Case study on Queensland Health Payroll Debacle & Queensland Governm.pdf

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Case study on Queensland Health Payroll Debacle & Queensland Government Bans IBM from IT contracts. Questions: 7. What could IBM have done to prevent the release of its defective payroll system? Solution The queensIsland had banned IBM for further contracts, following report about payroll system it was blamed towards vendorers. In 2010, government started introduction of IBM based Payroll system which makes million bollers over budget and indulge payments to thousands of people.projects initiall expectations is 6.19 million dollars but current estimation cost about 1.2 billion dollars. Early the qeens government has launced enquiry. the report that resluts out from enqury had given in the week.the report gaves a critcal of all members including IBM.The problms which makes project comes dowm, in major failure is by multiple peoape to loose up their responsibility properly,” Richard who led enquery had said that.. then IBM is not allowed to any of the contracts with the state .which was said by Queeens Government. But IBM bought an application to claim up the damage, argue that in agrment signed in 2010 the Government has waved its rights to sue. This afternoon supreme court judge ruled IBM favour.it means that state government cannot go for current damage some defect are found in payroll system when it is launced in 2010 which produce a result that many of the employees in the queen state had remaind withe unpaid and underpaid..

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one human resource manager recently got a thank you note on her iPhone that said “Thx 4 the Iview! Wud to wrk 4 u!!!” The manager had liked the candidate’s interview, but after getting the note, she put him in the reject pile. Do you think it was fair for the manager to automatically reject the candidate? Should “textspeak” be considered acceptable workplace communication? Discuss. Solution as per the above explanation we can not say that what the manager has did and because of the text mail only he rejected the candidate. and some times it may not possible to select the candidate even he did best in the interview. there would some one else who performed best than the other. in those times, he may reject the good one to choose the best. and based on the above explanation it can be observed that the interviwer is a lady and the interviewee is a gent or boy. she may not like expressing his ability and interest to work in the firm. and she have not decided and discolsed it to the guy that he is selected or rejected. she may put it hold and mean while by looking at his message she may refused the application. but, because of the only message if she rejected him, it is not fair..

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Mutations rII1 and rII2 are known to fall into the rIIA and rIIB cistrons (genes) respectively. Mutations rII1 and rII2 are used to infect E. coli B at very low multiplicity of infection (m.o.i.) and the progeny phage collected. These phage are then used to infect strain K at very low m.o.i.. The resulting phage are diluted and plated on both E. coli B and E. coli K. Do you expect more plaques on strain B, on Strain K, an equal amount on both or is it impossible to tell? Explain Solution Strain B have more plaques because it supports the growth of all viruses,wild type T4 as well as rII mutants which are able to complete their lifecycle in this strain thus giving total number of viruses developed. Strain K have low number of plaques because in this strain only wild type T4viruses can grow and complete their lifecycle..

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Let X and Y be continuous r.v. with joint density function Please give two different formulas that show how to find cov(X,Y) **No need to do the actual integration** One formula should use the definition of cov (which should have one definite double integral) and the other uses the short cut formula for finding cov. (which should have difference between two definite double integrals). Thank you Solution Cov(x,y)= E(x)=x E(y)= E(xy)= Short cut formula is: Cov(x,y)=E(xy)-E(x)E(Y).

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Discuss the reasons why the Human Microbiome Project is a step forward in characterizing normal and pathogenic biota in humans. Solution The Human Microbiome is defined a the collection of microbes- bacteria, viruses and single cell eukaryotes that inhabits the human body. Microbes in adult human body are estimated to outnumber human cells by a number of ten to one. The Human Microbiome project is a United States National Institute of Health initiative with a goal of identifying and characterising the microorganisms which are found in association with both healthy and diseased humans. The main goals of Human Microbiome Project is a step forward in characterising normal and pathogenic biota due to its goals which are as follows:.

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Consider an extension of the queue abstract data type, called a minqueue, which supports operations ENQUEUE, DEQUEUE, and FINDMIN. Assume that the elements to be stored are integers (or any other totally ordered data type). FINDMIN simply returns the smallest element in the minqueue, but does not remove it. Using a standard implementation of a queue, the FINDMIN operation takes O(n) time in the worst case. However, consider a more clever data structure for this abstract data type which works as follows: There are two regular queues used. The first queue is called the ”real queue” and the second queue is called the ”helper queue”. When the ENQUEUE(x) operation is performed, the element x is enqueued in the regular way on the real queue. The element x is also enqueued at the end of the helper queue. However, if the element y immediately ”in front” of x on the helper queue is larger than x, then y is removed from the helper queue. This process of x annihilating the element immediately in front of it is repeated until the element immediately in front of x is less than or equal to it or x is at the front of the helper queue. Describe how the DEQUEUE and FINDMIN operation are implemented in this data structure. Give the worst case running time for each of ENQUEUE, DEQUEUE, and FINDMIN using this data structure. Give an argument using the potential method to show that any sequence of n ENQUEUE, DEQUEUE, and FINDMIN operations requires only O(n) time. Or namely, the amortized complexity of these operations is O(1). Consider an extension of the queue abstract data type, called a minqueue, which supports operations ENQUEUE, DEQUEUE, and FINDMIN. Assume that the elements to be stored are integers (or any other totally ordered data type). FINDMIN simply returns the smallest element in the minqueue, but does not remove it. Using a standard implementation of a queue, the FINDMIN operation takes O(n) time in the worst case. However, consider a more clever data structure for this abstract data type which works as follows: There are two regular queues used. The first queue is called the ”real queue” and the second queue is called the ”helper queue”. When the ENQUEUE(x) operation is performed, the element x is enqueued in the regular way on the real queue. The element x is also enqueued at the end of the helper queue. However, if the element y immediately ”in front” of x on the helper queue is larger than x, then y is removed from the helper queue. This process of x annihilating the element immediately in front of it is repeated until the element immediately in front of x is less than or equal to it or x is at the front of the helper queue. Describe how the DEQUEUE and FINDMIN operation are implemented in this data structure. Give the worst case running time for each of ENQUEUE, DEQUEUE, and FINDMIN using this data structure. Give an argument using the potential method to show that any sequence of n ENQUEUE, DEQUEUE, and FINDMIN operations requires only O(n) t.

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Can you give an example of a binary heap program Can you give an example of a binary heap program Solution /* * C Program to Implement a Heap & provide Insertion & Deletion Operation */ #include int array[100], n; main() { int choice, num; n = 0;/*Represents number of nodes in the heap*/ while(1) { printf(\"1.Insert the element \ \"); printf(\"2.Delete the element \ \"); printf(\"3.Display all elements \ \"); printf(\"4.Quit \ \"); printf(\"Enter your choice : \"); scanf(\"%d\", &choice); switch(choice) { case 1: printf(\"Enter the element to be inserted to the list : \"); scanf(\"%d\", &num); insert(num, n); n = n + 1; break; case 2: printf(\"Enter the elements to be deleted from the list: \"); scanf(\"%d\", &num); delete(num); break; case 3: display(); break; case 4: exit(0); default: printf(\"Invalid choice \ \"); }/*End of switch */ }/*End of while */ }/*End of main()*/ display() { int i; if (n == 0) { printf(\"Heap is empty \ \"); return; } for (i = 0; i < n; i++) printf(\"%d \", array[i]); printf(\"\ \"); }/*End of display()*/ insert(int num, int location) { int parentnode; while (location > 0) { parentnode =(location - 1)/2; if (num <= array[parentnode]) { array[location] = num; return; } array[location] = array[parentnode]; location = parentnode; }/*End of while*/ array[0] = num; /*assign number to the root node */ }/*End of insert()*/ delete(int num) { int left, right, i, temp, parentnode; for (i = 0; i < num; i++) { if (num == array[i]) break; } if (num != array[i]) { printf(\"%d not found in heap list\ \", num); return; } array[i] = array[n - 1]; n = n - 1; parentnode =(i - 1) / 2; /*find parentnode of node i */ if (array[i] > array[parentnode]) { insert(array[i], i); return; } left = 2 * i + 1; /*left child of i*/ right = 2 * i + 2; /* right child of i*/ while (right < n) { if (array[i] >= array[left] && array[i] >= array[right]) return; if (array[right] <= array[left]) { temp = array[i]; array[i] = array[left]; array[left] = temp; i = left; } else { temp = array[i]; array[i] = array[right]; array[right] = temp; i = right; } left = 2 * i + 1; right = 2 * i + 2; }/*End of while*/ if (left == n - 1 && array[i]) { temp = array[i]; array[i] = array[left]; array[left] = temp; } } Output:.

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4. Records of the missions of Baja California in the form of baptisms and burial records provide an ideal opportunity to examine the consequences of what is known as the \"Colombian Exchange,\" the movement of organisms large and small between the Old and New Worlds after the appearance of Columbus in 1492. The data below are two time series of the numbers of baptisms at two Central Baja missions, San Jose de Comondu and San Gertrudis. The data are for the decade 1770 - 1779. (a) How would you generally describe the trend of baptisms for the two missions over this period? Do the missions appear to show similar trends? Justify your response by appealing to the graph. (b) One of the influences of the meeting of the Old and New Worlds was the transmission of diseases to the native populations, who had developed no natural immunity from European diseases. If there were anecdotal evidence for an outbreak of disease during this time, which mission and what year seems to be the most likely candidate? Again, justify your response by appealing to the graph. (a) How would you generally describe the trend of baptisms for the two missions over this period? Do the missions appear to show similar trends? Justify your response by appealing to the graph. (b) One of the influences of the meeting of the Old and New Worlds was the transmission of diseases to the native populations, who had developed no natural immunity from European diseases. If there were anecdotal evidence for an outbreak of disease during this time, which mission and what year seems to be the most likely candidate? Again, justify your response by appealing to the graph. Solution.

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4. Cafazzo et al. studied a group of stray dogs in Rome, and ranked them from 1 (most dominant) to 27 (least dominant.) They split the results by gender . Ranks of males: 1, 2, 3, 4, 5, 6, 11, 12, 13, 14, 17, 18, 19, 20, 21 . Ranks of females: 7, 8, 9, 10, 15, 16, 22, 23, 24, 25, 26, 27 Is there any diference between males and Is there any difference between males and females? Find the two-tailed P-value for: (a) The Wilcoxon rank-sum test. (b) A van der Waerden test Solution.

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