Mba530R4Problemsolutiontemplatewk6 V2

In: Business and Management

Submitted By khwaja
Words 4909
Pages 20

Problem Solution: Riordan Manufacturing
Khwaja Shaik
University of Phoenix

Problem Solution: Riordan Manufacturing Riordan Manufacturing is a global plastics producer employing 550 people with projected annual earnings of $46 million. The company is wholly owned by Riordan Industries, a Fortune 1000 enterprise with revenues in excess of $1 billion. Production is divided among three plants: plastic beverage containers in Albany, Georgia; custom plastic parts in Pontiac, Michigan; and plastic fan parts in Hangzhou, China. Research and Development is conducted at corporate headquarters in San Jose, California. Riordan's major customers are automotive parts manufacturers, aircraft manufacturers, the Department of Defense, beverage makers and bottlers, and appliance manufacturers. This research paper will discuss the issues being faced by Riordan Manufacturing and provides solution based on various motivation, rewards and performance concepts

Situation Analysis
Issue and Opportunity Identification – Total reward system The current reward system is barely based on performance, instead recognizing cost-of-living increases, seniority and position. Faced with declining morale and work ethic, Riordan managers have been pressuring the CEO to "do something" about the rewards system. Riordan’s employees comprise three major demographic groups. Baby boomers make up the bulk of the managerial and about half of the manufacturing staff; GenXers make up the majority of the professional staff, as well as some of the manufacturing staff; and the GenY contingent are the newest hires, found primarily in manufacturing, engineering and IT. These three groups have radically different perspectives on rewards and motivation, valuing everything from interesting work to bigger paychecks.

Similar Documents


...labeled v1, v2, v3. v1 R1 v2 + Vs R3 R2 v3 R4 _ 6.071, Spring 2006. Chaniotakis and Cory 2 Figure 3. Circuit with assigned nodal voltages. For the next step we assign current flow and polarities, see Figure 4. v1 R1 + i1 _ v2 i3 + R3 + Vs + i2 _ i1 _ R2 _ v3 + _ R4 Figure 4. Example circuit with assigned node voltages and polarities. Before proceeding let’s look at the circuit shown on Figure 4 bit closer. Note that the problem is completely defined. Once we determine the values for the node voltages v1, v2, v3 we will be able to completely characterize this circuit. So let’s go on to calculate the node voltages by applying KCL at the designated nodes. For node n1 since the voltage of the voltage source is known we may directly label the voltage v1 as v 1 = Vs and as a result we have reduced the number of unknowns from 3 to 2. KCL at node n2 associated with voltage v2 gives: (4.1) i1 = i 2 + i 3 (4.2) The currents i1, i2, i3 are expressed in terms of the voltages v1, v2, v3 as follows. i1 = Vs- v 2 R1 v2 i2 = R2 v2 - v3 i3 = R3 (4.3) (4.4) (4.5) 6.071, Spring 2006. Chaniotakis and Cory 3 By combining Eqs. 4.2 – 4.5 we obtain Vs- v 2 v 2 v 2 - v 3 =0 R1 R2 R3 (4.6) Rewrite the above expression as a linear function of the unknown voltages v2 and v3 gives. 1 1 ⎞ 1 1 ⎛ 1 (4.7) v 2⎜ + + = Vs ⎟-v 3 R3 R1 ⎝ R1 R 2 R 3 ⎠ KCL at node n3 associated with voltage v3 gives: v2 -v......

Words: 3646 - Pages: 15


...Is the work positive, negative or zero? The boundary work is: P drops but does V go up or down? The process equation is: PVn = C P 1 2 W V W = ⌠P dV ⌡ 4W = πP∆x 4 × 2.5 kJ = 0.113 m π × 500 kPa × 0.5 m so we can solve for P to show it in a P-V diagram P = CV-n as n = 1.667 the curve drops as V goes up we see V2 > V1 giving dV > 0 and the work is then positive. Sonntag, Borgnakke and van Wylen 4.11 An ideal gas goes through an expansion process where the volume doubles. Which process will lead to the larger work output: an isothermal process or a polytropic process with n = 1.25? The process equation is: PVn = C The polytropic process with n = 1.25 drops the pressure faster than the isothermal process with n = 1 and the area below the curve is then smaller. P 1 n=1 2 W V 4.12 Show how the polytropic exponent n can be evaluated if you know the end state properties, (P1, V1) and (P2, V2). Polytropic process: PVn = C Both states must be on the process line: P2V2 = C = P1V1 Take the ratio to get: P1 V2 P2 = V1   n n n n and then take ln of the ratio P1 V2 V2 ln P  = ln V  = n ln V   1  1  2 now solve for the exponent n P1 V2 n = ln P  ln V   2  1 / Sonntag, Borgnakke and van Wylen 4.13 A drag force on an object moving through a medium (like a car through air or a submarine through water) is Fd = 0.225 A ρV2. Verify the unit becomes Newton. Solution: Fd = 0.225 A ρV2 Units = m2 × ( kg/m3 ) × ( m2/ s2 ) = kg m /......

Words: 15028 - Pages: 61

Week 4 Problems

...E1) and G2 = (V2, E2) be the loop-free undirected connected graphs in Fig. 11.42. a) Determine |V1|, |E1|, |V2|, and |E2|. Counting the vertices and edges in both graphs: |V1| = |V2| = 8 |E1| = |E2| = 14 11.4 (540-553) 3a. How many vertices and how many edges are there in the complete bipartite graphs K4,7, K7,11, and Km,n, where m, n, ∈ Z+? K4,7 = 11 vertices (4 + 7) = 28 edges (4 * 7) K 7,11 = 18 vertices (7 + 11) = 77 edges (7 * 11) Km,n = m + n vertices = m * n edges 11.5 (556-562) 2. Characterize the type of graph in which an Euler trail (circuit) is also a Hamilton path (cycle). A Euler trail is open and crosses each edge at least once without repeating. A Hamilton path passes through each vertex only once, and becomes a cycle when only the first and last vertices connect. According to these definitions, a shape can have a Euler trail and a Hamiltonian path if they have the same amount of edges and vertices. A triangle and square are both examples of a Euler trail and a Hamiltonian path because each uses each of its edges only once, and each only connects vertices at the same beginning and endpoint to close the shape. 12.1 (581-585) 2. Let T1 = (V1, E1), T2 = (V2, E2) be two trees where |E1| = 17 and |V2| = 2|V1|. Determine |V1|, |V2|, and |E2|. In every tree T = (V , E), |V | = |E| + 1. If T1 = (V1, 17), then V1 = 18, and T1 = (18, 17) If V2 = 2(V1), then V2 = 2(18) = 36 If V2 = 36, and V = E +1, then E2 = 37 V1 = 18; V2 = 36; and......

Words: 772 - Pages: 4

L Brands - Icr Xchange Conference - Final 1 14 14 V2

...2013 INVESTOR HANDOUT KEY MESSAGES     We have global, category-leading brands with high emotional content that generate substantial income and cash flow Our overriding focus is on the substantial growth opportunity in North America We believe there is substantial opportunity for international growth and expect it to be accretive to the total company operating margin We are targeting 10% annual operating income $ growth (or better) and an operating income rate in the high-teens We continue to emphasize maintaining a strong cash and liquidity position while optimizing our cost of capital We will continue to manage inventory, expenses and capital with discipline We remain committed to returning excess cash and generating superior returns for shareholders 2    SUPERIOR RETURNS TO SHAREHOLDERS 3-year Total Return 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 Fast Retailing TJX Home Depot L Brands Foot Locker Inditex Gap Giordano Ascena Retail Group Buckle Chico's Ralph Lauren Bed Bath & Beyond Wal-Mart Stores Walgreen Co. Tiffany & Co. H&M Ann Inc. Best Buy American Eagle Children's Place Target Coach bebe Staples Abercrombie & Fitch Esprit Holdings Li & Fung Aeropostale S&P 500 S&P Retail Index 51.7 43.7 35.9 35.5 31.4 30.9 23.7 20.7 19.6 19.4 18.8 17.9 17.9 16.3 16.3 16.1 14.7 12.2 8.9 6.0 4.5 4.2 2.4 0.1 (8.7) (15.8) (21.1) (21.7) (28.1) 15.8 24.0 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......

Words: 3479 - Pages: 14


...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 3, Problem 3. Find the currents i1 through i4 and the voltage vo in the circuit in Fig. 3.52. Figure 3.52 Chapter 3, Solution 3 Applying KCL to the upper......

Words: 19454 - Pages: 78

Unilateral Network input port and an output port. i1 + + i2 A C V1 Circuit V2 B D i1 ` - - i2` Block Diagram of a Two-port Network There are several restrictions in using this building block. i. There can be no energy stored within the circuit. ii. There can be no independent sources within the circuit, though dependent sources are permissible. iii. The current into the port must equal the current out of the port, ie. i1= i2` and i1`= I2. iv. All external connections must be made to either the input port or the output port; no external connections between ports are allowed, that is between terminals A & C and A & D or between B & C and B & D. The fundamental principle underlying two-port modeling of a system is that only the terminal variables, ie V1, V2, i1, i2, i1` and i2` are of interest. 4.2 Y-Parameter or Admittance Parameter The Y-parameter equations that describe the network are as followed: I1 = Y11V1 + Y12V2 I2 = Y21V1 + Y22V2 In matrix form, Y11 Y12 (Y( = Y21 Y22 These parameters can be determined by short circuiting one port at a time. With output port CD shorted (V2 = 0), I1 I2 Y11 = and Y21 = V1 V1 With the input port AB shorted (V1 = 0), I1 I2 Y12 = and Y22 = V2 V2 Due to this nature of obtaining the parameters, they are also called short-circuit admittance...

Words: 5013 - Pages: 21

Allstate Insurance Company V2

...Running head: BUS520 Assign 2 Allstate Insurance Company Allstate Insurance Company Myrna L. Hunt-Young In Partial fulfillment of BUS520 Professor Danette O’Neal 04/28/2011 Abstract This paper will discuss the strategy that the Allstate Insurance Company, located in Northbrook, Illinois uses for motivating their employees. There will be a discussion of the advantage that Allstate has by using the Diversity Index and Quality Leadership Measurement System (QLMS) program. Furthermore, there will be list of options that Allstate could use to motivate their employees to meet the goals set by the company. Lastly, I will discuss whether or not Allstate’s Diversity Index and QLMS program would motivate me to work for the company. Background The human resource team at Allstate started an affirmative action program during the late 1960’s. This program was considered innovative. The program consisted of 4 steps: Succession, Programming, Development, Measurement, and Accountability and Reward. Each step takes into affect the employee, managers, and the company. Allstate is an Insurance company located in Northbrook Illinois. This is a suburb located north of Chicago, Illinois. Using the model for goal setting, evaluate Allstate’s goal setting process to determine whether or not Allstate has an effective goal-setting program. Goal setting is documented in the book Organizational Ethics, by Hellriegel and Slocum (2010, page 194), “Goal Setting is the process of......

Words: 855 - Pages: 4

Microwave Engineering

...(ABCD) parameters  V1   Z1,1 V    Z  2   2,1 Z1,2   I1  Z 2,2   I 2    , ,  I1   Y1,1 Y1,2   V1   I   Y Y2,2   V2   2   2,1    V1   h1,1 I   h  2   2,1  I1   g1,1 V    g  2   2,1 h1,2   I1  h2,2   V2    g1,2   V1  g 2,2   I 2    , , , , , ,  V1   A B   V2   I    C D   I   2   1  EE 221 S. Gedney, University of Kentucky The Impedance (Z) Parameters The impedance parameters are quantified by the Zmatrix which relates the port currents to the port voltages  V1   Z1,1 V    Z  2   2,1 , Z1,2   I1  Z 2,2   I 2    The Z-parameters have the units of We can “measure” the Z-parameters of a linear twoport network in a very simple way: Open circuit port 2 (This sets = 0).  V1   Z1,1 Z1,2   I1  V    Z  Z 2,2   0   2   2,1   Drive port 1 with a current source and measure port voltages V1  Z1,1 I1  Z1,1  V1 / I1 V2  Z 2,1 I1  Z 2,1  V2 / I1 EE 221 S. Gedney, University of Kentucky Measuring the Z-parameters Repeat by open circuiting port 1, drive port 2 with current, and measure the port voltages  V1   Z1,1 V    Z  2   2,1 Z1,2   0  Z 2,2   I 2    V1  Z1,2 I 2  Z1,2  V1 / I 2 V2  Z 2,2 I 2  Z 2,2  V2 / I 2 , , is the input impedance of port 2 with port 1 open circuited is the trans-impedance with port 1 open circuited Similarly: , , is the......

Words: 3337 - Pages: 14

Retrospective Analysis of Personality-V2

...Retrospective Analysis of Personality Retrospective Analysis of Personality Reflection of My Life History Growing up in Gary, Indiana, I moved to Houston, TX when I was 10 years old when my parents got a divorce. Being in a new city away from all of my childhood friends, I was very shy and reserved. My mom made me go to Sunday school at church each weekend and I believe this was to allow me to meet new friends. I never liked Sunday school or going to church as a kid or even playing any sports. Then around my freshman year of high school I decided to try out for football, mainly because my best friend talked me into it. I was one to follow the crowd by the time I became a teenager and that got me into a lot of trouble. For example, sneaking out of my bedroom window with my mother’s car keys to go for a joyride with my buddies at the age of 14. There was no father figure in my home, so I would either go to my best friend or my coaches to fill that void. I was the exact opposite of a leader. Finally by the time I reached my senior year, my personality started to take a turn for the better. As a senior football player with 3 years of experience, I became a leader on the team. After high school, I joined the Army National Guard and that was another culture shock for me. Going through this experience taught me further leadership skills and proved to myself that I could accomplish anything that I put my mind to. Later in life I met my wife and she was a woman that......

Words: 914 - Pages: 4

A Report to Critically Compare a Number of Routing Protocols; Including Rip V2, Eigrp & Ospf

...Routing & Router Configuration A Report to Critically Compare a Number of Routing Protocols; Including RIP v2, EIGRP & OSPF Paul McDermott CCNA 2 Table of Contents 1.0 Abstract 3 2.0 Introduction 4 3.0 Protocol overview 5 3.1 RIP v2 Overview 5 3.2 EIGRP Overview 6 3.3 OSPF Overview 6 4.0 Protocol Comparison 10 4.1 Topology Overview 10 4.2 Protocol Types 10 4.3 Administration Distance 10 4.4 Protocol Tables 11 4.5 Algorithm 11 4.6 Metric 12 4.7 Periodic Updates 12 4.8 Hierarchical / Scalable 12 4.9 Load Balance 13 4.10 Comparison Table 14 5.0 Conclusion 15 6.0 References 16 Abstract The following report is a critical comparison of three routing protocols; RIPv2, EIGRP and OSPF, detailing the protocol features, as well as their similarities and differences. The report takes an in-depth look at the technical elements and algorithms used in these protocols, such as Bellman Ford, DUAL, and the Dijkstra Algorithm; and how these algorithms are used to calculate the routing metric. The report also discusses the fact that EIGRP is the most desirable protocol to use on Cisco based routers, while OSPF can be used across different router manufacturers. While looking at the technical considerations that are needed in choosing a routing protocol for a desired network the report will also look into the CPU/memory requirements, and how difficult the protocol is to install and......

Words: 4222 - Pages: 17

Db Iv - Oracle 11g V2 - Final Examination -Chapters 10 and 11 Student Version-Completed

...Multiple Choice Identify the choice that best completes the statement or answers the question. __A__ 1. All of the following would cause a trigger to fire, except ____. a. | BEFORE | c. | DELETE | b. | INSERT | d. | UPDATE | __A__ 2. Which of the following indicates that the trigger is fired only once, regardless of the number of rows affected by the DML statement? a. | Statement level | c. | DML level | b. | Event level | d. | Row level | __B__ 3. Which of the following statements is correct? a. | If multiple triggers exist on a table, the order in which the triggers will be fired can be set. | b. | A BEFORE statement level trigger will fire before a BEFORE row level trigger. | c. | If you have two statement level triggers, the order in which they are fired is dependent on the order in which they were written. | d. | Only one trigger can be constructed per table. | __D__ 4. Row level options are only applicable for ____ events. a. | CREATE | c. | DECLARE | b. | INSERT | d. | UPDATE and DELETE | __A__ 5. The default timing of a trigger is ____. a. | statement level | c. | row level | b. | system level | d. | header level | __A__ 6. Which of the following events will cause the trigger to fire? AFTER UPDATE OF orderplaced ON bb_basket a. | INSERT | c. | DELETE | b. | UPDATE | d. | AFTER | __D__ 7. CURSOR basketitem_curIS SELECT idproduct, quantity, option1 FROM bb_basketitem WHERE idbasket =......

Words: 1769 - Pages: 8


...capacitance, Ceq, from Q · 1/Ceq = ∆V = ∆V1 + ∆V2 = Q/C1 + Q/C2 = Q [1/C1+1/C2] 1/Ceq = 1/C1 + 1/C2 (2) (3) Parallel: In the parallel connection, the components are connected together at both ends as shown below: C1 C2 For a parallel connection, the voltage drops will be the same, but the charges will add. Then the equivalent capacitance can be calculated by adding the charges: Ceq ∆V = Q = Q1 + Q2 = C1∆V + C2∆V = [C1+C2] ∆V Ceq = C1 + C2 (5) (6) Procedure: Figure 2. AC power supply Figure 1. Multimeter 1. Turn on the power supply and set the AC voltage to 10 V. Measure the actual power supply voltage with the multimeter and record it below: VPS = ___________________V 2. Connect two 0.1 µF capacitors in series. Measure V2 (across C2) and record it below. V2 (measured) = ____________ V 3. Compute the expected value of V2 using VPS, the values of C1 and C2 with equations 1 and 3. Remember that Eq. 1 is true for each capacitor, including the combined C12. C2 C1 V V2 (expected) = ____________ V % difference = |measured - expected| / measured x 100 % = __________ 4. Connect a third 0.1 µF capacitor in parallel with C2. Compute their equivalent capacitance C23. ∼ C2 C23 = _________ µF. Measure and compute the voltage across C2. [Hint: is this the same as the voltage across the equivalent capacitor C23? You may want to compute the total equivalent capacitance seen by the power supply, C123] C3 V2 (measured) = ________ V, V V2 (expected) = ________ V, %......

Words: 577 - Pages: 3

Managing Across Cultue

...between the volume, pressure and temperature of gasses. If you decrease the volume of a gas (compress it), the pressure and the temperature will go up. If you know volume (V1), Temperature (T1) and Pressure (P1) of the gas before you compress it and the then measure the Pressure (P2) and Temperature (T2) of the gas after it is compressed, then the we can calculate the new compressed volume (V2). So if we know the volume, temperature and pressure of a gas at any one time, we can calculate what the volume of the gas would be if the temperature and pressure were changed to any other given values. Most gas meters measure the volume of gas flowing through them. So if we measure the pressure and temperature at the same time, we can determine condition 1 (V1, P1 and T1) as above. If we know the V1, T1 and P1 in the first condition at the meter, we can calculate the volume in an imaginary second condition (P2, V2 and T2). If we say that the temperature T2 is 288.15 K (15oC) and the pressure P2 is 101 325 Pa then by using the V1, P1 and T1 data at the meter, we can calculate what the volume V2 would be at the second imaginary condition. This is the volume that is normally reported by the gas suppliers. A simple example of this is attached as Appendix 3. The amount of energy available in a gas is dependent on the mass of gas and is usually declared as J per kg (J/kg) However, if the gas is normalised to a known condition, as described above, then the mass of gas in......

Words: 2660 - Pages: 11

Video Transcript

...Video 1 – Why V2 is better than the other electronic cigarette V2 is the most advanced electronic cigarette in the market today. Don’t be fooled by the other electronic cigarette brands that employ outdated or sub-standard technologies. V2 ‘s breakthrough two-piece design was developed to correct the problems commonly experienced with the early 3-piece each cigarette models. Three piece electronic cigarettes are unsanitary and messy. Some models not only re-use the atomizer but also the nicotine cap which is annually refilled with vaccine liquid. This makes it nearly impossible to avoid getting sticky nicotine liquid on your fingers and your mouth. This liquid can stain your clothes and is irritating to your skin. Because the nicotine cap is fully exposed, the nicotine liquid inside these caps dries out quickly leading to inconsistent vapor thicknesses and a dried burnt flavor. These cheap plastic caps only accommodate a small amount of vapor liquid and must be replaced often which leads to increased costs for you. The reusable atomizer in these models must be cleaned regularly to maintain optimal performance and avoid closure. When replacing the nicotine cap or cleaning the atomizer it is easy to disturb the sensitive heating element which can cause the device to stop working altogether and even lead to fire risks. 3-piece electronic cigarettes sacrifice freshness, convenience, economy and safety. V2’s easy 2-piece design addresses all of these problems. The V2 electronic......

Words: 1548 - Pages: 7

It Managers Book

...of the ITIL Advisory Group Jennifer Zaino 12 15 The Evolution of Service Management Philosophy Troy DuMoulin Key Differences Between ITIL v2 and v3 Mike Tainter and Martin Likier 12 15 The IT Manager's Guide to ITIL v3, an IT Management eBook. © 2008, Jupitermedia Corp. 1 [ The IT Manager's Guide to ITIL v3 ] ITIL Version 3 is All Business By Drew Robb Shrek III” was better. So were "Rocky III" and "Mission Impossible III." But while ITIL III is no blockbuster, it is getting decent ratings from the IT community overall. In particular, it is earning kudos for attempting to bridge the chasm between IT and the various business units within organizations. IT Infrastructure Library (ITIL) made its debut in 1989 and since then has become the de facto standard for IT service management best practices. Since the release of v2 at the start of the millennium, there have been new business regulations and mandates, technology advancements, and general shifts in how IT is valued. In light of this, the UK's Office of Government and Commerce (OGC) has been working with industry leaders to update ITIL for the third time in two decades even as many organizations are in varying stages of ITIL adoption. With the release of ITIL v3 come many questions from IT managers and CIOs. Most common among them: "What's the difference between v2 and v3?" and "Should I wait to start my ITIL project?” and, "What “ does this have to do with the data center?" "For those worried......

Words: 8563 - Pages: 35