Summer CH3 Solutions

# Summer CH3 Solutions - MIME 310 ENGINEERING ECONOMY...

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1 M I M E 3 1 0 E N G I N E E R I N G E C O N O M Y SOLUTIONS TO PROBLEM EXERCISES – THE TIME VALUE OF MONEY 1. Monetary flow diagrams illustrate inflows and outflows of cash over time. Downward- pointing arrows represent disbursements, i.e. outflows of cash, and upward-pointing arrows, receipts, i.e. inflows of cash. These diagrams are very useful in solving time value problems. i) ii) iii) 1 2 3 4 5 0 6 7 8 \$700 9 10 1 2 3 4 5 0 \$1000 \$600 \$800 1 2 3 4 0 \$200 \$300 \$500 \$400 \$500

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2 iv) Notes : Part ii – Because the payments are received at the beginning of each year, there is no payment at time 10, i.e., the end of year 10. Part iv – Because the payments are received at the beginning of each year, there is no payment at time 8, i.e., the end of year 8. 2. The amount borrowed is \$6000 and the interest rate is 12 percent compounded annually. The table below gives the repayment schedules for the four proposed schedules. Values are rounded to the nearest dollar. Schedule Year Balance owed at beginning of year Interest owed at end of year End-of-year payment Balance owed at end of year i 1 6000 720 720 6000 2 6000 720 720 6000 3 6000 720 6720 0 ii 1 6000 720 2720 4000 2 4000 480 2480 2000 3 2000 240 2240 0 iii 1 6000 720 0 6720 2 6720 806 0 7526 3 7526 903 8429 0 iv 1 6000 720 2498 4222 2 4222 507 2498 2230 3 2230 268 2498 0 Note : The annual end-of-year payment under schedule iv is determined as follows: 6000 (A/P,12%,3) = 6000 (0.4163) = 2498 Using a financial calculator with N=3, I/Y=12% and PV=6000, compute PMT to obtain -2098.09. 1 2 3 4 5 0 6 7 8 \$1200 \$1260 \$1323 \$1389 \$1459 \$1532 \$1608 \$1689
3 This payment contains the interest owed at the end of the year as well as an amount to repay the principal. For instance, from the end-of-year payment of \$2498 in year 1, an amount of \$720 covers the interest due on the balance owed at the beginning of the year (i.e. 12% of \$6000), and the remainder, i.e. [2498 - 720] or \$1778, is applied to the reimbursement of the principal. The balance owed is thus [6000 - 1778] or \$4222. 3. The amount to be deposited today is the present value equivalent (PV) of the 8 withdrawals. i) PV = 1000 (P/A,6%,8) = 1000 (6.2098) = \$6210 ii) PV = 1000 (P/A,10%,8) = 1000 (5.3349) = \$5335 iii) PV = 1000 (P/A,15%,8) = 1000 (4.4873) = \$4487 Using a financial calculator with N=8, PMT=1000 and I/Y=6%, 10% and 15%, compute PV to obtain -6209.79, -5334.93 and -4487.32, respectively. As the interest rate increases, the amount to be deposited today decreases. 4. The amount that can be withdrawn in nine years' time is the future value equivalent (FV) of the three deposits at that time. FV = 1000 (F/P,10%,9) + 1000 (F/P,10%,6) + 1000 (F/P,10%,3) = 1000 (2.3579 + 1.7716 + 1.3310) = \$5461 There are many ways of solving this problem with a financial calculator. For instance, with N=3, I/Y=10% and PV=1000, compute FV to obtain -1331. Change the sign, add 1000 to obtain 2331, and enter this value in PV. Compute FV to obtain -3102.56. Change the sign, add 1000 to

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## This note was uploaded on 10/06/2009 for the course MIME 310 taught by Professor Bilido during the Summer '08 term at McGill.

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Summer CH3 Solutions - MIME 310 ENGINEERING ECONOMY...

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