The Dangers of Social Networking free essay sample

Remember when you were ten years old, and you had hundreds of friends from various parts of the world who all interacted with you over Facebook? No? That scenario doesn’t exactly describe my childhood, either, but it does ring true to many of today’s children. In our state, in our country, and throughout the world, an increasing number of children are using social networking websites. We like to think that this is safe for children to be doing. After all, social networking websites have privacy settings. However, many children are either unaware of such settings, or simply don’t use them. As a matter of fact, a large number of children lie about their ages on their social networking profiles, deliberately avoiding the age-based privacy settings that these websites have so cleverly implemented. Of course, we also assume that the parents of these children are monitoring their social networking. We will write a custom essay sample on The Dangers of Social Networking or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page This is true to a certain extent; most parents try to regulate their child’s internet usage. However, the rules are often poorly reinforced, and many parents are not as aware of their child’s internet activity as they believe themselves to be. As a result of the increase in social networking, many of today’s children have already experienced cyberbullying. Most of us have heard of Megan Meier, the thirteen-year-old girl who committed suicide after having been bullied over Myspace a few years ago. While this certainly is an extreme case of cyberbullying, it is something that we can expect to see more and more of, especially with the increase in young social networkers. In less extreme cases, a child who is cyberbullied is more likely to experience depression, anxiety, bedwetting, headaches, and low self-esteem. These effects may even be long-term, following the child into his or her adult life. Unfortunately, most of the children who are cyberbullied are also bullied in school. As a result, the internet becomes an inescapable means of harassment to these children, forcing them to cope with bullying both in and out of school. Another issue with children using social networking websites is their ability to â€Å"tal k to strangers.† Many children today are using social networking as a means of making new friends. This is a scary concept because in reality, there are no guarantees that the people that children are befriending are the people that their profiles portray. Just imagine how easy it would be for a sex offender to make a fake profile, befriend an overly trusting child, and lure him or her to meet in person. In addition to this potential danger, we have to consider that many recent internet-based sex crimes have been committed in a more straightforward manner than we anticipate. In fact, most of these sex offenders were upfront with the child about their ages and their intentions to have sex with the child. Furthermore, a majority of the children who met the offender in person did so on more than one occasion. Although this disturbing information is difficult to face, it demonstrates that many children are not mature enough to recognize the danger in their online activities. Wit h all of the risks that accompany children using social networking websites, why does it continue to be a problem? It seems like we should be taking action to prevent such dangers, yet more and more children are using social networking websites every day. Something must be done; no benefits of social networking can possibly outweigh the safety of a child.

A Study of Prejudice in Harper Lees To Kill a Mockingbird Research Paper Example

A Study of Prejudice in Harper Lees To Kill a Mockingbird Research Paper Example A Study of Prejudice in Harper Lees To Kill a Mockingbird Paper A Study of Prejudice in Harper Lees To Kill a Mockingbird Paper Essay Topic: A Woman Killed With Kindness Literature To Kill a Mockingbird We see in the novel To Kill a Mockingbird, by Harper Lee an assortment of racial, social, religious and gender prejudice. The narrative is set in the small town of Maycomb, located in Alabama. The era is the early 1930s, a very prejudiced time in the southern states of America. This period of history was also the time of the Great Depression that occurred due to the Wall Street Crash in 1929. This novel is based upon a court case of a black man that is accused of the rape of a white woman. This tale is split up into two different parts. Part one introduces the main characters and portrays the several different prejudices which they both feel and experience. The second part of this novel presents the case of Tom Robinson, the black man. To Kill a Mockingbird focuses predominantly on the subject of racial prejudice throughout its entirety. There were an excessive amount of prejudice people at this time in the southern states. The Civil War ended in 1876, giving the blacks their deserved freedom from slavery. Even though the war had come to a close so long before the story takes place, in the 1930s, racial tension is still very high. There is strain between the blacks and the whites because the blacks legally are not subject to the whites anymore, yet the whites do not want to change their ways of living above the blacks, with the blacks under their authority. Even the small town of Maycomb was greatly overruled by the prejudices of whites; Atticus, his family, and Miss Maudie, are shown as the only unprejudiced people throughout the whole of the Maycomb County. Mr Ewell, the man who accuses Tom Robinson of raping his daughter, is by far the most prejudiced man in the whole novel. We see this right after the trial when Bob Ewell stops Atticus on the post office corner, spits in his face and says, according to Miss Stephanie Crawford, Hed get him if it took the rest of his life. The reader might consider that this doesnt show prejudice, but we see that throughout the novel Bob Ewell hates the unprejudice white people like Atticus. We see later on in the book that Tom Robinson is probably innocent and falsely accused. Because of Mr Ewell accusing Tom and causing his conviction, this eventually brings about the death of Tom. The black community is presented throughout the story in a very positive light. The Finchs maid is a loving, tender-hearted woman, who cares for Jem and Scout, bringing them up as if they were her own children, disciplining them at appropriate times, and giving them a mothers love. Tom Robinson is a very hard-working family man, praised by his employer, that That boys worked for me for eight years and I aint had a speck o trouble outa him. Not a speck. His kindness is shown in his feelings for Mayella Ewell, he felt right sorry for her. After the trial, when Calpurnia brings Atticus into the kitchen the next morning, the kitchen table was loaded with enough food to bury the family, brought by the black community, because they preciate what you did, Mr Finch. Gender prejudice is focused on in this book a great deal less than racial prejudice. Gender prejudice is defined as attributing certain characteristics to one sex or other which is not based on reason or experience. Jem told me I was being a girl, that girls always imagined things. We see that women were treated more as servants and maids than as wives and helpers in the 1930s. They were not allowed to sit on a jury because they were considered too frail. At the beginning of the book, Scout relates, I sat at the little table in the dining room; Jem and Francis sat with the adults at the dinning table. Scout feels that Aunty dislikes her because she isnt enough of a lady. We believe Scouts aunt is unreasonable in her demand on her to sit alone, because at such a young age it is abnormal for a young girl to sit still and try to be a lady. Throughout the novel, Scout is constantly embarrassed by, or embarrasses her aunty because of her continuous tom-boy attitude. The second strongest discrimination that is shown in this story is the social prejudice of Maycomb County. The Ewells are perceived as low down people, because they are poor and dirty and never going to school. They are despised by the middle class almost as much as blacks of Maycomb. Another poor family that is depicted, is the Cunninghams. When Walter Cunningham comes over for dinner, he poured syrup on his vegetables, and meatand probably would have poured it into his milk glass if (Scout) hadnt asked him what the sam hill he was doing. The book displays the Cunninghams as honest and upright people, they never took anything they couldnt pay back but because of their poverty they are looked down upon in the society. One of the Cunninghams sat on the jury during Tom Robinsons trial, in the beginning he was rarin for an outright acquittal. The Radleys, Jem and Scouts next door neighbour, have a son named Arthur Radley but all the children in the neighbourhood call him Boo. There is social prejudice shown here, in that he is not accepted because he is different from most people. The reason the children call him Boo is because for years time no one had seen him and the kids imagined him to be a ghost. There was a incident when Boo was thirty years of age he stabbed his father in the leg with a pair of scissors. Boo was imprisoned in the basement of the county jail until one day his father took him home and he was never seen again. Inside the house lived a malevolent phantom Any stealthy crimes committed in Maycomb were his work. The pecans from the Radley tree fall into the school yard, but they lay untouched because Radley pecans would kill youfolks say he pizened em and put em over on the school side of the fence. This shows how intolerant the county is against Boo, just for one crime he is sent to jail and is never seen again after his father takes him home. No one really knows Boo Radley but he is suspected by all that he is a crook and a killer. This is displayed, one September afternoon, Scout nearly smashes a roly-poly but is stopped by her brother. She asks, Why couldnt I mash him? Jem says, Because they dont bother you. Jem is learning to empathize with others, including Boo. You can shoot all the blue jays you want, but it is a sin to kill a mockingbird. Atticus told the children when they received air rifles. Miss Maudie explained Your father is right, mockingbirds dont do one thing but make music for us to enjoy. They dont eat up peoples gardens, dont nest in corncribs, they dont do one thing but sing their hearts out for us. That is way its a sin to kill a mockingbird. Mockingbirds here symbolize both Tom Robinson and Boo Radley. These two persons are harmless to society, yet because of the time and age, they are looked down upon and despised. This is the real sin, because they are not only harmless, but defenceless. Religious prejudice plays a small part in the book, when Cal, the familys black maid, took the children to a black church named First Purchase African M.E. Church. There was a big, tall Black woman called Lula and she objected to Cal bringing Scout and Jem to the black church. Lula wants to know why you bringin white chillum to a nigger church. This shows that the churches are segregated from each other, and that the Black church does not accept white people and the white church does not accept the Black people. We see later on that the children are asked to stay; this is because their father Atticus is not prejudiced and is defending Tom Robinson. The discrimination and intolerance that Harper Lee is really trying to expose is racial prejudice. This book has caused millions of people to consider the effect that racial prejudice really has. Tom Robinson is killed trying to escape from jail, because although Atticus proved that he was innocent, the jury still convicted him. The story ends with Boo finally coming out of his house, to save Jem and Scout from Mr Ewell, who tries to kill them, showing the children that the way that had imagined Boo was wrong. Over-all, this book teaches everyone something important about prejudice and how to live in the world today.

Applying Theory to a Practice Problem of Nursing Essay

Applying Theory to a Practice Problem of Nursing - Essay Example Nursing has developed differently through the years, with the efforts of the theorists of the different nursing eras; they have helped turn it into a respectable and reputable profession. It cannot be questioned for it has been backed up the different nursing theories which helped carve what nursing practice is today. Providing care is one of the main responsibilities that a nurse has to give to a patient. In fact, nursing has almost come to be defined as synonymous to caring, because nurses include care into their daily interventions. Caring for people in the field of nursing involves the simplest of things. Indeed, listening attentively (and hearing the message between the words) to what patients say about themselves, about their environment, about their current situation, about matters concerning the mind, heart, and soul, and different other things are part of caring that is inherent in the nursing profession (Bernick, 2004). Caring in nursing also manifests itself in the simple aspects of therapeutic communication, in a touch that can transcend the barriers of age and race, and even in the nurse’s simple presence at the bedside, giving the patient the feeling of peace and security. This process of caring has been maintained for all these years by both contemporary and pioneer nurses. But with the appearance of technologies that can diagnose a patient without having to ask them any questions aother than their personal information and lessen the time of poking and proding, the idea of spending a quiet tim e with the patient has lost some touch to the nurses nowadays. The idea of saving time and alloting them to other chores rather than giving the patients the care that they needed runs within the nurse’s mind. Time spent with the patient decreases which means the care given to the patient also diminishes. The application of nursing theories actualy loses its touch on the nurse, without the nurse even noticing. Caring is one of the essential component of nursing, one they tend to overlook and disregard. To be unable to perform this task means that the nurse is not performing the responsibility placed upon their shoulders. The only time that nurses realize that providing more than the average activities and giving the caring needed not only improves the healing ability of the patient is through the presentation of evidences of improvement when they are given; one of which is presented by situations where the cases are treated with the application of the different theories develo ped through the years.This paper aims to determine the importance of application of theories on the everyday job of a nurse. How with the guidance of the theories developed years ago can improve the ability of a nurse to provide proper care for their patients. The application of nursing theories in practice has improved the nurse’s ability to provide for all the needs of the patient. Whether it be his physical, emotional, psychological or spiritual need; a nurse can help the patient with the proper application of a specific theory needed by the situation. The theories used in practice were identified and classified according to three general divisions: The grand theories, mid-range theories and the borrowed theories. Application of Grand theory to the Problem Nursing theory is an important aspect of nursing that

Discussion of the theories on Optimal Capital Structure Essay

Discussion of the theories on Optimal Capital Structure - Essay Example The study by Modigliani and Miller was based on the following assumptions: 1. There are no brokerage costs. 2. There are no taxes. 3. There are no bankruptcy costs. 4. Investors can borrow at the same rate as corporations. 5. All investors have the same information as management about the firm’s future investment opportunities. 6. EBIT is not affected by the use of debt. This theory says that if these assumptions hold true, the value of the firm is not affected by the capital structure. This situation is expressed as follows: VL = VU = SL + D. Here VL is the value of a levered firm, VU is the value of an identical, unlevered firm, SL is the value of the levered firm’s stock and D is the value of its debt. As we know that WACC is a combination of cost of debt and cost of equity. The cost of debt is lower than the cost of equity. As a company raises capital through debt, the weight of debt increases and hence, it drives up the cost of equity as equity gets riskier. According to the assumptions by Modigliani and Miller, the cost of equity increases by an amount to keep the WACC constant. In other words, under these assumptions it does not matter whether the firm uses debt or equity to raise capital. So, capital structure decisions are irrelevant in such conditions. Modigliani and Miller: The Effect of Corporate Taxes In 1963, Modigliani and Miller relaxed the assumption that there are no corporate taxes. The corporate tax laws favour debt financing over equity financing because the tax laws allow companies to deduct interest payments as expense and on the other hand dividends are not deductible. So this treatment encourages debt financing. Interest payments reduce the amount the firm pay s to the government in the form of taxes and more of its cash is available for its investors. Hence, tax deductibility of the interest payments acts as a shield for the firm’s income before tax. Modigliani and Miller presented this concept as follows: VL = VU + Value of side effects = VU + PV of tax shield. They further simplified the concept as: VL = VU + TD. Here T is the corporate tax rate and D is the amount of debt. This relationship is expressed in the graph below. If the corporate tax rate is 40%, then this formula implies that every dollar of debt will increase the value of the firm by 40 cents. Hence, the optimal capital structure is 100% debt. Under this theory, the cost of equity increases as the amount of debt increases but it does not increase as fast as it does under the assumption that there are no taxes. As a result, under this theory the WACC falls as the amount of debt increases. This relationship is shown in the following graph. Miller: The Effect of corpor ate and personal taxes Later Miller brought in the aspect of personal taxes in this model. He said that income from the bonds is considered as interest which is taxed as personal income at a particular rate (Td). On the other hand, income from stocks comes in the form of dividends and capital gains. The tax on long-term capital gains is deferred until the stock is sold and the gain is realized. Of the stock is held until the owner dies no capital gains tax is paid. So he concluded that the returns on stock are taxed at a lower effective tax rate (Ts) than returns on debt. Looking gat this, Miller argued

The Definition of Nationalism In Different Regions and Countries Essay

The Definition of Nationalism In Different Regions and Countries - Essay Example It was Indian peasantry that formed a major force in the national liberation movement in the interwar period. To involve a broad amount of population into this movement, it was needed to take into account socio-psychological characteristics of the Indian peasantry and urban workers, yesterday’s peasants. A prominent role in the organization of mass non-violent campaign of the resistance to the colonial regime in the 20 - 40’s of the 20th century belonged to Mahatma Gandhi. He was the one to shape India national ideology. Thanks to Gandhi, the idea of the absolute independence of India was put forward. In this respect, Indian nationalism acquired the traits of Gandhi’s ideas.Gandhism resonated with wide layers of the peasantry and the urban poor because it was connected with the social ideal of the belief that the struggle for independence from British rule, is a struggle for justice. Gandhi drew upon his appeals from cultural, historical, and religious traditions that were familiar to peasants. Therefore, the demands for independence and the transformation of the society were explained in the traditional way and were clear to the dozens of millions of Indians. The tactical method of Gandhism was marked with the understanding of the traditions and psychology of the peasants. It was a method of nonviolent resistance. Gandhi’s active protest was combined with the tolerance to the enemy. Nonviolent resistance was considered as the only possible form of struggle with the colonial regime.... Prominent role in the organization of mass non-violent campaign of the resistance to the colonial regime in the 20 - 40’s of the 20th century belonged to Mahatma Gandhi. He was the one to shape India national ideology. Thanks to Gandhi, the idea of the absolute independence of India was put forward. In this respect, Indian nationalism acquired the traits of Gandhi’s ideas. Gandhism resonated with wide layers of peasantry and the urban poor, because it was connected with the social ideal of the belief that the struggle for independence from British rule, is a struggle for justice. Gandhi drew upon his appeals from cultural, historical, and religious traditions that were familiar to peasants. Therefore, the demands for independence and the transformation of the society were explained in the traditional way and were clear to the dozens of millions of Indians. The tactical method of Gandhism was marked with the understanding of the traditions and psychology of the peasants. It was a method of nonviolent resistance (boycott, peaceful marches, denial of cooperation). Gandhi’s active protest was combined with the tolerance to the enemy. Nonviolent resistance was considered as the only possible form of struggle with the colonial regime. Gandhi rejected the class struggle as a destabilizing factor the separates the nation and makes the liberation from the foreign rule impossible (Spielvogel 927). The Indian nation was united according to the primordial racial idea. So, there was a clear understanding the colonizers were the odd element in the national state structure. Gandhism tied together peasants, artisans, and the national bourgeoisie. Finally,

Features of Goodpastures Syndrome

Features of Goodpastures Syndrome Introduction Goodpastures syndrome, a rare autoimmune disease is characterized by anti-GBM (anti-glomerular basement membrane) antibodies attacking glomerular and alveolar basement membranes of the kidneys and lungs respectively. It was first reported by Dr. Ernest William Goodpasture in 1919 and first used by Stanton and Tange in 1957 in their case studies involving nine patients with the pulmonary-renal syndrome. [1, 2] Clinical Features   Ã‚   The onset of this disease ranges from the ages of 20-30 and 60-70 especially in young men in their late twenties or in men and women over sixty years of age study. [3] The diagnostic techniques involved in detection of Goodpastures syndrome include i) urine analysis that detects kidney damage by presence of high number of red blood cells or protein in the urine sample ii) blood tests showing the presence of anti-GBM antibodies iii) x-rays that can show anomalies in lung anatomy or iv) biopsies that involve imaging of a kidney tissue sample to demonstrate glomeruli characterised by crescent-shaped structures and lines of antibodies attached to the GBM. [4] While Goodpastures syndrome constitutes the representation of clinical features like rapidly progressive glomerulonephritis (RPGN) and pulmonary hemorrhage from any cause, Goodpasture disease also includes the presence of anti-GBM antibodies in addition to the other characteristics. The term anti-GBM disease constitutes a patient with the typical autoantibodies, irrespective of clinical symptoms and characteristic features. [1,5] The clinical manifestations associated with Goodpastures syndrome include acute renal failure resulting from rapidly progressive glomerulonephritis along with pulmonary hemorrhage that might prove fatal. The symptoms in relation to it consist of bleeding of lungs, kidney failure, hematuria, proteinuria, general malaise, fatigue, and weight loss. [1,6,7,8,9] The exact etiology of this syndrome is not known however there seem to be genetic and environmental risk factors. The factors being i) exposure to organic solvents or hydrocarbons ii) smoking and drugs iii) infection iv) exposure to metal particulate matter v) lymphocyte-depletion therapy. [1,5,10] The characteristic pathology in individuals experiencing the Goodpastures Syndrome can be detected by immunofluorescence staining technique of the IgG on the GBM that shows smooth diffuse linear patterns. [11] Hemodialysis, plasma exchange, cyclophosphamide drugs and immunosuppressive agents like methylprednisolone pulse therapy or oral administration of prednisolone are possible treatments for Goodpastures syndrome. [12,13,14] Basic Cellular and Molecular Mechanisms The localization of immunoglobulin IgG deposits at sites of inflammation within the pulmonary and renal basement membranes shows Goodpastures syndrome (a form of the anti-GBM disease) to be an antibody-mediated autoimmune disease. The pathogenic role of these antibodies has been confirmed by  transplantation of circulating or kidney-eluted anti-GBM antibodies to Rhesus monkey or human kidney allografts that result in the development of the disease.  A type II hypersensitivity reaction occurs when antibodies are targeted against extracellular matrix (ECM) specific antigens. [15]   The hypersensitivity response affects all organs in the body of which collagen is a constituent but the alveolar and glomerular basement membranes are more prone to the effect. This discrepancy is a result of increased accessibility of epitopes (antigen molecules facilitating attachment to a matching antibody) linked to overexpression of ÃŽÂ ±3 collagen chains in the respective basement membranes allowing access and formation of antibodies. [16] While ÃŽÂ ±3NC1 antibodies are the most common in patients with Goodpastures syndrome, ÃŽÂ ±5NC1 antibodies are less prevalent. Sometimes antineutrophil cytoplasmic antibody [ANCA] can also be present. [5,17] The disorder develops antibodies that target ÃŽÂ ±3 chain of basement membrane collagen (type IV collagen) present in alveoli in lungs and in the glomeruli that form the filtering units of the kidneys within the nephrons. These structures contain the basement membrane with collagen as its essential component that differentiates the epithelia from the underlying tissue. The conformational epitopes of the Goodpasture antigen are localized within 2 regions in the carboxyl terminal, noncollagenous (NC1) domain of a type IV collagen chain, ÃŽÂ ±3(IV)NC1. [1, 5, 18]. Upon interaction of the anti-GBM antibodies with the conformational epitope of the GBM glycoproteins, the complement pathway of the immune system gets activated. This results in infiltration by polymorphonuclear leukocytes (PMNs) and monocytes. The severely damaged GBM induces reflux of fibrinogen into the Bowman space, fibrinogen polymerizes to fibrin through the proliferation of procoagulant factors from activated mono cytes, leading to a crescent formation.[19] Goodpastures syndrome is linked with specific HLA types. Both positive (HLA-DR15) and negative (HLA-DR7) associations are defined and being used to develop an understanding of antigen presentation, tolerance and autoimmunity. [20,21,22] Recent Developments Recent developments like the plasmapheresis technique, steroidal drugs, and immunosuppressive therapy have drastically ameliorated the course of the medical condition in comparison to yesteryears, in which Goodpasture syndrome was deemed fatal. [23] Zhao et al., demonstrate the significant role of ÃŽÂ ±5NC1-specific antibodies in pathogenesis of Goodpastures disease and also re-confirm ÃŽÂ ±345 collagen IV molecule as the original GP autoantigen. [17] The invention of a drug, now patented, with its active element containing boron  that constitutes inhibitors of arginase activity has claimed remedial effects in the pathological state of Goodpastures Syndrome. [24] A recently developed, patented prophylaxis for glomerulonephritis resulting from Goodpastures syndrome comprises of administration of a therapeutically effective amount of an IL-6 antibody that binds with or regulates the expression or activity of a mammalian IL-6 polypeptide. [25] Conclusions Goodpastures Syndrome is an autoimmune disease characterized by anti-GBM antibodies attacking glomerular and alveolar basement membranes. The innate immune response comprises of (i) cell death; (ii) polymorphonuclear cell releasing neutrophils, basophils, eosinophils, antigens and monocytes to infiltrate the glomerulus. The adaptive immune response triggers the classical pathway of complement activated by antigen-antibody complex formation, and type II hypersensitivity reaction. Here antigens are targeted against cell- specific and tissue specific antigens (chiefly the connective tissue). Unanswered Questions Currently, there is a lot of research focusing on deciphering the causative agents of the harmful antibodies that lead to the development of Goodpastures syndrome. Evidence from this research can lead to novel drug discovery, eventually leading to a potential definitive cure for Goodpastures syndrome. [17] The exact the genetic determinants that constitute the etiology of Goodpastures syndrome are yet to be found. Bibliography Salama AD, Pusey CD. Goodpasture syndrome and other antiglomerular basement membrane diseases. In: Gilbert SJ, Weiner DE, eds. National Kidney Foundations Primer on Kidney Diseases. 6th ed. Philadelphia, PA: Elsevier Saunders; 2014: chap 21. Benoit, F. L., D. B. Rulon, G. B. Theil, P. D. Doolan, and R. H. Watten. Goodpastures syndrome: a clinicopathologic entity. The American journal of medicine 37, no. 3 (1964): 424-444. Hudson B, Tryggvason K, Sundaramoorthy M, Neilson E. Alport syndrome, goodpasture syndrome, and type IV Collagen. New Engl J Med 2003; 348:2543-56. Fervenza, Fernando C. Goodpasture Syndrome | NIDDK National Institute of Diabetes and Digestive and Kidney Diseases. https://www.niddk.nih.gov/health-information/kidney-disease/glomerular-diseases/goodpasture-syndrome (accessed March 1, 2017). Phelps RG, Turner AN. Anti-glomerular basement membrane disease and Goodpasture disease. In: Johnson RJ, Feehally J, Floege J, eds. Comprehensive Clinical Nephrology. 5th ed. Philadelphia, PA: Elsevier Saunders; 2015: chap 24. Lahmer T, Heemann U. Anti-glomerular basement membrane antibody disease: a rare autoimmune disorder affecting the kidney and the lung. Autoimmun Rev 2012;12:169-73. Pedchenko V, Bondar O, Fogo AB, Vanacore R, Voziyan P, Kitching AR, et al. Molecular architecture of the Goodpasture autoantigen in anti-GBM nephritis. N Engl J Med2010;363:343-54. Salant David J. Goodpastures disease new secrets revealed. N Engl J Med 2010; 363:388-91. Dammacco F, Battaglia S, Gesualdo L, Racanelli V. Goodpastures disease: a report of ten cases and a review of the literature. Autoimmun Rev 2013;12:1101-8. Jones, Joanne L., Sara AJ Thompson, Priscilla Loh, Jessica L. Davies, Orla C. Tuohy, Allison J. Curry, Laura Azzopardi et al. Human autoimmunity after lymphocyte depletion is caused by homeostatic T-cell proliferation. Proceedings of the National Academy of Sciences 110, no. 50 (2013): 20200-20205. MD, Edward. Renal Pathology http://library.med.utah.edu/WebPath/RENAHTML/RENAL093.html (accessed March 1, 2017). Greco, Antonio, Maria Ida Rizzo, Armando De Virgilio, Andrea Gallo, Massimo Fusconi, Giulio Pagliuca, Salvatore Martellucci, Rosaria Turchetta, Lucia Longo, and Marco De Vincentiis. Goodpastures syndrome: a clinical update. Autoimmunity reviews 14, no. 3 (2015): 246-253. Bolton, W. Kline. Goodpastures syndrome. Kidney international 50, no. 5 (1996): 1753-1766. Johnson, John P., Walter Whitman, William A. Briggs, and Curtis B. Wilson. Plasmapheresis and immunosuppressive agents in anti-basement25] membrane antibody-induced Goodpastures syndrome. The American journal of medicine 64, no. 2 (1978): 354-359. Rutgers A, Meyers KEC, Canziani G, Kalluri R, Lin J, Madaio MP. High affinity of anti-GBM antibodies from Goodpasture and transplanted Alport patients to 3 (IV) NC1 collagen. Kidney Int. 2000;58:115-122. Kelly, Patrick T., and Edward F. Haponik. Goodpasture syndrome: molecular and clinical advances. Medicine 73, no. 4 (1994): 171-185. Zhao J, Cui Z, Yang R, et al. Anti-glomerular basement membrane autoantibodies against different target antigens are associated with disease severity. Kidney Int 2009; 76:1108. Borza, Dorin-Bogdan, Eric G. Neilson, and Billy G. Hudson. Pathogenesis of Goodpasture syndrome: a molecular perspective. In Seminars in nephrology, vol. 23, no. 6, pp. 522-531. WB Saunders, 2003. Morita, Takashi, Yasunosuke Suzuki, and Jacob Churg. Structure and development of the glomerular crescent. The American journal of pathology 72, no. 3 (1973): 349. Phelps, Richard G., and Andrew J. Rees. The HLA complex in Goodpastures disease: a model for analyzing susceptibility to autoimmunity. Kidney international 56, no. 5 (1999): 1638-1653. Phelps, Richard G., Victoria Jones, A. Neil Turner, and Andrew J. Rees. Properties of HLA class II molecules divergently associated with Goodpastures disease. International immunology 12, no. 8 (2000): 1135-1143. Turner AN, Rees AJ. Anti-glomerular basement membrane disease (Chapter 3.11). In: Cameron SDAM, Grunfeld JP, Kerr DNS, Ritz E, eds Oxford Textbook of Nephrology, 2nd edn. Oxford University Press, Oxford, 1997 Shah MK, Hugghins SY. Characteristics and outcomes of patients with Goodpastures syndrome. South Med J 2002;95:1411-8. Van Zandt, Michael, Adam Golebiowski, Min Koo Ji, Darren Whitehouse, Todd Ryder, and Raymond Paul Beckett. Inhibitors of arginase and their therapeutic applications. U.S. Patent 9,266,908, issued February 23, 2016. Marshall, Diane, and Stevan Shaw. Method for the treatment of glomerulonephritis by administering an IL-6 antibody. U.S. Patent 9,321,837, issued April 26, 2016.

The relationship between the concentration of Sodium Thiosulphate and the reaction rate with Hydrochloric acid :: GCSE Chemistry Coursework Investigation

The relationship between the concentration of Sodium Thiosulphate and the reaction rate with Hydrochloric acid Introduction The rate of reaction is measured by finding the quantity of product made in a certain time. The rate of reaction can be made faster by an increase of temperature, adding a catalyst, increasing concentration or pressure or making the reactants surface area larger. An increase in temperature causes the particles of the reactants to gain more energy and move faster. Collision theory states that this results in more collisions and more of these collisions will have enough energy to cause a reaction. By adding a catalyst the particles will stick to the sides of the catalyst and this increases the rate of reaction. By increasing the surface area there is more surface for the particles to collide with thus causing more collisions and increasing the rate of reaction. Finally by increasing concentration there are more particles moving around in the same space. This equals more reactions, which equals faster rate of reaction. More collisions = faster rate of reaction. In this experiment I shall be investigating the relationship between the concentration of sodium thiosulphate and the rate of its reaction with Hydrochloric acid. The formula for this reaction is: Hydrochloric acid + Sodium thiosulphate>>>>>>> Sodium chloride + Sulphur oxide + Sulphur + water 2HCL(aq)+Na2S2O3(aq)>>>>>>>>>>>>NaCl(aq)+SO 2(g)+S(s)+H2O(l) The reaction is considered to be over when sufficient cloudiness is built up in the reaction flask. This cloudiness is sulphur precipitated from the reactants. This therefore is a precipitation reaction. We are investigating the effect of varying the concentration of Sodium Thiosulphate and not the other variables that can affect rates because: We do not know the right catalyst for this reaction, we are not able to break the reactants down to increase their surface area and we decided not to do temperature because we felt that the most dramatic results would come from concentration. To make this test fair all of these other variables have to remain the same. Hypothesis ---------- I believe that the less concentrated the Sodium Thiosulphate the slower the rate of reaction will be. I say this because the less the concentration the less the amount of particles in the same volume and therefore the less collisions. This means that the rate of reaction will be slower. Plan ---- To obtain our results we will be using the following equipment: Conical flask (To contain the liquids whilst the reaction is taking place.) Pipette (To move reactants and measure out precise amounts.) Goggles (As we are using acid our eyes must be shielded from harmful materials) Apron (To protect the clothes and body. Stopwatch (To count the time taken for the reaction to take place.

Dai Park Textbook

Stochastic Manufacturing & Service Systems Jim Dai and Hyunwoo Park School of Industrial and Systems Engineering Georgia Institute of Technology October 19, 2011 2 Contents 1 Newsvendor Problem 1. 1 Pro? t Maximization 1. 2 Cost Minimization . 1. 3 Initial Inventory . . 1. 4 Simulation . . . . . . 1. 5 Exercise . . . . . . . 5 5 12 15 17 19 25 25 27 29 29 31 32 33 34 39 39 40 40 42 44 46 47 48 49 51 51 51 52 54 55 57 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Queueing Theory 2. 1 Introduction . . . . . . . 2. 2 Lindley Equation . . . . 2. 3 Tra? c Intensity . . . . . 2. 4 Kingman Approximation 2. 5 Little’s Law . . . . . . . 2. 6 Throughput . . . . . . . 2. 7 Simulation . . . . . . . . 2. 8 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Discrete Time Markov Chain 3. 1 Introduction . . . . . . . . . . . . . . . . . . . . 3. 1. 1 State Space . . . . . . . . . . . . . . . . 3. 1. 2 Transition Probability Matrix . . . . . . 3. 1. 3 Initial Distribution . . . . . . . . . . . . 3. 1. 4 Markov Property . . . . . . . . . . . . . 3. 1. 5 DTMC Models . . . . . . . . . . . . . . 3. 2 Stationary Distribution . . . . . . . . . . . . . 3. 2. 1 Interpretation of Stationary Distribution 3. 2. 2 Function of Stationary Distribution . . 3. 3 Irreducibility . . . . . . . . . . . . . . . . . . . 3. 3. 1 Transition Diagram . . . . . . . . . . 3. 3. 2 Accessibility of States . . . . . . . . . . 3. 4 Periodicity . . . . . . . . . . . . . . . . . . . . . 3. 5 Recurrence and Transience . . . . . . . . . . . 3. 5. 1 Geometric Random Variable . . . . . . 3. 6 Absorption Probability . . . . . . . . . . . . . . 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3. 7 3. 8 3. 9 3. 0 Computing Stationary Distribution Using Cut Method Introduction to Binomial Stock Price Model . . . . . . Simulation . . . . . . . . . . . . . . . . . . . . . . . . . Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CONTENTS . . . . . . . . . . . . . . . . . . . . 59 61 62 63 71 71 72 73 75 78 80 80 80 82 84 91 91 96 97 100 101 103 103 104 106 107 107 108 109 111 111 117 117 130 135 148 159 4 Poisson Process 4. 1 Exponential Distribution . . . . . . . 4. 1. 1 Memoryless Property . . . . 4. 1. 2 Comparing Two Exponentials 4. 2 Homogeneous Poisson Process . . . . 4. 3 Non-homogeneous Poisson Process . 4. Thinning and Merging . . . . . . . . 4. 4. 1 Merging Poisson Process . . . 4. 4. 2 Thinning Poisson Process . . 4. 5 Simulation . . . . . . . . . . . . . . . 4. 6 Exercise . . . . . . . . . . . . . . . . 5 Continuous Time Markov Chain 5. 1 Introduction . . . . . . . . . . . 5. 1. 1 Holding Times . . . . . 5. 1. 2 Generator Matrix . . . . 5. 2 Stationary Distribution . . . . 5. 3 M/M/1 Queue . . . . . . . . . 5. 4 Variations of M/M/1 Queue . . 5. 4. 1 M/M/1/b Queue . . . . 5. 4. 2 M/M/? Queue . . . . . 5. 4. 3 M/M/k Queue . . . . . 5. 5 Open Jackson Network . . . . . 5. 5. 1 M/M/1 Queue Review . 5. 5. 2 Tandem Queue . . . . . 5. 5. Failure Inspection . . . 5. 6 Simulation . . . . . . . . . . . . 5. 7 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Exercise Answers 6. 1 Newsvendor Problem . . . . . . . 6. 2 Queueing Theory . . . . . . . . . 6. 3 Discrete Time Markov Chain . . 6. 4 Poisson Process . . . . . . . . . . 6. 5 Continuous Time Markov Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Chapter 1 Newsvendor Problem In this course, we will learn how to design, analyze, and manage a manufacturing or service system with uncertainty. Our ? rst step is to understand how to solve a single period decision problem containing uncertainty or randomness. 1. 1 Pro? t Maximization We will start with the simplest case: selling perishable items. Suppose we are running a business retailing newspaper to Georgia Tech campus. We have to order a speci? c number of copies from the publisher every evening and sell those copies the next day.One day, if there is a big news, the number of GT people who want to buy and read a paper from you may be very high. Another day, people may just not be interested in reading a paper at all. Hence, you as a retailer, will encounter the demand variability and it is the primary un certainty you need to handle to keep your business sustainable. To do that, you want to know what is the optimal number of copies you need to order every day. By intuition, you know that there will be a few other factors than demand you need to consider. †¢ Selling price (p): How much will you charge per paper? Buying price (cv ): How much will the publisher charge per paper? This is a variable cost, meaning that this cost is proportional to how many you order. That is why it is denoted by cv . †¢ Fixed ordering price (cf ): How much should you pay just to place an order? Ordering cost is ? xed regardless of how many you order. †¢ Salvage value (s) or holding cost (h): There are two cases about the leftover items. They could carry some monetary value even if expired. Otherwise, you have to pay to get rid of them or to storing them. If they have some value, it is called salvage value. If you have to pay, it is called 5 6 CHAPTER 1.NEWSVENDOR PROBLEM holding cost. Hence , the following relationship holds: s = ? h. This is per-item value. †¢ Backorder cost (b): Whenever the actual demand is higher than how many you prepared, you lose sales. Loss-of-sales could cost you something. You may be bookkeeping those as backorders or your brand may be damaged. These costs will be represented by backorder cost. This is per-item cost. †¢ Your order quantity (y): You will decide how many papers to be ordered before you start a day. That quantity is represented by y. This is your decision variable. As a business, you are assumed to want to maximize your pro? t. Expressing our pro? t as a function of these variables is the ? rst step to obtain the optimal ordering policy. Pro? t can be interpreted in two ways: (1) revenue minus cost, or (2) money you earn minus money you lose. Let us adopt the ? rst interpretation ? rst. Revenue is represented by selling price (p) multiplied by how many you actually sell. The actual sales is bounded by the realized dema nd and how many you prepared for the period. When you order too many, you can sell at most as many as the number of people who want to buy. When you order too few, you can only sell what you prepared. Hence, your revenue is minimum of D and y, i. . min(D, y) or D ? y. Thinking about the cost, ? rst of all, you have to pay something to the publisher when buying papers, i. e. cf +ycv . Two types of additional cost will be incurred to you depending on whether your order is above or below the actual demand. When it turns out you prepared less than the demand for the period, the backorder cost b per every missed sale will occur. The amount of missed sales cannot be negative, so it can be represented by max(D ? y, 0) or (D ? y)+ . When it turns out you prepared more, the quantity of left-over items also cannot go negative, so it can be expressed as max(y ? D, 0) or (y ? D)+ .In this way of thinking, we have the following formula. Pro? t =Revenue ? Cost =Revenue ? Ordering cost ? Holding c ost ? Backorder cost =p(D ? y) ? (cf + ycv ) ? h(y ? D)+ ? b(D ? y)+ (1. 1) How about the second interpretation of pro? t? You earn p ? cv dollars every time you sell a paper. For left-over items, you lose the price you bought in addition to the holding cost per paper, i. e. cv + h. When the demand is higher than what you prepared, you lose b backorder cost. Of course, you also have to pay the ? xed ordering cost cf as well when you place an order. With this logic, we have the following pro? t function. Pro? t =Earning ?Loss =(p ? cv )(D ? y) ? (cv + h)(y ? D)+ ? b(D ? y)+ ? cf (1. 2) 1. 1. PROFIT MAXIMIZATION 7 Since we used two di? erent approaches to model the same pro? t function, (1. 1) and (1. 2) should be equivalent. Comparing the two equations, you will also notice that (D ? y) + (y ? D)+ = y. Now our quest boils down to maximizing the pro? t function. However, (1. 1) and (1. 2) contain a random element, the demand D. We cannot maximize a function of random element if we all ow the randomness to remain in our objective function. One day demand can be very high. Another day it is also possible nobody wants to buy a single paper. We have to ? ure out how to get rid of this randomness from our objective function. Let us denote pro? t for the nth period by gn for further discussion. Theorem 1. 1 (Strong Law of Large Numbers). Pr g1 + g2 + g3 +  ·  ·  · + gn = E[g1 ] n>? n lim =1 The long-run average pro? t converges to the expected pro? t for a single period with probability 1. Based on Theorem 1. 1, we can change our objective function from just pro? t to expected pro? t. In other words, by maximizing the expected pro? t, it is guaranteed that the long-run average pro? t is maximized because of Theorem 1. 1. Theorem 1. 1 is the foundational assumption for the entire course.When we will talk about the long-run average something, it involves Theorem 1. 1 in most cases. Taking expectations, we obtain the following equations corresponding to (1. 1) and ( 1. 2). E[g(D, y)] =pE[D ? y] ? (cf + ycv ) ? hE[(y ? D)+ ] ? bE[(D ? y)+ ] =(p ? cv )E[D ? y] ? (cv + h)E[(y ? D)+ ] ? bE[(D ? y)+ ] ? cf (1. 4) (1. 3) Since (1. 3) and (1. 4) are equivalent, we can choose either one of them for further discussion and (1. 4) will be used. Before moving on, it is important for you to understand what E[D? y], E[(y? D)+ ], E[(D ? y)+ ] are and how to compute them. Example 1. 1. Compute E[D ? 18], E[(18 ? D)+ ], E[(D ? 8)+ ] for the demand having the following distributions. 1. D is a discrete random variable. Probability mass function (pmf) is as follows. d Pr{D = d} 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 Answer: For a discrete random variable, you ? rst compute D ? 18, (18 ? D)+ , (D ? 18)+ for each of possible D values. 8 d CHAPTER 1. NEWSVENDOR PROBLEM 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 Pr{D = d} D ? 18 (18 ? D)+ (D ? 18)+ 10 8 0 15 3 0 18 0 2 18 0 7 18 0 12 Then, you take the weighted average using corresponding Pr{D = d} for each possible D. 1 1 1 1 1 125 (10) + (15) + (18) + (18) + (18) = 4 8 8 4 4 8 1 1 1 1 1 19 + E[(18 ?D) ] = (8) + (3) + (0) + (0) + (0) = 4 8 8 4 4 8 1 1 1 1 1 + E[(D ? 18) ] = (0) + (0) + (2) + (7) + (12) = 5 4 8 8 4 4 E[D ? 18] = 2. D is a continuous random variable following uniform distribution between 10 and 30, i. e. D ? Uniform(10, 30). Answer: Computing expectation of continuous random variable involves integration. A continuous random variable has probability density function usually denoted by f . This will be also needed to compute the expectation. In this case, fD (x) = 1 20 , 0, if x ? [10, 30] otherwise Using this information, compute the expectations directly by integration. ? E[D ? 18] = ? 30 (x ? 18)fD (x)dx (x ? 18) 10 18 = = 10 18 1 dx 20 1 20 dx + 30 (x ? 18) x 10 dx + 18 30 (x ? 18) 1 20 dx 1 20 dx = = x2 40 1 20 + 18 x=18 x=10 18x 20 18 x=30 x=18 The key idea is to remove the ? operator that we cannot handle by separating the integration interval into two. The other two expectations can 1. 1. PROFIT MAXIMIZATION be computed in a similar way. 9 ? E[(18 ? D)+ ] = 30 (18 ? x)+ fD (x)dx (18 ? x)+ 10 18 = = 10 18 1 dx 20 1 20 1 20 +0 30 (18 ? x)+ (18 ? x) 10 x2 2 x=18 dx + 18 30 (18 ? x)+ 0 18 1 20 dx = dx + 1 20 dx 18x ? = 20 x=10 ? E[(D ? 18)+ ] = 30 (18 ? x)+ fD (x)dx (x ? 8)+ 10 18 = = 10 18 1 dx 20 1 20 30 (x ? 18)+ 0 10 x2 2 dx + 18 30 (x ? 18)+ 1 20 dx 1 20 dx = =0 + 1 20 dx + 18 x=30 (x ? 18) ? 18x 20 x=18 Now that we have learned how to compute E[D? y], E[(y? D)+ ], E[(D? y)+ ], we have acquired the basic toolkit to obtain the order quantity that maximizes the expected pro? t. First of all, we need to turn these expectations of the pro? t function formula (1. 4) into integration forms. For now, assume that the demand is a nonnegative continuous random variable. 10 CHAPTER 1. NEWSVENDOR PROBLEM E[g(D, y)] =(p ? cv )E[D ? y] ? (cv + h)E[(y ? D)+ ] ? bE[(D ? y)+ ] ? f ? =(p ? cv ) 0 (x ? y)fD (x)dx ? ? (cv + h) 0 ? (y ? x)+ fD (x)dx ?b 0 (x ? y)+ fD (x)dx ? cf y ? =(p ? cv ) 0 xfD (x)dx + y y yfD (x)dx ? (cv + h) 0 ? (y ? x)fD (x)dx ?b y (x ? y)fD (x)dx ? cf y y =(p ? cv ) 0 xfD (x)dx + y 1 ? 0 y y fD (x)dx xfD (x)dx ? (cv + h) y 0 y fD (x)dx ? 0 y ? b E[D] ? 0 xfD (x)dx ? y 1 ? 0 fD (x)dx ? cf (1. 5) There can be many ways to obtain the maximum point of a function. Here we will take the derivative of (1. 5) and set it to zero. y that makes the derivative equal to zero will make E[g(D, y)] either maximized or minimized depending on the second derivative.For now, assume that such y will maximize E[g(D, y)]. We will check this later. Taking the derivative of (1. 5) will involve di? erentiating an integral. Let us review an important result from Calculus. Theorem 1. 2 (Fundamental Theorem of Calculus). For a function y H(y) = c h(x)dx, we have H (y) = h(y), where c is a constant. Theorem 1. 2 can be translated as follows for our case. y d xfD (x)dx =yfD (y) dy 0 y d fD (x)dx =fD (y) dy 0 (1. 6) (1. 7) Also remember the relationship between cd f and pdf of a continuous random variable. y FD (y) = fD (x)dx (1. 8) 1. 1. PROFIT MAXIMIZATION Use (1. 6), (1. 7), (1. ) to take the derivative of (1. 5). d E[g(D, y)] =(p ? cv ) (yfD (y) + 1 ? FD (y) ? yfD (y)) dy ? (cv + h) (FD (y) + yfD (y) ? yfD (y)) ? b (? yfD (y) ? 1 + FD (y) + yfD (y)) =(p + b ? cv )(1 ? FD (y)) ? (cv + h)FD (y) =(p + b ? cv ) ? (p + b + h)FD (y) = 0 If we di? erentiate (1. 9) one more time to obtain the second derivative, d2 E[g(D, y)] = ? (p + b + h)fD (y) dy 2 11 (1. 9) which is always nonpositive because p, b, h, fD (y) ? 0. Hence, taking the derivative and setting it to zero will give us the maximum point not the minimum point. Therefore, we obtain the following result. Theorem 1. 3 (Optimal Order Quantity).The optimal order quantity y ? is the smallest y such that FD (y) = p + b ? cv ? 1 or y = FD p+b+h p + b ? cv p+b+h . for continuous demand D. Looking at Theorem 1. 3, it provides the following intuitions. †¢ Fixed cost cf does not a? ect the o ptimal quantity you need to order. †¢ If you can procure items for free and there is no holding cost, you will prepare as many as you can. †¢ If b h, b cv , you will also prepare as many as you can. †¢ If the buying cost is almost as same as the selling price plus backorder cost, i. e. cv ? p + b, you will prepare nothing. You will prepare only upon you receive an order.Example 1. 2. Suppose p = 10, cf = 100, cv = 5, h = 2, b = 3, D ? Uniform(10, 30). How many should you order for every period to maximize your long-run average pro? t? Answer: First of all, we need to compute the criterion value. p + b ? cv 10 + 3 ? 5 8 = = p+b+h 10 + 3 + 2 15 Then, we will look up the smallest y value that makes FD (y) = 8/15. 12 1 CHAPTER 1. NEWSVENDOR PROBLEM CDF 0. 5 0 0 5 10 15 20 25 30 35 40 D Therefore, we can conclude that the optimal order quantity 8 62 = units. 15 3 Although we derived the optimal order quantity solution for the continuous demand case, Theorem 1. applies to t he discrete demand case as well. I will ? ll in the derivation for discrete case later. y ? = 10 + 20 Example 1. 3. Suppose p = 10, cf = 100, cv = 5, h = 2, b = 3. Now, D is a discrete random variable having the following pmf. d Pr{D = d} 10 1 4 15 1 8 20 1 8 25 1 4 30 1 4 What is the optimal order quantity for every period? Answer: We will use the same value 8/15 from the previous example and look up the smallest y that makes FD (y) = 8/15. We start with y = 10. 1 4 1 1 3 FD (15) = + = 4 8 8 1 1 1 1 FD (20) = + + = 4 8 8 2 1 1 1 1 3 FD (25) = + + + = 4 8 8 4 4 ? Hence, the optimal order quantity y = 25 units.FD (10) = 8 15 8 < 15 8 < 15 8 ? 15 < 1. 2 Cost Minimization Suppose you are a production manager of a large company in charge of operating manufacturing lines. You are expected to run the factory to minimize the cost. Revenue is another person’s responsibility, so all you care is the cost. To model the cost of factory operation, let us set up variables in a slightly di? erent way. 1. 2. COST MINIMIZATION 13 †¢ Understock cost (cu ): It occurs when your production is not su? cient to meet the market demand. †¢ Overstock cost (co ): It occurs when you produce more than the market demand.In this case, you may have to rent a space to store the excess items. †¢ Unit production cost (cv ): It is the cost you should pay whenever you manufacture one unit of products. Material cost is one of this category. †¢ Fixed operating cost (cf ): It is the cost you should pay whenever you decide to start running the factory. As in the pro? t maximization case, the formula for cost expressed in terms of cu , co , cv , cf should be developed. Given random demand D, we have the following equation. Cost =Manufacturing Cost + Cost associated with Understock Risk + Cost associated with Overstock Risk =(cf + ycv ) + cu (D ? )+ + co (y ? D)+ (1. 10) (1. 10) obviously also contains randomness from D. We cannot minimize a random objective itself. Instead, based on Theorem 1. 1, we will minimize expected cost then the long-run average cost will be also guaranteed to be minimized. Hence, (1. 10) will be transformed into the following. E[Cost] =(cf + ycv ) + cu E[(D ? y)+ ] + co E[(y ? D)+ ] ? ? =(cf + ycv ) + cu 0 ? (x ? y)+ fD (x)dx + co 0 y (y ? x)+ fD (x)dx (y ? x)fD (x)dx (1. 11) 0 =(cf + ycv ) + cu y (x ? y)fD (x)dx + co Again, we will take the derivative of (1. 11) and set it to zero to obtain y that makes E[Cost] minimized.We will verify the second derivative is positive in this case. Let g here denote the cost function and use Theorem 1. 2 to take the derivative of integrals. d E[g(D, y)] =cv + cu (? yfD (y) ? 1 + FD (y) + yfD (y)) dy + co (FD (y) + yfD (y) ? yfD (y)) =cv + cu (FD (y) ? 1) + co FD (y) ? (1. 12) The optimal production quantity y is obtained by setting (1. 12) to be zero. Theorem 1. 4 (Optimal Production Quantity). The optimal production quantity that minimizes the long-run average cost is the smallest y such tha t FD (y) = cu ? cv or y = F ? 1 cu + co cu ? cv cu + co . 14 CHAPTER 1. NEWSVENDOR PROBLEM Theorem 1. can be also applied to discrete demand. Several intuitions can be obtained from Theorem 1. 4. †¢ Fixed cost (cf ) again does not a? ect the optimal production quantity. †¢ If understock cost (cu ) is equal to unit production cost (cv ), which makes cu ? cv = 0, then you will not produce anything. †¢ If unit production cost and overstock cost are negligible compared to understock cost, meaning cu cv , co , you will prepare as much as you can. To verify the second derivative of (1. 11) is indeed positive, take the derivative of (1. 12). d2 E[g(D, y)] = (cu + co )fD (y) dy 2 (1. 13) (1. 13) is always nonnegative because cu , co ? . Hence, y ? obtained from Theorem 1. 4 minimizes the cost instead of maximizing it. Before moving on, let us compare criteria from Theorem 1. 3 and Theorem 1. 4. p + b ? cv p+b+h and cu ? cv cu + co Since the pro? t maximization problem solved previously and the cost minimization problem solved now share the same logic, these two criteria should be somewhat equivalent. We can see the connection by matching cu = p + b, co = h. In the pro? t maximization problem, whenever you lose a sale due to underpreparation, it costs you the opportunity cost which is the selling price of an item and the backorder cost.Hence, cu = p + b makes sense. When you overprepare, you should pay the holding cost for each left-over item, so co = h also makes sense. In sum, Theorem 1. 3 and Theorem 1. 4 are indeed the same result in di? erent forms. Example 1. 4. Suppose demand follows Poisson distribution with parameter 3. The cost parameters are cu = 10, cv = 5, co = 15. Note that e? 3 ? 0. 0498. Answer: The criterion value is cu ? cv 10 ? 5 = = 0. 2, cu + co 10 + 15 so we need to ? nd the smallest y such that makes FD (y) ? 0. 2. Compute the probability of possible demands. 30 ? 3 e = 0. 0498 0! 31 Pr{D = 1} = e? 3 = 0. 1494 1! 32 ? Pr{D = 2} = e = 0. 2241 2! Pr{D = 0} = 1. 3. INITIAL INVENTORY Interpret these values into FD (y). FD (0) =Pr{D = 0} = 0. 0498 < 0. 2 FD (1) =Pr{D = 0} + Pr{D = 1} = 0. 1992 < 0. 2 FD (2) =Pr{D = 0} + Pr{D = 1} + Pr{D = 2} = 0. 4233 ? 0. 2 Hence, the optimal production quantity here is 2. 15 1. 3 Initial Inventory Now let us extend our model a bit further. As opposed to the assumption that we had no inventory at the beginning, suppose that we have m items when we decide how many we need to order. The solutions we have developed in previous sections assumed that we had no inventory when placing an order.If we had m items, we should order y ? ? m items instead of y ? items. In other words, the optimal order or production quantity is in fact the optimal order-up-to or production-up-to quantity. We had another implicit assumption that we should order, so the ? xed cost did not matter in the previous model. However, if cf is very large, meaning that starting o? a production line or placing an order i s very expensive, we may want to consider not to order. In such case, we have two scenarios: to order or not to order. We will compare the expected cost for the two scenarios and choose the option with lower expected cost.Example 1. 5. Suppose understock cost is $10, overstock cost is $2, unit purchasing cost is $4 and ? xed ordering cost is $30. In other words, cu = 10, co = 2, cv = 4, cf = 30. Assume that D ? Uniform(10, 20) and we already possess 10 items. Should we order or not? If we should, how many items should we order? Answer: First, we need to compute the optimal amount of items we need to prepare for each day. Since cu ? cv 1 10 ? 4 = , = cu + co 10 + 2 2 the optimal order-up-to quantity y ? = 15 units. Hence, if we need to order, we should order 5 = y ? ? m = 15 ? 10 items. Let us examine whether we should actually order or not. . Scenario 1: Not To Order If we decide not to order, we will not have to pay cf and cv since we order nothing actually. We just need to conside r understock and overstock risks. We will operate tomorrow with 10 items that we currently have if we decide not to order. E[Cost] =cu E[(D ? 10)+ ] + co E[(10 ? D)+ ] =10(E[D] ? 10) + 2(0) = $50 16 CHAPTER 1. NEWSVENDOR PROBLEM Note that in this case E[(10 ? D)+ ] = 0 because D is always greater than 10. 2. Scenario 2: To Order If we decide to order, we will order 5 items. We should pay cf and cv accordingly. Understock and overstock risks also exist in this case.Since we will order 5 items to lift up the inventory level to 15, we will run tomorrow with 15 items instead of 10 items if we decide to order. E[Cost] =cf + (15 ? 10)cv + cu E[(D ? 15)+ ] + co E[(15 ? D)+ ] =30 + 20 + 10(1. 25) + 2(1. 25) = $65 Since the expected cost of not ordering is lower than that of ordering, we should not order if we already have 10 items. It is obvious that if we have y ? items at hands right now, we should order nothing since we already possess the optimal amount of items for tomorrow’s op eration. It is also obvious that if we have nothing currently, we should order y ? items to prepare y ? tems for tomorrow. There should be a point between 0 and y ? where you are indi? erent between order and not ordering. Suppose you as a manager should give instruction to your assistant on when he/she should place an order and when should not. Instead of providing instructions for every possible current inventory level, it is easier to give your assistant just one number that separates the decision. Let us call that number the critical level of current inventory m? . If we have more than m? items at hands, the expected cost of not ordering will be lower than the expected cost of ordering, so we should not order.Conversely, if we have less than m? items currently, we should order. Therefore, when we have exactly m? items at hands right now, the expected cost of ordering should be equal to that of not ordering. We will use this intuition to obtain m? value. The decision process is s ummarized in the following ? gure. m* Critical level for placing an order y* Optimal order-up-to quantity Inventory If your current inventory lies here, you should order. Order up to y*. If your current inventory lies here, you should NOT order because your inventory is over m*. 1. 4. SIMULATION 17 Example 1. 6.Given the same settings with the previous example (cu = 10, cv = 4, co = 2, cf = 30), what is the critical level of current inventory m? that determines whether you should order or not? Answer: From the answer of the previous example, we can infer that the critical value should be less than 10, i. e. 0 < m? < 10. Suppose we currently own m? items. Now, evaluate the expected costs of the two scenarios: ordering and not ordering. 1. Scenario 1: Not Ordering E[Cost] =cu E[(D ? m? )+ ] + co E[(m? ? D)+ ] =10(E[D] ? m? ) + 2(0) = 150 ? 10m? 2. Scenario 2: Ordering In this case, we will order.Given that we will order, we will order y ? ?m? = 15 ? m? items. Therefore, we will start tomorrow with 15 items. E[Cost] =cf + (15 ? 10)cv + cu E[(D ? 15)+ ] + co E[(15 ? D)+ ] =30 + 4(15 ? m? ) + 10(1. 25) + 2(1. 25) = 105 ? 4m? At m? , (1. 14) and (1. 15) should be equal. 150 ? 10m? = 105 ? 4m? ? m? = 7. 5 units (1. 15) (1. 14) The critical value is 7. 5 units. If your current inventory is below 7. 5, you should order for tomorrow. If the current inventory is above 7. 5, you should not order. 1. 4 Simulation Generate 100 random demands from Uniform(10, 30). p = 10, cf = 30, cv = 4, h = 5, b = 3 1 p + b ? v 10 + 3 ? 4 = = p + b + h 10 + 3 + 5 2 The optimal order-up-to quantity from Theorem 1. 3 is 20. We will compare the performance between the policies of y = 15, 20, 25. Listing 1. 1: Continuous Uniform Demand Simulation # Set up parameters p=10;cf=30;cv=4;h=5;b=3 # How many random demands will be generated? n=100 # Generate n random demands from the uniform distribution 18 Dmd=runif(n,min=10,max=30) CHAPTER 1. NEWSVENDOR PROBLEM # Test the policy where we order 15 it ems for every period y=15 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 33. 4218 # Test the policy where we order 20 items for every period y=20 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 44. 37095 # Test the policy where we order 25 items for every period y=25 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 32. 62382 You can see the policy with y = 20 maximizes the 100-period average pro? t as promised by the theory. In fact, if n is relatively small, it is not guaranteed that we have maximized pro? t even if we run based on the optimal policy obtained from this section.The underlying assumption is that we should operate with this policy for a long time. Then, Theorem 1. 1 guarantees that the average pro? t will be maximized when we use the optimal ordering policy. Discrete demand case can also be simulated. Suppose the demand has the following distribution. All other parameters remain same. d Pr{D = d} 10 1 4 15 1 8 20 1 4 25 1 8 30 1 4 The theoretic optimal order-up-to quantity in this case is also 20. Let us test three policies: y = 15, 20, 25. Listing 1. 2: Discrete Demand Simulation # Set up parameters p=10;cf=30;cv=4;h=5;b=3 # How many random demands will be generated? =100 # Generate n random demands from the discrete demand distribution Dmd=sample(c(10,15,20,25,30),n,replace=TRUE,c(1/4,1/8,1/4,1/8,1/4)) # Test the policy where we order 15 items for every period y=15 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 19. 35 # Test the policy where we order 20 items for every period y=20 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 31. 05 # Test the policy where we order 25 items for every period 1. 5. EXERCISE y=25 mean(p*pmin(Dmd,y)-cf-y*cv-h*pmax(y-Dmd,0)-b*pmax(Dmd-y,0)) > 26. 55 19There are other distributions such as triangular, normal, Poisson or binomial distributions available in R. When you do your senior project, for example, you will observe the demand for a departm ent or a factory. You ? rst approximate the demand using these theoretically established distributions. Then, you can simulate the performance of possible operation policies. 1. 5 Exercise 1. Show that (D ? y) + (y ? D)+ = y. 2. Let D be a discrete random variable with the following pmf. d Pr{D = d} Find (a) E[min(D, 7)] (b) E[(7 ? D)+ ] where x+ = max(x, 0). 3. Let D be a Poisson random variable with parameter 3.Find (a) E[min(D, 2)] (b) E[(3 ? D)+ ]. Note that pmf of a Poisson random variable with parameter ? is Pr{D = k} = ? k e . k! 5 1 10 6 3 10 7 4 10 8 1 10 9 1 10 4. Let D be a continuous random variable and uniformly distributed between 5 and 10. Find (a) E[max(D, 8)] (b) E[(D ? 8)? ] where x? = min(x, 0). 5. Let D be an exponential random variable with parameter 7. Find (a) E[max(D, 3)] 20 (b) E[(D ? 4)? ]. CHAPTER 1. NEWSVENDOR PROBLEM Note that pdf of an exponential random variable with parameter ? is fD (x) = ? e x for x ? 0. 6. David buys fruits and vegetables wholesal e and retails them at Davids Produce on La Vista Road.One of the more di? cult decisions is the amount of bananas to buy. Let us make some simplifying assumptions, and assume that David purchases bananas once a week at 10 cents per pound and retails them at 30 cents per pound during the week. Bananas that are more than a week old are too ripe and are sold for 5 cents per pound. (a) Suppose the demand for the good bananas follows the same distribution as D given in Problem 2. What is the expected pro? t of David in a week if he buys 7 pounds of banana? (b) Now assume that the demand for the good bananas is uniformly distributed between 5 and 10 like in Problem 4.What is the expected pro? t of David in a week if he buys 7 pounds of banana? (c) Find the expected pro? t if David’s demand for the good bananas follows an exponential distribution with mean 7 and if he buys 7 pounds of banana. 7. Suppose we are selling lemonade during a football game. The lemonade sells for $18 per g allon but only costs $3 per gallon to make. If we run out of lemonade during the game, it will be impossible to get more. On the other hand, leftover lemonade has a value of $1. Assume that we believe the fans would buy 10 gallons with probability 0. 1, 11 gallons with probability 0. , 12 gallons with probability 0. 4, 13 gallons with probability 0. 2, and 14 gallons with probability 0. 1. (a) What is the mean demand? (b) If 11 gallons are prepared, what is the expected pro? t? (c) What is the best amount of lemonade to order before the game? (d) Instead, suppose that the demand was normally distributed with mean 1000 gallons and variance 200 gallons2 . How much lemonade should be ordered? 8. Suppose that a bakery specializes in chocolate cakes. Assume the cakes retail at $20 per cake, but it takes $10 to prepare each cake. Cakes cannot be sold after one week, and they have a negligible salvage value.It is estimated that the weekly demand for cakes is: 15 cakes in 5% of the weeks, 1 6 cakes in 20% of the weeks, 17 cakes in 30% of the weeks, 18 cakes in 25% of the weeks, 19 cakes in 10% of the weeks, and 20 cakes in 10% of the weeks. How many cakes should the bakery prepare each week? What is the bakery’s expected optimal weekly pro? t? 1. 5. EXERCISE 21 9. A camera store specializes in a particular popular and fancy camera. Assume that these cameras become obsolete at the end of the month. They guarantee that if they are out of stock, they will special-order the camera and promise delivery the next day.In fact, what the store does is to purchase the camera from an out of state retailer and have it delivered through an express service. Thus, when the store is out of stock, they actually lose the sales price of the camera and the shipping charge, but they maintain their good reputation. The retail price of the camera is $600, and the special delivery charge adds another $50 to the cost. At the end of each month, there is an inventory holding cost of $25 fo r each camera in stock (for doing inventory etc). Wholesale cost for the store to purchase the cameras is $480 each. (Assume that the order can only be made at the beginning of the month. (a) Assume that the demand has a discrete uniform distribution from 10 to 15 cameras a month (inclusive). If 12 cameras are ordered at the beginning of a month, what are the expected overstock cost and the expected understock or shortage cost? What is the expected total cost? (b) What is optimal number of cameras to order to minimize the expected total cost? (c) Assume that the demand can be approximated by a normal distribution with mean 1000 and standard deviation 100 cameras a month. What is the optimal number of cameras to order to minimize the expected total cost? 10.Next month’s production at a manufacturing company will use a certain solvent for part of its production process. Assume that there is an ordering cost of $1,000 incurred whenever an order for the solvent is placed and the solvent costs $40 per liter. Due to short product life cycle, unused solvent cannot be used in following months. There will be a $10 disposal charge for each liter of solvent left over at the end of the month. If there is a shortage of solvent, the production process is seriously disrupted at a cost of $100 per liter short. Assume that the initial inventory level is m, where m = 0, 100, 300, 500 and 700 liters. a) What is the optimal ordering quantity for each case when the demand is discrete with Pr{D = 500} = Pr{D = 800} = 1/8, Pr{D = 600} = 1/2 and Pr{D = 700} = 1/4? (b) What is the optimal ordering policy for arbitrary initial inventory level m? (You need to specify the critical value m? in addition to the optimal order-up-to quantity y ? . When m ? m? , you make an order. Otherwise, do not order. ) (c) Assume optimal quantity will be ordered. What is the total expected cost when the initial inventory m = 0? What is the total expected cost when the initial inventory m = 700? 22 CHAPTER 1. NEWSVENDOR PROBLEM 11.Redo Problem 10 for the case where the demand is governed by the continuous uniform distribution varying between 400 and 800 liters. 12. An automotive company will make one last production run of parts for Part 947A and 947B, which are not interchangeable. These parts are no longer used in new cars, but will be needed as replacements for warranty work in existing cars. The demand during the warranty period for 947A is approximately normally distributed with mean 1,500,000 parts and standard deviation 500,000 parts, while the mean and standard deviation for 947B is 500,000 parts and 100,000 parts. (Assume that two demands are independent. Ignoring the cost of setting up for producing the part, each part costs only 10 cents to produce. However, if additional parts are needed beyond what has been produced, they will be purchased at 90 cents per part (the same price for which the automotive company sells its parts). Parts remaining at the end of the warr anty period have a salvage value of 8 cents per part. There has been a proposal to produce Part 947C, which can be used to replace either of the other two parts. The unit cost of 947C jumps from 10 to 14 cents, but all other costs remain the same. (a) Assuming 947C is not produced, how many 947A should be produced? b) Assuming 947C is not produced, how many 947B should be produced? (c) How many 947C should be produced in order to satisfy the same fraction of demand from parts produced in-house as in the ? rst two parts of this problem. (d) How much money would be saved or lost by producing 947C, but meeting the same fraction of demand in-house? (e) Is your answer to question (c), the optimal number of 947C to produce? If not, what would be the optimal number of 947C to produce? (f) Should the more expensive part 947C be produced instead of the two existing parts 947A and 947B. Why? Hint: compare the expected total costs.Also, suppose that D ? Normal( µ, ? 2 ). q xv 0 (x?  µ)2 1 e? 2? 2 dx = 2 q (x ?  µ) v 0 q (x?  µ)2 1 e? 2? 2 dx 2 + µ =  µ2 v 0 (q?  µ)2 (x?  µ)2 1 e? 2? 2 dx 2 t 1 v e? 2? 2 dt +  µPr{0 ? D ? q} 2 2 where, in the 2nd step, we changed variable by letting t = (x ?  µ)2 . 1. 5. EXERCISE 23 13. A warranty department manages the after-sale service for a critical part of a product. The department has an obligation to replace any damaged parts in the next 6 months. The number of damaged parts X in the next 6 months is assumed to be a random variable that follows the following distribution: x Pr{X = x} 100 . 1 200 . 2 300 . 5 400 . 2The department currently has 200 parts in stock. The department needs to decide if it should make one last production run for the part to be used for the next 6 months. To start the production run, the ? xed cost is $2000. The unit cost to produce a part is $50. During the warranty period of next 6 months, if a replacement request comes and the department does not have a part available in house, it has to buy a part from the spot-market at the cost of $100 per part. Any part left at the end of 6 month sells at $10. (There is no holding cost. ) Should the department make the production run? If so, how many items should it produce? 4. A store sells a particular brand of fresh juice. By the end of the day, any unsold juice is sold at a discounted price of $2 per gallon. The store gets the juice daily from a local producer at the cost of $5 per gallon, and it sells the juice at $10 per gallon. Assume that the daily demand for the juice is uniformly distributed between 50 gallons to 150 gallons. (a) What is the optimal number of gallons that the store should order from the distribution each day in order to maximize the expected pro? t each day? (b) If 100 gallons are ordered, what is the expected pro? t per day? 15. An auto company is to make one ? al purchase of a rare engine oil to ful? ll its warranty services for certain car models. The current price for the engine oil is $1 per g allon. If the company runs out the oil during the warranty period, it will purchase the oil from a supply at the market price of $4 per gallon. Any leftover engine oil after the warranty period is useless, and costs $1 per gallon to get rid of. Assume the engine oil demand during the warranty is uniformly distributed (continuous distribution) between 1 million gallons to 2 million gallons, and that the company currently has half million gallons of engine oil in stock (free of charge). a) What is the optimal amount of engine oil the company should purchase now in order to minimize the total expected cost? (b) If 1 million gallons are purchased now, what is the total expected cost? 24 CHAPTER 1. NEWSVENDOR PROBLEM 16. A company is obligated to provide warranty service for Product A to its customers next year. The warranty demand for the product follows the following distribution. d Pr{D = d} 100 . 2 200 . 4 300 . 3 400 . 1 The company needs to make one production run to satisfy the wa rranty demand for entire next year. Each unit costs $100 to produce; the penalty cost of a unit is $500.By the end of the year, the savage value of each unit is $50. (a) Suppose that the company has currently 0 units. What is the optimal quantity to produce in order to minimize the expected total cost? Find the optimal expected total cost. (b) Suppose that the company has currently 100 units at no cost and there is $20000 ? xed cost to start the production run. What is the optimal quantity to produce in order to minimize the expected total cost? Find the optimal expected total cost. 17. Suppose you are running a restaurant having only one menu, lettuce salad, in the Tech Square.You should order lettuce every day 10pm after closing. Then, your supplier delivers the ordered amount of lettuce 5am next morning. Store hours is from 11am to 9pm every day. The demand for the lettuce salad for a day (11am-9pm) has the following distribution. d Pr{D = d} 20 1/6 25 1/3 30 1/3 35 1/6 One lettu ce salad requires two units of lettuce. The selling price of lettuce salad is $6, the buying price of one unit of lettuce is $1. Of course, leftover lettuce of a day cannot be used for future salad and you have to pay 50 cents per unit of lettuce for disposal. (a) What is the optimal order-up-to quantity of lettuce for a day? b) If you ordered 50 units of lettuce today, what is the expected pro? t of tomorrow? Include the purchasing cost of 50 units of lettuce in your calculation. Chapter 2 Queueing Theory Before getting into Discrete-time Markov Chains, we will learn about general issues in the queueing theory. Queueing theory deals with a set of systems having waiting space. It is a very powerful tool that can model a broad range of issues. Starting from analyzing a simple queue, a set of queues connected with each other will be covered as well in the end. This chapter will give you the background knowledge when you read the required book, The Goal.We will revisit the queueing the ory once we have more advanced modeling techniques and knowledge. 2. 1 Introduction Think about a service system. All of you must have experienced waiting in a service system. One example would be the Student Center or some restaurants. This is a human system. A bit more automated service system that has a queue would be a call center and automated answering machines. We can imagine a manufacturing system instead of a service system. These waiting systems can be generalized as a set of bu? ers and servers. There are key factors when you try to model such a system.What would you need to analyze your system? †¢ How frequently customers come to your system? > Inter-arrival Times †¢ How fast your servers can serve the customers? > Service Times †¢ How many servers do you have? > Number of Servers †¢ How large is your waiting space? > Queue Size If you can collect data about these metrics, you can characterize your queueing system. In general, a queueing system can be denoted as follows. G/G/s/k 25 26 CHAPTER 2. QUEUEING THEORY The ? rst letter characterizes the distribution of inter-arrival times. The second letter characterizes the distribution of service times.The third number denotes the number of servers of your queueing system. The fourth number denotes the total capacity of your system. The fourth number can be omitted and in such case it means that your capacity is in? nite, i. e. your system can contain any number of people in it up to in? nity. The letter â€Å"G† represents a general distribution. Other candidate characters for this position is â€Å"M† and â€Å"D† and the meanings are as follows. †¢ G: General Distribution †¢ M: Exponential Distribution †¢ D: Deterministic Distribution (or constant) The number of servers can vary from one to many to in? nity.The size of bu? er can also be either ? nite or in? nite. To simplify the model, assume that there is only a single server and we have in? ni te bu? er. By in? nite bu? er, it means that space is so spacious that it is as if the limit does not exist. Now we set up the model for our queueing system. In terms of analysis, what are we interested in? What would be the performance measures of such systems that you as a manager should know? †¢ How long should your customer wait in line on average? †¢ How long is the waiting line on average? There are two concepts of average. One is average over time.This applies to the average number of customers in the system or in the queue. The other is average over people. This applies to the average waiting time per customer. You should be able to distinguish these two. Example 2. 1. Assume that the system is empty at t = 0. Assume that u1 = 1, u2 = 3, u3 = 2, u4 = 3, v1 = 4, v2 = 2, v3 = 1, v4 = 2. (ui is ith customer’s inter-arrival time and vi is ith customer’s service time. ) 1. What is the average number of customers in the system during the ? rst 10 minutes? 2 . What is the average queue size during the ? rst 10 minutes? 3.What is the average waiting time per customer for the ? rst 4 customers? Answer: 1. If we draw the number of people in the system at time t with respect to t, it will be as follows. 2. 2. LINDLEY EQUATION 3 2 1 0 27 Z(t) 0 1 2 3 4 5 6 7 8 9 10 t E[Z(t)]t? [0,10] = 1 10 10 Z(t)dt = 0 1 (10) = 1 10 2. If we draw the number of people in the queue at time t with respect to t, it will be as follows. 3 2 1 0 Q(t) 0 1 2 3 4 5 6 7 8 9 10 t E[Q(t)]t? [0,10] = 1 10 10 Q(t)dt = 0 1 (2) = 0. 2 10 3. We ? rst need to compute waiting times for each of 4 customers. Since the ? rst customer does not wait, w1 = 0.Since the second customer arrives at time 4, while the ? rst customer’s service ends at time 5. So, the second customer has to wait 1 minute, w2 = 1. Using the similar logic, w3 = 1, w4 = 0. E[W ] = 0+1+1+0 = 0. 5 min 4 2. 2 Lindley Equation From the previous example, we now should be able to compute each customerâ€℠¢s waiting time given ui , vi . It requires too much e? ort if we have to draw graphs every time we need to compute wi . Let us generalize the logic behind calculating waiting times for each customer. Let us determine (i + 1)th customer’s waiting 28 CHAPTER 2. QUEUEING THEORY time.If (i + 1)th customer arrives after all the time ith customer waited and got served, (i + 1)th customer does not have to wait. Its waiting time is 0. Otherwise, it has to wait wi + vi ? ui+1 . Figure 2. 1, and Figure 2. 2 explain the two cases. ui+1 wi vi wi+1 Time i th arrival i th service start (i+1)th arrival i th service end Figure 2. 1: (i + 1)th arrival before ith service completion. (i + 1)th waiting time is wi + vi ? ui+1 . ui+1 wi vi Time i th arrival i th service start i th service end (i+1)th arrival Figure 2. 2: (i + 1)th arrival after ith service completion. (i + 1)th waiting time is 0.Simply put, wi+1 = (wi + vi ? ui+1 )+ . This is called the Lindley Equation. Example 2. 2. Given the f ollowing inter-arrival times and service times of ? rst 10 customers, compute waiting times and system times (time spent in the system including waiting time and service time) for each customer. ui = 3, 2, 5, 1, 2, 4, 1, 5, 3, 2 vi = 4, 3, 2, 5, 2, 2, 1, 4, 2, 3 Answer: Note that system time can be obtained by adding waiting time and service time. Denote the system time of ith customer by zi . ui vi wi zi 3 4 0 4 2 3 2 5 5 2 0 2 1 5 1 6 2 2 4 6 4 2 2 4 1 1 3 4 5 4 0 4 3 2 1 3 2 3 1 4 2. 3. TRAFFIC INTENSITY 9 2. 3 Suppose Tra? c Intensity E[ui ] =mean inter-arrival time = 2 min E[vi ] =mean service time = 4 min. Is this queueing system stable? By stable, it means that the queue size should not go to the in? nity. Intuitively, this queueing system will not last because average service time is greater than average inter-arrival time so your system will soon explode. What was the logic behind this judgement? It was basically comparing the average inter-arrival time and the average serv ice time. To simplify the judgement, we come up with a new quantity called the tra? c intensity. De? nition 2. 1 (Tra? Intensity). Tra? c intensity ? is de? ned to be ? = 1/E[ui ] ? =  µ 1/E[vi ] where ? is the arrival rate and  µ is the service rate. Given a tra? c intensity, it will fall into one of the following three categories. †¢ If ? < 1, the system is stable. †¢ If ? = 1, the system is unstable unless both inter-arrival times and service times are deterministic (constant). †¢ If ? > 1, the system is unstable. Then, why don’t we call ? utilization instead of tra? c intensity? Utilization seems to be more intuitive and user-friendly name. In fact, utilization just happens to be same as ? if ? < 1.However, the problem arises if ? > 1 because utilization cannot go over 100%. Utilization is bounded above by 1 and that is why tra? c intensity is regarded more general notation to compare arrival and service rates. De? nition 2. 2 (Utilization). Utilization is de? ned as follows. Utilization = ? , 1, if ? < 1 if ? ? 1 Utilization can also be interpreted as the long-run fraction of time the server is utilized. 2. 4 Kingman Approximation Formula Theorem 2. 1 (Kingman’s High-tra? c Approximation Formula). Assume the tra? c intensity ? < 1 and ? is close to 1. The long-run average waiting time in 0 a queue E[W ] ? E[vi ] CHAPTER 2. QUEUEING THEORY ? 1 c2 + c2 a s 2 where c2 , c2 are squared coe? cient of variation of inter-arrival times and service a s times de? ned as follows. c2 = a Var[u1 ] (E[u1 ]) 2, c2 = s Var[v1 ] (E[v1 ]) 2 Example 2. 3. 1. Suppose inter-arrival time follows an exponential distribution with mean time 3 minutes and service time follows an exponential distribution with mean time 2 minutes. What is the expected waiting time per customer? 2. Suppose inter-arrival time is constant 3 minutes and service time is also constant 2 minutes. What is the expected waiting time per customer?Answer: 1. Tra? c intensity is ? = 1/E[ui ] 1/3 2 ? = = = .  µ 1/E[vi ] 1/2 3 Since both inter-arrival times and service times are exponentially distributed, E[ui ] = 3, Var[ui ] = 32 = 9, E[vi ] = 2, Var[vi ] = 22 = 4. Therefore, c2 = Var[ui ]/(E[ui ])2 = 1, c2 = 1. Hence, s a E[W ] =E[vi ] =2 ? c2 + c2 s a 1 2 2/3 1+1 = 4 minutes. 1/3 2 2. Tra? c intensity remains same, 2/3. However, since both inter-arrival times and service times are constant, their variances are 0. Thus, c2 = a c2 = 0. s E[W ] = 2 2/3 1/3 0+0 2 = 0 minutes It means that none of the customers will wait upon their arrival.As shown in the previous example, when the distributions for both interarrival times and service times are exponential, the squared coe? cient of variation term becomes 1 from the Kingman’s approximation formula and the formula 2. 5. LITTLE’S LAW 31 becomes exact to compute the average waiting time per customer for M/M/1 queue. E[W ] =E[vi ] ? 1 Also note that if inter-arrival time or service time distribution is deterministic, c2 or c2 becomes 0. a s Example 2. 4. You are running a highway collecting money at the entering toll gate. You reduced the utilization level of the highway from 90% to 80% by adopting car pool lane.How much does the average waiting time in front of the toll gate decrease? Answer: 0. 8 0. 9 = 9, =4 1 ? 0. 9 1 ? 0. 8 The average waiting time in in front of the toll gate is reduced by more than a half. The Goal is about identifying bottlenecks in a plant. When you become a manager of a company and are running a expensive machine, you usually want to run it all the time with full utilization. However, the implication of Kingman formula tells you that as your utilization approaches to 100%, the waiting time will be skyrocketing. It means that if there is any uncertainty or random ? ctuation input to your system, your system will greatly su? er. In lower ? region, increasing ? is not that bad. If ? near 1, increasing utilization a little bit can lead to a disaster. Atl anta, 10 years ago, did not su? er that much of tra? c problem. As its tra? c infrastructure capacity is getting closer to the demand, it is getting more and more fragile to uncertainty. A lot of strategies presented in The Goal is in fact to decrease ?. You can do various things to reduce ? of your system by outsourcing some process, etc. You can also strategically manage or balance the load on di? erent parts of your system.You may want to utilize customer service organization 95% of time, while utilization of sales people is 10%. 2. 5 Little’s Law L = ? W The Little’s Law is much more general than G/G/1 queue. It can be applied to any black box with de? nite boundary. The Georgia Tech campus can be one black box. ISyE building itself can be another. In G/G/1 queue, we can easily get average size of queue or service time or time in system as we di? erently draw box onto the queueing system. The following example shows that Little’s law can be applied in broade r context than the queueing theory. 32 CHAPTER 2. QUEUEING THEORY Example 2. 5 (Merge of I-75 and I-85).Atlanta is the place where two interstate highways, I-75 and I-85, merge and cross each other. As a tra? c manager of Atlanta, you would like to estimate the average time it takes to drive from the north con? uence point to the south con? uence point. On average, 100 cars per minute enter the merged area from I-75 and 200 cars per minute enter the same area from I-85. You also dispatched a chopper to take a aerial snapshot of the merged area and counted how many cars are in the area. It turned out that on average 3000 cars are within the merged area. What is the average time between entering and exiting the area per vehicle?Answer: L =3000 cars ? =100 + 200 = 300 cars/min 3000 L = 10 minutes ? W = = ? 300 2. 6 Throughput Another focus of The Goal is set on the throughput of a system. Throughput is de? ned as follows. De? nition 2. 3 (Throughput). Throughput is the rate of output ? ow from a system. If ? ? 1, throughput= ?. If ? > 1, throughput=  µ. The bounding constraint of throughput is either arrival rate or service rate depending on the tra? c intensity. Example 2. 6 (Tandem queue with two stations). Suppose your factory production line has two stations linked in series. Every raw material coming into your line should be processed by Station A ? rst.Once it is processed by Station A, it goes to Station B for ? nishing. Suppose raw material is coming into your line at 15 units per minute. Station A can process 20 units per minute and Station B can process 25 units per minute. 1. What is the throughput of the entire system? 2. If we double the arrival rate of raw material from 15 to 30 units per minute, what is the throughput of the whole system? Answer: 1. First, obtain the tra? c intensity for Station A. ?A = ? 15 = = 0. 75  µA 20 Since ? A < 1, the throughput of Station A is ? = 15 units per minute. Since Station A and Station B is linked in series, the throughput of Station . 7. SIMULATION A becomes the arrival rate for Station B. ?B = ? 15 = = 0. 6  µB 25 33 Also, ? B < 1, the throughput of Station B is ? = 15 units per minute. Since Station B is the ? nal stage of the entire system, the throughput of the entire system is also ? = 15 units per minute. 2. Repeat the same steps. ?A = 30 ? = = 1. 5  µA 20 Since ? A > 1, the throughput of Station A is  µA = 20 units per minute, which in turn becomes the arrival rate for Station B. ?B =  µA 20 = 0. 8 =  µB 25 ?B < 1, so the throughput of Station B is  µA = 20 units per minute, which in turn is the throughput of the whole system. 2. 7 SimulationListing 2. 1: Simulation of a Simple Queue and Lindley Equation N = 100 # Function for Lindley Equation lindley = function(u,v){ for (i in 1:length(u)) { if(i==1) w = 0 else { w = append(w, max(w[i-1]+v[i-1]-u[i], 0)) } } return(w) } # # u v CASE 1: Discrete Distribution Generate N inter-arrival times and service times = sample( c(2,3,4),N,replace=TRUE,c(1/3,1/3,1/3)) = sample(c(1,2,3),N,replace=TRUE,c(1/3,1/3,1/3)) # Compute waiting time for each customer w = lindley(u,v) w # CASE 2: Deterministic Distribution # All inter-arrival times are 3 minutes and all service times are 2 minutes # Observe that nobody waits in this case. 4 u = rep(3, 100) v = rep(2, 100) w = lindley(u,v) w CHAPTER 2. QUEUEING THEORY The Kingman’s approximation formula is exact when inter-arrival times and service times follow iid exponential distribution. E[W ] = 1  µ ? 1 We can con? rm this equation by simulating an M/M/1 queue. Listing 2. 2: Kingman Approximation # lambda = arrival rate, mu = service rate N = 10000; lambda = 1/10; mu = 1/7 # Generate N inter-arrival times and service times from exponential distribution u = rexp(N,rate=lambda) v = rexp(N,rate=mu) # Compute the average waiting time of each customer w = lindley(u,v) mean(w) > 16. 0720 # Compare with Kingman approximation rho = lambda/mu (1/mu)*(rho/(1-rho)) > 16. 33333 The Kingman’s approximation formula becomes more and more accurate as N grows. 2. 8 Exercise 1. Let Y be a random variable with p. d. f. ce? 3s for s ? 0, where c is a constant. (a) Determine c. (b) What is the mean, variance, and squared coe? cient of variation of Y where the squared coe? cient of variation of Y is de? ned to be Var[Y ]/(E[Y ]2 )? 2. Consider a single server queue. Initially, there is no customer in the system.Suppose that the inter-arrival times of the ? rst 15 customers are: 2, 5, 7, 3, 1, 4, 9, 3, 10, 8, 3, 2, 16, 1, 8 2. 8. EXERCISE 35 In other words, the ? rst customer will arrive at t = 2 minutes, and the second will arrive at t = 2 + 5 minutes, and so on. Also, suppose that the service time of the ? rst 15 customers are 1, 4, 2, 8, 3, 7, 5, 2, 6, 11, 9, 2, 1, 7, 6 (a) Compute the average waiting time (the time customer spend in bu? er) of the ? rst 10 departed customers. (b) Compute the average system time (waiting time plus service time) of the ? st 10 departed customers. (c) Compute the average queue size during the ? rst 20 minutes. (d) Compute the average server utilization during the ? rst 20 minutes. (e) Does the Little’s law of hold for the average queue size in the ? rst 20 minutes? 3. We want to decide whether to employ a human operator or buy a machine to paint steel beams with a rust inhibitor. Steel beams are produced at a constant rate of one every 14 minutes. A skilled human operator takes an average time of 700 seconds to paint a steel beam, with a standard deviation of 300 seconds.An automatic painter takes on average 40 seconds more than the human painter to paint a beam, but with a standard deviation of only 150 seconds. Estimate the expected waiting time in queue of a steel beam for each of the operators, as well as the expected number of steel beams waiting in queue in each of the two cases. Comment on the e? ect of variability in service time. 4. The arrival rate of customers to an ATM machi ne is 30 per hour with exponentially distirbuted in- terarrival times. The transaction times of two customers are independent and identically distributed.Each transaction time (in minutes) is distributed according to the following pdf: f (s) = where ? = 2/3. (a) What is the average waiting for each customer? (b) What is the average number of customers waiting in line? (c) What is the average number of customers at the site? 5. A production line has two machines, Machine A and Machine B, that are arranged in series. Each job needs to processed by Machine A ? rst. Once it ? nishes the processing by Machine A, it moves to the next station, to be processed by Machine B. Once it ? nishes the processing by Machine B, it leaves the production line.Each machine can process one job at a time. An arriving job that ? nds the machine busy waits in a bu? er. 4? 2 se? 2? s , 0, if s ? 0 otherwise 36 CHAPTER 2. QUEUEING THEORY (The bu? er sizes are assumed to be in? nite. ) The processing times fo r Machine A are iid having exponential distribution with mean 4 minutes. The processing times for Machine B are iid with mean 2 minutes. Assume that the inter-arrival times of jobs arriving at the production line are iid, having exponential distribution with mean of 5 minutes. (a) What is the utilization of Machine A?What is the utilization of Machine B? (b) What is the throughput of the production system? (Throughput is de? ned to be the rate of ? nal output ? ow, i. e. how many items will exit the system in a unit time. ) (c) What is the average waiting time at Machine A, excluding the service time? (d) It is known the average time in the entire production line is 30 minutes per job. What is the long-run average number of jobs in the entire production line? (e) Suppose that the mean inter-arrival time is changed to 1 minute. What are the utilizations for Machine A and Machine B, respectively?What is the throughput of the production system? 6. An auto collision shop has roughly 10 cars arriving per week for repairs. A car waits outside until it is brought inside for bumping. After bumping, the car is painted. On the average, there are 15 cars waiting outside in the yard to be repaired, 10 cars inside in the bump area, and 5 cars inside in the painting area. What is the average length of time a car is in the yard, in the bump area, and in the painting area? What is the average length of time from when a car arrives until it leaves? 7. A small bank is sta? d by a single server. It has been observed that, during a normal business day, the inter-arrival times of customers to the bank are iid having exponential distribution with mean 3 minutes. Also, the the processing times of customers are iid having the following distribution (in minutes): x Pr{X = x} 1 1/4 2 1/2 3 1/4 An arrival ? nding the server busy joins the queue. The waiting space is in? nite. (a) What is the long-run fraction of time that the server is busy? (b) What the the long-run average waiting tim e of each customer in the queue, excluding the processing time? c) What is average number of customers in the bank, those in queue plus those in service? 2. 8. EXERCISE (d) What is the throughput of the bank? 37 (e) If the inter-arrival times have mean 1 minute. What is the throughput of the bank? 8. You are the manager at the Student Center in charge of running the food court. The food court is composed of two parts: cooking station and cashier’s desk. Every person should go to the cooking station, place an order, wait there and pick up ? rst. Then, the person goes to the cashier’s desk to check out. After checking out, the person leaves the food court.The coo