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Cash_Flow_Fundamentals_Interest_Factors

# Cash_Flow_Fundamentals_Interest_Factors - IE 168 Production...

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Unformatted text preview: IE 168 Production Analysis < ;6—&Jlm_. Exam 1 Fall 2004 1...(10 pts.) One dollar is to be compounded quarterly at 12% nominal annual interest rate for 3 years. Determine an equivalent annual interest rate if the dollar is compounded continuously over those 3 years. (l+%>ﬂ)(n: 67>? /,2/@m(1.as) = Er y =,//g15 a ”3.23% Z" 2...(15 pts.) Determine the uniform end-of—year payments that would be equivalent at an interest rate of 15% to the cash flow diagram, below, where payments have an annual spacinn. 2500 ........ . 1500 2000 3000 ,8qu_ '73-4' .éS7S ,5-7/% 'F: ——3cwo (P/F)'(S,|\+ISUO(P/F).l§32) ~2m(P/F).i\$”)\$\ +200v (P/F).IS)‘+) ,‘(4'7JL o‘f3313 ‘- 0 : '— .. _ 2,233 "9323 [aw (i /( ).Ih,53 +3500 (?/F,"°>") “l“ 800“) ,2472 m 1 . is, 3)(P/F),,5;2/ ~I=900(1’/F,.i-:)zo) +/00'D = 3mg: A = ? (A/?,.ib‘)zo\ = 356,25 (4943} =71 N (\ It 3 I H I OJ N 04 L” F: moo +500 (”P/evazzsrﬁ) +1500 [P/A 325 g) + )' \ /Uoo(P/Q),322§,33] (F/P) .I5, I) + 300 (”F/’4 A :r(A/F,.i§>10> [' 30‘” (P/A;,3zzs‘) 5\ + i . ' =7; 33(WFJ .15) é)—/50:>(r/i§,l§‘ Name 3) (25 pts) You wish to start a retirement account to supplement your pension. Starting next year (one year from now = year 1), you place \$1000 in an account that earns 5% interest per year. Each year after that your deposit increases an additional \$100 for a total of20 deposits (the second year deposit is \$1 100, the third year deposit is \$1200 ...). In years 21 through 30 you deposit a constant amount of\$4000. The 30th deposit is the last. How much is in the account at the end ofthe ﬂ year. )(z' 7 3’0 5—76 1 Liam How (3 “0 Au: 0 — _[,w0+ 1000‘} /G 5% Hzo):((P/ﬁ 970 2:5)w “LIUOq/F/ﬁfw/Qrﬁijzo 7 ‘tﬂcﬁo 12 L(«ﬁzz 77217 '37“? + x (10/5575 L70) v/J/ r’Li-Zlk 1'12le e [loabﬁ‘msj (12 than) + {1001/ 325” =- WHIM 1* .1‘12151 :— W 33‘79—2 :‘I’E’g 33130 V lue at year 40 = 2-7363, q go # wrong IE 124 ‘ Engineering Economy and Decision Analysis Spring '94 Quiz 1 Prof. Wilson 4. The effective interest rate in each odd month is 3% and in each even month is 5%. What is the equivalent annual effective interest rate? (+1; = [(Mnﬂmsﬁy IE 124 Engineering Economy & Decision Analysis Spring 1996 Quiz 1 1 ...... Find the present worth of the following cash ﬂow diagram if i = 18%. IE 124 g, Engineering Economy & Decision Analysis Spring 1996 Quiz 1 2 ...... For the cash ﬂow diagram that follows and an effective annual interest rate of 18%, ﬁnd the value of X so that equivalent receipts equal equivalent disbursements. 17= §o+§o[?/A,: 13 ﬂ— (IX)[A'/?).I2 21(_PIA.nw][P/ﬁmz€1 =0 124.5os = (1)0 (3439) =- HST-t? X g m 124 ﬁg Engineering Economy & Decision Analysis Spring 1996 Quiz 1 3 ...... An investment of \$100,000 is to be made for a period of 10 years. Over the ﬁrst 6 years, the inﬂation rate is expected to be 8% annually, and over the last 4 years, it is expected to be 12%. Expenses are expected to be \$10,000 in constant dollars annually. What amount in actual dollars should the investment be sold for in 10 years such that the inﬂation-free return on investment is 15% compounded annually? ““6 IE 124 Engineering Economy and Decision Analysis Spring '94 Quiz 2 Prof. Wilson 2. You have just won \$2,000,00011 You plan to quit your job and live on the interest earned on the \$2,000,000 when invested at 12%. You want to have a “constant dollar" income over the next 50 years, anticipating an average 6% inflation rate over that period. What will your "constant dollar“ annual income be (paid at the end of each year) over the next 50 years without spending any of the original \$2, 000, 000? 2-11.33“ WJW.1 ”M ————~——’——"“ . canoe W4: is ,, E=WWM4a¢uc M1maujoquaé . . Wf‘zgwﬁ 1 J’MWWW alt-£41. (canary; WA) A = ,i. can: (.IZ)(Z)0013|¢mo) : Z‘fo’ow Pad [MU-Lt W 1.15:! 31m] = Pw Lt‘aMIut Jada-a fd-Iabui'M-l 7 3°” (4 54:7 2.40 ,(TUOIP/A .lZ 501:: [ID/46,501 41:.“ Hf- F I: I+.lz - Imwo ’ ,T': 0‘" 5"” 3 M. ' = .0564. F = I: 14 ( ‘ I 1 P/A=((.asu)53( (WWJW aft=o I (-osu)(l.os¢qs' 1:: F,[P/F).oo,n1 “6-5"? =(1211o1)(.44zn : 1261297 IE 124 Engineering Economy & Decision Analysis Spring 1996 Makeup Exam 1 ...... Interest Factors: (A) given: geometric gradient, discrete compounding, discrete payments, P = 1,000,000, F1: 1000, i = .1, n =inf1nity. mi: g (B) given: uniform gradient, continuous compounding, discrete payments, G'= 100, n =10, wa uniform series with P = 1000, n =10. Find: r that will make both series have the same annual equivalent value. (C) giiven: compounding periods per year = 42 and the effective interest rate over a 1.65 year period must be a value such that an investment triples in 5 of these periods. Fin : r \A £0, _ Marv r r _ lave e -' e_‘ (or e " -—(or l—e [0199f e—Ior) [ova (he—[0T3 Cﬂ| " [or ‘ /m(€t'\ e ——| {61%- ”°" IoV‘ -1") T" _ If 7(6 -ﬁrlmﬂe (‘9 3” cl" -(oVr ’— /w(e —-|~‘+e )#(W(er— eqr~l*e IDY‘ T:.¥227 K I ,L: ([+‘EM\ W‘_( Nut; 3: [HP x 4cm: , 1.244. = [+4 i=4“. 2 1.02% 41+ J52) (.0032 = I’r—‘C— "f'Z (MSW‘IZW ...
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