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ISEN Formula Sheet - Quiz Problem 1 if an engineer borrows...

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Unformatted text preview: Quiz Problem 1 if an engineer borrows $1000 from the company credit union at 5% per year compound interest, compute the total amount due after 4 years. H Solution Quiz Problem ow long will it take for $1000 to quadruple if the interest rate is 5% per year? If the solution is not integer, round it up. Solution $1000 (1 + 0.05)‘1 : $1215.51 The n value can be determined using either the F / P or P/ F factor. Using the P/ F fa Quiz Problem 2 A group of engineers estimated that a total of $500,000 will be deposited at the end of next year into an account for the repair of old and safetyrquestionable bridges throughout the area. Further, they estimate that the deposits will increase by $100,000 per year for only 8 years thereafter, then cease. Determine the equivalent present worth if county flmds earn interest at a rate of 5% per year. Solution PT : $500(P/A, 5%, 9) + $100(P/G, 5%, 9) : $500(7.1078) + $100(26.1268) : $6166.59 Quiz Problem Jacki obtained a new credit card from a national bank, M'BNA, with a stated rate of 12% per year, compounded monthly. For a $1000 balance at the beginning of the year, find the effective annual rate and the total amount owed to M'BNA after 1 year, provided no payments are made during the year. Solution There are 12 compounding periods per year. Thus, an : 12 and 1' : 12%/12 : 1% per month. For a $1000 balance that is not reduced during the year, the effective armual rate and the total owed to MBNA after 1 year are calculated as: id = (14.0.01)12 71:1.1268371=D.12683 F : $1000(1.12683) : $1126.83 Jacki will pay 12.683%. or $126.83 plus the $1000 balance. for the use ofthe bank’s money during the year. Quiz Problem Engineers Marci and Suzanne both invest $5,000 for 10 years at 8% per year. Compute the future worth for both individuals if Marci receives annual compounding and Suzanne receives continuous compounding. Solution Marci: For annual compounrhng, the future worth is F : P(F fP.8%,10) : $5.000(1+.08)m : $5.000(2.158925) : $10,794.62 Suzanne: Using equation 1?: e’ —1, first find the effective 1' per year for use in the HP factor. Effective i: em 71: 0.083287 = 8.3287% F = P(F fP. 8.3287%,10) = $5,000(1+.083287)1° = $5,000(2.22554) = $11,127.? Continuous compounding causes a $333 increase in earnings. ctor, P:F(P/F,i,n) $1000 : $4000(P / 17.5%, n) (P/ F,5%,n) : 0.25 1/105” :025 1.05” :4 nlnl.05 : 1114 n : ln4/lnl.05 : 28.4134 By rounding up, the answer is 29 years. Quiz Problem Suppose you plan to purchase a car and carry a loan of $12,500 at 12% per year, compounded monthly. Payments Will be made monthly for 4 years. Determine the monthly payment. Solution A monthly series A is sought; the PP and CP are both a month. Use the steps for PP = CP when a uniform series is present. The effective interest per month is 12% / 12 = 1%, and the number of payments is (4 years)(12 months per year} = 48. The monthly payment is calculated as A = $12,500(A/P,1%,48) = $12,500(0.02633} = $329125 Quiz Problem A new software system will be used for the indefinite future. The system has an installed cost of $150,000 and an additional cost of $50,000 after 10 years. The annual software maintenance contract cost is $5000 for the first 4 years and $8000 thereafter. In addition. there is expected to be a recturing major upgrade cost of $15000 every 13 year. Assume that i = 5% per year for cormty funds. Find the capitalized cost for the recurring major upgade cost. Solution Convert the recurring cost of $15,000 every 13 years into an annual worth A] for the first 13 years. A1 = —$15,000(A/F, 5%, 13) = -$847 The same value, A1 : i$847, applies to all the other 13—year periods as well. This annual cost series are converted into a capitalized cost CC1. A. 7 7$847 1? _ 0.05 CCl : —$ 16. 940 Qu iz Problem Wells Fargo Bank lent a newly graduated engineer $1000 at i = 10% per year for 4 years to buy home office equipment. From the bank’s perspective (the lender), the investment Quiz Problem Detamineflrewmpositemteofreflnnforthecashflowsmfllefiflowingtabkiflhe reinvesmlentrateistheMARRof 15%peryear. Istheprtjectjustified‘? in this young engineer is expected to produce an equivalent net cash flow of $3 15.47 for Cash Flow, 3 each of 4 years. Year Net Cumulative A = $1000(A-"‘P.10%,4) = $315.47 0 0 0 Compute the amount of unrecovered investment for each of the 4 years using the rate of 1 200 +200 retum on the unrecovered balance (the correct basis) and fill the blanks of the table below. i 1:3 72% + Beginning Interest on Ending 4 —1000 "650 Um'ecovered Unreeovered Recovered Um'ecovered 5 600 60 Year Balance Balance Cash Flow Amount Balance 0 --- --- -$1_.000.00 --- $1,000.00 1 , 1 $1,000.00 $315.47 $215.47 m 2 $78453 $31547 $23102 To findtheone i’ value, develop theret—investment serieng, througth usingEquation 3 $547.51 $315.47 $250.72 FI—E_1(1-t-i)+C,andc=15% 4 45286.79 $3l5.47 $285.79 F0=0 F1=$200 (F150,u5€‘3) Solution F; = 2000.15) + 100 = $330 (F2 > 0, use e) F;=330(1.15]+50=$429.50 LFg>0,usee) (5)44), E.=429.50(1.15)—1000=— $500.03 LF4<0,usei’] (1) (2) (3) = 0710 " (2) (4) (3) (5) = (2) t (5) F5 2 5135-93077!) + 500 Beginning SetF5=0andsolvefiIf directly Unrecovered Inmrest onUmecovered Recovered Ending Unrecovered 000 Year Balance Balance Cash Flow Amount Balance 1+i' = =1.135595 — $1,000.00 —— $1,000.00 505-03 1 $1,000.00 0.1 >< $1,000.00 = $100.00 $315.42 $215.41 $1,000.00 + $215.42 i‘=18.56% =$7s453 Sinoei’Ht/IARILIheprojectisjustified. 2 —$784.53 0 1 N $784.53 = $78.45 $315.47 $237.02 $784.53 + $237.02 = $547.51 3 3547.51 0.1 N $547.51 2 $54.75 $315.47 $260.72 -$547.51+ $260.72 = $236.79 4 4286.79 0.1 " $286.79 = $28.68 $315.47 $286.79 $286.79 + $286.79 = 0 Quiz Problem ln2000,BellAflantic andG'IE mgedtoformagiantteleotmmmicationscorporafion named Verizon Comnmnications. As expected, some equipment incompaubilities hadto be rectified, especially for long distance and international Wireless and vitko services. One itemhadtwo suppliers—aUS. firm (A) andanAsian firm (B). ApproximatelySOflD units ofthis equipmentwereneeded. Estimates tbrvendors AandBaregivenforeaeh Quiz Problem unit. A fiber optics testing device is to be DDB depreciated. It has a first cost of $25,000 and an estimated salvage of $2,500 after 12 years. (a) Calculate the depreciation and book value for years 1 and 4. (b) Calculate the implied salvage value after 12 years. Solution (a) The DDB fixed depreciation rate is d = 2/ n = 2/ 12 = 0.1667 per year. Use equations D: = 6130—61)” and BVt = 13(1—0?)f . Year 1: D1 = (0.1667)(25,000)(1.0166711-1 = $4,167 3V1 = 25,000(1—0.1667)1 = $20,833 Year 4: D4 = (0.1667)(25,000)(1.01667141 = $2,411 3V4 = 25,000(140.1667)4 = $12,054 (b) From Equation ”implied S = BVI. = Bard)“ ”, the implied salvage value after 12 years is Implied S = $25,000(140.166'fl12 = $2,803 Since the estimated S = $2,500 is less than $2,803, the asset is not fully depreciated when its 12—year expected life is reached. A B Initial 0051, $ 3,001) 43,001) Annual costs, $ 6,500 4,000 Salvage value, 3 0 2,000 Life, years 10 5 Show all the steps, including incremental cash flow tabulation and incremental cash flow diagram, to determine which vendor should be selected it the NIARR is 15%pervear. ShowtherateofreturnequationbasedmtleWofirmementaloashflowsbut do NOT solve it. Given a unique rate of return i;_A calculated from the equation, how do we determine which vendor should be selected? Solution Sinceallcashflowsarecosgusetheprocedurebelowtodetermine 1";_,,: 1. AlternativesA andB are correctlyr ordered with thehigher first—cost alternativeincolumn (2). 2. ThecashflowsfortheICMoflUyearsaretabulatedinthetablebelow Year Cash Flow A Cash Flow B Incremental Cash Flow (1) (2) (3) = (2) — (1) 0 $0,000 $13,000 $5,000 1.5 $3,500 $1,600 41,900 5 0 $2,000$13,000 $11,000 5.10 $3,500 $1,600 41,900 10 0 $2,000 42,000 3. The incrementalcash flowdiag‘ramis shown in theflgmebelow: 0 = —5000 + 1900M}; r;_, ,10) — 11,000(P/1=, {H ,5) + mfP/F, 17,440) 5. [file calculated 1;). valueislessthan 15% MARRflielower—{DstvendorA is selected. Ifitisgreater than 15% MARK, thehigher—oost vendorB is selected. ...
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