6 Pages

Park - FUNDAMENTALS OF ENGINEERING ECONOMICS, 2nd Edition_Solution Manual_im13

Course: ENGR ENG106, Winter 2010
School: UC Davis
Rating:
 
 
 
 
 

Word Count: 1057

Document Preview

13 Chapter Understanding Financial Statements 13.1) (2) Income statement; (1) balance sheet; (3) cash flow statement; (4) operating activities; (5) investing activities, and (6) financing activities; (7) capital account (paid-in capital) 13.2) (7), (8), (1), (11), (3), (9) 13.3) (a) Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 = $580,000 Current liabilities = $50,000 + $100,000 + $80,000...

Register Now

Unformatted Document Excerpt

Coursehero >> California >> UC Davis >> ENGR ENG106

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
13 Chapter Understanding Financial Statements 13.1) (2) Income statement; (1) balance sheet; (3) cash flow statement; (4) operating activities; (5) investing activities, and (6) financing activities; (7) capital account (paid-in capital) 13.2) (7), (8), (1), (11), (3), (9) 13.3) (a) Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 = $580,000 Current liabilities = $50,000 + $100,000 + $80,000 = $230,000 Working capital = $580,000 - $230,000 = $350,000 Shareholders equity = $100,000 + $150,000 + $150,000 + $70,000 = $470,000 (b) EPS = $500,000/10,000 = $50 per share (c) Par value = $15; capital surplus = $150,000/10,000 = $15; Market price = $15 + $15 = $30 per share 13.4) (a) Working capital = Current assets Current liabilities; Working capital requirements = Changes in current assets Changes in current liabilities; WC req. = (+$100,000 - $20,000) (+$30,000 - $40,000) = $90,000, indicating that additional financing is needed to fund the increase in current assets. (b) Taxable income = $1,500,000 - $650,000 - $150,000 - $20,000 = $680,000 (c) Net income = $680,000 - $272,000 = $408,000 (d) Net cash flow: Operating activities = net income + depreciation WC = $408,000 + $200,000 - $90,000 = $518,000 Investing activities = equipment purchase = ($400,000) Financing activities = borrowed fund = $200,000 Net cash flow = $518,000 - $400,000 + $200,000 = $318,000 13.5) (b) 13.6) (b) 13.7) (d) Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. 13.8) (b) ROE= Profit margin Asset turnover Financial leverage (a) 0.1668 (b) 0.1900 (c) 0.1447 (d) 0.1152 13.9) Income Statement: A B C D E F $900,000 $585,000 $315,000 $270,000 $108,000 $162,000 Balance Sheet: 0 $160,000 1 $120,000 2 $320,000 3 $600,000 4 $900,000 5 $1,500,000 6 $450,000 7 $700,000 8 $100,000 9 $700,000 $800,000 From Current ratio Total current assets = 2.4 $250,000 = $600,000 ----------------------------------- Plant and equipment, net = $1,500,000-$600,000=$900,000----------------------- 4 From Quick ratio Inventory = $600,000 - (1.12 $250,000) = $320,000 ----------------------------- From Inventory Turnover Net Revenue = (($320,000 +$280,000)/2) 6.0 =$1,800,000 Cost of goods sold = $1,800,000- $900,000= $900,000 ------- A From DSO Accounts receivable = 24.3333 ($1,800,000 365) = $120,000 ------------------ Cash = -(+) = $160,000 ----------------------------------------------------------- From interest expense of income statement Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage 2 in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. Bond = $450,000 ----------------------------------- 250,000 + = $700,000 -------------------------- From Debt to Equity ratio Total Equity = $700,000 0.875 =$ 800,000 ------------- Total assets or Total liabilities and equity = + = $1,500,000 ------ From Return on total assets Net income F = 14% ($1,350,000) - ($45,000)(0.6)=$162,000 From F, D =F 0.6 = $270,000, E = D (0.4) =$108,000 C = D+45,000 = $315,000 B=$900,000-C = $585,000 From EPS Stock Outstanding = F 4.05 = 40,000 shares Common stock = $2.50 40,000 = $100,000 ------------------------------- Retained Earnings = - = $700,000 ------------------------------------9 13.10) Given Olsons EPS = $8 per share; Cash dividend = $4 per share; Book value per share = $80; Changes in the retained earnings = $24 million; Total debt = $240 million; Find debt ratio = total debt/total assets Net Income = $8 X Where X = the number of outstanding shares EPS = Total shareholders' equity = $80 X Retained earnings = Net income Cash dividend; Net income = 8X from EPS relationship and the total cash dividend = 4X, so we rewrite 8X 4X = $24 million, or X = 6 million shares Book value = From the book value per share, we know that total shareholders equity = 80X, or $480 million; Total assets Total = liabilities + Total shareholders equity = $240 million + $480 million = $720 million Debt ratio = $240 million/$720 million = 0.33 Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage 3 in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. 13.11) (a) Debt ratio = Total debt $2,047,599 = = 123.46% Total assets $1,658,528 EBIT -$1,038,770 = = 1, 622.27% (b) Times-interest-earned ratio= Interest expense $64,032 Current assets $617,266 (c) Current ratio = = = 70.54% Current liabilities $875,065 Current assets - Inventory $617,266-34,502 = = 66.60% Current liabilities $875,065 Net sales $637,235 (e) Inventory turnover = = = 26.14 times avg. Inventory ($34,502+$14,256)/2 (d) Quick ratio = (f) DSO = A/R $71,014 = = 40.68 days avg. sales/day $637,235/365 Net sales $637,235 = = 0.3842 times Total assets $1,658,528 (g) Total assets turnover ratio = (h) Profit margin on sales = Net Income -$1,104,867 = = 173.38 % Net sales $637,235 (i) Return on total assets = Net Income+Interst expense(1-tax rate) -$1,104,867+64,032(0.6) = = 56.97% avg. total assets ($1,658,528+2,085,362)/2 with an assumption of 40% tax rate. (j) Return on common equity = Net Income -$1,104,867 = = 34.47% avg. shareholder's equity (-$389,701+$324,968)/2 Price per share $2.83 (k) Price-earning ratio = = = 3.59 Earnings per share $0.7877 Net income $1,104, 867 = Earnings per share= Avg. number of shares outstanding 1, 402, 619 = $0.7877 (l) Book value per share = total stockholders' equity-preferred stock -$389,071 = = 0.277 Shares outstanding 1, 402, 619 Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage 4 in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. 13.12) Total debt $1,118,455 = = 79.39% Total assets $1,408,785 EBIT $290,235 = = 946.69% (b) Times-interest-earned ratio= Interest expense $30,658 Current assets $515,173 (c) Current ratio = = = 179.88% Current liabilities $286,394 (a) Debt ratio = Current assets - Inventory $515,173-$311,464 = = 71.13% Current liabilities $286,394 Net revenue $3,198,084 (e) Inventory turnover = = = 16.37 times avg. Inventory ($311,464+$79,181)/2 (d) Quick ratio = (f) DSO = A/R $126,634 = = 14.45 days avg. sales/day $3,198,084/365 Net revenue $3,198,084 = = 2.27 times Total assets $1,408,785 (g) Total assets turnover ratio = (h) Profit margin on sales = Net Income $157,368 = = 4.92 % Net sale $3,198,084 (i) Return on total assets = Net Income+Interst expense(1-tax rate) $157,368+30,658(0.6) = = 15.94% avg. total assets ($1,408,785+758,780)/2 We assume a tax rate of 40%. (j) Return on common equity = Net Income $157,368 = = 55.23% avg. share holder's equity ($290,330+$279,493)/2 Price per share $65 (k) Price-earning ratio = = = 19.30 Earnings per share $3.37 $157, 368 Earnings per share= 46, 738 = $3.37 (l) Book value per share = Total stockholders' equity - Preferred stock $290,330 = = 6.21 Shares outstanding 46, 738 13.13) Not provided Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage 5 in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. 13.14) Not provided. Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage 6 in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

UC Davis - ENGR - ENG106
Instructors ManualFUNDAMENTALS OF ENGINEERING ECONOMICS, Second EditionChan S. Park Auburn UniversityInstructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. 2008 Pearson E
UC Davis - ENGR - ENG106
Chapter 1 Engineering Economic DecisionsThere are no end-of-chapter questions in this introductory chapter. However, the following questions could be added as a part of instruction: 1. Ask students to review the contents of The Wall Street Journal for th
UC Davis - ENGR - ENG106
Chapter 2: Time Value of Money2.1) 2.2) Simple interest:F = P (1 + iN ) $4, 000 = $2, 000(1 + 0.08 N ) N = 12.5 years (or 13 years)I = iPN = (0.09)($3, 000)(5) = $1, 350Compound interest:$4, 000 = $2, 000(1 + 0.07) N 2 = 1.07 N log 2 = N log 1.07 N =
UC Davis - ENGR - ENG106
Chapter 3 Understanding Money Management3.1) (a)r = 1.5% 12 = 18%ia = (1 + 0.015)12 1 = 19.56%(b)3.2) Nominal interest rate: r = 0.95% 12 = 11.40% Effective annual interest rate: ia = (1 + 0.0095)12 1 = 12.01%3.3)Assume a continuous compounding:r
UC Davis - ENGR - ENG106
Chapter 4 Equivalence Calculations under Inflation4.1)1.1(1 + f )11 = 3.15 f = 10.04% 100(1 + 0.1004)11 = 286.454.2)(a) 144.5(1 + f )5 = 170.6 f = 3.3766%(b)170.6(1 + 0.033766) 2 = 182.324.3)100(1 + 0.05)(1 + 0.08) = 113.40 100( F / P, f ,2) = 113
UC Davis - ENGR - ENG106
Chapter 5 Present-Worth Analysis5.1) (a), (b), (c)n 0 1 2 3 4 5 6 7 8 Inflow $0 $215,500 $215,500 $215,500 $215,500 $215,500 $215,500 $215,500 $215,500 Outflow $65,000 $53,000 $53,000 $53,000 $53,000 $53,000 $53,000 $53,000 $53,000 Net Cash Flow -$65,00
UC Davis - ENGR - ENG106
Chapter 6 Annual Equivalence Method6.1)AE(9%) = $20, 000( A / P, 9%, 5) = $51, 4206.2)AE (10%) = A( P / A,10%, 3) = $100, 000 A = $40, 210.706.3)AE(12% =$25, 000(A/ P,12% 6) ) ,$4, 000(P/ F,12%,1) +$13, 000(P/ F,12%, 2) +$13, 000(P/ F,12%, 3) + (A/
UC Davis - ENGR - ENG106
Chapter 7 Rate of Return AnalysisNote: Symbol convention-The symbol i* represents the breakeven interest rate that makes the PW of the project equal to zero. The symbol IRR represents the internal rate of return of the investment. For a simple (or pure)
UC Davis - ENGR - ENG106
Chapter 8 Accounting for Depreciation and Income TaxesNote: For the most up-to-date depreciation and income tax information, consult thebooks website at http:/www.prenhall.com/park and click on Tax Information 8.1) 8.2) 8.3) Total property value with th
UC Davis - ENGR - ENG106
Chapter 9 Project Cash Flow Analysis9.1) 9.2) a) 6 b) 11 c) 5 (Note: It is tempting to select 1, but the graphs are drawn on cumulative basis) d) 4 e) 2 f) 10 g) 3 h) 7 i) 9 9.3) a) Incremental cost Description Soldering operation Direct materials Direct
UC Davis - ENGR - ENG106
Chapter 10 Handling Project Uncertainty10.1) (a)AEC(10%) = (25, 000 5, 000)( A / P,10%, 6) + 0.1(5, 000) + 3, 000 = $8, 092 (b) AEC(10%) = (25, 000 5, 000)( A / P,10%, 5) + 0.1(5, 000) + 3, 000 = $8, 776 (c) AEC(10%) = (25, 000 5, 000)( A / P,10%, 6) +
UC Davis - ENGR - ENG106
Chapter 11 Replacement Decisions11.1)Tax Rate(%) = MARR(%) = 0Income Statement0.00% 10.00% 1 2PW(i) = AE(%) = 3($20,065)($6,329.8)4Revenues (savings) Expenses: O&M Depreciation Taxable Income Income Taxes (%) Net IncomeCash Flow Statement$2,500
UC Davis - ENGR - ENG106
Chapter 12 Benefit-Cost Analysis12.1) B = $117, 400( P / A, 6%, 5) = $494, 535.76 C = $5, 000 + $48, 830( P / A, 6%, 5) = $210, 691.49 BC(6%) = $494, 535.76 $210, 691.49 = 2.35 > 1This project is justifiable based on the benefit-cost analysis. 12.2) B =
UC Davis - ENGR - ENG104
tJ ~ 0vt IS 1h,1 j6ah~- pa ,-tk~ ~ eQ IltJ P~J rJ~~(fedP(+/5'tuv ~ : ;:w~1- ;')pAJdU b ~=~IJ7.\[ (S~(l:r f~, , f s~fy : <ul~IJ~ leJ ~ fu /f(b"I'- crl ~o,cfw_ lutAlJ;k C; d@J f(upe~eJuP-~ 7dsifV 5S'dr ~cfw_,~r:,t'l~ILIle
UC Davis - ENGR - ENG104
UC Davis - ENGR - ENG104
Lecture#1:Jan5,2010ENG 104 Mechanics of Materials - Winter 2010READING ASSIGNMENT Textbook Sections 1.1-1.13 HOMEWORK ASSIGNMENT Problems: Due: 1.3, 1.13, 1.40, 1.47, 1.53, 1.68 Jan 14ADMINISTRATION (HOUSE-KEEPING) Instructor: Telephone: Email: Office
UC Davis - ENGR - ENG104
UC Davis - ENGR - ENG104
UC Davis - ENGR - ENG104
UC Davis - ENGR - ENG104
UC Davis - ENGR - ENG104
L~-,' (if : qJ 16) 24-) R~( '~ : ,1 - 61-+_.' \w;#- 7.l,\~-jr_I,-.1-"" -IIrI~.I\I\. \ I .L ~'I-I'I 'If),_l .~,~-'If t1-",\1"',1 1j".J't' " . I.J-II~I' .\ .1 I'I \.I \'j. \ ~".\.Ir ~.L .\ ( \,IIt
UC Davis - ENGR - ENG104
+.I'Jfh~Cl, \,Cc.tCfl, -4' '( (tIe> .,- cf, I h 1/\ 0 ( 'n f D I (\ t0 ')(Jrerel'-' , ll6?0L\(- (:.JL(iIJ ,(',.( '<'0('1 , E'( IJ(cn\f2 () ,enCQI~tlcfw_-fX\ fa ) ]e 1(EsO) vea.X/.~. rrolc-t' -~i1l)\c cfw_'1 "t=I\l_~L,
UC Davis - ENGR - ENG104
\~p l~ It U :k ;" U(j\r\~IC\ g -tlttx\1 ,:;t' ste;s-E-S i (: ~()J.c; j (-'If ( O fJ r1(Cl~ (\te J-:>ll' t-1C~'G I~J\ '~(J 0 , \ \ 1 i i B:X\./ (; 01r;~ , I +f j~ . 1:7<81 c>,~\IeTt,ll G (ILlm.'IO 'l SIc1ey(\hy(c Y A .,' Y('flfl BIt 0 cfw_- .
UC Davis - ENGR - ENG104
I~~ '=z :=.Alg, +V\,g., +1f'Vj,A I -+ lAz. -I1A-s2(7s)Oo)(cfw_J) (cfw_G)! ~-+ Z (w)(io)4c + Z(25)~o)5( ;>0(10; -+ 2 ( 6 0 ) (!o) +-i!"-<0-<- l=5 fr-h()(\ G.-c~ _~ LV(\., = 6>- 1 Of1sC, 0 )(3C;>,S;) =- l:xLf 6~' /O~ t'Y\'(Y)~ ),71,'Cf( 1 0)
UC Davis - ENGR - ENG104
2,/ lt /IOI-/w7 :b) fD )2.- Lf ) 32) l D6 )11)/ 2. I fAnc I,p/:S -I. '50UJe e x) S ec(l()nS0 '( s+re~ ~~ .Q.aJ:/(1\9 ~ 7 ; i -/3sir(/. )/1~vh,~ve sfrrJ i ed~"1~ ClAY' ( ()Ordin'Jm QXlG)".stresses- ~sfmlns->--Hiifl-2(5~2.a)(ldl"f/O(J
UC Davis - ENGR - ENG104
/ Ccfw_~(1 c\./1 i rc1y\ (l~ I/' -r(~Ure W~cfw_ f-= I M~d)(~t I'~(\Shwr ~(U;(~ H.l1'fhtc ln0Sl I t:L 1<';2OH~?()Y\X, alh,x-~b1~uJn(J ,'" ff6-X- ~'Y'C'r \ ~<.,s (t iJ , Kero.\- _l._ _&=-2 QA. _ T v -LG6c~cr(ea\ f 0 1 c.A..I~>J z-
UC Davis - ENGR - ENG104
2-/23 10Jcfuacfw_!I+dll!.~WiJr1-r 7 . 1~()Reo.J.t'flj-<(:3'f/_~ : / - if-I7IIJe&tie-Cdl/~ C~ I-Ws l ( a.v.J ~ CdMfl de~ rt~r-) h kfLJi(Ie- vt'srf6a (~We-dd'ltr!.cluJ i I\Cj~~ 4.ad-erl oJa.y :, Pe J 07Frllis jC~-kr-?CAfeloo)
UC Davis - ENGR - ENG104
UC Davis - ENGR - ENG104
DeF(~ G t i<t1lC)wple rofe aM5.4-:frAte Bc,1[ d '"Il!) - 7 5 f r et;tj~7 D~fluj-'i611, - .-' - . M - P ELf=-rod jtJ.5 aFcur-ifrJJuveGhQ.fl-e rr:V=SA eo. r :q t1: F(X)dM d-X oprovlJeJ +1t aJ :5 cVJ;1 f Ve JI. OM!- ~ f r;/tCI fd. 1
UC Davis - ENGR - ENG104
RuoJii)1.we-,Jei j v~dE- I.-:; hoseJOnk :- J1(~)ct'f1w~CAM.~fl.rl/~ r/t b~ Jcfw_,(~ ~l1~it(jCtsf,gMlf rl;~vJ(L~JiFfe-r ~ah' ~ ~ f (~ Ii t'\() l J ' i ~ :.- d";- f- I t! eLI i 001 ~ O~ffere.h~ ()~r;e.k-=;>;t~ ~ Vcj)di-1-: ~ l
UC Davis - ENGR - ENG104
(drrecA1 JV15p re0/[P(<2- vi dcfw_)SLec~'t-CD5 ufe' fJ -i ft'J I\ XfyJ ~ O ll/fc(LTo ~JcG1 )s~_-2:-I-1 ~ (/+L1J e- ~ 0 R,-e v; 0 WS'kft'(dyJ f;~ Jele0t;~ Jf-~ Sk~cdfy In rj~kcfw_j . vJe +b~ Jef1e t~tt1 P-eUleWs t-aJ7 ca)!yk'efe,fJt(~ (
UC Davis - ENGR - ENG106
Eng 106Homework #8 SolutionWinter 20109.47 Given: Beta = 1.7, market return rm = 14%, risk-free rate rf = 7%. Find: cost of equity, ie. Cost of equity, ie = rf + Beta(rm rf) = 7% + 1.7(14% - 7%) = 18.9%9.48 Given: a.t. cost of equity ie = 25%, b.t. co
UC Davis - ENGR - ENG106
Eng 106Homework #1 SolutionWinter 2010Problem 1. Youve been asked to analyze the economic feasibility of installing various sizes of photovoltaic systems on a commercial distribution center. a) What might be one or more possible objectives (wants)? Min
UC Davis - ENGR - ENG106
Eng 106Homework #2 SolutionWinter 20102.40 Given: Cash deposits and withdrawals (negative amounts) below, i=12%/yr. EOY: 0 1 2 3 4 5 6 7 8 9 10 Flow: 1,000 800 600 400 200 0 -C -2C -3C -4C -5C Find: the one of four values of C given in the book that ma
UC Davis - ENGR - ENG106
ENG 106 Present WorthHomework #3 SolutionWinter 20105.13 Given: Four projects, each has a three-year life, i = 13%/yr EOY A B C D 0 -5000 -2000 4500 -3500 1 5800 -4400 -6000 $1000 2 12400 7000 2000 $5000 3 8200 3000 4000 $6000 a) Find: NPWs of the proj
UC Davis - ENGR - ENG106
ENG 106 Problem 12.6Homework #4 SolutionWinter 2010Given: MARR, i: 10% three M.E. alternatives for parking structures, 30-yr need Design: A B C Site cost, $ 240 180 200 Bldg cost, $ 2200 700 1400 Annual income, $ 830 750 600 Annual maint cost, $ 410 36
UC Davis - ENGR - ENG106
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54A ENG106BCDE HW#5FGH Winter, 2006IProblem 1 Knowns: (Read the problem statement) Two M.E. alt
UC Davis - ENGR - ENG106
ENG 106Homework #6 SolutionWinter 2010Problem 9.34 All dollar values in k$ Assumptions for Both Alternatives Marginal tax rate, % 40% MARR, %/year 15% Inflation rate, %/year 0 A) Buy Option Inputs/Assumptions (all dollar amounts in year 0$) Equip Purch
UC Davis - ENGR - ENG106
ENG 106 Homework #7 Solution Winter 2010 Problem 11.13 Given: Company considering replacing an existing machine. For the challenger: Purchase cost, $ 31000 Economic service life (ESL), years 10 Salvage value @ EOY 10, $ 2500 Operating costs, $/year, EOY 1
Irvine Valley College - MGMT - 12A
Chapter 11. The stability and predictability created by the law is essential to businessactivities. True2. Law is a body of enforceable rules governing relationships among individuals andbetween individuals and their society. True3. The natural law p
Middle East Technical University - EE - ee446
DM54LS181 DM74LS181 4-Bit Arithmetic Logic UnitJune 1992DM54LS181 DM74LS181 4-Bit Arithmetic Logic UnitGeneral DescriptionThe LS181 is a 4-bit Arithmetic Logic Unit (ALU) which can perform all the possible 16 logic operations on two variables and a va
Middle East Technical University - EE - ee446
SN54/74LS181 4-BIT ARITHMETIC LOGIC UNITThe SN54 / 74LS181 is a 4-bit Arithmetic Logic Unit (ALU) which can perform all the possible 16 logic, operations on two variables and a variety of arithmetic operations. Provides 16 Arithmetic Operations Add, Sub
Middle East Technical University - EE - ee446
Freescale Semiconductor, Inc.M68HC11K/D68HC11M6 HC11M68HC 1M68HC11MM68HC11K Family Technical DataHCMOS Microcontroller UnitFreescale Semiconductor, Inc.blankFreescale Semiconductor, Inc.For More Information On This Product, Go to: www.freescale.co
Middle East Technical University - EE - ee446
DatasheetArchiveRequest For Quotation.comOrder the parts you need from our real-time inventory database. Simply complete a request for quotation form with your part information and a sales representative will respond to you with price and availability.
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #3Deadline: April 10 2009 @ 23:59 (No late submission is allowed.) Posted on April 03 2009.Question # 1Consider a local area network (LAN) which operates over 10Mbits/s CSMA/CD with 8 statio
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #4Deadline: April 24 2009 @ 23:59 (No late submission is allowed.) Posted on April 17 2009.Question # 1Given the message 1010001101 and the CRC polynomial P(x) = x5+x4+x2+1.(a)(b)Compute
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #4Deadline: April 24 2009 @ 23:59 (No late submission is allowed.) Posted on April 17 2009.Question # 1Given the message 1010001101 and the CRC polynomial P(x) = x5+x4+x2+1. (a) (b) Compute
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #5Deadline: May 8 2009 @ 23:59 (No late submission is allowed.) Posted on May 1 2009.Question # 1The round-trip propagation delay for a 10Mbps link is 200msec. Total probability of error (pr
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #5Deadline: May 8 2009 @ 23:59 (No late submission is allowed.) Posted on May 1 2009.Question # 1The round-trip propagation delay for a 10Mbps link is 200msec. Total probability of error (pr
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKS Spring 2009 HOMEWORK #6Deadline: May 31 2009 (Sunday) @ 23:59 (No late submission is allowed.) Posted on May 15 2009.Question # 1B 2 A 1 C Consider the network given above with each link associated with its cost
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKS Spring 2009 HOMEWORK #6Deadline: May 31 2009 (Sunday) @ 23:59 (No late submission is allowed.) Posted on May 15 2009.Question # 1B 2 A 1 C Consider the network given above with each link associated with its cost
Middle East Technical University - EE - ee444
EE444 Term Project Guidelines (Spring 2009) Part IDue Date: April 20th, 2009 Consider the network configuration given below: 1. There are 5 stations sharing an Ethernet local area network (LAN) (i.e., S1, S2, S3, S4, and S5) with a capacity of 10 Mb/s wi
Middle East Technical University - EE - ee444
EE444 Term Project Guidelines (Spring 2009)Phase 2: Due Date: May 29th, 2009 You are required to construct the network topology with the link capacities and delays given on the diagram. Turksat2 Mbps, 100 ms(at t=180s) 1 Mbps, 1000msBU2 Mbps, 50 ms 1
Middle East Technical University - EE - ee444
Lecture Set 3, EE 679, Copyright 2000, Sanjay K. Bose1Basic Queuing TheoryKendall's Notation for Queues:This is a useful way to represent different types of queues in a compact and easily understood fashion. Kendall's Notation describes the nature of
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #1Deadline: March 13 2009 @ 23:59 (No late submission is allowed.) Posted on March 06 2009. Please follow the submission method as described over the course webpage (any other submission, e.g.
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKS2009 SpringHomework #1 with AnswersDeadline: March 13 2009 @ 23:59 (No late submission is allowed) Posted on March 06 2009.Question # 1Consider a car dealer that is also responsible for repairing the cars they
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #2Deadline: March 27 2009 @ 23:59 (No late submission is allowed.) Posted on March 20 2009.Question # 1Is it possible to apply Littles Law in a finite buffer M/M/1/m queue? If it is possible
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #2 AnswersDeadline: March 27 2009 @ 23:59 (No late submission is allowed.) Posted on March 20 2009.Question # 1Is it possible to apply Littles Law in a finite buffer M/M/1/m queue? If it is
Middle East Technical University - EE - ee444
EE 444 INTRODUCTION TO COMPUTER NETWORKSSpring 2009HOMEWORK #3Deadline: April 10 2009 @ 23:59 (No late submission is allowed.) Posted on April 03 2009.Question # 1Consider a local area network (LAN) which operates over 10Mbits/s CSMA/CD with 8 statio
Middle East Technical University - EE - ee446
Date: 19980205From: Tito Smailagich <epetrozl@eunet.yu>Organization: ENICTo: Mot-68HC11-Apps@freeware.mcu.motsps.comSubject: Re: MC68HC11 instruction set Message sent by TITO SMAILAGICH <epetrozl@EUnet.yu> to the mot-68hc11-apps Mailing List.-ENIC
Middle East Technical University - EE - ee446
"" File:C:\FNDTN\ACTIVE\PROJECTS\EXPPRE2\pre2.abl" created:03/23/06 08:27:09" from:'C:\FNDTN\ACTIVE\PROJECTS\EXPPRE2\pre2.asf'" by:fsm2hdl - version: 2.0.1.49"module pre2Title 'pre2'Declarations"clocksCLK PIN;"input portsENDCNT PIN;RST PI