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Industrial Engineering Chapter 1 Excel Problem Time Value Money Rationale See Rule The Four Discounted Cash Flow
FE PROBLEM 1.1 a. TIME VALUE OF MONEY rationale: SEE RULE #2 OF THE FOUR DISCOUNTED CASH FLOW RULES FE PROBLEM 1.2 b. COLLECTIVELY EXHAUSTIVE rationale: DEFINITION OF COLLECTIVELY EXHAUSTIVE FE PROBLEM 1.3 d. POSITIVELY rationale: SEE PRINCIPLE #9 OF […]
Industrial Engineering Chapter 1 Homework July Overview Engineering Economic Analysis Sample Responses Discussion Questions What The
July 22, 2013 Chapter 1 – An Overview of Engineering Economic Analysis Sample Responses to Discussion Questions 1) What do the preceding two examples have in common? Sample Response: We have two entities – one large (with $419 billion in […]
Industrial Engineering Chapter 10 Excel Problem Answer Since The Cash Flow Thencurrent Dollars Combined Discount Rate Must
FE PROBLEM 10.1 ANSWER: d SINCE THE CASH FLOW IS IN THEN-CURRENT DOLLARS, A COMBINED DISCOUNT RATE MUST BE USED. FE PROBLEM 10.2 ANSWER: c SINCE EACH SUBSEQUENT CASH FLOW IS 6% GREATER THAN THE PREVIOUS CASH FLOW, IT IS […]
Industrial Engineering Chapter 10 Excel Problem Basis Loan Princ Real Btcf Tax Rate Infl Rate Est Salv
PROBLEM 10.65 BASIS $190,000.00 LOAN PRINC $100,000.00 REAL BTCF $45,000.00 TAX RATE 40.00% INFL RATE= 3.90% EST SALV $5,000.00 LOAN PERIOD 6 BTCF INCR $0.00 USEFUL LIFE 12 MARRr=10.00% RCVD SALV $5,000.00 LOAN INT 13.00% METHOD 1 PERIODS 12 MARRc=14.29% […]
Industrial Engineering Chapter 10 Excel Problem Basis Real Btcf Infl Rate Est Salv Btcf Incr Tax Rate
PROBLEM 10.50 BASIS $1,400,000.00 REAL BTCF $400,000.00 INFL RATE= 4.20% EST SALV $200,000.00 BTCF INCR $0.00 TAX RATE 40.00% RCVD SALV $200,000.00 USEFUL LIFE 8 MARRr=10.00% PERIODS 8 MARRc=14.62% EOY REAL BTCF REAL DWO TI T REAL ATCF MACRS(5) 0 […]
Industrial Engineering Chapter 10 Excel Problem Inflation Rate Eoy Thencurrent Constant Worth Problem Growth Rate Inflation Rate
PROBLEM 10.30 INFLATION RATE 5.00% EOY THEN-CURRENT CONSTANT WORTH 1$10,000.00 $10,000.00 2$12,000.00 $11,428.57 3$15,000.00 $13,605.44 4$17,500.00 $15,117.16 PROBLEM 10.31 GROWTH RATE 12.00% INFLATION RATE 5.00% EOY THEN-CURRENT CONSTANT 0$420,000.00 $420,000.00 1$470,400.00 $448,000.00 2$526,848.00 $477,866.67 3$590,069.76 $509,724.44 4$660,878.13 $543,706.07 5$740,183.51 $579,953.15 […]
Industrial Engineering Chapter 10 Excel Problem Year Rate Index Index Indexindex Indexindexindex Solver Target Cell Solver Changed
PROBLEM 10.12 aYEAR 2000 2001 2002 2003 2004 2005 RATE -14.84% 13.75% 10.09% 18.87% 18.79% INDEX 165.9 141.3 160.7 176.9 210.3 249.8 b ((1+f)^4)-1 = ((INDEX2005 – INDEX2001)/INDEX2001) (INDEX2005-INDEX2001)/INDEX2001= 76.82816% ((1+f)^4)-1 = 76.82814% SOLVER TARGET CELL f= 15.31554% SOLVER CHANGED […]
Industrial Engineering Chapter 10 Homework August Inflation Sample Responses Discussion Questions Caterpillar Generated Percent Its Sales
August 19, 2013 Chapter 10 – Inflation Sample Responses to Discussion Questions 1) In 2011, Caterpillar generated 70 percent of its sales and revenue from outside the United States. Why is this relevant to our study of inflation? Sample Response: […]
Industrial Engineering Chapter 11 Excel Problem Cash Flows Are Expressed Costs Positive Values Represent Cost Not Revenue
PROBLEM 11.52 a CASH FLOWS ARE EXPRESSED AS COSTS (I.E., POSITIVE VALUES REPRESENT COST NOT REVENUE) ALTERNATIVE A WILL BE PREFERRED IF PW(COSTS A) < PW(COSTS B) OR IF PW(COSTS A-B) < 0 VARIANCE = (STANDARD DEVIATION)^2 SALVAGE VALUE IS […]
Industrial Engineering Chapter 11 Excel Problem Ecfj Pwcfj Vcj Varpwj Epw Varpw Sdpw Prpwgt Przgtepwsdpw Prpwgt Prpwgt
PROBLEM 11.44 j E(CFj) (P|F 10%,,j) PW(CFj) V(Cj) (P|F 10%,,j)2 Var(PWj) 0 -$22,500 1.00000 -$22,500 6,250,000 1.00000 6250000 1 $4,000 0.90909 $3,636 160,000 0.82645 132231.405 2 $5,000 0.82645 $4,132 250,000 0.68301 170753.364 3 $6,000 0.75131 $4,508 360,000 0.56447 203210.615 4 […]
Industrial Engineering Chapter 11 Excel Problem For Breakeven Answer Problem The Line With The Steepest Slope Positive
FE PROBLEM 11.1 TC = 8,000 + 0.75X TR = 4.00X FOR BREAKEVEN, TR = TC 4.00X = 8,000 + 0.75X 3.25X = 8,000 X = 8,000/3.25 X = 2,461.54 ANSWER: c FE PROBLEM 11.2 THE LINE WITH THE STEEPEST […]
Industrial Engineering Chapter 11 Excel Problem Let The Number Days Snow Removal Required Cost Contracting Xpa Cost
PROBLEM 11.11 LET X BE THE NUMBER OF DAYS SNOW REMOVAL IS REQUIRED PW OF COST OF CONTRACTING PW = ($400)(X)(P|A 12%,6) PW OF COST OF PURCHASING PW = $25,000 + $5,000(P|A 12%,6) FOR BREAKEVEN THE COST OF CONTRACTING AND […]
Industrial Engineering Chapter 11 Excel The Analysis Most Sensitive Changes Initial Investment Since This Line Has The
THE ANALYSIS IS MOST SENSITIVE TO CHANGES IN INITIAL INVESTMENT SINCE THIS LINE HAS THE STEEPEST (LARGEST ABSOLUTE VALUE) SLOPE. THE ANALYSIS IS LEAST SENSITIVE TO CHANGES SALVAGE VALUE SINCE THIS LINE HAS THE FLATEST (SMALLEST ABSOLUTE VALUE) SLOPE. PERCENT […]
Industrial Engineering Chapter 11 Homework August Breakeven Sensitivity And Risk Analysis Sample Responses Discussion Questions The
August 22, 2013 Chapter 11 – Break-Even, Sensitivity, and Risk Analysis Sample Responses to Discussion Questions 1) Is the risk experienced by BP limited to financial risk? If not, what other risks are present? Sample Response: BP has more at […]
Industrial Engineering Chapter 12 Excel Optimal Portfolio Percent Total Total Investment See Below For Details Investment Opportunity
e OPTIMAL PORTFOLIO 6 1 7 9 4 5 10 PERCENT 100% 100% 100% 100% 100% 100% 50.00% TOTAL PW $1,058,608.60 TOTAL INVESTMENT $2,500,000.00 SEE BELOW FOR DETAILS INVESTMENT OPPORTUNITY 1 2 3 4 5 6 7 8 9 10 […]
Industrial Engineering Chapter 12 Excel Portfolio Cap Constraint Marr See Below For One Setup Investment Opportunity Initial
PORTFOLIO PW g CAP CONSTRAINT $280,000.00 1, 2 $46,000.00 $350,000.00 2, 5 $56,590.58 $420,000.00 1, 2, 3 $70,281.85 h MARR 9.60% 1, 5 $100,637.36 12.00% 2, 5 $56,590.58 14.40% 1, 2 $24,062.77 *** SEE BELOW FOR ONE SETUP *** INVESTMENT […]
Industrial Engineering Chapter 12 Excel Present Worth Internal Rate Return Value Portfolio Investment Portfolio Present Worth Marr
PRESENT WORTH $379.08 $947.70 $189.54 $2,937.86 $2,843.09 $3,222.17 INTERNAL RATE OF RETURN 10.40% 10.71% 10.17% 11.94% 11.25% 11.70% VALUE OF x 1 0 0 1 1 1 PORTFOLIO INVESTMENT $25,000.00 $0.00 $0.00 $40,000.00 $60,000.00 $50,000.00 $175,000.00 PORTFOLIO PRESENT WORTH $379.08 […]
Industrial Engineering Chapter 12 Excel Problem Answer Since The Printer Should Not Purchased The Computer Not Purchased
FE PROBLEM 12.1 ANSWER: b SINCE THE PRINTER SHOULD NOT BE PURCHASED IF THE COMPUTER IS NOT PURCHASED, THE PURCHASE OF THE PRINTER IS CONTINGENT ON THE PURCHASE OF THE COMPUTER. FE PROBLEM 12.2 ANSWER: a MUTUALLY EXCLUSIVE MEANS BOTH […]
Industrial Engineering Chapter 12 Homework August Capital Budgeting Sample Responses Discussion Questions The Announced Spinoff The
August 22, 2013 Chapter 12 – Capital Budgeting Sample Responses to Discussion Questions 1) The announced spin-off of the pharmaceutical portion of Abbott’s business is meant to reduce risk for investors. What other benefits might be realized? Sample Response: The […]
Industrial Engineering Chapter 2 Excel Notes Solutions Following Are Some Thoughts About The Problems And Solutions
NOTES – CHAPTER 2 SOLUTIONS 1 2 3 4 5 6 7 FOLLOWING ARE SOME THOUGHTS ABOUT THE PROBLEMS AND SOLUTIONS IN THIS CHAPTER THAT MAY BE OF HELP TO FACULTY AND STUDENTS. THIS IS, PERHAPS, THE MOST CRITICAL CHAPTER […]
Industrial Engineering Chapter 2 Excel Problem Apa Gpg Using Electronic Interest Tables This Using Excels Solver Tool
PROBLEM 2.147 i= 1.25% G= $60.00 n= 12 P= $5,000.00 $5,000 = A*(P|A 1.25%,12)+G*(P|G 1.25%,12)) $130.17 USING ELECTRONIC INTEREST TABLES THIS IS =($5,000-$60*(P|G 1.25%,12))/(P|A 1.25%,12)) =($5,000-60*59.296701)/11.079312 USING EXCEL’S SOLVER TOOL THIS IS EOM CF 1 $130.17 SOLVER CHANGE CELL 2 […]
Industrial Engineering Chapter 2 Excel Problem Bfp Ifa Biini Using Using Interest Tables This Using Using Compound
1 2 3 4 5 6 7 8 9 10 11 12 13 14 A B C D E F G H I PROBLEM 2.78 aF= B*(F|P i%,1)*(F|A i%,n) bF= B*(1+i)*((1+i)^n-1)/i cB= $1,000.00 n= 5 i= 10.00% USING (a) F= […]
Industrial Engineering Chapter 2 Excel Problem Four Dcf Rules Money Has Time Value Money Cannot Added Subtracted
1 2 3 4 5 6 7 8 A B PROBLEM 2.4 FOUR DCF RULES: 1. MONEY HAS A TIME VALUE 2. MONEY CANNOT BE ADDED OR SUBTRACTED UNLESS IT OCCURS AT THE SAME POINT(S) IN TIME 3. TO MOVE […]
Industrial Engineering Chapter 2 Excel Problem Initial Using Interest Tables This See Below For Tabulated Cash Flows
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 A B C D E PROBLEM 2.95 i= 10.00% n= 8 INITIAL CF= -$50,000.00 […]
Industrial Engineering Chapter 2 Excel Problem Ltsolver Changed Cell Ltsolver Target Cell Using Excels And Npv Functions
1 2 3 4 5 6 7 8 9 19 20 21 22 23 24 25 26 27 28 29 =PV(6%,4,,-NPV(6%,X,X+100,X+400,2*X)) 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 […]
Industrial Engineering Chapter 2 Excel Problem Notice Using Compound Interest Formulas This Using Excels Npv Function And
PROBLEM 2.113 i= 12.00% j= 12.00% NOTICE i=j n= 5 P= $75,000 P|A1=4.46429 USING COMPOUND INTEREST FORMULAS THIS IS =5/(1+0.12) A1|P= 0.2240000 =(1+0.12)/5 A1 = $75,000(A1|P 18%,10%,5) A1 = $16,800.00 =C7*C10 USING EXCEL’S NPV FUNCTION AND SOLVER TOOL A1 = […]
Industrial Engineering Chapter 2 Excel Problem Using Compound Interest Formulas This Fpin Solving For Nlnfplni Lnln Use
1 2 3 4 5 F= $2.00 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 […]
Industrial Engineering Chapter 2 Excel Problem Using Compound Interest Formulas This Using Electronic Interest Tables This Using
1 2 3 4 5 6 7 8 9 10 11 12 A B C D E F G H PROBLEM 2.59 i= 7.00% n= 35 F= $2,000,000.00 A= $14,467.92 USING COMPOUND INTEREST FORMULAS THIS IS =2000000*0.07/((1+0.07)^35-1) A= $14,467.92 USING […]
Industrial Engineering Chapter 2 Excel Problem Using Electronic Interest Tables This See Below For Tabulated Cash Flows
1 2 3 4 5 6 7 8 9 10 11 14 15 16 17 18 19 23 24 25 26 27 28 31 32 33 34 37 38 39 40 41 42 43 44 48 49 50 51 52 […]
Industrial Engineering Chapter 2 Homework June Time Value Money Calculations Sample Responses Discussion Questions What You
June 10, 2013 Chapter 2 – Time Value of Money Calculations Sample Responses to Discussion Questions 1) What do you suppose Kellie is attempting to do with her investment portfolio by selecting multiple forms of investments? Sample Response: Kellie is […]
Industrial Engineering Chapter 3 Excel Problem Beginning Loan Interest Principal Ending Per Year Month Balance Payment Payment
PROBLEM 3.10 P= $25,000.00 BEGINNING LOAN INTEREST PRINCIPAL ENDING i= 12.000% PER YEAR MONTH BALANCE PAYMENT PAYMENT PAYMENT BALANCE m= 12 MONTHS n= 4 YEARS 1 $25,000.00 $658.35 $250.00 $408.35 $24,591.65 2 $24,591.65 $658.35 $245.92 $412.43 $24,179.22 aA= $658.35 PER […]
Industrial Engineering Chapter 3 Excel Problem Cfa Cfb Future Worth Future Worth Future Worth Equivalent Fwa Fwb
FE PROBLEM 3.1 i = 8% t CF(A) CF(B) 1$0 $5,000 2$0 $6,000 3$0 $7,000 4 X $8,000 FUTURE WORTH OF B = [$5,000 + $1,000(A|G 8%,4)](F|A 8%,4) FUTURE WORTH OF B = [$5,000 + $1,000(1.40396)](4.50811) = $28,869.76 FUTURE WORTH […]
Industrial Engineering Chapter 3 Excel Problem Facepar Value Bond Rate Per Year Paymentsyr Annual Years Purchase Price
PROBLEM 3.26 FACE/PAR VALUE= $100,000.00 BOND RATE= 8.00% PER YEAR PAYMENTS/YR= 1 ANNUAL n= 6 YEARS PURCHASE PRICE= $95,250.00 EARNED RATE= 7.0000% PER YEAR EXCEL FW= $85,718.24 =FV(C10,C8,C3*C5/C7,-C9) TABLES FW= $85,718.21 = $95,250(F|P 7%,6) – $100,000(0.08)(F|A 7%,6) PROBLEM 3.27 FACE/PAR […]
Industrial Engineering Chapter 3 Homework July Equivalence Loans And Bonds Sample Responses Discussion Questions Discuss The
July 22, 2013 Chapter 3 – Equivalence, Loans, and Bonds Sample Responses to Discussion Questions 1) Discuss the quantitative (economic) tradeoffs that Samuel should consider when he decides how much money to use from his personal savings versus borrowing money […]
Industrial Engineering Chapter 4 Excel Problem Problem Rationale Problem Less Than Rationale For This Type Investment The
FE PROBLEM 4.1 a. 0.0 FE PROBLEM 4.2 d. $1,082,000 rationale: $200,000(P|A 8%,10) – $10,000(P|G 8%,10) = $1,082,248 FE PROBLEM 4.3 a. LESS THAN $10,000 rationale: FOR THIS TYPE OF INVESTMENT, THE PW DECREASES WITH INCREASING i; THEREFORE, LESS THAN […]
Industrial Engineering Chapter 4 Excel Problem Solution Using Individual Cash Flows And Factors Alternative Eoy Solution Using
PROBLEM 4.33 SOLUTION USING INDIVIDUAL CASH FLOWS AND P|F FACTORS ALTERNATIVE W 11.00% EOY CF (P|F i%,n) PW OF CF 0 -$100,000.00 1.00000 -$100,000.00 1 $20,000.00 0.90090 $18,018.02 2 $20,000.00 0.81162 $16,232.45 3 $50,000.00 0.73119 $36,559.57 4 $80,000.00 0.65873 $52,698.48 […]
Industrial Engineering Chapter 4 Excel Problem Solution Using Individual Cash Flows And Factors Eoy Net Solution Using
PROBLEM 4.6 a SOLUTION USING INDIVIDUAL CASH FLOWS AND P|F FACTORS 5.00% EOY + CF – CF NET CF (P|F i%,n) PW OF CF 0 -$135,000.00 -$135,000.00 1.00000 -$135,000.00 1$12,000.00 -$2,000.00 $10,000.00 0.95238 $9,523.81 2$12,000.00 -$2,000.00 $10,000.00 0.90703 $9,070.29 3$12,000.00 […]
Industrial Engineering Chapter 4 Excel Problem Solution Using Individual Cash Flows And Factors Project Eoy Net Solution
PROBLEM 4.23 a SOLUTION USING INDIVIDUAL CASH FLOWS AND P|F FACTORS PROJECT A 12.00% EOY + CF – CF NET CF (P|F i%,n) PW OF CF 0 -$50,000.00 -$50,000.00 1.00000 -$50,000.00 1$20,000.00 -$12,500.00 $7,500.00 0.89286 $6,696.43 2$20,000.00 -$12,500.00 $7,500.00 0.79719 […]
Industrial Engineering Chapter 4 Excel Wins Winner Problem Incremental East West West Over East First Cost Salvage
B-C -$17,520,000.00 ############ -$12,860,000.00 SB WINS SB IS WINNER PROBLEM 4.45 INCREMENTAL EAST WEST WEST OVER EAST FIRST COST $3,500,000.00 $5,000,000.00 $1,500,000.00 SALVAGE $0.00 $0.00 $0.00 MAINTENANCE $120,000.00 $90,000.00 -$30,000.00 BENEFITS (RE $0.00 $220,000.00 $220,000.00 SERVICE LIF 20 INTEREST RA […]
Industrial Engineering Chapter 4 Homework July Present Worth You Think That Small Organizations Rely Present Worth
July 31, 2013 Chapter 4 – Present Worth 1) Do you think that small organizations rely on present worth analysis as well, or is this type of analysis important only for extremely large firms such as ConocoPhillips? Sample Response: Using […]
Industrial Engineering Chapter 5 Excel Decision Purchase Preferred Since Has The Highest Problem Light Bar Solution Based
DECISION: PURCHASE IS PREFERRED SINCE IT HAS THE HIGHEST AW. PROBLEM 5.23 LIGHT BAR SOLUTION BASED ON INDIVIDUAL CASH FLOWS AND P|F FACTORS THEN A|P FACTOR 10.00% EOY BAR REVENUE NET CF (P|F i%,n) PW OF CF 0 -$6,000.00 -$6,000.00 […]
Industrial Engineering Chapter 5 Excel Problem Rationale Problem Problem Both And Rationale Since Pwawpa And Fwawfa And
FE PROBLEM 5.1 b. $31,050 rationale: $150,000(A|P 8%,30) – $25,000(A|F 8%,30) + $17,500 = $30,599.32 FE PROBLEM 5.2 c. 0.0 FE PROBLEM 5.3 c. BOTH I AND II rationale: SINCE PW=AW(P|A i%,n) AND FW=AW(F|A i%,n) AND AW=0 THEN PW=0 AND […]
Industrial Engineering Chapter 5 Excel Problem Solution Based Individual Cash Flows And Factors Then Factor Eoy Net
PROBLEM 5.10 SOLUTION BASED ON INDIVIDUAL CASH FLOWS AND P|F FACTORS THEN A|P FACTOR 10.00% EOY + CF – CF NET CF (P|F i%,n) PW OF CF 0 -$1,000.00 –$1,000.00 1.00000 -$1,000.00 1 $600.00 -$300.00 $300.00 0.90909 $272.73 2 $600.00 […]
Industrial Engineering Chapter 5 Excel Problem Solution Using Excels Function Solution Using Excels Functions Pmtfv Solution Using
PROBLEM5.30 a F = $2,000(F|A 10%,42) = $2,000(537.63699) = $1,075,273.98 SOLUTION USING EXCEL’S FV FUNCTION FW = $1,075,273.98 =FV(10%,42,-2000) b A = $1,075273.98(A|P 10%,15) = $1,075273.98(0.13147) = $141,366.27 SOLUTION USING EXCEL’S FUNCTIONS A = $141,370.33 =PMT(10%,15,-FV(10%,42,-2000)) c A = $1,075273.98(A|P […]
Industrial Engineering Chapter 5 Excel Problem Solution Using Individual Cash Flows And Factors Polisher Eoy Net Int
PROBLEM 5.46 a SOLUTION USING INDIVIDUAL CASH FLOWS AND F|P FACTORS POLISHER 1 15.00% EOY + CF – CF NET CF (F|P i%,n-t) FW OF CF 0 -$20,000.00 -$20,000.00 4.04556 -$80,911.15 1 $4,465.00 $4,465.00 3.51788 $15,707.32 2 $4,465.00 $4,465.00 3.05902 […]
Industrial Engineering Chapter 5 Homework July Annual Worth And Future Worth Thus Far Joshs Annual Investments
July 31, 2013 Chapter 5 – Annual Worth and Future Worth 1) Thus far, Josh’s annual investments have been in the stock market. What other investment options might you suggest to Josh? What factors should Josh consider in his stock […]
Industrial Engineering Chapter 6 Excel Felike Problem Answer Rationale Xpa Pmt Felike Problem Answer Both And Rationale
FE-LIKE PROBLEM 6.1 ANSWER: a. $246 rationale: 0 = -$400 + X(P|A 15%,2); X = $246.05 $246.05 =PMT(15%,2,-400) FE-LIKE PROBLEM 6.2 ANSWER: d. BOTH I AND II rationale: SINCE THE IRR = MARR, THEN PW EVALUATED AT MARR WILL BE […]
Industrial Engineering Chapter 6 Excel Problem Irr Comparisons Must Done Incrementally This Solution Assumes That Null Not
PROBLEM 6.53 IRR COMPARISONS MUST BE DONE INCREMENTALLY THIS SOLUTION ASSUMES THAT NULL IS NOT AN OPTION; THE PROBLEM CAN ALSO BE WORKED ASSUMING THAT NULL IS AN OPTION. MARR IS 10%; ALL CASH FLOWS ARE IN $K FOR AN […]
Industrial Engineering Chapter 6 Excel Problem One Sign Change Therefore One Positive Real Root Eoy Cum One
PROBLEM 6.28 a ONE SIGN CHANGE, THEREFORE, ONE POSITIVE REAL ROOT b EOY CF CUM CF 0 -$100.00 -$100.00 1$25.00 -$75.00 2$25.00 -$50.00 3$25.00 -$25.00 4$25.00 $0.00 5$25.00 $25.00 6$25.00 $50.00 ONE SIGN CHANGE, THEREFORE, UNIQUE POSITIVE REAL ROOT c […]
Industrial Engineering Chapter 6 Excel Problem Order Comparison Null Marr Current Best Challenger Increment Irr Increment Winner
PROBLEM 6.37 ORDER OF COMPARISON: NULL, C, D, A, B a MARR = 50% CURRENT BEST CHALLENGER INCREMENT IRR OF INCREMENT WINNER NULL C C-NULL 39.20% NULL NULL D D-NULL 36.00% NULL NULL A A-NULL 44.00% NULL NULL B B-NULL […]
Industrial Engineering Chapter 6 Excel Problem Solution Using Excels Irr Function Eoy Net Irr Irrnetcfcolumn Irr Decision
PROBLEM 6.8 a SOLUTION USING EXCEL’S IRR FUNCTION EOY + CF – CF NET CF 0 -$12.00 -$12.00 1 -$1.00 -$1.00 2$5.00 $5.00 3$2.00 $2.00 4$5.00 $5.00 5$5.00 $5.00 6$2.00 $2.00 7$5.00 $5.00 IRR = IRR(NET_CF_COLUMN) IRR = 15.70% b […]
Industrial Engineering Chapter 6 Excel Problem Solutions Uses Goal Seek Determine Err Marr Eoy Net Int Marrnt
PROBLEM 6.66 a SOLUTIONS USES GOAL SEEK TO DETERMINE ERR MARR IS 13.5% 14.07% 13.50% EOY – CF + CF NET CF (F|P i’,n-t) (F|P MARR,n-t) FW OF -CF FW OF +CF FW OF CF 0 -$1,400.00 -$1,400.00 7.20501 6.68248 […]
Industrial Engineering Chapter 6 Homework August Rate Return What Factors Would Determine The Rate Return Values
August 19, 2013 Chapter 6 – Rate of Return 1) What factors would determine the rate of return values that Motorola Solutions would use for their investment decisions? Sample Response: The type of investment, its duration, and risks involved in […]
Industrial Engineering Chapter 7 Excel Problem First Cost Oampm First Year Oampm Each Addn Salvage Decline Marr
PROBLEM 7.30 FIRST COST = $14,000.00 O&M FIRST YEAR = $3,500.00 O&M EACH ADDN YR = $600.00 SALVAGE % DECLINE = 30% MARR = 15.00% LIFE StCR O&M EUAO&MC EUAC 0 1 $9,800.00 $6,300.00 $3,500.00 $3,500.00 $9,800.00 2 $6,860.00 $5,420.93 […]
Industrial Engineering Chapter 7 Excel Problem First Cost Oampm First Year Oampm Each Addn Salvage Decline Min
PROBLEM 7.39 FIRST COST = $250,000.00 O&M FIRST YEAR = $60,000.00 O&M EACH ADDN YR = $12,000.00 SALVAGE % DECLINE = 100.00% MIN SALVAGE = $0.00 MARR = 10.00% LIFE StCR O&M EUAO&MC EUAC 0 1 $0.00 $275,000.00 $60,000.00 $60,000.00 […]
Industrial Engineering Chapter 7 Excel Problem Insiders Viewpoint Keep Old Turret Lathe And Contract Out Replace With
PROBLEM 7.10 INSIDER’S VIEWPOINT a KEEP OLD TURRET LATHE AND CONTRACT OUT REPLACE WITH NEW TURRET LATHE MARKET VALUE = $18,000.00 FIRST COST = $65,000.00 ANNUAL CONTRACT COST = $13,000.00 ANNUAL O&M COST = $10,000.00 ANNUAL O&M COST = $8,000.00 […]
Industrial Engineering Chapter 7 Excel Problem Sunk Cost The Defenders Quotsalvage Valuequot The Investment Cost For The
FE PROBLEM 7.1 $60,000 IS A SUNK COST $80,000 IS THE DEFENDER’S “SALVAGE VALUE” $450,000 IS THE INVESTMENT COST FOR THE CHALLENGER WITH THE INSIDER’S OR CASH FLOW APPROACH, THE NET FIRST COST REQUIRED TO REPLACE THE DEFENDER IS $450,000 […]
Industrial Engineering Chapter 7 Homework August Replacement Analysis You Suppose That Jbht Makes Replacement Decisions For
August 19, 2013 Chapter 7 – Replacement Analysis 1) Do you suppose that JBHT makes replacement decisions for its equipment on an item-by-item level or for a fleet of equipment? What might be some of the advantages and disadvantages to […]
Industrial Engineering Chapter 8 Excel Est Salv Useful Life Rcvd Salv Periods Eoy Exc Func Exc Func
51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 5$15,271.35 $76,356.74 6$12,726.12 $63,630.62 7$10,605.10 $53,025.51 8$9,605.10 […]
Industrial Engineering Chapter 8 Excel Felike Problem Answer Must Have Basis Greater Than Felike Problem Answer Rationale
FE-LIKE PROBLEM 8.1 ANSWER: b. IT MUST HAVE A BASIS GREATER THAN $1,000 FE-LIKE PROBLEM 8.2 ANSWER: c. $17,500.00 rationale: ($200,000 – $25,000) / 10 = $17,500 FE-LIKE PROBLEM 8.3 ANSWER: d. FOR MACRS-GDS AN ESTIMATE OF THE SALVAGE VALUE […]
Industrial Engineering Chapter 8 Excel Problem Basis Macrs Est Salv Year Rcvd Salv Eoy Macrsgds Sum Depr
PROBLEM 8.23 BASIS $180,000.00 MACRS CL 3 5 7 10 15 20 EST SALV $5,000.00 10 YEAR RCVD SALV $5,000.00 12 1 33.33% 20.00% 14.29% 10.00% 5.00% 3.750% 2 44.45% 32.00% 24.49% 18.00% 9.50% 7.219% EOY dt Bt MACRS-GDS 3 […]
Industrial Engineering Chapter 8 Excel Problem Macrsgds Basis Est Salv Useful Life Rcvd Salv Periods Eoy Macrsgds
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 […]
Industrial Engineering Chapter 8 Homework August Depreciation Sample Responses Discussion Questions What Impact Does Depreciation Have
August 19, 2013 1 Chapter 8 – Depreciation Sample Responses to Discussion Questions 1) What impact does depreciation have on capital investment decision making at Schneider? 2) What will be the impact of these capital investment decisions on after-tax income […]
Industrial Engineering Chapter 9 Excel Felike Problem Answer Rationale See Table Felike Problem Answer Rationale Felike Problem
FE-LIKE PROBLEM 9.1 ANSWER: d. 34.0% rationale: SEE TABLE 9.1 FE-LIKE PROBLEM 9.2 ANSWER: b. 20.5% rationale: ($13,750 + 0.34($87,000 – $75,000)) / $87,000 = 0.2049 = 20.5% FE-LIKE PROBLEM 9.3 ANSWER: c. $4,450,000 rationale: $3,400,000 + 0.35($13,000,000 – $10,000,000) […]
Industrial Engineering Chapter 9 Excel Problem Basis Btcf Tax Rate Est Salv Use Life Marr Rcvd Salv
PROBLEM 9.26 BASIS $1,350,000.00 BTCF $300,000.00 TAX RATE 40.00% EST SALV $0.00 USE LIFE 12 MARRAT = 9.00% RCVD SALV $900,000.00 PERIODS 4 a SLN EOY BTCF DWO TI T ATCF 0 -$1,350,000.00 -$1,350,000.00 1 $300,000.00 $112,500.00 $187,500.00 $75,000.00 $225,000.00 […]
Industrial Engineering Chapter 9 Excel Problem Basis Loan Princ Btcf Tax Rate Basis Loan Princ Btcf Tax
PROBLEM 9.58 BASIS $58,500.00 LOAN PRINC $40,000.00 BTCF ########## TAX RATE 40.00% BASIS $87,500.00 LOAN PRINC $40,000.00 BTCF ########## TAX RATE 40.00% EST SALV $0.00 LOAN PERIOD 4 USE LIFE 9 MARRAT = 10.00% EST SALV $0.00 LOAN PERIOD 4 […]
Industrial Engineering Chapter 9 Excel Problem Tax Rate Expenses Btlcf Dwo Ipmt Ppmt Tibtlcfdwoipmt Ttax Rateti Atcfbtlcfipmtppmtt
PROBLEM 9.44 TAX RATE 40.00% GI $500,000.00 EXPENSES $120,000.00 BTLCF $380,000.00 DWO $171,480.00 IPMT $60,500.00 PPMT $0.00 a TI=BTLCF-DWO-IPMT= $148,020.00 b T=TAX RATE*TI= $59,208.00 c ATCF=BTLCF-IPMT-PPMT-T= $260,292.00 PROBLEM 9.45 BASIS $850,000.00 TAX RATE 40.00% GI $750,000.00 EXPENSES $475,000.00 BTLCF $275,000.00 […]
Industrial Engineering Chapter 9 Excel Problem Tax Totals Tax Totals Problem Tax Totals Tax Totals Problem Base
PROBLEM 9.9 a$50,001.00 b TI= $83,928.580 TI TAX $50,000.00 $7,500.00 $25,000.00 $6,250.00 $8,928.58 $3,035.72 $0.00 $0.00 $0.00 $0.00 $0.00 $83,928.580 $16,785.72 20.0000014% TOTALS c TI= $119,642.86 TI TAX $50,000.00 $7,500.00 $25,000.00 $6,250.00 $25,000.00 $8,500.00 $19,642.86 $7,660.72 $0.00 $0.00 $0.00 $0.00 […]
Industrial Engineering Chapter 9 Homework August Income Taxes Sample Responses Discussion Questions Intel Hightech Company With
August 19, 2013 Chapter 9 – Income Taxes Sample Responses to Discussion Questions 1) Intel is a high-tech company with significant R&D. What income tax considerations would such a company have as compared to a “lower–tech” company? Sample Response: As […]