CommunicationsSystems have four basic properties (holism, equi-fi.pdf

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Communications: Systems have four basic properties (holism, equi-finality, negative entropy). List the principle behind each property. What does it mean that “a system is more than the sum of its parts?” How does this concept apply to organizations? Solution Four basic properties of systems are: Principle of Holism: It emphasizes that system is more than sum of its part due to interdependence upon each other. Principle of Equi-finality: It states that outcome can be achieved through many different paths due to interdependence of components. Principle of Negative Entropy: It is based upon openness of the system that says that system owns the ability to prevent deterioration and thrive. Principle of Requisite Variety: It is also based upon openness of the system and it says that internal complexity should be maintained to deal with external complexity. Statement of “a system is more than the sum of its parts” means that system achieves more when all of its parts complement each other and work together. Each component / part of the system cannot do much in isolation but they can achieve big goals when grouped together as a team. It is very important in organizational perspective as it consists of many different functions as components. When, these functions work together, organizational objective is achieved. If not, organization falls apart..

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Chapter 4: Systems and Critical Approaches to Organizational Communication Introduction Consideration of the organismic metaphor Understanding organizations from a systems perspective Use a radical frame of reference to look at how critical approaches view organizations as sites of domination Systems Approaches Systems approaches take a more macro view of organizations, emphasizing the permeable nature of organizational boundaries Systems approaches to understanding organizations use this general idea to examine the component parts of an organization and how they work together to keep an organization “alive” General Systems Theory Ludwig von Bertalanffy suggested systems concepts for a variety of disciplines “Novel developments” include: Hierarchical order Entropy Control mechanisms Open systems http://isss.org/conferences/madison2008/20080714_1130_ISSS_Drack.pdf 4 Cybernetics Norbert Wiener’s contribution, from the Greek word for “steersman” Control and communication theory that examines both machine and animal Especially important to studying organizational communication: System goals Feedback Entropy Gestalt (nonsummativity) The Process of Organizing Karl Weick’s definition of organizations includes: Equivocality Interdependence In his book Making Sense of the Organization (2001) Weick states that he views “organizations as collections of people trying to make sense of what is happening around them” (p. 5) Chaos and Complexity Theories Used in mathematics to help explain random occurrences Represents the collapse of order in the status quo Chaotic behavior can appear random but can be defined mathematically Complexity theory suggests nature is incredibly complex but is a result of underlying simplicity The Organization as a System The component parts and environment of a system interact in a way that can be studied and understood Parts are interdependent or naturally reliant on one another Inputs: incoming information Throughputs: assimilation of that information Outputs: end results of throughput decisions Critical Thinking Questions – Systems Characteristics In what ways do you use feedback to make changes in your life? Does the quality of input affect the quality of the output in an organizational system? Organizational System Characteristics General terms of systems properties include: Holism: system is more than just the sum of its parts Nonsummative: results are more than simply adding together each piece Equifinality: reaching the same final state from differing initial conditions; variety of paths Entropy: process of degeneration Typical of closed systems Negative entropy: flow of information from external environment; leads to growth and sustainability Typical of open systems Subsystems and Suprasystems Departments in an organization need to be aware of what is happening in other departments There are 10 common characteristics of open systems highlighted by Katz and Kahn (1978) Feedback: re.

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Problem 1: Create Node class (or use what you have done in Lab4) • Add methods to get and set, Data and Link • Here the data is an object. (java object) your Linked List can hold any java class. a) Create MyLinkedList2 class and write methods to • Methods to insert nodes at the beginning, at the end and at a specific position. insertFirst(Node x), insertLast(Node x) , insertLast(Node x, int pos) • Methods to delete node from the beginning and from the end. delFirst(), delLast() • Method to search the Linked List for a specific object. searchList(x Object) • Method to Sort the Linked List in ascendant or descendant order according to the object value(s). for example sorted by name, by age, by gpa. sortAsc(),sortDesc(). • Method to display all nodes objects (printing an object is related to its toString() method). • Method to return the largest and smallest object in the list – maxList(), minList() according to some specific object members. . • Method to return the number of nodes – listSize() • Method to merge two linked Lists into one. MyLinkedList2 list3= mergeList(MyLinkedList2 list1, MyLinkedList2 list2) b) Create Main class : MyLinkedList2Demo • Create a linked List LL1 and insert some objects to it. Use any a student java class. • Fill LL1 by adding students from an input file. • Display all the objects (students) in the List LL1 • Search for a particular object (Student by name or by Id or both) in LL1 and print true or false • Sort LL1 and display it ( using asc and desc on multiple members) • Perform all the delete methods on LL1. • Perform all the insert methods on LL1. • Call and run all the implemented methods in a sequential order. NB: Organize a nice output for all the questions above. Problem 2: Same as Pb1 using double linked lists. Create Nod3e and MyLinkedList3 classes. -------------------------------------------------------------------------------------------------------------------- --------------------------------------------------- The input file : id fname lname age gpa year college cuCourses totalCourses 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201112345 Amna chamisi 17 2.1 2014 ENG 31 32 201412345 Amna chamisi 17 2.1 2014 COE 5 29 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 201412345 Amna chamisi 17 2.1 2014 CIT 5 30 Solution /** * Node.java * @author * */ public class Node { public Object getObj() { return obj; } public Node setObj(Object obj) { this.obj = obj; return this; } public Node getNext() { return next; } public Node setNext(Node next) { this.next = next; return this; } private Object obj; private Node next; } import java.util.Collections; import java.util.Comparator; import java.util.LinkedList; /** * MyLinkedList2.java * @author * */ public class MyLinkedList2 { public void insertFirst(Node x) { if (head == null) { head = x; } else { x.setNext(head); } } public void insertLast(Node x) { if (head == null) { head = x; r.

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For Programming Embedded Systems Q-07 *Whcih variable and condition is sued to detect unfinished tasks (interrupted interrupt)? A) Line 28 B) Line 36 C) Line 25 D) Line 22 I. #include \"rims.h\" 2 3 4. typedef struct task state BL LedOff; break; default: int state; unsigned long period; unsigned long elapsedTime: int (*TickFct) (int); taski state ) I Transitions switch (state) case BL Ledoff: 11. task tasks [2] break; case BL Ledon: 13. const unsigned char tasksNum 2; 14. const unsigned long periodBlinkLed = 1500; 15. const unsigned long periodThreeLeds 500; break; default: break; ) // State actions 17, const unsigned long tasksPeriodGCD = 500; return state 19. int TickFct BlinkLed (int state) 20. int TickFct ThreeLeds (int state) 22. unsigned char process!ngRdyTasks = 0; 23. void TimerISR 98. enum TL States TL TO, TL T1, TL T2 TL State; 100 99. int TickFct ThreeLeds (int state) ( 101 VARIABLES MUST BE DECLARED STATIC/ unsigned char i; if (processingRdyTasks) /*e.g., static int x = 0; */ /*variables for state machine go here. printf (\"Period too short \ \"): 103 switch (state) case -1: process!ngRdyTasks = 1; for (1 = 0; 1 = 107 108 109 tasks [i].period) case TL TO: tasks [1]. state = 1 tasks[i].TickFct (tasks[i].state); state = TL T1; tasks[i].elapsedTime0 break; case TL T1: 112 113 tasks [1].elapsedTime += tasksPeriodGCD; processingRdyTasks - 0; state = TL T2 115 116 38. int main) break; case TL T2: unsigned char i=0; tasks [1]. state =-1; tasks [1].period = periodBlinkLed; tasks [1] .elapsedTime tasks [1] ·period; tasks [1].TickFct 6T1ckFct BlinkLed; 118 119 120 121 122 123 1 state = TL T0; break; default: state --1; tasks [1] , state = -1; tasks [1].period periodThreeLeas; tasks [1].elapsedTime = tasks [1].period; tasks [1].TICkFct = &TICkfct; ThreeLeas; 125 switch (state) 127 128 129 130 case TL TO: get Timerset (tasksPeriodGCD); break case TL T1: TimerOn () while (1) Sleep return 0 133 135 137 139 h break case TL T2: 61. enum BL States BL Ledoff, BL Ledon BL State; 62. int TickFct BlinkLed (int state) *VARIABLES MUST BE DECLARED STATIC break; 64. /*e.g., static Int X = 0; */ 65. /*variables for state machine go here/ default: break switch (state) case-1: return state state = BL LedOff break; 146 case BL Ledoff: state = BL Ledon; break; case BL Ledon: 1 Solution c) Line 25 is correct Line 25 contains if condition and processingRdyTasks variable is used to detect the unfinished tasks.

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Abraham Shine recently opaned his own accounting fiem on April 1, which he operates as a corporation. The name of the new entity is Abraham Shine CPA Shine xparienced the following events during the organizing phase of the new business and ies Srst month of operations in 2018 (Cick the icon to view the transactions) Read the teaments Requirement 1. Analyze the effects of the events on the accounšing equation of Abraham Shine, CPA Anal ze the events chronolo caly one transaction at a rme egning with the transaction on the 6th caklate the balance in each account after analyzing the efect of the transaction on the account g equation Aher calculating the ending balance of each account on the 30m caloulate total assets and totall Babilities and equity (Complete only the nacessary answer boxes for your transaction ines Do not enter any zeros for your transaction lines 1 Car down all balances to the al line including zero balance accounts entering \"Q for any zero balances Enter decrease in ar account with 4 minus sign or pare theses Abbreviations used Ap Accounts Payable A Accounts Receivable. Com Common Contr “Conttuted. Div . Dividends. Exp . Expense. Fum . Fumitu, Liab. Liablies Rev-Revenue Sup . Supplies, Uni-Uilties) ASSETS LIAB+ EQUITY Contr. Retained Earnings ServiceRent Exp Cap Cash + AR + Ofice + Furn AP + Com Die UtI Sup Stock Rev Exp Enter any inumber in the edit fhelds and then click Check Answer 13-taring Clear Al Check Answer Solution Answer Assets = Liabilities + Equity Contributed Capital + Retained earnings Date Cash + A/R + Office Sup + Furn = A/P + Com Stock - Div + Service Revenue - Rent Exp - Util Exp Apr-05 80000 + + + = + 80000 - + - - Apr-06 -200 + + 200 + = + - + - - Apr-07 + + + 7500 = 7500 + - + - - Apr-10 3500 + + + = + - + 3500 - - Apr-11 -390 + + + = + - + - - 390 Apr-12 + 12000 + + = + - + 12000 - - Apr-18 -1600 + + + = + - + - 1600 - Apr-25 12000 + -12000 + + = + - + - - Apr-27 -7500 + + + = -7500 + - + - - Apr-30 -4000 + + + = + - 4000 + - - TOTAL 81810 0 200 7500 0 80000 4000 15500 1600 390 Income Service Revenue 15500 Expenses Rent Expense 1600 Utilities expense 390 Total expense 1990 Net Income 13510 Opening Bal 0 Add: Net Income 13510 Less: Dividend paid 4000 9510 Closing balance 9510 A. Owner\'s Equity Common Stock 80000 Retained Earnings 9510 89510 B. Liabilities Accounts payable 0 Total Owners\' Equity and Liabilities 89510 C. Assets Cash 81810 Accounts Receivables 0 Office Supplies 200 Furniture 7500 Total Assets 89510 Assets = Liabilities + Equity Contributed Capital + Retained earnings Date Cash + A/R + Office Sup + Furn = A/P + Com Stock - Div + Service Revenue - Rent Exp - Util Exp Apr-05 80000 + + + = + 80000 - + - - Apr-06 -200 + + 200 + = + - + - - Apr-07 + + + 7500 = 7500 + - + - - Apr-10 3500 + + + = + - + 3500 - - Apr-11 -390 + + + = + - + - - 390 Apr-12 + 12000 + + = + - + 12000 - - Apr-18 -1600 + + + = + - + - 1600 - Apr-25 12000 + -12000 + + = + - + - - Apr-27 -7500 + + + = -7500 + - + - - Apr-3.

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13. What is the difference between post-consumer recycled materials and post- industrial recycled content materials? Solution Postconsumer material means a material or finished product that has served its intended use and has been diverted or recovered from waste destined for disposal, having completed its life as a consumer item. Post-consumer recycled content is defined as waste materials generated by households or by commercial, industrial and institutional facilities in their role as end-users of a product, which can no longer be used for its initial purpose..

13. What is the difference between post-consumer recycled materials a.pdf

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