...Chapter 2 ------------------------------------------------- Managerial Accounting and Cost Concepts Solutions to Questions 2-1 The three major elements of product costs in a manufacturing company are direct materials, direct labor, and manufacturing overhead. 2-2 a. Direct materials are an integral part of a finished product and their costs can be conveniently traced to it. b. Indirect materials are generally small items of material such as glue and nails. They may be an integral part of a finished product but their costs can be traced to the product only at great cost or inconvenience. c. Direct labor consists of labor costs that can be easily traced to particular products. Direct labor is also called “touch labor.” d. Indirect labor consists of the labor costs of janitors, supervisors, materials handlers, and other factory workers that cannot be conveniently traced to particular products. These labor costs are incurred to support production, but the workers involved do not directly work on the product. e. Manufacturing overhead includes all manufacturing costs except direct materials and direct labor. Consequently, manufacturing overhead includes indirect materials and indirect labor as well as other manufacturing costs. 2-3 A product cost is any cost involved in purchasing or manufacturing goods. In the case of manufactured goods, these costs consist of direct materials, direct labor, and manufacturing overhead. A period cost is a cost that is taken directly...
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...Chapter 1 27. If they get married in 2010 and file jointly they will pay $38,643.50 [(10% x $16,750) + (15% x $51,250) + (25% x $69,300) + (28% x $42,700)]. If they wait until 2011 to get married and Conrad files as a single individual he will pay $44,516.75 [(10% x $8,375) + (15% x $25,625) + (25% x $48,400) + (28% x $89,450) + (33% x $8,150)]. It would be to their advantage to marry before the end of 2010 because they would save $5,873.25 in income taxes. If they each made $90,000 and both filed individually, they would pay in a combined total of $37,818.50 = 2 x $18,909.25 [(10% x $8,375) + (15% x $25,625) + (25% x $48,400) + (28% x $7,600)]. In this case, it would make more sense to wait until 2011 to get married and each file individually in 2010 because it will save them an additional $825. It would make no difference if they got married in 2010 and filed jointly or separately $38,643.50 = 2 x $19,321.75 [(10% x $8,375) + (15% x $25,625) + (25% x $34,650) + (28% x $21,350)]. 28. John: $4,081.25 [(10% x $8,375) + (15% x $21,625)] William: $11,181.25 [(10% x 8,375) + (15% x $25,625) + (25% x $26,000)] William’s taxable income is twice that of John’s, but William’s taxes are 2.74 times that of John’s. This illustrates vertical equity as well as a progressive tax system where each tax bracket pays a different percentage of their income in taxes – the more money you earn, the higher your tax rate; and, just because you earn twice as much as someone else, it...
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...Chapter 2 Transaction Analysis Short Exercises (5 min.) S 2-1 The transaction had a financial impact on the business and should be recorded. The payment for the computer was not an expense. The payment related to the purchase of an asset, “Equipment,” because the computer is an economic resource of the business. The computer will provide benefit over more than one fiscal period. (5 min.) S 2-2 a. $12,000 (Cash $10,000–$5,000; Supplies $2,000, Computer $5,000) b. $2,000 Accounts Payable Chapter 2 Transaction Analysis Copyright © 2012 Pearson Canada Inc. 65 (5-10 min.) Cash 25,000 2,000 23,000 Supplies 9,000 S 2-3 4,000 Accounts Receivable 6,000 Bal. Accounts Payable 9,000 Rent 4,000 Service Revenue 8,000 Common Shares 25,000 (5 min.) S 2-4 Increased total assets: May 1 (Cash) May 1 (Medical supplies) May 3 (Cash, Accounts receivable) Decreased total assets: May 2 (Cash) 66 Financial Accounting Fourth Canadian Edition Instructor’s Solutions Manual Copyright © 2012 Pearson Canada Inc. (10 min.) S 2-5 CREDIT Journal DATE ACCOUNT TITLES AND EXPLANATION DEBIT June 15 Cash ................................................. Note Payable ............................... Borrowed money from the bank. 25,000 25,000 22 Accounts Receivable ...................... 9,000 Service Revenue ......................... Delivered portrait to be paid on account. 28 Cash .................................................
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...Chapter 2 Review Questions Solutions 1. Describe and compare the six sources of software. The six sources of software identified in the textbook are: (1) information technology services firms, (2) packaged software providers, (3) vendors of enterprise solution software, (4) application service providers and managed service providers, (5) open-source software, and (6) in-house development. IT services firms help companies develop custom information systems for internal use; they develop, host, and run applications for customers; or they provide other services. An IT services firm may be chosen if the system can’t be developed internally or requires customer support. Packaged software providers are companies that produce software exclusively, like Microsoft or Intuit, and are preferable if the task needing the system is generic. Vendors of enterprise solution software create a system that is composed of a series of integrated modules. Each module supports a business function, such as accounting, or human resources. ERP systems may be appropriate if a complete system is required that can cross functional boundaries. A more intense option for larger, more customizable solutions are Managed service providers who can provide more services than application service providers. ASPs and MSPs may be appropriate when instant access to an application is desired, and in the case of ASPs, when the task is generic. Open-source software is a type of software that is developed by...
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...Chapter 2 The Cost Function LEARNING OBJECTIVES Chapter 2 addresses the following questions: Q1 What are different ways to describe cost behavior? Q2 What is a learning curve? Q3 What process is used to estimate future costs? Q4 How are the engineered estimate, account analysis, and two-point methods used to estimate cost functions? Q5 How does a scatter plot assist with categorizing a cost? Q6 How is regression analysis used to estimate a mixed cost function? Q7 What are the uses and limitations of future cost estimates? These learning questions (Q1 through Q7) are cross-referenced in the textbook to individual exercises and problems. COMPLEXITY SYMBOLS The textbook uses a coding system to identify the complexity of individual requirements in the exercises and problems. Questions Having a Single Correct Answer: |No Symbol |This question requires students to recall or apply knowledge as shown in the textbook. | |e |This question requires students to extend knowledge beyond the applications shown in the textbook. | Open-ended questions are coded according to the skills described in Steps for Better Thinking (Exhibit 1.10): ( Step 1 skills (Identifying) ( Step 2 skills (Exploring) ( Step 3 skills (Prioritizing) ( Step 4 skills (Envisioning) QUESTIONS 2.1 This function has both fixed costs and variable costs. ...
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...CHAPTER 2 ECONOMISTS’ VIEW OF BEHAVIOR CHAPTER SUMMARY This chapter uses the cheating scandal at Merrill Lynch to illustrate how a manager’s view of behavior can affect decision making. It summarizes the economic view of behavior and contrasts it with other views. The chapter presents a graphical analysis of utility maximization and decision making under uncertainty. The concepts in this chapter are an important foundation for subsequent material in the book. CHAPTER OUTLINE ECONOMIC BEHAVIOR: AN OVERVIEW Economic Choice Marginal Analysis Managerial Application: Marginal Analysis of Customer Profitability Opportunity Costs Managerial Application: Opportunity Costs and V-8 Creativity of Individuals Managerial Application: Creative Gaming of the System GRAPHIC TOOLS Individual Objectives Indifference Curves Constraints Individual Choice Changes in Choice MOTIVATING HONESTY AT MERRILL LYNCH MANAGERIAL IMPLICATIONS Managerial Application: Medicare Creates Perverse Incentives for Doctors ALTERNATIVE MODELS OF BEHAVIOR Only-Money-Matters Model Happy-Is-Productive Model Managerial Application: Happy-Is-Productive versus Economic Explanations of the Hawthorne Experiments Good-Citizen Model Managerial Application: Culture and Behavior Product-of-the-Environment Model WHICH MODEL SHOULD MANAGERS USE? ...
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...Strategic Marketing Problems Cases and Comments V11. Chapter 2 1a) Contribution per CD unit = Unit Selling Price – Unit Variable Cost = $9.00 – ($1.25 + $0.35 + $1.00) = $6.40 b) Break-even volume in CD units Total Fixed Costs = $275,000 + $250,000 = $525,000 Unit Break-even Volume = Total Fixed Costs/Contribution per unit = $525,000 - $6.40 = 82,031.25units Break-even volume in dollars Contribution Margin = (Unit selling price – unit variable cost) / unit selling price = ($9.00 – $2.60) / $9.00 = 0.7111 = 71.111% Break-even Dollar Volume = Total Fixed Costs / Contribution Margin = $525,000 / 0.7111 = $738,282.40 c) Net profit if 1 million CDs are sold = Total Contribution – Total Fixed Costs = ($6.40 X 1,000,000) - $525,000 = $6,400,000 - $525,000 = 5,875,000 d) Necessary CD unit volume to achieve a $200,000 profit = (Total Dollar Fixed Costs + Dollar Profit Goal) / Contribution Per Unit = ($525,000 + $200,000) / $6.40 = $725,000 /$6.40 = 113,281.25 2a) VCI’s Unit contribution = $20.00 – [(4000/1000) + (500/1000) + (500/1000) = $20.00 – ($4.00 + $0.50 + 0.50) = $20.00 - $5.00 = $15.00 Contribution margin = ($20.00 - $5.00) / $20.00 = $0.75 b) Break-even point in units Total Fixed Costs = $150,000 + $125,000 + $5,000 + $10,000 + $35,000 = $325,000 Unit Break-even Volume = $325,000 / $15.00 = $21,666,67 Break-even point in dollars = $325,000 / $0.75 = $433,333.33 c) Share of the market would the film have to achieve to earn a 20% return on VCI’s...
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...1. How do we create a partnership between the project manager and line managers when project manager focuses only on the best of his/her project and the line manager is expected to make impartial company decision? I think the most important point is that project manager and line manager should stand on the same level, which means each of them should have same formal authority, line manager should manage right people to project manager, and project manager should arrange right people to the right position, for the best interest of company, this is a partnership they should establish. 2. Who should have more of a say during negotiations for the resource manager or the line manager. I think it depends on what size the company is, for case one, line manager have more of a say during negotiations for resource, during the context, line manager control several key staff, while project manager have to work with the one he does not want to work with. Ideally, project manager should have a say in the assembly of his project team and his resource, which will help him to gain their personal commitment, support and required quality of service. 3. How should irresolvable conflicts over staffing between the project and line managers be handle? Basically, for this kind of conflict should be handle by sponsor, it is the sponsor who support the project, if there is a conflict between the project manager and line manager, sponsor should evaluate the situation to decide who should go...
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...Chapter Solutions Operations Management.pdf DOWNLOAD HERE CHAPTER 12: INVENTORY MANAGEMENT – Suggested Solutions to ... http://users.ipfw.edu/khamaljn/P301/Class_Notes/Solutions-Chapter12.pdf 1 BUS P301:01 CHAPTER 12: INVENTORY MANAGEMENT – Suggested Solutions to Selected Questions Summer II, 2009 Question 12.5 This is EOQ with D = 19,500 units/yr; H = $ ... CHAPTER 11 Operations Management http://www.eng.uwi.tt/depts/mech/ugrad/courses/meng3006/Week09b.pdf Operations Management, ... • To decouple operations ... CHAPTER 11 11-39 Inventory Management Economic Production Quantity I n v en t o r y L ev el CHAPTER 10 Operations Management - Academic Resources at ... http://academic.missouriwestern.edu/mlewis14/MGT%20416/Lectures/My%20Chap010.pdf Operations Management 8th edition 10-2 Quality Control CHAPTER 10 Quality Control McGraw-Hill/Irwin Operations Management, Eighth Edition, by William J. Stevenson PRODUCTION/OPERATIONS MANAGEMENT - … http://faculty-staff.ou.edu/B/Samir.Barman-1/POMsp02.doc PRODUCTION/OPERATIONS MANAGEMENT. BAD 5262 . SPRING 2002. Course Instructor: Samir Barman, Ph.D. ... Apr 02 Chapter 3 All Example Problems. Operations in a PR: 3.1-3.6 Chapter 7 Accounting for Financial Management http://harbert.auburn.edu/~yostkev/teaching/finc3630/notes/Chapter7solutions.pdf Chapter 7 Accounting for Financial Management ANSWERS TO END-OF-CHAPTER QUESTIONS 7-3 No, because the $20 million of retained earnings would probably not be held as cash...
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...Business Decision Making Session – IV Chapter 11: Linear Programming Chapter 12: Transportation, Transshipment, and Assignment Problems Chapter 13: Decision Analysis Faculty Pankaj Dutta Chapter – 11 Linear Programming LPP solution through Excel Solver: The Steps in Implementing an LP Model in a Spreadsheet: 1. 2. 3. 4. Organize the data for the model on the spreadsheet. Reserve separate cells in the spreadsheet for each decision variable in the model. Create a formula in a cell in the spreadsheet that corresponds to the objective function. For each constraint, create a formula in a separate cell in the spreadsheet that corresponds to the left-hand side (LHS) of the constraint. PANKAJ DUTTA IMTCDL Chapter – 11 Linear Programming LPP solution through Excel Solver: Max Z = 350X1 + 300X2 Subject To 1X1 + 1X2 ≤ 200 9X1 + 6X2 ≤ 1566 12X1+16X2 ≤ 2880 X1 , X 2 ≥ 0 1. Organize the data for the model on the spreadsheet. PANKAJ DUTTA IMTCDL Chapter – 11 Linear Programming LPP solution through Excel Solver: Max Z = 350X1 + 300X2 Subject To 1X1 + 1X2 ≤ 200 9X1 + 6X2 ≤ 1566 12X1+16X2 ≤ 2880 X1 , X 2 ≥ 0 1. Organize the data for the model on the spreadsheet. Changing cells Target cell PANKAJ DUTTA IMTCDL Constraint cells Chapter – 11 Linear Programming Max Z = 350X1 + 300X2 Subject To 1X1 + 1X2 ≤ 200 9X1 + 6X2 ≤ 1566 12X1+16X2 ≤ 2880 X1 , X 2 ≥ 0 LPP solution through Excel Solver: 1. Organize the...
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...Chapter 5, Problem 1. The equivalent model of a certain op amp is shown in Fig. 5.43. Determine: (a) the input resistance. (b) the output resistance. (c) the voltage gain in dB. 8x104vd Figure 5.43 for Prob. 5.1 Chapter 5, Solution 1. (a) (b) (c) Rin = 1.5 MΩ Rout = 60 Ω A = 8x104 Therefore AdB = 20 log 8x104 = 98.0 dB Chapter 5, Problem 2 The open-loop gain of an op amp is 100,000. Calculate the output voltage when there are inputs of +10 µV on the inverting terminal and + 20 µV on the noninverting terminal. Chapter 5, Solution 2. v0 = Avd = A(v2 - v1) = 105 (20-10) x 10-6 = 1V PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 5, Problem 3 Determine the output voltage when .20 µV is applied to the inverting terminal of an op amp and +30 µV to its noninverting terminal. Assume that the op amp has an open-loop gain of 200,000. Chapter 5, Solution 3. v0 = Avd = A(v2 - v1) = 2 x 105 (30 + 20) x 10-6 = 10V Chapter 5, Problem 4 The output voltage of an op amp is .4 V when the noninverting input is 1 mV. If the open-loop gain of the op amp is 2 × 106, what is the inverting...
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... FOR MORE CLASSES VISIT www.fin534tutor.com FIN 534 Week 1 Chapter 1 Solution FIN 534 Week 1 Chapter 2 Solution FIN 534 Week 2 Chapter 3 Solution FIN 534 Week 3 Chapter 4 Solution FIN 534 Week 3 Chapter 5 Solution FIN 534 Week 4 Chapter 6 Solution FIN 534 Week 4 Chapter 7 Solution FIN 534 Week 5 Chapter 8 Solution FIN 534 Week 5 Chapter 9 Solution FIN 534 Week 6 Chapter 10 Solution FIN 534 Week 6 Chapter 11 Solution FIN 534 Week 7 Chapter 12 Solution FIN 534 Week 7 Chapter 13 Solution FIN 534 Week 8 Chapter 14 Solution FIN 534 Week 8 Chapter 15 Solution FIN 534 Week 9 Chapter 16 Solution FIN 534 Week 10 Chapter 17 Solution Fin 534 Week 1 Quiz 1 FIN 534 Week 3 Quiz 2 FIN 534 Week 4 Quiz 3 FIN 534 Week 5 Quiz 4 FIN 536 Week 6 Quiz 5 FIN 534 Week 7 Quiz 6 FIN 534 Week 8 Quiz 7 FIN 534 Week 9 Quiz 8 FIN 534 Week 10 Quiz 9 FIN 534 Week 11 Quiz 10 FIN 534 Week 1 DQ 1 FIN 534 Week 1 DQ 2 FIN 534 Week 2 DQ 1 FIN 534 Week 2 DQ 2 FIN 534 Week 3 DQ 1 FIN 534 Week 3 DQ 2 FIN 534 Week 4 DQ 1 FIN 534 Week 4 DQ 2 FIN 534 Week 5 DQ 1 FIN 534 Week 6 DQ 1 FIN 534 Week 7 DQ 1 FIN 534 Week 7 DQ 2 FIN 534 Week 8 DQ 1 FIN 534 Week 8 DQ 2 FIN 534 Week 9 DQ 1 FIN 534 Week 9 DQ 2 FIN 534 Week 10 DQ 1 FIN 534 Week 10 DQ 2 FIN 534 Week 11 DQ 1 FIN 534 Week 11 DQ 2 ------------------------------------------------------------------------------------ FIN 534 Week 1 Chapter 1 Solution (Str Course) FOR MORE CLASSES VISIT www.fin534tutor.com 1. Which...
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...Investment Analysis, 7th Edition Solutions to Text Problems: Chapter 1 Chapter 1: Problem 1 A. Opportunity Set With one dollar, you can buy 500 red hots and no rock candies (point A), or 100 rock candies and no red hots (point B), or any combination of red hots and rock candies (any point along the opportunity set line AB). then: Algebraically, if X = quantity of red hots and Y = quantity of rock candies, 0.2 X + 1Y = 100 That is, the money spent on candies, where red hots sell for 0.2 cents a piece and rock candy sells for 1 cent a piece, cannot exceed 100 cents ($1.00). Solving the above equation for X gives: X = 500 − 5Y which is the equation of a straight line, with an intercept of 500 and a slope of −5. Elton, Gruber, Brown, and Goetzmann Modern Portfolio Theory and Investment Analysis, 7th Edition Solutions to Text Problems: Chapter 1 1-1 B. Indifference Map Below is one indifference map. The indifference curves up and to the right indicate greater happiness, since these curves indicate more consumption from both candies. Each curve is negatively sloped, indicating a preference of more to less, and each curve is convex, indicating that the rate of exchange of red hots for rock candies decreases as more and more rock candies are consumed. Note that the exact slopes of the indifference curves in the indifference map will depend an the individual’s utility function and may differ among students. Chapter 1: Problem 2 A. Opportunity Set The individual...
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...Adventures in Debentures c Copyright ⃝ 2004 by Michael R. Gibbons Adventures in Debentures we can solve for y to obtain Solutions to The Grammar of Fixed Income Securities 1. Part Part Part Part Part Part a. b. c. d. e. f. $100/(1.20)3 = $57.87. $100/(2)3 = $12.50. $100/(1)3 = $100. $100/(1.10)6 = $56.45. $100/(1.05)12 = $55.68. $100/e.20×3 = $54.88. y= So yB = and yA = 2nLN nSN 1 + c/2 −1 . P/100 2 · 184 1 + .09125/2 − 1 = 2.991185% 105 103.6776495/100 2 · 184 1 + .09125/2 − 1 = 2.778233% . 105 103.7401495/100 2. Remember we are trying to find r such that the growth in our initial investment is the ˙ same as the growth from an alternative investment providing a rate of r(m). That is, find r ˙ Part a. Part b. Part c. Part d. ˙ ˙ such that er = (1 + r(m)/m)m so r = m ln(1 + r(m)/m). m = 1: ln(1.04) = 3.922%. m = 1: ln(1.20) = 18.232%. m = 4: 4 ln(1.05) = 19.516%. m = 1: ln(2) = 69.315%. In addition, we have Part b. Ignoring for the moment any restrictions on lot size, we note that a cash flow of $100 to be received on 12/31/92 and purchased on 9/17/92 is available either in the form of $100 par amount of the 12/31/92 bill or in the form of $100/(1+.09125/2) par amount of the 9 1/8’s of 12/31/92. (Indeed, since $100 par of the 9 1/8’s will pay off $(100 + 9.125/2) on 12/31/92, $100/(1 + .09125/2) par amount of the 9 1/8’s will pay off $100.) Can we buy this cash flow low through one instrument and sell it high through the other? We must see if the asked price of one of the...
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...Chapter 3, Problem 1. Determine Ix in the circuit shown in Fig. 3.50 using nodal analysis. 1 kΩ Ix 9V + _ 2 kΩ + _ 6V 4 kΩ Figure 3.50 For Prob. 3.1. Chapter 3, Solution 1 Let Vx be the voltage at the node between 1-kΩ and 4-kΩ resistors. 9 − Vx 6 − Vx Vk + = 1k 4k 2k Vx Ix = = 3 mA 2k ⎯⎯ Vx = 6 → PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter 3, Problem 2. For the circuit in Fig. 3.51, obtain v1 and v2. Figure 3.51 Chapter 3, Solution 2 At node 1, − v1 v1 v − v2 − = 6+ 1 10 5 2 At node 2, 60 = - 8v1 + 5v2 (1) v2 v − v2 = 3+ 6+ 1 4 2 Solving (1) and (2), v1 = 0 V, v2 = 12 V 36 = - 2v1 + 3v2 (2) PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Chapter...
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