2007 ECN 801 Sample test 2

2007 ECN 801 Sample test 2 - ECN801, V. Bardis Sample Test...

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Unformatted text preview: 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 2. Consider a project with the following cash flows of income and expenses: Year 0-3 4-6 7-12 Income 0 2,000 5,000 Expense 3,000 2,000 2,500 Net Income - 3000 0 2500 (a) (3 marks) Fill in the net-income column. (b) (5 marks) Find the present worth of the project if the interest rate is 20% per year. (Factor values for i = 20% appear below.) P = -3000(P/A,20%,4) (F/P,20%,1) +2500(P/A,20%,6)(P/F,20%,6) = -3000(2.5887)(1.2) + 2500( 3.3255)(.3349) = -6,535.045125 (c) (2 marks) If you had to choose whether to undertake the project or not, 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. n 1 2 3 4 5 6 7 8 9 10 11 12 (F/P, i, n) 1.2 1.44 1.728 2.0736 2.4883 2.9860 3.5832 4.2998 5.1598 6.1917 7.4301 8.9161 (P/F, i, n) 0.8333 0.6944 0.5787 0.4823 0.4019 0.3349 0.2791 0.2326 0.1938 0.1615 0.1346 0.1122 (A/F, i, n) 1 0.45455 0.27473 0.18629 0.13438 0.10071 0.07742 0.06061 0.04808 0.03852 0.0311 0.02526 (F/A, i, n) 1 2.2 3.64 5.368 7.4416 9.9299 12.9159 16.4991 20.7989 25.9587 32.1504 39.5805 (A/P, i, n) 1.2 0.65455 0.47473 0.38629 0.33438 0.30071 0.27742 0.26061 0.24808 0.23852 0.23110 0.22526 (P/A, i, n) 0.8333 1.5278 2.1065 2.5887 2.9906 3.3255 3.6046 3.8372 4.031 4.1925 4.3271 4.4392 (P/G, i, n) 0.6944 1.8519 3.2986 4.9061 6.5806 8.2551 9.8831 11.4335 12.8871 14.233 15.4667 (A/G, i, n) 0.4545 0.8791 1.2742 1.6405 1.9788 2.2902 2.5756 2.8964 3.0739 3.2893 3.4841 3 3. Shane bought a house today for $120,000. The financing terms are as follows: 30-year fixed rate of 12% per year interest, 5% downpayment, zero origination fee. Interest is compounded monthly. (a) (6 marks) If Shane agrees to make monthly payments to pay for the principal plus interest by the end of the term period, how much will he have to pay the bank each month? The effective rate per payment period (month) is i_12 = 12%/12 = 1%. Then we simply find the equivalent monthly amount given the present value of the loan is 120,000 - .05(120,000) (the amount actually borrowed after the 5% downpayment). A = (1-.05)120,000(A/P,1%, 360) = (114,000)(0.01029)=1,173.06 (b) (4 marks) Given the above, what will be the total amount of interest paid in the first 10 years? Denominated in year 10 dollars, he borrowed an amount of 114,000(F/P,1%,120)= 114,000(3.3004)=376,245.6 and he made payments over the 10 years whose total is equivalent to the following amount in year-10 dollars: 1,173.06(F/A,1%,120) = 1,173.06 (230.0387)=269,849.1974 Therefore the balance on the loan is 376,245.6-269,849.1974=106,396.4026 that is, he still owes 106,396.4026 of the principal. This means that over the 10 year period only 114,000-106.396.4026 = 7,603.5974 of the principal was paid. Since the sum of payments made is 120(1,173.06)=140,767.2, the total interest amount paid over the first 10 years is 140,767.2 - 7,603.5974 = 133,163.6026 i = 1% n (F/P, i, n) 120 3.3004 240 10.8926 360 35.9496 i = 2% n (F/P, i, n) 120 10.7652 240 115.8887 360 1247.56 (P/F, i, n) 0.3030 0.0918 0.0278 (P/F, i, n) 0.0929 0.0086 0.0008 (A/F, i, n) 0.00435 0.00101 0.00029 (A/F, i, n) 0.00205 0.00017 0.00002 (F/A, i, n) 230.0387 989.2554 3494.96 (F/A, i, n) 488.2582 5744.44 62328 (A/P, i, n) 0.01435 0.01101 0.01029 (A/P, i, n) 0.02205 0.02017 0.02002 (P/A, i, n) 69.7005 90.8194 97.2183 (P/A, i, n) 45.3554 49.5686 49.9599 (P/G, i, n) 3334.11 6878.6 8720.43 (P/G, i, n) 1710.42 2374.88 2482.57 (A/G, i, n) 47.8349 75.7393 89.6995 (A/G, i, n) 37.7114 47.911 49.7112 4 4. Use the least-common-multiple method two compare the following two projects given the applicable interest rate is 20%. (Factor values are given below.) 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, i, n) 1.2 1.44 1.728 2.0736 2.4883 2.9860 3.5832 4.2998 5.1598 6.1917 7.4301 8.9161 10.6993 12.8392 15.407 (P/F, i, n) 0.8333 0.6944 0.5787 0.4823 0.4019 0.3349 0.2791 0.2326 0.1938 0.1615 0.1346 0.1122 0.0935 0.0779 0.0649 (A/F, i, n) 1 0.45455 0.27473 0.18629 0.13438 0.10071 0.07742 0.06061 0.04808 0.03852 0.0311 0.02526 0.02062 0.01689 0.01388 (F/A, i, n) 1 2.2 3.64 5.368 7.4416 9.9299 12.9159 16.4991 20.7989 25.9587 32.1504 39.5805 48.4966 59.1959 72.0351 Project 1 -10,000 -4,000 1,000 3 (A/P, i, n) 1.2 0.65455 0.47473 0.38629 0.33438 0.30071 0.27742 0.26061 0.24808 0.23852 0.23110 0.22526 0.22062 0.21689 0.21388 Project 2 -20,000 -1,000 5,000 5 (P/A, i, n) 0.8333 1.5278 2.1065 2.5887 2.9906 3.3255 3.6046 3.8372 4.031 4.1925 4.3271 4.4392 4.5327 4.6106 4.6755 (P/G, i, n) 0.6944 1.8519 3.2986 4.9061 6.5806 8.2551 9.8831 11.4335 12.8871 14.233 15.4667 16.5883 17.6008 18.5095 (A/G, i, n) 0.4545 0.8791 1.2742 1.6405 1.9788 2.2902 2.5756 2.8964 3.0739 3.2893 3.4841 3.6597 3.8175 3.9588 LCM = 15 => project 1 will have to be undertaken 5 times and project 2 will be undertaken 3 times. Project 1: The "first project" has present value P_1f = -10,000 + (-4,000)(P/A, 20%,3)+1000(P/F,20%,3) = = -10,000 + (-4,000)(2.1065)+1000(0.5787) = = -17,847.3 Therefore the present value over a period of 15 years with 5 such identical projects is P_1 = [ 1 + (P/F,20%,3)+(P/F,20%,6)+(P/F,20%,9)+(P/F,20%,12)](P_1f) = [ 1+ 0.5787 + 0.3349+ 0.1938+ 0.1122 ] (-17,847.3) = 19.7899(-16,698) =-39,613.86708 Project 2: Using the same steps as above we get P_2 = [ 1 + (P/F,20%,5) + (P/F,20%,10)] (-20,000 - 1,000(P/A,20%,5) +5,000(P/F,20%,5)) = [ 1 + 0.4019 + 0.1615 ] (-20,981.1)= = -32,801.85174 Since P_2 > P_1, project 2 is better. 5 5. Jerry bought a company 5 years ago for 10 million dollars. 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. This trend is expected to continue for the foreseeable future. If Jerry wishes to make an annual rate of return equal to 20%, should he sell the company today for 25 million dollars or sell it 5 years from today for 35 million dollars? (The factor values below are for i = 20%.) 32 n 5 6 10 (F/P, i, n) 2.4883 2.9860 6.1917 (P/F, i, n) 0.4019 0.3349 0.1615 (A/F, i, n) 0.13438 0.10071 0.03852 (F/A, i, n) 7.4416 9.9299 25.9587 (A/P, i, n) 0.33438 0.30071 0.23852 (P/A, i, n) 2.9906 3.3255 4.1925 (P/G, i, n) 4.9061 6.5806 12.8871 (A/G, i, n) 1.6405 1.9788 3.0739 Offer 1's future value (period 0) in millions F_1 = -10 (F/P,20%,5) + (-2)(F/A,20%,5)+(1)(F/G,20%,5)+25 = -10(2.4883)+(-2)(7.4416)+(1)(2.4883)(4.9061)+25 = -2.5584 Offer 2's future value (period 5) in millions F_2 = -10(F/P,20%,10)+(-2)(F/A,20%,10)+(1)(F/G,20%,10)+35 = 0.9586 > 0 Therefore choose offer 2. 6 ...
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This note was uploaded on 02/06/2012 for the course PHYSICS 125 taught by Professor Carloss during the Spring '11 term at Ryerson.

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