ECN801, V.Bardis Sample Test 2 1. Suppose a bank charges a 16% semiannual rate, compounded quartely. (a) (3 marks) What is the nominal annual rate? What is the nominal quarterly rate? r_1 = 2 (16%) = 32% r_4 = (2/4) r_2 = 8% or r_4 = (1/4)r_1 = 8% (b) (3 marks) What is the effective annual rate? The effective rate per compounding period (quarterly) is i_4 = r_4 = 8% The effective annual rate is i_1 = (1+i_4)^(4)-1 = (1+.08)^4 - 1 = 36.04 % (c) (4 marks) If instead the bank compounds interest continuously, what are the effective annual rate and the effective monthly rate (given the semiannual rate remains 16%)? i_1 = e^(2 (.16))-1 = 37.7127% i_12 = e^(.16(2/12))-1 = 2.7025% 2 3349 0.65455 0.2902 2.26061 0.1598 6.8871 14. what would be your ‘best choice’ ? Why? If the "do nothing" alternative has zero PW (as is implied) then it is best to not undertake the project since its PW is less than zero.44 1.031 4. i.9860 3.4883 2.5887)(1.2791 0.1346 0. (b) (5 marks) Find the present worth of the project if the interest rate is 20% per year.4019 0.3271 4.4301 8.1917 7.38629 0.9061 6.3349) = -6.20%.27473 0.2986 4.23110 0.2998 5.3255)(.0739 3.8964 3.6046 3.1) +2500(P/A. n) 0.23852 0.233 15. (Factor values for i = 20% appear below.2742 1.5806 8.8333 1.8372 4.2.10071 0. n) (A/G.728 2.6) = -3000(2.20%.000 5.7989 25.45455 0.2893 3.20%. i.2 1.22526 (P/A.4392 (P/G.1065 2. n) 1. n) 0.2 0.500 Net Income .8831 11.18629 0.2551 9.5278 2.4545 0. i.) P = -3000(P/A.000 2.535.0736 2.1925 4. i. i.9159 16.4) (F/P.4667 0.045125 (c) (2 marks) If you had to choose whether to undertake the project or not.06061 0. i.2326 0.9299 12.47473 0.8519 3.9906 3. n 1 2 3 4 5 6 7 8 9 10 11 12 (F/P. n) 1 0.4841 .30071 0.2) + 2500( 3.5887 2.1122 (A/F.4823 0.8333 0.04808 0.5832 4.5805 3 (A/P.33438 0.9788 2.3255 3.13438 0.02526 (F/A.4335 12.4991 20.0311 0. n) 0.000 2.1615 0.6944 1.1938 0.368 7.5787 0. i.64 5.6405 1.6)(P/F.3000 0 2500 (a) (3 marks) Fill in the net-income column.1504 39.24808 0.9161 (P/F. Consider a project with the following cash flows of income and expenses: Year 0-3 4-6 7-12 Income 0 2.8791 1. i.27742 0.5756 2.4416 9.03852 0.6944 0.2 3. n) 1.9587 32. n) 1 2.20%.000 Expense 3.07742 0. 01435 0. Since the sum of payments made is 120(1.43 (A/G.603. i.000.849.00435 0.0929 0.6-269.396. 5% downpayment. n) 47. n) 0.00101 0.8887 360 1247.1974 Therefore the balance on the loan is 376.9599 (P/G.173. i. Then we simply find the equivalent monthly amount given the present value of the loan is 120.4026 = 7. n) 0.8194 97.05(120.3004)=376. i.3.2 .173.396. i.396. n) 0.8926 360 35.3004 240 10.173.5974 of the principal was paid. the total interest amount paid over the first 10 years is 140.00002 (F/A. (a) (6 marks) If Shane agrees to make monthly payments to pay for the principal plus interest by the end of the term period.06)=140.44 62328 (A/P.02002 (P/A. zero origination fee.911 49. i. i.0387)=269. n) 488. Shane bought a house today for $120. n) 69. i.2.173.000(3.767.767. i. n) 0.06(F/A.05)120.0918 0. n) 3334. i.4026 of the principal.7005 90.120)= 114.5686 49. n) 0.02017 0.603.7114 47.00205 0. 360) = (114. n) 1710.000 . i.120) = 1.01029 (P/A.01029)=1.3030 0.7112 4 .02205 0.7393 89.7652 240 115. he borrowed an amount of 114. This means that over the 10 year period only 114.00029 (F/A. Interest is compounded monthly. n) 0. what will be the total amount of interest paid in the first 10 years? Denominated in year 10 dollars.06 (b) (4 marks) Given the above.96 (A/P.0008 (A/F. n) 230. how much will he have to pay the bank each month? The effective rate per payment period (month) is i_12 = 12%/12 = 1%.000) (the amount actually borrowed after the 5% downpayment).6 8720.5974 = 133.3554 49.245. n) 45.1%.0278 (A/F.000(F/P. i.8349 75. i.2183 (P/G. i.1%.11 6878.000)(0.0387 989.56 (P/F.6026 i = 1% n (F/P.6 and he made payments over the 10 years whose total is equivalent to the following amount in year-10 dollars: 1.4026 that is.1%.88 2482.. n) 120 3.000(A/P. i. i.2582 5744.06 (230. The financing terms are as follows: 30-year fixed rate of 12% per year interest.01101 0. A = (1-.00017 0.849.1974=106.000-106.7.57 (A/G.163. he still owes 106. n) 120 10.9496 (P/F.6995 i = 2% n (F/P.0086 0.2554 3494. n) 37.245.42 2374. i. 3)+(P/F.22526 0.1122 0.20%.9)+(P/F.4392 4.) Type of Cost First Cost Annual Operating Cost Salvage Value Life in Years n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 (F/P. (Factor values are given below.5756 2.5327 4.12)](P_1f) = [ 1+ 0.4335 12.000 5.2551 9.18629 0.8333 0.26061 0.847. i.000(P/F.45455 0.20%.000 5 (P/A.000 + (-4.2326 0.1615 0. n) 0.3) = 19.3)+1000(P/F.8175 3.9161 10.4991 20.1938+ 0.8333 1. i.85174 Since P_2 > P_1.64 5.000 -4.000)(2.20%.000 1.44 1.01388 (F/A.6944 1.1598 6.9860 3. i. n) 1.0351 Project 1 -10.20%.0779 0.13438 0.6993 12.2902 2.1959 72.5787 + 0. Project 1: The "first project" has present value P_1f = -10.5) +5.03852 0.613.06061 0.9299 12.9587 32.1)= = -32.8392 15.3 Therefore the present value over a period of 15 years with 5 such identical projects is P_1 = [ 1 + (P/F.000 .8964 3. project 2 is better.2998 5.1615 ] (-20.2986 4.4841 3.1938 0.1.0935 0.1065 2.24808 0.3271 4.4966 59.21689 0.20%.2 3.1917 7.698) =-39.4019 0. 5 .4823 0.86708 Project 2: Using the same steps as above we get P_2 = [ 1 + (P/F.031 4.5)) = [ 1 + 0. n) 1 0.8519 3.7989 25.8791 1.1925 4.2742 1.6008 18. i.9788 2.4301 8.1065)+1000(0.4883 2.20%. i.000 3 (A/P.8871 14.07742 0.01689 0.10071 0.0649 (A/F.27742 0.000 -1.47473 0. 20%.0311 0.5805 48.9159 16. Use the least-common-multiple method two compare the following two projects given the applicable interest rate is 20%.38629 0.368 7.3255 3.981.5787 0.801.4.20%.5787) = = -17.8372 4.4667 16.23110 0.5806 8.1346 0.3349 0.2 1.1122 ] (-17.8831 11.4416 9.2791 0. i.30071 0.5095 0.3349+ 0.0739 3.1504 39.0736 2. n) (A/G.5883 17.20%. n) 1 2.04808 0.6597 3.27473 0.6106 4.6944 0.65455 0.7899(-16.33438 0.5) + (P/F. n) 1.6046 3.3) = = -10.407 (P/F. n) 0.23852 0.6405 1.9061 6.6755 (P/G.22062 0.233 15.000)(P/A.4545 0.6)+(P/F.5887 2.2893 3.02062 0.02526 0.728 2.000(P/A. i.000 + (-4.20%. n) 0.10)] (-20.2 0.5278 2. i.5832 4.21388 Project 2 -20.4019 + 0.9588 LCM = 15 => project 1 will have to be undertaken 5 times and project 2 will be undertaken 3 times.847.9906 3. n) 2.20%.23852 Offer 1's future value (period 0) in millions F_1 = -10 (F/P. n) 0.1615 (A/F. If Jerry wishes to make an annual rate of return equal to 20%. n) 0.3255 4.5584 Offer 2's future value (period 5) in millions F_2 = -10(F/P. The company’s profit was -2 million (a loss) in the first year of ownership and has increased since then by 1 million each year.1925 (P/G.5806 12. i.6405 1.9061 6.9061)+25 = -2. n) 1.5)+25 = -10(2.4019 0. n) 4.10)+(-2)(F/A.20%.5.3349 0.9299 25.4883)+(-2)(7. i. i.10)+(1)(F/G. Jerry bought a company 5 years ago for 10 million dollars. i.10)+35 = 0.9906 3.5) + (-2)(F/A. This trend is expected to continue for the foreseeable future.03852 (F/A.20%.9586 > 0 Therefore choose offer 2. n) 7.9860 6. i.20%.13438 0. should he sell the company today for 25 million dollars or sell it 5 years from today for 35 32 million dollars? (The factor values below are for i = 20%.4883 2.20%. n) 0.4416 9.5)+(1)(F/G.4883)(4. n) 2.0739 .20%.9587 (A/P.1917 (P/F. 6 (P/A. i. i. i.8871 (A/G.9788 3.) n 5 6 10 (F/P.4416)+(1)(2.10071 0.33438 0.30071 0.