...Task 1- Explain the resources available to support the project Manager Information Understand client’s requirements It essential that the project manager meet with the customers to talk about the necessities of the system they are going to outline and the system director have to listen to the customer requirement on what they have to do and it is important that the system administrator make a list of the number of details he has gained from the customers. Facilities Equipment’s The equipment’s that are used should be available so that the project manager can use them and that it can support them and it is also important for the team so that they can use the correct equipment’s for the project they work on. Technology For a project manager technologies are important and they have to take into consideration of these when doing a project they will have to depend on a variety of technologies for different projects and the team working on the project should also make sure that they choose the technology that is need for the project they are doing. Finance The manager has to take into consideration of the finance that are used in the project so when doing the project they will have to carefully make sure that the budget they make for the project is good and they don’t overspend and the team who is working on the project should also make sure that they buy whatever they need for the project according to the budget. People Designer When doing a project...
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...ANATOMY AND PHYSIOLOGY P2, P3 4/27/2015 Khadra Ali | P1 – Outline the functions of the main cell components The human body is made up of millions of tiny cells that can only be seen under a microscope, cell also vary in shape and size. Cells are the basic structural of all living things. The human body is poised of trillions of cells. They give structure for the body, take in nutrients from food, convert those nutrients into energy, and carry out specialized functions. Cells also contain the body’s hereditary material and can make copies of them. Cells all have different sizes, shapes, and jobs to do. Each cell has a different function. The actual definition of cells is the smallest structural unit of the body that is capable of independent functioning, it consisting of one or more nuclei; it has a cytoplasm, and various organelles which are all surrounded by a cell membrane. There are four main parts to a cell; Plasma/Cell membrane, Cytoplasm, Nucleus and Cell Organelles. Plasma/Cell membrane: The plasma/cell membrane is a phospho-lipid-protein bi-layer; the lipids are small fatty molecules in two layers (bi-layer) with larger protein molecules inserted at intervals partly or completely through the bi-layer. The lipid molecules are phospholipids, the two lipid chains are insoluble in water and the phosphate head is water soluble. The fluid which surrounds the cells and the cytoplasm are watery environments next to the phosphate heads. Protein molecules create channels...
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...P1: Explain concepts of equality, diversity and rights in relation to health and social care Equality, diversity and rights are the core elements in health and social care. It affects every single person using the health and social care sector as well as those working within them. It is important for health and social care workers to understand the importance of treating all individuals equally no matter their ethnicity, gender, race, beliefs, sexuality, education, language, background or skin colour. Individuals must all be treated equally; Equality in terms of rights, status or opportunities. This has become an important focus as there are laws and policies in all organisations regarding it to ensure that everyone in the organisation has same opportunities and choices, and no one is discriminated. The new equality act came into force in 2010; it consists of over 116 separate pieces of legislation into one act to protect the rights of individuals and to advance a fair and more equal opportunity for all. In health and social care sector this means everyone using or working within it should have equal opportunities, this can range from job opportunities to getting medical treatment options and that all individuals must receive same high-quality service. Diversity is the term used to describe the differences between individuals. It is important that individual and group diversities are recognised to ensure that everybody’s needs and requirements are understood and responded...
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...Unit 2: Equality, Diversity and Rights in Health and Social Care Assessment Criteria: Explain the Concept of Equality, Diversity and Rights in Relation to Health and Social Care. Describe Discriminatory Practise in Health and Social Care. Describe the Potential Effects of Discriminatory Practice on Those Who Use Health or Social Care Services. Assess the Effects on Those Using the Service of Three Discriminatory Practices in Health and Social Care Settings. Equality: - The term equality according to Dictionary.com (2015) “Is the state or quality of being equal; correspondence in quantity, degree, value, rank and or ability” however I believe that this definition can establish misconception on the view of an individual, and so the more realistic concept behind ‘Equality’ in regards to the health and social industry; is for all individuals to be provided with equal opportunities to access of the concerning services, and for these users to be treated fairly (not equally) accordingly to the approaches used to meet their individual needs. Diversity: - Is a termed used to describe the promotion and celebrating of differences which an individual may have on the aspects of your own identity as well as others. Rights: - "They are the legal entitlements which we are all privileged to own, with the aim of allowing equal opportunities for all citizens in certain aspects of an individuals' life. There are a number of rights entitled under the enforcement of The Human Rights Act. 2000...
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...number of possible different schedules for 3 processes on one processor is: 3! == 6. Gantt Charts: 1. P1 | P2 | P3 | 2. P1 | P3 | P2 | 3. P2 | P1 | P3 | 4. P2 | P3 | P1 | 5. P3 | P1 | P2 | 6. P3 | P2 | P1 | General Formula: For “N” number of processes, the number of possible, different schedules will be N! I.e. N*N-1*N-2*….2*1 Q2 (ANS). (A). 1. SJF Gantt Chart: P4 | P3 | P5 | P1 | P2 | 2. Non Pre-emptive Priority: P2 | P5 | P1 | P3 | P4 | 3. Round Robin Gantt Chart with quantum 4: P1 | P2 | P3 | P4 | P5 | P1 | P2 | P3 | P4 | P5 | P1 | P2 | P3 | P5 | (B). Turnaround Time for all process, in all schedules: Process | SJF | Non Pre-emptive Priority | Round Robin | P1 | 35 | 31 | 41 | P2 | 47 | 12 | 45 | P3 | 16 | 40 | 46 | P4 | 7 | 47 | 35 | P5 | 25 | 21 | 47 | (C). Waiting time for all processes, in all schedules: Process | SJF | Non Pre-emptive Priority | Round Robin | P1 | 26 | 21 | 31 | P2 | 38 | 0 | 33 | P3 | 7 | 31 | 37 | P4 | 0 | 38 | 28 | P5 | 16 | 12 | 38 | (D).The waiting time is least in the SJF (Shortest Job First) scheduling. This is because the long-time taking jobs are pushed back and the shortest time taking jobs are done first. Q3 (ANS). (A). Waiting time for all in FCFS scheduling algorithm: 1. P1 = 0 2. P2 = 7.5 3. P3 = 11 Avg. Turnaround Time = (waiting Time + burst Time for n processes)/n Avg. Turnaround Time = ((0 + 8) + (7.5 +...
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...ps->age+3); struct Student *ps1; // allocate memory by malloc() dynamically. ps1 = malloc(sizeof(struct Student)); ps1->name = malloc(20*sizeof(char)); scanf("%s%d",ps1->name,&(ps1->age)); printf("4:%s %d\n",ps1->name,ps1->age+3); printf("5:%s %d\n", (*ps1).name, (*ps1).age); } Student s 0028FF08 malloc 003C1110 name name age age ps malloc 0028FF00 003C1188 ps1 malloc #include #include struct S1 { int a1; char b1; }; struct S2{ int a2; char b2; struct S1 *p2; }; struct S3{ int a3; char b3; struct S1 *p3a; struct S2 *p3b; }; Linked structures // Self referencing structure. Also called: recursive structure. struct S4{ int a4; char b4; struct S4 *p4; }; int main(){ struct S1 s1; struct S2 s2; struct S3 s3; struct S4 s4; s1.a1=2; s1.b1='A'; s2.a2=20; s2.b2='B'; s2.p2 = &s1; // Creates a list: s2 ---> s1 printf("First:%d %c %d %c\n",s1.a1,s1.b1,s2.a2,s2.b2); printf("Second:%d %c %d %c \n", (*(s2.p2)).a1,...
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... and Evaluation: Beating the Market ABSTRACT During the period of 2005 to 2010, the market portfolio (P1) and one suggested portfolio (P3) post a positive absolute return of 0.80% and 0.82% respectively which underperformed the active fund portfolio (P2) 0.91%. This report follows various modeling methods in order to back test the performance of the active fund portfolio and compare its performance with that of two other portfolios. The findings indicate that, even though P2 achieves the highest return on the overall performance, the limitations such as the macro environment, the assumptions set, and the Shrinkage method used that accidentally downsizes some valuable stocks in out-‐samples as they are closely correlated are being ignored. By contrast, P3 will probably offer a “middle-‐choice” which will bring a promising and more stable return. 1 Portfolio Modeling and...
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...overtime between 6 PM and 7 PM. There will be seven sets of part-time employees who begin their work day at hour j=j␣1,2,...,7,withP1beingthenumberofworkers beginning at 9 AM, P2 at 10 AM, . . . , P7 at 3 PM. Note that because part-time employees must work a minimum of four hours, none can start after 3 PM because the entire operation ends at 7 PM. Similarly, some number of part-time employees, Qj, leave at the end of hour j, j 4, 5, . . . , 9. The workforce requirements for the first two hours, 9 AM and 10 AM, are: F P1 14 F P1 P2 25 At 11 AM half of the full-time employees go to lunch; the remaining half go at noon. For those hours: 0.5F P1 P2 P3 26 0.5F P1 P2 P3 P4 38 Starting at 1 PM, some of the part-time employees begin to leave. For the remainder of the straight-time day: F P1 P2 P3 P4 P5 −Q4 55 F P1 P2 P3 P4 P5 P6 −Q4 −Q5 60 F P1 P2 P3 P4 P5 P6 P7 −Q4 −Q5 −Q6 51 F P1 P2 P3 P4 P5 P6 P7 −Q4 −Q5 −Q6 −Q7 29 For the two overtime hours: F P1 P2 P3 P4 P5 P6 P7 −Q4 −Q5 −Q6 −Q7 −Q8 14 F P1 P2 P3 P4 P5 P6 P7 −Q4 −Q5 −Q6 −Q7 −Q8 −Q9 9 If the left-hand sides of these 10 constraints are added, one finds that 7F hours of full-time...
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...Dallas | c) Load the following data into the project table: project_no | project_name | budget | p1 | Apollo | 120000.00 | p2 | Gemini | 95000.00 | p3 | Mercury | 186500.00 | d) Load the following data into the works_on table: emp_no | project_no | job | enter_date | 10102 | p1 | Analyst | 2006.10.1 | 10102 | p3 | Manager | 2008.1.1 | 25348 | p2 | Clerk | 2007.2.15 | 18316 | p2 | NULL | 2007.6.1 | 29346 | p2 | NULL | 2006.12.15 | 2581 | p3 | Analyst | 2007.10.15 | 9031 | p1 | Manager | 2007.4.15 | 28559 | p1 | NULL | 2007.8.1 | 28559 | p2 | Clerk | 2008.2.1 | 9031 | p3 | Clerk | 2006.11.15 | 29346 | p1 | Clerk | 2007.1.4 | Question 2 a) Update the following data in the employee table. emp_no | emp_fname | emp_lname | dept_no | 25348 | Matthew | Smith | d3 | 10102 | Ann | Jones | d3 | 18316 | John | Barrimore | d1 | 29346 | James | James | d2 | 9031 | Elsa | Bertoni | d2 | 2581 | Elke | Hansel | d2 | 28559 | Sybill | Moser | d1 | b) Increase the budget for all projects by 50%. c) Update the following data in the works_on table. emp_no | project_no | job | enter_date | 10102 | p1 | Analyst | 2006.10.1 | 10102 | p3 | Manager | 2008.1.1 | 25348 | p2 | Clerk | 2007.2.15 | 18316 | p2 | NULL Manager | 2007.6.1 | 29346 | p2 | NULL | 2006.12.15 | 2581 | p3 | Analyst |...
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...Assignment No 6: Submission date 19th November 1, 2012 LOAD FLOW CALCULATIONS y12 = -j2.5 P3, Q3 V1 2 V3 V2 z12 = j0.4 z13 = j0.4 Y13=-j5 P2, Q2 P1, Q1 3 1 Node | Vpu | δo | P | Q | Type | 1 | 1.0 | 0 | - | - | Reference | 2 | 1.0 | - | -1.0 | - | P-V | 3 | - | - | -0.6 | -0.3 | PQ | 6.1 Establish the equations for power injection (Pi,Qi) at all nodes. 6.2 Calculate the unknown variables in the table shown above by analytical solution. 6.3 Develop analytical expressions for Jacobian matrix elements. 6.4 Calculate the Jacobian matrix in power flow balance. Assume per unit values are given referred to Vref = 60 kV, line to line Sref = 100 MVA, 3-phase After analytical solution the voltage at node no. 3 will be V3 = 0.93 pu which is below the acceptable limit i.e. 57 kV, line to line. Two alternatives to bring V3 up to 57 kV are proposed: A: Reduce active power consumption P3 from 60 MW to 50 MW. B: Reduce reactive power consumption Q3 from 30 MVAr to 20 MVAr. In order to test these two alternatives we can use a linear approximation where the Jacobian matrix (in power flow balance) is applied: ∆Q3∆P2∆P3=∂Q3/∂V3∂Q3/∂δ2∂Q3/∂δ3∂P2/∂V3∂P2/∂δ2∂P2/∂δ3∂P3/∂V3∂P3/∂δ2∂P3/∂δ3∆V3∆δ2∆δ3 For alt A : ΔQ3 = 0, ΔP3 ≠ 0 as the proposed change. For alt B : ΔP3 = 0, ΔQ3 ≠ 0 as the proposed change. 6.5 Test both alternatives and choose the best. 6.6 Check the best solution found under 6.5 by exact analytical solution of the equations...
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...1. Deadlock (“deadly embrace”) is a system-wide tangle of resource requests that begins when 2 or more jobs are put on hold. • Each job is waiting for a vital resource to become available. • Needed resources are held by other jobs also waiting to run but can’t because they’re waiting for other unavailable resources. • The jobs come to a standstill. • The deadlock is complete if remainder of system comes to a standstill as well. • The resources can be categorized into physical and logical resources. The physical resources are printer, disk drive, cpu, memory, scanner etc. The logical resources are files. • Deadlock is more serious than indefinite postponement or starvation because it affects more than one job. • Because resources are being tied up, the entire system (not just a few programs) is affected. • Requires outside intervention (e.g., operators or users terminate a job) to resolved the deadlock. 2. Seven Cases of Deadlocks Case 1 Deadlocks on file requests Case 2 Deadlocks in databases Case 3 Deadlocks in dedicated device allocation Case 4 Deadlocks in multiple device allocation Case 5 Deadlocks in spooling Case 7 Deadlocks in disk sharing Case 8 Deadlocks in a network Case 1: Deadlocks on File Requests | |If jobs can request and hold files for duration of their...
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...General: 1. What devices do you own with operating systems (smartphones, tablets, computers). For each (up to 3): (5 points) a. What OS does it run? What version? What is the update mechanism? Answer) my computer OS is windows 8 and my smartphone OS is IOS 7. Windows update mechanism is Windows Updates and IOS update mechanism is ITunes. Chapter 2 2. What is the main advantage for an operating-system designer of using a virtual-machine architecture? What is the main advantage for a user? (10 points) Answer) OS designers can solve debug, network problem and security problem easily. Users also can undergo many different OS on one physical system. It is easy to research and develop. Chapter 3 3. Discuss three major complications concurrent processing adds to an OS. (10 points) Answer) 1. Processes and system resources should protected from each other. 2. More than one process executing concurrently for protect performance degradation 3. It occurs deadlocks. So you can m Chapter 4: 4. Provide a programming example where multithreading does not provide better performance than a single-threaded solution. (5 points) Answer) Sequential program is not good at threaded. So example is Shell program. 5. Can a multithreaded solution using multiple user-level threads achieve better performance on a multiprocessor system than on a single-processor system? How does this change if kernel-level threads are used? Explain. (10 points) ...
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...Case Study: Chase Manhattan Bank Chase Manhattan Bank The workload in many areas of bank operations has the characteristics of a nonuniform distribution with respect to time of day. For example, at Chase Manhattan Bank, the number of domestic money transfer requests received from customers, if plotted against time of day, would appear to have the shape of an inverted-U curve with the peak reached around 1 P.M. For efficient use of the resources, the manpower available should, therefore, also vary correspondingly. A variable capacity can effectively be achieved by employing part-time personnel. Since part-timers are not entitled to all fringe benefits, they are often more economical than full-time employees. However, other considerations may limit the extent to which part-time people can be hired in a given operating department. The problem is to find an optimum workforce schedule that would meet manpower requirements at any given time and also be economical. Some of the factors affecting personnel assignments can be listed 1. By corporate policy, part-time personnel hours are limited to a maximum of 40 percent of the day's total requirement. 2. Full-time employees work for eight hours (one hour for lunch included) per day. Thus, a full-timer's productive time is 35 hours per week. 3. Part-timers work for at least four hours but less than eight hours and are not allowed any lunch break. 4. Of the full-timers, 50 percent go out to lunch between 11 A.M. and 12...
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...The Experimental Determination of Heat Capacity Ratios of Gases, N2, CO2 and Ar, by the method of Adiabatic Expansion ABSTRACT The heat capacity ratios [Cp/Cv] of Nitrogen, Carbon Dioxide and Argon were obtained by the method of adiabatic reversible expansion. Four studies were conducted for each gas using an 18.0 L carboy and phthalate manometer at 297K. The initial and final pressures were obtained and used to calculate the experimental heat capacity ratios. Based on the results, Argon has an experimental heat capacity ratio of 1.66 and a Cv value of 1.56, which agrees with the theoretical Cp/Cv and Cv values obtained from the classical equipartition theorem. Based on the results, the diatomic molecule, N2 and triatomic molecule, CO2 have heat capacity ratios ...
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...Chapter 6: CPU Scheduling • • • Basic Concepts Scheduling Criteria Scheduling Algorithms Operating System Concepts 6.1 Basic Concepts • Maximum CPU utilization obtained with multiprogramming. • CPU–I/O Burst Cycle – Process execution consists of a cycle of CPU execution and I/O wait. – Example: Alternating Sequence of CPU And I/O Bursts – In an I/O – bound program would have many very short CPU bursts. – In a CPU – bound program would have a few very long CPU bursts. Operating System Concepts 6.2 1 CPU Scheduler • The CPU scheduler (short-term scheduler) selects from among the processes in memory that are ready to execute, and allocates the CPU to one of them. • A ready queue may be implemented as a FIFO queue, priority queue, a tree, or an unordered linked list. • CPU scheduling decisions may take place when a process: 1. Switches from running to waiting state (ex., I/O request). 2. Switches from running to ready state (ex., Interrupts occur). 3. Switches from waiting to ready state (ex., Completion of I/O). 4. Terminates. • Scheduling under 1 and 4 is nonpreemptive; otherwise is called preemptive. • Under nonpreemptive scheduling, once the CPU has been allocated to a process, the process keeps the CPU until it releases the CPU either by terminating or by switching to the waiting state. Operating System Concepts 6.3 Dispatcher • Dispatcher module gives control of the CPU to the process selected by the short-term scheduler;...
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