ࡱ> MOLG Jbjbjَ F]84Ltn (P R R R C P P5 $$Y Y ||||FP P ||P P $M5YZ"P 15.415, Spring 1999 S.C. Myers Problem Set 1 Solutions Prepared by Wes Chan Numerical answers are in bold. 1. Annuity Problem. This is an annuity of which we know the present value, the annual payment, and the discount rate. We are asked for the life of the annuity. The formula for the present value of an annuity (see B&M p. 40) is given by:  EMBED Equation.2  In the first case with an annual interest rate of 8%, we can substitute the following values:  EMBED Equation.2  This equation can be solved algebraically for t. The solution is t=20.9 years. In the second case with an annual interest rate of 10%, there is no solution. The payment of $10,000 covers only the interest from the loan. The principal will never be repaid (you can verify this by plugging the relevant numbers into the formula above and seeing that only t= infinity will work). 2. Annuity Problem. Step 1: Determine the monthly payment that will be required on the loan. We can use the formula for the present value of an annuity:  EMBED Equation.2  This can be solved algebraically for C. The solution is C=$1,028.61. Step 2: Determine the effective annual interest rate of the above-calculated payment on the actual loan amount of $98,000. We can use the same formula again:  EMBED Equation.2  This could be solved algebraically for r=1.02%, but it is difficult to do so. It is better to go straight to a calculator or a computer. Step 3: Convert the monthly rate of 1.02% to an effective annual interest rate. This can be done with the following formula:  EMBED Equation.2  where n equals the number of periods. In this case we have:  EMBED Equation.2  which gives r=12.95%. 3. Annuity Problem. The first step is to convert the annual discount rate to a semiannual rate:  EMBED Equation.2  The above formula can be solved algebraically to get rsemiannual=3.92%. With this rate in hand we can go back to our annuity formula, with $100,000 paid over 9 periods:  EMBED Equation.2  This gives the present value of the nine future payments as $746,251. To get the total present value of all payments we must add the initial payment of $100,000 (not discounted), which gives our answer of $846,251. Note: As illustrated by this problem, it is very important to match discounting periods to cash flows. It would not be correct to assume that we have annual cash flows of $200,000 and that these can be discounted at 8% a year. We must discount whenever the cash flow is received. 4. Growing Perpetuity Problem. We will want to use the following formula (see B&M p. 39):  EMBED Equation.2  EMBED Equation.2  We know r = 8%. We need to determine the appropriate growth rate, g. We do this by multiplying the two factors (price and quantity) that affect the value of the cash flow stream. Production (quantity) will increase by -4% per year and prices will increase by 2% per year. This gives: g = (1 + -0.04)(1 + 0.02) 1 = (0.96)(1.02) 1 = 0.9792 1 = -0.0208 Note: The perpetuity formula requires that we use the next years cash flow in the numerator! C0 (current cash flow) is 100,000 times the current price of oil, which we will let be p. To obtain C1 we must multiply C0 by 1 plus the growth rate. Therefore: C1 = (100,000p)(1 + -0.0208) = (100,000p)(0.9792) = 97,920 This gives us all the pieces we need, so we can get our solution with:  EMBED Equation.2  When calculated this gives the present value of the wells production at 971,429p. 5a. Discounting Problem. The first step is to find the future value of $10,000,000 in 15 years when increased by 5% per year. This future cost is found by: $10,000,000 * (1.05)15 = $20,789,282 The second step is to discount the future cost back at a rate of 11% to find its present value:  EMBED Equation.2  This equation gives the solution of PV = $4,345,050. Alternatively, we could have used the simple 1-step formula:  EMBED Equation.2  b. Annuity Problem. For this part, we can go back to the formula for the present value of an annuity. The only trick is that we know the future value of the annuity to be $20,789,282 (which we derived previously), but the formula calls for the present value of the annuity. But we can find the present value of the annuity by discounting back at the rate at which this fund will be growing, 6.5% (note we can rearrange the following equation to show this):  EMBED Equation.2  Now we are ready to use the standard formula:  EMBED Equation.2  This equation can be solved algebraically for C. The solution tells us that the oil company will have to put $859,695 in the trust fund each year in order to cover the scrapping cost in 15 years. Note: We discount by the T-bond yield even though the trust fund grows at that rate because we are trying to get C using a present value ($8,083,424). This can be verified using a spreadsheet by inputting the annual payments and having them each grow by 6.5% a year. The final total value will equal the future cost. 6. Annuity Problem. Solving the first part of this question is a matter of calculating the present value of the stream of 19 payments. Each payment will be $9,420,713/19 = $495,827. We should be willing to pay up to the present value of the stream, discounted at 8%. Once again we can use the formula for the present value of an annuity:  EMBED Equation.2  Thus we should be prepared to bid up to $4,761,724 for the prize. In the second part of the question, we use the same formula, but this time we know the present value and were looking for the appropriate interest rate.  EMBED Equation.2  As we noted previously, when r is the unknown in this equation it is much more difficult to solve. Better to head straight for your calculator to get the solution of r=9.8%. In other words, Enhance was looking for a return of approximately 9.8% (instead of 8%) on their bid. 7. Annuity Problem. Lets start by finding the monthly payment that will be required under each of the options. Once again we will use the PVA formula. Dealership A will finance $18,000 over 36 months (after the down payment) at 8/12% per month. How much will we have to pay each month?  EMBED Equation.2  We solve this equation for C=$564.05. This is the required monthly payment under option A. Doing the same thing for dealership B, we have $15,500 over 36 months at 10/12% per month:  EMBED Equation.2  We solve this equation for C=$500.14. This is the required monthly payment under option B. Now lets compare the stream of outflows under the two options: Time0123. . .36Option A-$2,000-$564.05-$564.05-$564.05-$564.05-$564.05Option B-$4,000-$500.14-$500.14-$500.14-$500.14-$500.14 To determine which deal to choose, we simply need to discount back each stream of flows to find the present value of the cost of each option. We should choose the option with the lower present value. The only problem is that we are not given a discount rate to use. Our answer will depend on the discount rate we choose. So the official answer is it depends on the discount rate. But to learn more from the problem you might want to calculate the present values at varying discount rates and see which answer is better. You should be able to see just from looking at the flows that the higher the discount rate the more likely it is that option A will be preferred. That is because option A has relatively more cash payment deferred to the future. With a given discount rate the present value can be computed, once again, with the PVA formula. For a given rate r, we plug in $564.05 for C under option A, and $500.14 for C under option B. Once we have this present value we add to it the initial outflow ($2,000 for A, $4,000 for B) to get the total present value cost of the loan. The following table gives computed values for the two options under various discount rates: Discount RatePV Option APV Option BBest Choice6%$20,541$20,440B8%$20,000$19,960B10%$19,481$19,500A12%$18,982$19,058A If we wanted to, we could calculate the break-even discount rate, or the rate at which we would be indifferent between the two options. From the above table we can see that it should be somewhere between 8% and 10%. The question didnt ask you for this specifically, but if you did it, the rate is 9.33%. That is, with a discount rate below 9.33% we should choose option B, and with a rate above 9.33% we should choose option A. 8. Compounding Problem. We need to calculate the effective annual yields for each offer and see which rate is higher: Bank A  EMBED Equation.2  Bank B  EMBED Equation.2  Because it offers the higher effective annual yield, we should choose Bank A. 9a. Growing Perpetuity Problem. We are given r = 0.1, g = 0.4, and C1 = $400,000 * (1.04) = $416,000 (remember we need next years cash flow for the perpetuity!). We plug these into our formula:  EMBED Equation.2  Thus we should be willing to pay $6,933,333 for the complex. Let us redo the calculations using real cash flows and the real discount rate. We need the real discount rate, which can be calculated from the following formula (see B&M p. 47): (1 + rreal)(1 + inflation) = (1 + rnominal) We know the inflation rate (4%) and the nominal interest rate (10%), so by rearranging the formula we get rreal = 5.77%. Now we apply the same growing perpetuity formula using real terms (note g = 0% here since we have eliminated the inflationary growth by using real values):  EMBED Equation.2  This is the same as our previous answer except for a small error caused by the rounding of rreal. Note: We have two options in our calculations: 1. Forecast nominal and discount nominal. 2. Forecast real and discount real. We should not mix the two! b. Growing Annuity Problem. For this part we need to use the formula for the present value of a growing annuity (PVGA). It is similar to the PVA formula we have been using (and can be derived from the growing perpetuity formula by following the method in B&M, footnote on page 39):  EMBED Equation.2  We need to plug in the values from this problem, and then add the present value of the site:  EMBED Equation.2  The present value of the site is $5,000,000/(1.1)20 = $743,218. Adding this to the PVGA gives our final answer of $5,418,389 as the amount we should be willing to pay for the apartment complex. In real terms, the cash flows are not growing over time. We plug the real discount rate into the standard PVA formula:  EMBED Equation.2  To this we add the present value of the site to get $5,418,094, which is the same as the previous answer in nominal terms except for the rounding error. 10. Compounding Problem. If we withdraw the money before 5 years, we will receive only 4%/year over the time we left it in the CD. We will get no further interest unless we reinvest the money somewhere else. We need to compare two alternatives. Our first option is to not withdraw our money. If we leave the money in the current CD, we will earn 6%, compounded monthly, for 5 years:  EMBED Equation.2  If we dont withdraw the money we will have $13,489 after 5 years. Our second option is to withdraw the money after the first year. We would only receive 4%, compounded monthly for that year:  EMBED Equation.2  So after 1 year we would have $10,407. In order to be enticed into withdrawing our money, then, we must be able to grow this $10,407 to at least $13,489 in the following 4 years (the remaining 48 months). We are looking for the minimum rate that allows us to do so. We can find that rate with this equation:  EMBED Equation.2  Solving this equation algebraically for r gives our answer of 6.50%. That is, if the rates on new 4-year CDs were 6.50% or higher, we should be willing to move our money. 11. Compounding Problem. If the annual rate is fixed at 5%, banks can increase the effective annual yield by increasing the frequency of compounding. The maximum effective annual yield would be achieved with continuous compounding. The continuously compounded yield is given by (B&M p. 44):  EMBED Equation.2  Where r is the annual rate and t is the number of years. So in this case we have:  EMBED Equation.2  So 5.13% is the highest effective annual yield banks could offer their depositors. 12. Growing Annuity Problem. This is another problem that calls for the PVGA formula. In order to simplify things, let us assume that Ms. Pools income grows at a 2% real rate. This is not strictly true, since (from previous): (1 + rreal)(1 + inflation) = (1 + rnominal) and it is obvious from rearranging that (1.02 + inflation)/(1 + inflation) does not equal (1 + rreal) in general! We also know that the discount rate is 5%. To use the PVGA formula, we need the present value of Ms. Pools desired retirement savings by year 30 (70-40). This is $500,000/(1.05)30 = $115,689. Finally we need to know C1, the cash flow with only 1 years growth (for use in the annuity formula). It is ($40,000)(1.02) = $40,800. Given these figures, we can calculate the fraction X of income (C1) that she needs to set aside. Use the PVGA formula:  EMBED Equation.2  Solving, we find X = 14.64%. This is the percentage of income each year that must be set aside. 13a. Growing Perpetuity Problem. Let us start by computing the present values in the standard way, assuming that the cash flows are received at yearend. This will be a benchmark. Using the formula for a growing perpetuity, with r = 0.05, g = 0.02, and C1 = ($20,000,000)(1.02) = $20,400,000, we have:  EMBED Equation.2  Now, given that we are using the midyear convention, which assumes that cash flows are received at the middle of the year, we need to redo the above with modified cash flows and interest rates. By the middle of the year, our projected cash flow has not grown to $20,400,000, but is instead ($20,000,000)(1.02)0.5 = $20,199,010. The second adjustment we need to make concerns the discount rate. The normal formula assumes that we discount for complete years. But since we are discounting at 1/2 year, 3/2 years, 5/2 years, etc., we need to discount each flow by 1/2 year less (remember this is a perpetuity, so there are no problems with the ending data). We accomplish this by multiplying the result by (1.05)0.5. Combining these elements, we have:  EMBED Equation.2  As one would expect, the PV of revenues has increased, to $689,927,537. This is due to the de facto earlier compounding. b. Growing Perpetuity Problem. We would expect that continuous compounding would increase the PV of revenues even further. In this calculation the cash flow in the numerator is the initial value, $20,000,000, because they are received continuously through the year beginning immediately (unlike annual compounding, which is once a year at the end of the year). We need the appropriate discount rate. We need to ask what rate, when compounded continuously, would give us an annual rate of 5%? In the problem we are told that the (effective/real/actual) annual interest rate is 5% (this is given and already in "actual" terms). We want to find the equivalent rate that will be continuously compounded. From B&M p. 44, we know that rActual = ert - 1. Plugging in rActual = 0.05 and t = 1, we get r = 4.88%  EMBED Equation.2  The present value of revenues has increased even further to $694,444,444. 14. T-Bill formula. This problem is different from all the other in this problem set, in that we must use something more than the basic present value formulas. The problem would more properly be placed later in the course, but is illustrates some of the mechanics with which we have been working. Specifically, we need to know the convention for calculating T-Bill prices and yields! Since this will be covered later in the course, we need not be too concerned about the methodology, but a useful reference is Bodie, Kane and Marcus pp. 40-41 (or B&M chapter 23). The formula for obtaining the true market price of a T-Bill is:  EMBED Equation.2  where P is the price, $10,000 is par value, rbd is the stated discount yield, n is the days to maturity, and 360 is the convention for days/year. We plug in the given values for the first part:  EMBED Equation.2  The bill traded at a price of $9,591.11. To compute the true yield, we first calculate the rate of return over the 230 days to maturity:  EMBED Equation.2  Then we annualize the return as follows (using the actual 365-day year):  EMBED Equation.2  The true yield offered on the bill is 6.84%. 15. Annuity Problem. Here we need to calculate the present value of payment option 2 and see if it is greater than $1,000,000. In payment option 1 it seems clear that the $1,000,000 is paid out immediately. For payment option 2, it doesnt say specifically, but I will assume that the first payment of $80,000 comes not immediately, but at the end of the first year. (If you assumed that the first payment comes immediately, thats okay. Your answer will be somewhat different, but equally valid.) Once again, we can use the PVA formula:  EMBED Equation.2  In order to come up with a single answer we would need to know what discount rate is appropriate for Stewart C. Myers. This is not given, so let us try a couple of values. If we let r = 8%, the PVA = $853,982. Then option 1 would be better (it has a larger present value- indeed, the entire $1,000,000). etc. We can use a spreadsheet solver or calculator to compute the breakeven discount rate (the rate at which both options have equal present values). By setting the PVA equal to $1,000,000, we can solve for r = 6.24%. Thus, it depends. If his discount rate is higher than 6.24%, he should choose option 1, and if it is lower, option 2. 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FMicrosoft Equation 2.0 DS Equation Equation.2@EGDxGD PV=C 1 r-g L=4@CompObj?AZObjInfoBEquation Native \_916642821EF0sKJ5YPJ5YOle PIC DGLMETA CompObjFHZ=4N P .1   & & MathTypeP &  FMicrosoft Equation 2.0 DS Equation Equation.2ObjInfoIEquation Native <_916643712LF0WJ5YPI_J5YOle  EGD GD aGW',  xL  : T .1    &`  & MathType-W Times New Roman- 2 `@PVPIC KNLMETA CompObjMOZObjInfoP  2 k pSymbol- 2 `= 2 - 2 -Times New Roman- 2 k97 2 k920 2 -08 2 0208 2 k*,` 2 .` 2 ( 2 w.` 2  ) & "System- FMicrosoft Equation 2.0 DS Equation Equation.2`E GDGD PV=97,920p.08-(-.0208)L DEquation Native |_916644389<fSFpfJ5Y@9pJ5YLOle PICObj RULMETAfo CompObjNativeTVZObjInfo0WFEquation Native |w   #%&'()*+,-./012478;=>?@ABCDEFGHIJKLNQRUWXYZ[\]^_`abcdefgilmnqstuvwxyz{|}~ : j .1    &`  & MathType-` Times New Roman- 2 `@PVSymbol- 2 `=Times New Roman- 2 k$20 2 k,` 2 k,` 2 ( 2 .` 2 +) 2 ks789 2 k 282 2 1 2 11@Times New RomanSy- 2 15`` & "System-',#F FMicrosoft Equation 2.0 DS Equation Equation.2`EGD$GD PV=$20,789,282(1.11) 15L D : j .1    &`  & MathType-` _980012373ZF`J5Y)J5YMOleObj  PICnfo Y\ LMETAon Native  Times New Roman- 2 `@PVSymbol- 2 `=Times New Roman- 2 k$20 2 k,` 2 k,` 2 ( 2 .` 2 +) 2 ks789 2 k 282 2 1 2 11@Times New RomanSy- 2 15`` & "System-',#F FMicrosoft Equation 3.0 DS Equation Equation.39qL PV=$10,000,000*(1.05) 15 (1.11) 15 =CompObj[]fObjInfo^Equation Native _916645140aF`DJ5Y^J5YZ$4,345,050L& &>   .1   & & MathType- Times New Roman- 2 `@PVAOle !PIC `c"LMETAbj $CompObjbd3ZSymbol- 2 `= 2 ` =Times New Roman- 2 k$20 2 k9,` 2 k ,` 2 9( 2 A.` 2  ) 2 `"$8 2 `,` 2 `$,` 2 k789 2 k, 282 2 1 2 065 2 `083 2 `~424@Times New Roman- 2 S 15`` & "System- FMicrosoft Equation 2.0 DS Equation Equation.2ՠE (GDh,GD PVA=$20,789,282(1.065) 15 ObjInfoe5Equation Native 6_916645333_mhFJ5YJ5YOleion Native 9=$8,083,424$8 2 `,L8VN  .1  &7 & MathType-= = =/=Times New RomanSy-PIC gj:LMETA <CompObjikMZObjInfolO 2 ($8 2 ,` 2 *,` 2  .` 2 7.` 2 ( 2 .` 2 7) 2 083 2 424 2  1 2 j 065 2 1 2 065 2 %1 2  065@Times New Roman- 2 15``Symbol- 2 ,= 2  - 2 t  2 t  2 t  2  2  2 Times New Roman- 2 fC & "System-Times New Roman- FMicrosoft Equation 2.0 DS Equation Equation.2Equation Native FP_916646304oF`J5YȮJ5YOle SPIC nqTLՠE&GD\%GD $8,083,424=C1.065-1.065(1.065) 15 []2 %1L     .1  &7 &META  VH CompObj&'()*,/0pr56789hZ@ObjInfoHILNOPsUVWXYjEquation Nativeijklmnop vyk MathType-= = =t=Times New RomanSy- 2 @PVASymbol- 2 = 2 G - 2 m  2 m  2 m  2  2  2  2 =Times New RomanSy- 2 $495 2 ,` 2  .` 2 |.` 2 D( 2 L.` 2 ) 2 $4 2 y,` 2 ,` 2 827 2  1 2 c 08 2 81 2 08 2 1 2 08 2 761 2 O724@Times New Roman- 2 19`` & "System-Times New Roman FMicrosoft Equation 2.0 DS Equation Equation.2E&GDh%GD PVA=$495,8271.08-1.08(1.08) 19 []=$4,761,724$495 2 _9166467745vFJ5YK5YOle oPIC uxpLMETA rLN  .1  @&7 & MathType-==X==NTimes New Roman- 2 ($4 2 ,` 2 B,` 2 f$495 2 T ,` 2 ( 2 ) 2 200 2 000 2  827 2 1 2 =1 2  1@Times New Roman\- 2 P15``Symbol- 2 8= 2 - 2 + 2   2   2   2 e 2 e 2 eTimes New Roman- 2 r 2 r 2 $r & "System-  8 l  FMicrosoft Equation 2.0 DS Equation Equation.2ՠE$'GDx%GD $4,200,000=$495,8271r-1rCompObjwyZObjInfoNativezEquation Native F_916647711}F+K5Y65K5Y(1+r) 15 []Times New L "L "M % .1  &7 & MathType-=@= =7=Times New Roman-Ole PIC |LMETAbj hCompObj~Z 2 ($18 2 b,` 2 T(.` 2  /k 2  ) 2 K(.` 2 /k 2  )( 2 (.` 2 /k 2 )) 2 000 2  1 2 " 08 2  12 2 <1 2 08 2 12 2 1 2 08 2 J12@Times New Roman- 2 36``Symbol- 2 X= 2  - 2 + 2  2  2  2   2   2  Times New Roman- 2 C & "System-"System- FMicrosoft Equation 2.0 DS Equation Equation.2ObjInfoEquation Native _916648219{FkAK5YGK5YOleion Native wE<'GD(GD $18,000=C1(.08/12)-1(.08/12)(1+(.08/12)) 36 [] 2 LPIC LMETA hCompObjZObjInfo  " .1  &7 & MathType-=(= =/=Times New Roman- 2 ($15 2 V,` 2 <(.` 2  /k 2  ) 2 C(.` 2 /k 2 )( 2 (.` 2 /k 2 )) 2 500 2  1 2 1 2  12 2 D1 2 1 2 12 2 1 2 1 2 b12@Times New Roman- 2  36``Symbol- 2 @= 2  - 2 + 2  2  2  2 % 2 % 2 %Times New Roman- 2 zC & "System-=(= =/ FMicrosoft Equation 2.0 DS Equation Equation.2E'GDh%GD $15,500=C1(.1/12)-1(.1/12)(1+(.1/12)) 36 []Equation Native _916657039F@PK5Y@ ]K5YwOleion Native PIC95970 FL= =/L M  .1  @`&  & MathType-= = Times New Roman- 2 ?rMETAbj (CompObjZObjInfoNativeEquation Native F Times New RomanD-2  EffectiveAIIcc>>pc2 nnualpppc>PSymbol- 2 = 2  + 2 d 2 d 2 Zd 2   2   2 Z  2  - 2 =Times New Roman- 2 1 2 ] 09 2 $ 12 2 1 2 U9 2 ]38%@ Times New RomanD- 2 12ppTimes New Roman- 2  .` 2  .` & "System-EMP\~WRF0000.tmp FMicrosoft Equation 2.0 DS Equation Equation.2ѠWEDx D r EffectiveAnnual =1+.0912[] 12 -1=9.38%""/L&M  .1  @`&  &_916657228FjK5Y`pK5YOleion Native PIC43712 FLMETA H MathType-= = Times New Roman:- 2 ?r Times New Roman-2  EffectiveAIIcc>>pc2 nnualpppc>PSymbol- 2 = 2  + 2 d 2 d 2 Wd 2   2   2 W  2 ;- 2 I=Times New Roman- 2 1 2 ] 085 2 4 365 2 >1 2 {8 2 x87%@ Times New Roman:- 2 m365pppTimes New Roman- 2  .` 2 (.` & "System-CompObj  Z ObjInfo&'()*+,-./078@Equation NativeIJKLN UWXY`_980014212gilmnF`}K5Y =K5Y~ FMicrosoft Equation 2.0 DS Equation Equation.2WE0DD r EffectiveAnnual =1+.085365[] 365 -1=8.87%= = Ole PIC LMETAbj (CompObjfLh N  .1  @@& & MathType-== Times New Roman:- 2 0$10 2 l,` 2 .` 2 , $13 2 W,` 2 000 2 1    "#&()*+,-./0123457:;<?ABCDEFGHIJKLMNPSTUVWZ\]^_`abcdefghiknopqtvwxyz{|}~2 06 2 12 2 489 Times New Roman- 2 x 60ppPSymbol- 2 + 2 " 2 " 2 Z" 2   2   2 Z  2  = & "System-365 2  FMicrosoft Equation 3.0 DS Equation Equation.39qL PV=C 1 r"g=$416,0000.1"0.04=$6,933,333Lh ObjInfo Equation Native  _980015082FK5Y K5YhOleObj PICObj LMETAfo (CompObjNativefObjInfo3F N  .1  @@& & MathType-== Times New Roman:- 2 0$10 2 l,` 2 .` 2 , $13 2 W,` 2 000 2 1 2 06 2 12 2 489 Times New Roman- 2 x 60ppPSymbol- 2 + 2 " 2 " 2 Z" 2   2   2 Z  2  = & "System-365 2  FMicrosoft Equation 3.0 DS Equation Equation.39qL PV=Cr=$400,0000.0577=$6,932,409L& &>   .Equation Native !_980016048F@HK5Y ܪK5YOleion Native $PIC45140 F%LMETAon Native F'CompObj4F6fObjInfo8Equation Native 91   & & MathType- Times New Roman- 2 `@PVASymbol- 2 `= 2 ` =Times New Roman- 2 k$20 2 k9,` 2 k ,` 2 9( 2 A.` 2  ) 2 `"$8 2 `,` 2 `$,` 2 k789 2 k, 282 2 1 2 065 2 `083 2 `~424@Times New Roman- 2 S 15`` & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qL, PVGA=C 1 1r"g"(1+g) t (r"g)(1+r) t []L& _980016054F̻K5Y0mK5YOle =PIC >LMETA @&>   .1   & & MathType- Times New Roman- 2 `@PVASymbol- 2 `= 2 ` =Times New Roman- 2 k$20 2 k9,` 2 k ,` 2 9( 2 A.` 2  ) 2 `"$8 2 `,` 2 `$,` 2 k789 2 k, 282 2 1 2 065 2 `083 2 `~424@Times New Roman- 2 S 15`` & "System- FMicrosoft Equation 3.0 DS EqCompObjOfObjInfoQEquation Native RF_980016379FK5Y6K5Yuation Equation.39qL* PVGA=$416,00010.1"0.04"(1.04) 20 (0.1"0.04)(1.1) 20 []=$4,675,171Olebj XPICObj YLMETAfoNative [CompObjNativeFjfL& &>   .1   & & MathType- Times New Roman- 2 `@PVASymbol- 2 `= 2 ` =Times New Roman- 2 k$20 2 k9,` 2 k ,` 2 9( 2 A.` 2  ) 2 `"$8 2 `,` 2 `$,` 2 k789 2 k, 282 2 1 2 065 2 `083 2 `~424@Times New Roman- 2 S 15`` & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qL PVA=$4,00010.0577"1(0.0577)(1.0577ObjInfolEquation Native m_916657842FkK5Y0 K5Y(OleObj r) 20 []=$4,674,876Lh N  .1  @@& & MathType-== PIC  sL META$%&'()*+,-./0 56789u(@CompObjFGHIJKNSiXVWYZZcObjInfoghjlkmnoqpuvxzyTimes New Roman:- 2 0$10 2 l,` 2 .` 2 , $13 2 W,` 2 000 2 1 2 06 2 12 2 489 Times New Roman- 2 x 60ppPSymbol- 2 + 2 " 2 " 2 Z" 2   2   2 Z  2  = & "System-365 2  FMicrosoft Equation 2.0 DS Equation Equation.2рWED`D $10,0001+.0612[] 60 =$13,489Equation Native _916658150tiveXF`|QL5YtL5Y Ole15082 FPICObj LLh N  .1  @@& & MathType-== Times New Roman:- 2 0$10 2 l,` 2 .` 2  $10METAbj (CompObjZObjInfoNativeEquation Native F 2 Y,` 2 000 2 1 2 04 2 12 2 407 Times New Roman- 2 i 12ppPSymbol- 2 + 2 " 2 " 2 Z" 2   2   2 Z  2  = & "System-365 2  FMicrosoft Equation 2.0 DS Equation Equation.2рWEDD $10,0001+.0412[] 12 =$10,407L _916658368tiveF@zL5YL5Y!Ole16048 FPICion Native LMETA5140 FhL  .1  @& & MathType-==A Times New Roman- 2 0$10 2 l,` 2  $13 2 ,` 2 407 2 1 2 12 2 O489 Times New Roman- 2  48ppPSymbol- 2 + 2  2  2 Z 2 X  2 X  2 ZX  2  =Times New Roman- 2 ;r & "System- FMicrosoft Equation 2.0 DS EqCompObjNativeFZObjInfo4FEquation Native _980605277tiveFL5YpL5Y9uation Equation.2рWE(D)D $10,4071+r12[] 48 =$13,489PSymbolL Olenfo PICion Native LMETA5333 FCompObjNativef nL 6 .1   & $ & MathTypepTimes New Roman- 2 ?r 2 e Times New Roman- 2 Contpp> 2 Comppp 2 rtJ> 2 y.8 2 .8PSymbol- 2 = 2 -Times New Roman- 2  1 & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qEQ r Actual =e rt "1ObjInfo   Equation Native)*+,-./0 57m_980605303GHIJKLMNPFhL5YҴL5Y^_`Oledefghiknop vwxyLT M  .1   && & MathTypepTimes New Roman- 2 ?r 2 e Times New Roman- 2 Contpp> 2 PICnfo LMETAon Native CompObj2FfObjInfoComppp 2 y.8 2 .8 2 (.J8 2 " )(1)JJpJTimes New Roman- 2 .`PSymbol- 2 = 2  - 2  = Times New Roman- 2 805ppTimes New Roman- 2  1 2 95 2 13%@ & "System-PSymbol- 2  FMicrosoft Equation 3.0 DS Equation Equation.39qEP r Actual =e (.05)(1) "1=5.13%Equation Native   _980018002'()*+,-./0FL5YL5Yu(@OleObjFGHIJKNSi XVWYZcPICnfoghjlkmnoqp uvxzyLL& &>   .1   & & MathType- Times New Roman- 2 `@PVASymbol- 2 `METAbj CompObjfObjInfoNativeEquation Native F6= 2 ` =Times New Roman- 2 k$20 2 k9,` 2 k ,` 2 9( 2 A.` 2  ) 2 `"$8 2 `,` 2 `$,` 2 k789 2 k, 282 2 1 2 065 2 `083 2 `~424@Times New Roman- 2 S 15`` & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qL $115,689=$40,800X10.05"0.02"(1.02) 30 (0.05"0.02)(1.05) 30 []L  : T .1    &`  & MathType-W _980018302tiveFpL5Y9L5Y!Ole16048 FPICion Native LMETA5140 F  "#$')*+,-./0123589:=?@ABCDEFGHIKNORTUVWXYZ[\]^`cdgijklmnopqrsuxy|~Times New Roman- 2 `@PV 2 k pSymbol- 2 `= 2 - 2 -Times New Roman- 2 k97 2 k920 2 -08 2 0208 2 k*,` 2 .` 2 ( 2 w.` 2  ) & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qL PV=$20,400,0000.05"0.02=$680,000,000CompObjNativeF fObjInfo4F Equation Native  _980019139tiveF 7L5Y M5Y9Olenfo    PICion Native)*+,-./0 57LMETA5303GHIJKLMNP F`CompObjfghiknopvwxyfL  : T .1    &`  & MathType-W Times New Roman- 2 `@PV 2 k pSymbol- 2 `= 2 - 2 -Times New Roman- 2 k97 2 k920 2 -08 2 0208 2 k*,` 2 .` 2 ( 2 w.` 2  ) & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfo Equation Native !_980603509Fj'M5Y+M5YfOlenfo %L PV=$20,199,0100.05"0.02(1.05) 0.5 =$689,927,537L PICbj &LMETAbj (CompObjNative4fObjInfoNativeF6 : T .1    &`  & MathType-W Times New Roman- 2 `@PV 2 k pSymbol- 2 `= 2 - 2 -Times New Roman- 2 k97 2 k920 2 -08 2 0208 2 k*,` 2 .` 2 ( 2 w.` 2  ) & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qE PV=$20Equation Native  7 _980020069'()*+,-./0F?@OleEFGHIJKNSi XVWYZ;cPICfeghjlkmnoqp uvxzy<L,000,0000.0488"0.02=$694,444,444L  : T .1    &`  &META6054 F>CompObjJfObjInfoLEquation Native M MathType-W Times New Roman- 2 `@PV 2 k pSymbol- 2 `= 2 - 2 -Times New Roman- 2 k97 2 k920 2 -08 2 0208 2 k*,` 2 .` 2 ( 2 w.` 2  ) & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qLv P=$10,0001"r bd (n/360)[]_980020240F-iM5YpHoM5YOle PPIC QLMETA S    #$ !"%&()*+,-./0L  : T .1    &`  & MathType-W Times New Roman- 2 `@PV 2 k pSymbol- 2 `= 2 - 2 -Times New Roman- 2 k97 2 k920 2 -08 2 0208 2 k*,` 2 .` 2 ( 2 w.` 2  ) & "System- FMicrosoft Equation 3.0 DS EqCompObjNative_fObjInfo8FaEquation Native b_980020369 FPM5Y0M5YKLuation Equation.39qLx< P=$10,0001"0.064(230/360)[]=$9,591.11L Oleion Native ePIC44389  FfLMETA hCompObj  tf : T .1    &`  & MathType-W Times New Roman- 2 `@PV 2 k pSymbol- 2 `= 2 - 2 -Times New Roman- 2 k97 2 k920 2 -08 2 0208 2 k*,` 2 .` 2 ( 2 w.` 2  ) & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qL $10,00ObjInfo vEquation Native w_980020418F̍M5YpM5YOleion Native zw0"$9,591.11$9,591.11=4.26%L  : T .1    &`  & MathType-W PICion Native {LMETA8150tive F}CompObj2FfObjInfoTimes New Roman- 2 `@PV 2 k pSymbol- 2 `= 2 - 2 -Times New Roman- 2 k97 2 k920 2 -08 2 0208 2 k*,` 2 .` 2 ( 2 w.` 2  ) & "System- FMicrosoft Equation 3.0 DS Equation Equation.39qLv d  r=1.0426() 365/230 =6.84%Equation Native F_980020530F)M5YM5Y 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