is 8%. How much money will you have (to the nearest penny) if this investment is a. compounded annually at 8% for 4 years? b. compounded semi-annually for 4 years? c. compounded quarterly for 7.5 years? d. compounded continuously for 4 years? 3. Consider the cash flow stream (C 0, C 1, C 2, C 3, C 4, C 5, C 6, C 7, C 8,, C 40, C 41 ) = (100, 110, 100, 110, 100, 110,, 100, 110). (The pattern (100, 110) repeats itself.) The appropriate per-period rate of interest is 10%. Determine the present value of this cash flow stream. 4. Ms. Jones financed her home purchase with a fixed-rate 20-yr mortgage at 6%. The original loan balance was 400,000.00. With her monthly mortgage just paid her current loan balance is 301,903.98. a. What is Jones s monthly payment to the bank? b. How many months remain until the loan is paid off? c. Jones would like to pay off her loan sooner. She has decided that she would like to pay off her loan in 10 years, and is willing to add $A per month to her payment. What is the value for A? 5. Jones financed his home purchase with a conventional fixed-rate 30-yr mortgage at 9%. The original loan balance was 200,000.00. With his monthly mortgage just paid his current loan balance is 173,719.16. a. What is Jones s monthly payment to the bank? b. How many months remain until the loan is paid off? c. Jones would like to pay off his loan sooner. He has decided that he can afford an extra 50 per month. How many months will it take to pay off his loan? 6. A company has two options for a machine it must purchase. The manufacturer s discount rate is 10%. Ignore taxes and depreciation. Type I: Initial cost = 250,000. Annual cost = 40,000. 4-yr lifetime. Salvage value = 80,000. Type II: Initial cost = 400,000. Annual cost = 20,000. 7-yr lifetime. Salvage value = 50,000. Use the annual equivalent method to determine which machine to purchase. 7. A company must purchase a piece of equipment. Two types are being considered: TYPE A: initial cost = 1000, annual cost = 400, salvage value = 300, 5-yr lifetime. TYPE B: initial cost = 2000, annual cost = 200, salvage value = 800, 8-yr lifetime. Discount rate is 10%. Use the annual equivalent method to recommend which equipment type to purchase. Ignore taxes and use annual compounding. 41

8. A manufacturer must acquire equipment to produce a product for a customer for a period of exactly two years. Relevant data are: Revenues are projected to be 600,000 in the first year and 800,000 in the second year. The manufacturer s discount rate is 15% and its marginal tax rate is 30%. For tax purposes straight-line depreciation is used. The equipment s market value depreciates at a rate of 30% per year, i.e., its value at the end of the year is 70% of the value at the beginning of the year. The equipment will be classified as 7-year property. The equipment initially costs 350,000, of which 150,000 will be financed by a 2-year loan at 12% interest, with principal repaid in two equal annual installments. First year annual operating costs will be 250,000. Each successive year these costs rise by 10%. Use annual compounding. Obtain the after-tax cash flows by filling out the following table. Revenues Expenses EBITDA Interest Depreciation EBT Taxes Net Income Adjustments Loan principal cash flow Depreciation Investment Free cash flow Net present value Project cash flow analysis 0 1 2 9. A manufacturer must acquire equipment to produce a product for a customer for a period of exactly two years. Relevant data are: Revenues are projected to be 500,000 in the first year and 700,000 in the second year. The manufacturer s discount rate is 12% and its marginal tax rate is 30%. For tax purposes straight-line depreciation is used. The equipment s market value depreciates at a rate of 40% per year, i.e., its value at the end of the year is 60% of the value at the beginning of the year. The equipment will be classified as 5-year property. The equipment initially costs 250,000, of which 100,000 will be financed by a 2-year loan at 10% interest, with principal repaid in two equal annual installments. First year annual operating costs will be 400,000. Each successive year these costs rise by 25%. Use annual compounding. Obtain the after-tax cash flows by filling out the following table. Revenues Expenses EBITDA Project cash flow analysis 0 1 2 42

Interest Depreciation EBT Taxes Net Income Adjustments Loan principal cash flow Depreciation Investment Free cash flow Net present value 10. Company X is considering investing in one of two mutually exclusive projects, A and B. The Company s cost of capital is 12% (annual compounding). The after tax cash flows for projects A and B are as follows: Project 0 1 2 A -18,000 5,650 16,600 B -6,000 4,600 2,645 The IRR for project A is 13% and the IRR for project B is 15%. Which project (if any) do you recommend for investment? 11. At time 0 the cash flows associated with a financial instrument are as shown in the table. Time 0.5 1.0 1.5 2.0 2.5 3.0 Cash flow 37.50 37.50 37.50 37.50 37.50 1037.50 How would the market describe this type of financial instrument? 12. The price of a 6% coupon bond with semi-annual payments and a face value of 100 is 102.65. Is its YTM higher or lower than 6%? Explain. 13. The current annual spot rate curve s = (s 1, s 2, s 3, s 4, s 5 ) = (5.00, 5.25, 5.50, 5.00, 4.75). Determine the forward rate f 24. 14. The current annual short rate curve r = (r 0, r 1, r 2 ) = (4.4, 4.8, 5.5). Determine the current annual spot rate curve s = (s 1, s 2, s 3 ). 15. In this problem all bonds have face values of 100 and all coupon payments are annual. Bond A is a 2-year, zero coupon bond. Its price is 87.34. Bond B is a 3-year, 6% coupon bond. Its price is 96.20. Bond C is a 2-year, 8% coupon bond. Its price is 101.843. Determine the annual spot rate curve s = (s 1, s 2, s 3 ). 16. In this problem all bonds have face values of 1000 and all coupon payments are annual. Bond A is a 1-year, zero coupon bond. Its price is 934.58. Bond B is a 2-year, 6% coupon bond. Its price is 977.61. Bond C is a 4-year, 9% coupon bond. Its price is 1043.45. Bond D is a 4-year, 8% coupon bond. Its price is 1009.94. Consider a liability stream (L 1, L 2, L 3, L 4 ) for the next 4 years. Precisely explain how it is possible to use the 4 bonds above to obtain an immunization that will exhibit absolutely no interest risk over the next 4 years. (Do not solve.) 43

CASH FLOW ANALYSIS SAMPLE PROBLEM SOLUTIONS 1. a. (1000)(1.04) 4 = 1169.86. b. 1000(1.02) 8 = 1171.66. c. 1000e 0.16 = 1173.51. 2. a. 1000(1.08) 4 = 1360.49. b. 1000(1.04) 8 = 1368.57. c. 1000(1.02) 30 = 1811.36. d. 1000e.32 = 1377.13. 3. The PV with r = 10% of (C 0, C 1 ) = (100, 110) at time 0 is, of course, 200. The cash flow stream is therefore equivalent to (200, 0, 200, 0, 200,, 200, 0), where the last 200 occurs at time 40. This cash flow stream is equivalent to 200 at each time t = 0, 1, 2,..., 20, where the period length is 2 years. Since (1.1) 2 1 = 0.21, the PV equals 200[1 + (1 (1.21) -20 )/0.21] = 1131.34. 4. a. M = (0.005)(400,000)/[1 (1.005) -240 ] = 2865.72. b. The ratio LB(t)/LB(0) = 301,903.48/400,000. We know that LB(t)/LB(0) = [1 (1.005) (240-t) ]/[1 (1.005) -240 ]. Thus, the number of months remaining is (240-t) = 150. Alternatively, 301,903.48 = [2865.72/0.005][1 (1.005) n ], which gives n = 150. c. We seek the value of A such that [(2865.72 + A)/0.005][1 (1.005) -120 ] = 301,903.48. 5. a. M = 200,000(0.09/12)/[1 (1 + 0.09/12) -360 ] = 1609.25. b. 173,719.16/200,000 = [1 (1.0075) -n ]/[1 (1.0075) -360 ] and so n = 222. c. Seek n such that 173,719.16 = (1659.25/0.0075) [1 (1.0075) -n ] and so n = 206. 6. PV I = 250,000 + 40,000[1 (1.1) -4 ]/(0.1) 80,000/(1.1) 4 = 322,154. AE I = [0.1PV I ]/[1 (1.1) -4 ] = 101,630. PV II = 400,000 + 20,000[1 (1.1) -7 ]/(0.1) 50,000/(1.1) 7 = 471,710. AE II = [0.1PV II ]/[1 (1.1) -7 ] = 96,892. Recommend Type II machine. 7. PV A = 1000 + (400/0.1)[1 (1.1) -5 ] - 300(1.1) -5 = 2330.04. AE A = 0.1(2330.04)/[1 (1.1) -5 ] = 614.66. PV B = 2000 + (200/0.1)[1 (1.1) -8 ] - 800(1.1) -8 = 2693.78. AE B = 0.1(2693.78)/[1 (1.1) -8 ] = 504.93. Recommend equipment type B. 44

8. Project cash flow analysis 0 1 2 Revenues 600,000 800,000 Expenses 250,000 275,000 EBITDA 350,000 525,000 Interest 18,000 9,000 Depreciation 50,000 50,000 EBT 282,000 466,000 Taxes 84,600 139,800 Net Income 197,400 326,200 Adjustments Loan principal cash flow 150,000-75,000-75,000 Depreciation 50,000 50,000 Investment -350,000 195,050 Free cash flow -200,000 172,400 496,250 Net present value 325,149 603,922 496,250 SV = (0.7) 2 (350,000) = 171,500. BV = 250,000. Taxes = 0.3(171,500 250,000) = -23,550. SV taxes = 171,500 (-23,550) = 195,050. 9. Project cash flow analysis 0 1 2 Revenues 500,000 700,000 Expenses 400,000 500,000 EBITDA 100,000 200,000 Interest 10,000 5,000 Depreciation 50,000 50,000 EBT 40,000 145,000 Taxes 12,000 43,500 Net Income 28,000 101,500 Adjustments Loan principal cash flow 100,000-50,000-50,000 Depreciation 50,000 50,000 Investment -250,000 108,000 Free cash flow -150,000 28,000 209,500 Net present value 42,012 215,054 209,500 SV = (0.6) 2 (250,000) = 90,000. BV = 150,000. Taxes = 0.3(90,000 150,000) = -18,000. SV taxes = 90,000 (-18,000) = 108,000. 10. NPV A = -18,000 + 5,560/1.12 + 16,600/(1.12) 2 = 278.06. NPV B = -6,000 + 4,600/1.12 + 2,645/(1.12) 2 = 215.72. Since NPV A > NPV B > 0, project A is recommended. (The IRR s are irrelevant.) 11. 7.5% level coupon bond, face value of 100, 3 yr maturity. 45

12. Bond sells at a premium, which implies its YTM must be less than 6%. 13. (1 + f 24 ) 2 = (1.05) 4 / 1.0525) 2. Thus, f 24 = 4.75%. {Note that [4(5) 2(5.25)]/2 = 4.75.} 14. s 1 = 4.4. (1.044)(1.048) = (1 + s 2 ) 2 and so s 2 = 4.6. (1 + s 2 ) 2 (1.055) = (1 + s 3 ) 3 and so s 3 = 4.9. 15. 87.34 = 100/(1 + s 2 ) 2 and so s 2 = 0.07. 101.843 = 8/(1 + s 1 ) + 108/(1.07) 2 and so s 1 = 0.065. Finally, 96.20 = 6/(1.065) + 6/(1.07) 2 + 106/(1 + s 3 ) 3 and so s 3 = 0.075. 16. The cash flows associated with these 4 bonds are: Bond 0 1 2 3 4 A -934.58 1000 B -977.61 60 1060 C -1043.45 90 90 90 1090 D -1009.94 80 80 80 1080 One can achieve an exact cash match by selecting the number of bonds of each type so that the following 4 equalities hold: 1000n A + 60n B + 90n C + 80n D = L 1. 1060n B + 90n C + 80n D = L 2. 90n C + 80n D = L 3. 1090n C + 1080n D = L 4. This linear system is easily solvable. The cost of the portfolio, which also equals the present value of the liabilities, is 934.58n A + 977.61n B + 1043.45n C + 1009.94n D. 46