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附件1.中山大学教师在线学习中心使用指南教师在线学习中心使用基本流程：注册会员1. 打开中山大学教师在线学习中心网页（），首先需进行会员注册。
（注意：首次注册必须使用中山大学IP地址登录）2. 点击【注册】，填写注册信息。
请填写真实姓名，以便培训认证使用；学校名称、院系请填写全称。
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选课，观看课程资源1.登录后，点击【在线培训课程】，可按个人需求选择课程。
2.选择一门课程，点击【查看课程】，进入课程介绍页面。
点击【我要报名】，核对个人信息后，点击提交。
3.报名后，在弹出的【个人中心】页面，可看到已报名的课程。
点击课程名，即可在课程页面进行视频观看，及配套PPT资源下载。
常见问题1.登录时提示“您不在授权IP范围内”或“非本校学员不能登录”怎么办？如学员登录时提示“您不在授权IP范围内”，则说明您当前未使用中山大学校园网IP 地址登录，首次注册必须使用校园网IP地址。
如学员登录时提示“非本校学员不能登录”，可以登录网培中心总部网站，在个人中心--个人信息修改【所在学校】名称为学校标准名称，然后就可以在校级中心网站登录。
2. 能否只观看课程视频，不提交作业？可以只观看课程视频获得学分学时。
但如果要打印电子证书（学时证明），则必须要提交作业，以满足相应的学时学分要求。
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4. 学员如何获得学分学时？学员参与以下培训内容均可获得学分学时：特别提示：同门课程重复学习、同一个教学活动重复参加学分不累计。
5. 如何查看我的学分、学时？学员进入个人中心，点击【学分与证书打印】，即可看到您的总学分、总学时和可打印学分。
点击【我的学分明细】可以查看您的学分明细记录。
6. 如何查询我的学况？学员进入个人中心，点击【我的学分】，即可看到【我的学况统计】。
中国临床肿瘤学会（CSCO）食管癌诊疗指南GUIDELINES OF CHINESE SOCIETY OF CLINICAL ONCOLOGY(CSCO) ESOPHAGEAL CANCER2020中国临床肿瘤学会指南工作委员会组织编写人民卫生出版社中国临床肿瘤学会指南工作委员会组长李进副组长（按姓氏汉语拼音排序）程颖郭军赫捷江泽飞梁军马军秦叔逵王绿化吴一龙徐瑞华中国临床肿瘤学会（CSCO）食管癌诊疗指南2020组长王绿化副组长黄镜韩泳涛李印傅剑华毛伟敏秘书王鑫执笔专家组成员（按姓氏汉语拼音排序）陈克能北京大学肿瘤医院胸外科方文涛上海市胸科医院胸外科樊青霞郑州大学第一附属医院肿瘤科傅剑华中山大学肿瘤防治中心胸外科韩泳涛四川省肿瘤医院胸外科胡兵四川大学华西医院消化内科黄镜中国医学科学院肿瘤医院内科梁军中国医学科学院肿瘤医院放疗科李印中国医学科学院肿瘤医院胸外科刘慧中山大学肿瘤防治中心放疗科毛伟敏浙江省肿瘤医院胸外科牟巨伟中国医学科学院肿瘤医院深圳医院胸外科束永前江苏省人民医院肿瘤科王贵齐中国医学科学院肿瘤医院内镜科王绿化中国医学科学院肿瘤医院深圳医院放疗科王鑫中国医学科学院肿瘤医院放疗科吴式琇杭州市肿瘤医院放疗科薛丽燕中国医学科学院肿瘤医院病理科袁响林华中科技大学同济医学院附属同济医院肿瘤科张述山东省肿瘤医院内科赵快乐复旦大学附属肿瘤医院放疗科祝淑钗河北医科大学第四医院放疗科庄武福建省肿瘤医院胸部肿瘤内科顾问专家组成员（按姓氏汉语拼音排序）白玉贤哈尔滨医科大学附属肿瘤医院消化内科包永星新疆医科大学第一附属医院肿瘤中心曹国春江苏省肿瘤医院内科曹建中山西省肿瘤医院放疗科陈椿福建医科大学附属协和医院胸外科陈俊强福建省肿瘤医院放疗科陈龙奇四川大学华西医院胸外科戴广海中国人民解放军总医院肿瘤内科邓艳红中山大学附属第六医院肿瘤内科樊祥山南京鼓楼医院病理科高树庚中国医学科学院肿瘤医院胸外科葛红河南省肿瘤医院放疗科龚新雷中国人民解放军东部战区总医院全军肿瘤中心肿瘤内科郭石平山西省肿瘤医院胸外科韩春河北医科大学第四医院放疗科韩大力山东省肿瘤医院放疗科何义富安徽省立医院肿瘤化疗科侯英勇复旦大学附属中山医院病理科胡春宏中南大学湘雅二医院肿瘤科黄晓俊兰州大学第二医院消化内科惠周光中国医学科学院肿瘤医院放疗科姬发祥青海大学附属医院肿瘤内科贾军北京大学肿瘤医院消化内科江浩蚌埠医学院第一附属医院放疗科姜宏景天津医科大学肿瘤医院食管肿瘤科姜慧卿河北医科大学第二医院消化内科康明强福建医科大学附属协和医院胸外科康晓征北京大学肿瘤医院胸外科李宝生山东省肿瘤医院放疗科李鹤成上海交通大学医学院附属瑞金医院胸外科李涛四川省肿瘤医院放疗科李媛复旦大学附属肿瘤医院病理科李志刚上海市胸科医院胸外科梁玮福建省立医院消化内科刘波山东省肿瘤医院内科刘俊峰河北医科大学第四医院胸心外科刘琳东南大学附属中大医院肿瘤科刘思德南方医科大学南方医院消化内科刘莺河南省肿瘤医院内科刘勇中国医学科学院肿瘤医院内镜科刘月平河北医科大学第四医院病理科柳硕岩福建省肿瘤医院胸部肿瘤外科路平新乡医学院第一附属医院肿瘤科罗素霞河南省肿瘤医院内科骆金华江苏省人民医院胸外科吕宁中国医学科学院肿瘤医院病理科马建群哈尔滨医科大学附属肿瘤医院胸外科马锴青岛大学附属医院胸外科毛友生中国医学科学院肿瘤医院胸外科庞青松天津医科大学肿瘤医院放疗科彭贵勇陆军军医大学第一附属医院消化内科彭林四川省肿瘤医院胸外科钱晓萍南京鼓楼医院肿瘤科秦建军中国医学科学院肿瘤医院胸外科屈东中国医学科学院肿瘤医院影像诊断科盛剑秋中国人民解放军陆军总医院消化内科宋岩中国医学科学院肿瘤医院内科隋红哈尔滨医科大学附属肿瘤医院消化内科孙明军中国医科大学附属第一医院消化内科孙新臣江苏省人民医院放疗科孙益峰上海市胸科医院胸外科谭锋维中国医学科学院肿瘤医院胸外科谭黎杰复旦大学附属中山医院胸外科田辉山东大学齐鲁医院胸外科王大力中国医学科学院肿瘤医院胸外科王峰郑州大学第一附属医院肿瘤科王晖湖南省肿瘤医院放疗科王澜河北医科大学第四医院放疗科王奇峰四川省肿瘤医院放疗科王实浙江省肿瘤医院内镜中心王维虎北京大学肿瘤医院放疗科王维威北京协和医院胸外科王哲中国医学科学院肿瘤医院深圳医院胸外科王铸中国医学科学院肿瘤医院影像诊断科郗彦凤山西省肿瘤医院病理科相加庆复旦大学附属肿瘤医院胸外科向锦中山大学肿瘤防治中心病理科肖菊香西安交通大学第一附属医院肿瘤内科肖泽芬中国医学科学院肿瘤医院放疗科徐红吉林大学白求恩第一医院内镜中心许洪伟山东省立医院消化内科许建萍中国医学科学院肿瘤医院内科于振涛天津医科大学肿瘤医院食管肿瘤科张百江山东省肿瘤医院胸外科张鹏天津医科大学总医院心胸外科张仁泉安徽医科大学第一附属医院胸外科张小田北京大学肿瘤医院消化内科张艳桥哈尔滨医科大学附属肿瘤医院肿瘤内科赵林北京协和医院肿瘤内科周平红复旦大学附属中山医院内镜中心周谦君上海市胸科医院肿瘤外科周炜洵北京协和医院病理科朱向帜江苏省肿瘤医院放疗科前言基于循证医学证据、兼顾诊疗产品的可及性、吸收精准医学新进展，制定中国常见癌症的诊断和治疗指南，是中国临床肿瘤学会（CSCO）的基本任务之一。
更多资讯请留意：中大青年网中山大学新生英语入学测试试卷College English Placement TestSeptember 7, 2008Part ⅠListening comprehension (25%)Part ⅡReading comprehension (1%,20%)Directions：There are 4 passages in this part. Each passages is followed by some questions or unfinished statements. For each of them there are four choices marked A, B, C, and D. You should decide on the best choice and mark the corresponding letter on the Answer Sheet with a single line through the center.Passage 1Questions 31 to 35 are based on the following passage：A third of Britons are overweight, states a report published in January by the Royal College of Physicians, the result of an 18-month-long study. About five per cent of the children weigh too much, and are likely to stay that way for life; in the mid-twenties age group the proportion of fat people rises to a third, and of the middle-age population half are overweight.Fat people risk severe health problem, says the report, including high blood pressure, breathlessness, and various forms of heart disease. Smoking is particularly risky for overweight people.The safest way to lose weight is to eat cereals, bread, fruit and vegetables, and cut down on fatty meats, butter and sweet foods. Fat diets do far more harm than good; slimming machines that vibrate muscles have not been proved useful; saunas merely remove a little body water, and health farms, says the report, serve as expensive holidays.Exercise is most important to health, the report emphasizes; though it doesn't necessarily reduce weight, it maintains the correct proportion of body fat to body muscle. And it isn't only for the young. From middle age a minimum of 20 minutes of gentle physical jerks should be practiced three times a week.The report advocates several public health measures to combat the high prevalence of overweight in this country. They include an increase of tax on alcohol to reduce its increasing, and dangerously fattening, consumption; and the provision of more sports facilities by local authorities. Britain's doctors, the report concludes, must learn to be more sympathetic and specific in their advice to the overweight, encouraging a change in eating habits on a long-term basis, and taking into account the many -often complex - reasons why fat people are fat31. The passage mainly talks about_______.A. Britons' overweight problemB. how to avoid getting overweight更多资讯请留意：中大青年网C. the cause of Britons overweightD. the relations between overweight and health problems32. According to the report, a person is most likely to stay fat for the whole of his life if he________.A. gets fat in his twentiesB. gets fat in his middle ageC. was born fatD. gets fat when he is a child33. The report suggests that exerciseA. is a way to reduce weightB. is a sure way to keep one healthyC. sometimes increases weightD. can convert fat to muscle34. The report points out that drinking too much alcoholA. will also cause a person to get overweightB. will cause a person to do less sportC. will make a person forget the fact that he is fatD. will lead a person to bad eating habits35. Britain’s doctors, when treating the overweight, shouldA. encourage a long-term diet for everyoneB. be more considerate and give detailed adviceC. first consider why so many people are overweightD. do all of the abovePassage TwoQuestions 36 to 40 are based on the following passage：Knowing ahead of time how to get out during a fire can save needed time. The best way out in a fire is the route you use to go in and out everyday. Yet in a fire this route may be blocked. Be sure to plan other escape routes.Practice your escape plan an night when it is dark. Be sure that your plan is good and will work. For instance, make sure that a child can actually open the window he is supposed to use for his escape. Teach children to close their bedroom doors. Tell them to will by an open window until someone can reach them from outside. If an adult cannot be wakened, children should understand that they must leave by themselves.Choose a meeting place outside. This way you can tell if everyone is safely out of the building. Know where nearby telephones or fire alarm boxes are found.If you live in an apartment, try to get everyone out. Learn where the fire alarm is in the building. Your family should know what the fire alarm bell or horn sounds like. They should know what to do when they hear it. Try to get the other families in your building together to have fire drills. Write down the telephone number of the fire department. Tape the number to each phone. Don’t forget to let the babysitter in on your plans. Tell your babysitter what to do in case of fire.Early warning is the key to a safe escape. It has been shown time and time again that a family can escape if warned early enough.36. Which of the following statements is true?A. You must take the route you use to go in and out in a fire.更多资讯请留意：中大青年网B. The route is always blocked in a fire.C. You should not rely on one route during a fire.D. During a fire, planning escape route can save you needed time.37. Which is the best advice for children during a fire?A. Children should awaken an adult before they leave.B. Children should close the bedroom doors and windows.C. Children should jump out of window even though they are in a high place.D. Children should open a window and wait there for help.38. People who live in apartment should______.A. practice the plan togetherB. make sure they know the fire chief.C. not have to pay for insuranceD. do fire drills separately39. If you are trapped in a bedroom during a fire, you should______.A. hide under the bed.B. stand in a cornerC. lie still.D. close the bedroom doors.40. When the best time to practice escape plan?A. At duskB. In the afternoonC. In the morningD. At nightPassage ThreeQuestions 41 to 45 are based on the following passage：Social change is more likely to occur in societies where there is a mixture of different kinds of people than in societies where people are similar in many ways. The simple reason for this is that there are more different ways of looking at things present in the first kind of society. There are more ideas, more disagreements in interest, and more groups and organizations with different beliefs. In addition, there is usually a greater worldly interest and greater tolerance in mixed societies. All these factors tend to promote social change by opening more areas of life to decision. In a society where people are quite similar in many ways, there are fewer occasions for people to see the need or the opportunity for change because everything seems to be the same. And although conditions may not be satisfactory, they are at least customary and undisputed.Within a society, social change is also likely to occur more frequently and more readily in the material aspects of the culture than in the non-material, for example, in technology rather than in values; in what has been learned later in life rather than what was learned early; in the less basic and less emotional aspects of society than in their opposites; in the simple elements rather than in the complex ones; in form rather than in substance; and in elements that are acceptable to the culture rather than in strange elements.Furthermore, social change is easier if it is gradual. For example, it comes more readily in human relations on a continuous scale rather than one with sharp dichotomies(一分为二). This is one reason why change has not come more quickly to Black Americans as compared to other American minorities, because of the sharp difference in appearance between them and their white counterparts.41. The passage is mainly discussing _______.更多资讯请留意：中大青年网A. two different societiesB. the necessity of social changeC. certain factors that determine the ease with which social changes occurD. certain factors that promote social change42. ______ is one of the factors that tend to promote social change.A. Joint interestB. Advanced technologyC. Different points of viewD. Less emotional people43. The expression greater tolerance（paragraph 1）refers to _____.A. greater willingness to accept social changeB. quicker adaptation to changing circumstancesC. more respect for different beliefs and behaviorD. greater readiness to agree to different opinions and ideas44. According to the passage，which of the following is NOT true？A. Social change tends to meet with more difficulty in basic and emotional aspects of society.B. Disagreement with and argument about conditions tend to slow down social change.C. Social change is more likely to occur in the material aspect of society.D. Social change is less likely to occur in what people learned when they were young.45. Why has change not come more quickly to Black Americans as compared to white people?A. Black people used to be slavesB. White people are of more different originsC. Black people are poorD. White people like to make shape changePassage fourQuestions 46 to 50 are based on the following passage：Most employers say that they wish to employ the right person for the right job. A recent report by Britain's Independent Institute of Manpower Studies，however，disagrees with this. The report states that most employers wish to avoid employing the wrong person. Rather than looking for the right person，they are looking for applicants to turn down.The report also suggests that in Britain and in many other parts of the world，the selection methods used to pick out the right person for the job certainly do not match up to those used to judge a piece of new equipment. Employers used three main selection methods: interviewing，checking resume or application forms and examining references. Most of the employers asked in this survey stated that these selection methods were used more for “weeding out” unsuitable applicants rather than for finding suitable ones.Interviews were considered to be more reliable than either resume checks or references from past employers. Research，however，proves otherwise. Interviewers’ decisions are often strongly influenced by their earlier judgment of the written application. Also，different employers view facts differently. One may consider applicants who have frequently changed jobs as people with broad and useful experience. Another will see such applicants as unreliable and unlikely to stay for long in the new job.更多资讯请留意：中大青年网Some employers place great importance on academic qualifications whereas the link between this and success in management is not necessarily strong. Some employers use handwriting as a standard. The report states that there is little evidence to support the value of the latter for judging working ability. References，also，are sometime unreliable as they are not very important，while checks on credit and security records and applicants' political opinions are often the opposite.The report is more favorable towards trainability tests and those which test personality and personal and mental skills. The report concludes by suggesting that interviewing could become more reliable if the questions were arranged in a careful，organized system and focused on the needs of the employing organization.46. According o the passage, when most employers want to hire workers, _______.A. they will try to find suitable peopleB. they will look for the right applicantC. the wrong applicants are to be turned downD. to turn down the wrong people is what they say they aim to do47. It is implied that ______.A. to evaluate a right person is more difficult than to evaluate equipmentB. employers are more successful in selecting the right equipment than the right personsC. criteria will be set up according to the real situation of the applicantsD. curriculum vitae means application forms48. Most of the recruiters ______.A. consult the applicantsB. can find suitable peoplesC. prefer curriculum vitae or referencesD. use different ways to sort out the unsuitable applicants49. Which of the following is True?A. Employers get different conclusions from the facts.B. Changing jobs frequently will reduce the change to be recruited.C. Academic qualifications will guarantee the applicant managing ability.D. Handwriting is a valid way to evaluate an applicant.50. It can be inferred from the passage that successful employees will be those who _____.A. have outstanding referencesB. are strong in emotional quotientC. take interviewing seriouslyD. have strong political leaningsPart ⅢVocabulary and Structure (35%Section A (1%, 30%)Directions: there are a number of incomplete sentences in this part. For each sentence there are fo ur choices marked A, B, C and D. Choose the ONE answer that best completes the sentence. Then mark the corresponding letter on the Answer Sheet with a single line through the centre.51. They had a fierce _____ as to whether their company should restore the trade relationship which was broken years ago.更多资讯请留意：中大青年网A. crashB. debateC. contestD. disagreement52. The early pioneers had to _____ many hardships to settle on the new land.A. go in forB. go back onC. go throughD. go on53. The local health organization is reported _____ twenty-five years ago when Dr.Audon became its first president.A. to be set upB. to have been set upC. being set upD. having been set up54. He invented a plough _____ any other in existence.A. far superior toB. more superior thanC. more superior toD. much superior than55. He is watching TV? He is _____ to be cleaning his room.A. knownB. consideredC. regardedD. supposed56. As scheduled, the communication satellite went into _____ round the earth.A. circleB. pathC. orbitD. course57. If this bill is not paid within five days, your gas supply will be _____.A. abandonedB. cut offC. turned downD. lost touch58. When no one answered the door, she _____ through the window to see if anyone was there.A. spottedB. inquiredC. scannedD. peered59. I don’t want to lend any money to him; he is already in debt _____ me.A. forB. toC. ofD. with60. Nowadays advertising cost money to him; he is already in debt _____ me.A. connectionB. reactionC. relationD. proportion61. A large part of human activity, particularly in relation to the environment, is _____ conditions or events.A. in contract toB. in favor toC. in response toD. in case of62. She said she liked dancing but was not in the _____ for it just then.A. mannerB. intentionC. desireD. mood63. The car _____ halfway for no reason and we could not move.A. broke offB. broke downC. broke upD. broke out64. In Britain today women _____ 44% of the workforce and nearly half the mothers with children are in paid work.A. build upB. stand forC. make upD. conform to65. The plan seem quite simple to me, but I just can’t _____ to my classmates.A. get it acrossB. get at itC. get it downD. get through it66. A man who is cruel to his children should be held in _____.A. contextB. contemptC. controversyD. convict67. The villagers were very _____ towards anyone who passed through.A. hospitalB. hospitalizedC. hospiceD. hospitable68. I meant _____ you about it, but I forgot to do so.A. to have toldB. to have been toldC. to tellD. telling更多资讯请留意：中大青年网 69. The American society is _____ an exceedingly shaky foundation of natural resources, which is connected with the possibility of a worsening environment.A. originated fromB. established onC. come fromD. turned out70. The French pianist who had been praised very highly _____ to be a great disappointment.A. turned upB. turned inC. turned downD. turned out71. Joe differs from his brother _____ he is very diligent while his brother is included to idleness.A. in whereB. in whichC. in whatD. in that72. –Can you come on Monday or Tuesday?–I’m afraid _____ day is possible.A. eitherB. neitherC. someD. both73. _____, he cannot succeed.A. Try as he mayB. As he may tryC. Try may he asD. How may he try74. Dr.Black comes from either Oxford or Cambridge, I can’t remember _____.A. whereB. whichC. thereD. that75. Only yesterday _____ that they had wronged their son.A. did the parent realizeB. the parent realizedC. the parent were realizedD. were the parent realized76. After I heard that I took a deferred pass in botany (植物学) , I was in a _____ emotional state.A. doubtfullyB. highlyC. greatlyD. nervously77. The staff proposed that wages _____.A. have been raisedB. be raisedC. have to be raisedD. were to be raised78. _____ the advance of science, the discomforts of old age will no doubt always be with us.A. As forB. BesidesC. ExceptD. Despite79. It sounds ridiculous but true that millions of people _____ to see the sun rise if they _____ pay for it.A. would have got up…had had toB. may get up…have toC. would get up…had toD. could get up…have to80. By using space technology China can predict natural disasters _____, Luan said.A. far more precisely than generally thoughtB. far precisely as generally thoughtC. more precisely than generally thinkingD. more precisely that of general thoughtSection B (0.5%, 10%)Directions: From the four choices given below the statements, choose the ONE that is closest in meaning to the underlined part in the sentence.81. Many cinema-goers were stunned by the film’s violent and tragic end.更多资讯请留意：中大青年网A. shockedB. disappointedC. indifferentD. angry82. The villages raked the district for a trace of the missing boy.A. exploredB. trampedC. exploitedD. combed83. Few countries today enjoy prosperous economy.A. staticB. assumedC. poorD. flourishing84. Collins was so seriously injured that he was unable to continue his career.A. resumedB. assumedC. consumedD. ensure85. Miss Baker’s respond to your request seemed gratifying.A. pleasingB. shrewdC. overwhelmingD. threatening86. You have to pay a premium for express delivery.A. extra tipB. extra changeC. extra bonusD. extra price87. The runaway had been in flight for a couple of days and was caught at the entrance to the library this morning.A. by planeB. in the long runC. on the runD. by air88. After many failures, his eventual success surprised all of us.A. finalB. outstandingC. exceptionalD. full89. Please express my good wish to your parents. I haven’t met them for years.A. informB. amuseC. denyD. convey90. There are so few bald eagles left in America that many people fear they are in danger of extinct.A. dyingB. dying outC. dying downD. dying awayPart ⅣClose (1%, 20%)Direction: There are 20 blanks in the following passage. For each blank there are four choices marked A, B, C and D. You should choose the ONE that best fits into the passage. Then mark the corresponding letter on the Answer Sheet with a single line through the center.Historians tend to tell the same joke when they are describing history education in America. It’s the one __91__ the teacher standing in the schoolroom door __92__ goodbye to students for the summer and calling __93__ them, “By the way, we won World War II.”The problem with the joke, of course, is that it’s __94__ funny. The recent surveys on __95__ illiteracy (无知) are beginning to numb (令人震惊): nearly one third of American 17-year-olds cannot even __96__ which countries the United States __97__ against in that war. One third have no __98__ when the Declaration of Independence was __99__. One third thought Columbus reached the New World after 1750. Two thirds cannot correctly __100__ the Civil War between 1850 and 1900. __101__ when they get the answers right, some are __102__ guessing.Unlike math or science, ignorance of history cannot be __103__ connected to loss of international __104__. But it does affect our future __105__ a democratic nation and as individuals.The __106__ news is that there is growing agreement __107__ what is wrong with the __108__ of history and what needs to be __109__ to fix it. The steps are tentative (尝试更多资讯请留意：中大青年网 性)__110__ yet to be felt in most classrooms.91. A. about B. in C. for D. by92. A. shaking B. waving C. nodding D. speaking93. A. in B. after C. for D. up94. A. rarely B. so C. too D. not95. A. historical B. educational C. cultural D. political96. A. distinguish B. acknowledge C. identify D. convey97. A. defeated B. attacked C. fought D. struck98. A. sense B. doubt C. reason D. idea99. A. printed B. signed C. marked D. edited100. A. place B. judge C. get D. lock101. A. Even B. Though C. Thus D. So102. A. hardly B. just C. still D. ever103. A. exclusively B. practically C. shortly D. directly104. A. competitiveness B. comprehensionC. communityD. commitment105. A. of B. for C. with D. as106. A. fine B. nice C. surprising D. good107. A. to B. with C. on D. of108. A. consulting B. coaching C. teaching D. instructing109. A. done B. dealt C. met D. reached110. A. therefore B. or C. and D. asPart Ⅴ. Oral CommunicationSection A Dialogue CompletionDirections: In the section, you will read 5 short incomplete dialogues between two speakers, each followed by four choices marked A, B, C and D. Choose the answer that appropriately suits the situation to complete the dialogue. Then mark the corresponding letter on the Answer Sheet a single line through the center.111. [A]: The chocolate cake is very good today.[B]: _____.A. Yes, I’ve to consider it carefullyB. No, I can’t afford itC. Yes, I’ll have it tomorrow, not todayD. No, thanks. I’ll have an apple pie with ice cream on the top112. [A]: _____[B]: I have only a bottle of Brandy and a painting.A. Do you have anything to declare?B. What have you bought abroad?更多资讯请留意：中大青年网C. What do you have to pay the tax?D. Anything you what me to know?113. [A]: Operator, I’d like to place a call to Beijing, please.[B]: _____A. OK. What are you?B. Just a moment, please.C. All right. How are youD. Sorry, you have to pay.114. [A]: Excuse me. When will the Flight 666 arrive?[B]: _____A. Sorry. I can’t tell you.B. Good. You have missed it.C. Well. It had been delayed.D. OK. Here you are.115. [A]: Have you got the book recommended by our teacher?[B]: _____A. No, I have got itB. No, it have been sold outC. Yes, they don’t sell it.D. Yes, I have one already.Section A Dialogue ComprehensionDirections: In the section, you will read 5 short conversations between a man and a woman. At the end of each conversation there is a question followed by four choices marked A, B, C and D. Choose the answer to the question from the four choices given. Then mark the corresponding letter on the Answer Sheet with a single line through the center.116. [Man]: This is Mr. Jones. My heater is not getting any power and the temperature is going to get down below freezing.[Woman]: This is our busiest time of the year, but I’ll speak to one of our men about getting over there sometime today.[Question]: What did the woman mean?A. She would come to fix the man’s heater soon.B. She would send someone to fix the man’s heater.C. She could not come and fix it.D. She was not happy about the heater.117. [Woman]: I feel like it’s only been a few weeks since school started.[Man]: Yes. How time flies! It’s almost time for our final exams.[Question]: What do the speakers imply?A. They didn’t like to study in their school.B. The school year seemed to go by very quickly.C. They’ve been in school for a few days only.D. Though final are over, they have to continue their study.118. [Man]: Are you sure this is the right way to get to the airport? My flight will depart in forty minutes.[Woman]: Sure. This is a shortcut. We’ll be there soon.[Question]: What does the woman mean?A. They have taken a wrong road.B. This way should take less time.更多资讯请留意：中大青年网11C. They will be late because of the bad traffic.D. The road is rough.119. [Woman]: Help yourself to some fish. I hope you ’ll like it.[Man]: My doctor told me to keep far from fish.[Question]: What does the man mean?A. He doesn ’t like eating fish.B. The woman is not good at cooking fish.C. He thinks the fish is too far from him to get.D. He shouldn ’t eat any fish as his doctor told him.120. [Man]: Will you come out this evening and go to cinema with me?[Woman]: That depends on what film the cinema is going to put on.[Question]: What does the woman ’s response mean?A. She does n’t like to go with the man.B. She is more interested in the film than the man.C. She knows there is a good film on.D. She wants the man to tell her where the cinema is.参考答案：31~35 ADBAB 36~40 DDADD 41~45 CCABA 46~50 DCDAD 51~55 BCAAD 56~60 CBDBD61~65 CDBCA 66~70 BDCBD 71~75 ABABA 76~80 BBDCA 81~85 ADDDA 86~90 BCADB 91~95 ABBDA 96~100 CCDBB 101~105 ABDCD 106~110 DCCAB 111~115 DCBCB116~120 BBBDB。
【关键字】方案、情况、方法、条件、前提、空间、领域、文件、质量、计划、运行、问题、系统、密切、充分、良好、执行、统一、发展、细化、建立、制定、提出、了解、研究、特点、基础、需要、环境、工程、资源、重点、体系、需求、方式、特色、作用、标准、规模、结构、任务、反映、关系、协商、分析、满足、管理、监督、保证、优先、维护、服务、指导、支持、解决、优化、改善、分工、完善、适应、实现、提高、协调、改进、规范、复杂性软件工程学第一次作业一、简答题1、需求分析的任务答：需求分析的任务主要包括以下几项：确定目标系统的综合要求，其中包括（目标系统的功能、性能、运行的环境及扩展性要求）；分析目标系统的数据要求，其中包括（系统平台需要哪些数据？数据间有什么关系？数据及数据结构？对数据的处理逻辑关系等）；导出目标系统的逻辑模型；修正系统流程图；修正系统开发计划、开发原型系统2、软件危机是指在计算机软件的开发和维护过程中所遇到的一系列严重问题，具体表现在哪些方面？答：1、无法开发复杂程度高的软件2、成本和进度估计不准3、无统一科学的规范，软件不可维护4、无质量保证，可靠性差5、软件常不能满足用户的需求6、无适当的文档资料7、软件生产率太低二、选择题1、从下列关于结构化程序设计的叙述中选出5条正确的叙述。
①程序设计比较方便，但比较难以维护。
②便于由多个人分工编制大型程序。
③软件的功能便于扩充。
④程序易于理解，也便于排错。
⑤在主存储器能够容纳得下的前提下，应使模块尽可能大，以便减少模块的个数。
⑥模块之间的接口叫做数据文件。
⑦只要模块之间的接口关系不变，各模块内部实现细节的修改将不会影响别的模块。
⑧模块间的单向调用关系叫做模块的层次结构。
⑨模块越小，模块化的优点越明显。
一般来说，模块的大小都在10行以下。
答：正确的叙述有②、③、④、⑦、⑧。
如果程序结构的模块化满足评价的标准（高内聚、低耦合），这样的结构是容易维护的，程序的功能也容易测试，容易理解、容易修改、容易维护的，程序的功能也容易扩充。
ERP系统应用与实践Course:Instructor: 张宏斌Homework: 第一次作业：MRP模块梁颖敏、陈丽平、温伟江、刘娜Name:Class: 08MLEDate:…20100413…本次作业MRP模块的操作主要包括以下几个部分：实验设计实验流程设计实验流程实验收获与问题实验设计某企业生产的一款蛋糕烤盘的产品结构如图所示款蛋糕烤盘生产的相关信息如下：◦TA-KP和TA-DP由企业生产◦TA-PP、TA-TL、TA-XJ、TA-GJ采购，采购价格及采购提前期详见下表：该款蛋糕烤盘生产的相关信息如下：◦TA-KP的售价为40元/个，预测4月的销售量为2000件，已接到一张需求量为400件的订单，要求发运时间为4月20该款蛋糕烤盘生产的相关信息如下：◦TA-DP经由自动喷涂工序（TA-PTA）生产，所需设备为归属于喷涂车间（TA-PTD）的自动喷涂生产线（TA-PTR）◦TA-KP经由装配工序（TA-ZPA）生产，所需设备为归属于装配车间（TA-ZPD）的装配生产线（TA-ZPR）◦TA-PTR和TA-ZPR的数量均为1，均采用一个白班的班次生产，生产能力均为每天100个实验流程设计1.创建物料2.创建物料清单3.创建资源4.创建部门5.创建工序6.创建工艺路线7.创建价目表8.创建销售订单9.创建预测10.定义和装载主需求计划11.定义和运行物料需求计划12.生成生产工单13.生成采购计划14.创建采购订单15.采购接收入库16.查找并释放生产工单17.生产子装配件18.子装配件完工入库19.生产产成品20.产成品完工入库实验流程责任：制造和分销经理组织：M1创建物料创建TA-KP复制自“成品”模板，分配给M1创建TA-DP复制自“子装配件”模板，分配给M1创建TA-PP、TA-TL、TA-XJ、TA-GJ复制自“采购物料”模板依据实验设计在“采购”属性的“价目表”栏输入价格，在“总计划”栏输入计划员分配给M1将上述物料的“MPS/MRP计划”属性中的“追溯”改为“总装/软反查”创建物料清单创建TA-DP的BOM在“物料”栏输入TA-DP在“组件”栏输入TA-PP、TA-TL，依据实验设计输入所需数量创建TA-KP的BOM在“物料”栏输入TA-KP在“组件”栏输入TA-DP、TA-XJ、TA-GJ，依据实验设计输入所需数量创建资源创建自动喷涂生产线TA-PTR在“类型”栏选择“设备”在“单位”栏选择“HR”在“费用类型”栏选择“WIP移动”创建装配生产线TA-ZPR在“类型”栏选择“设备”在“单位”栏选择“HR”在“费用类型”栏选择“WIP移动”创建部门创建喷涂车间TA-PTD在“资源”栏选择TA-PTR取消“24小时可用”，在“数量”栏输入1选择白班，在“数量”栏输入1创建装配车间TA-ZPD在“资源”栏选择TA-ZPR取消“24小时可用”，在“数量”栏输入1选择白班，在“数量”栏输入1创建工序创建自动喷涂工序TAPA在“资源”栏选择TA-PTR在“倒数”栏输入每小时产量12.5在“计划”栏选择“是”创建装配工序TAZA在“资源”栏选择TA-ZPR在“倒数”栏输入每小时产量12.5在“计划”栏选择“是”创建工艺路线创建TA-DP的工艺路线在“物料”栏选择TA-DP在“代码”栏选择TAPA创建TA-KP的工艺路线在“物料”栏选择TA-KP在“代码”栏选择TAZA创建价目表查找“Corporate”在“Corporate”的列表行新建一行在“产品属性”栏选择“项目编号”在“产品值”栏输入TA-KP在“价格”栏输入40创建销售订单在“客户”栏选择“Business World”在“行信息”的“订购项目”栏输入TA-KP在“日期”栏输入4月20日在“数量”栏输入400将“发运”中的仓库改为M1记录订单编号为64031创建预测在“预测集”栏输入TA-KPS在“预测”栏输入TA-KPC在“预测物料”的“物料”栏输入TA-KP在“详细资料”的“时段”栏选择“天”在“详细资料”的“日期”栏输入2010/4/10在“详细资料”的“数量”栏输入100定义和装载主需求计划定义主需求计划TA-MDS在“目标计划”栏输入TA-MDS在“来源类型”栏选择“特定预测”在“来源名称” 栏输入TA-KPS在“改写” 栏选择“所有分录”在“数量类型”栏选择“原始”在“冲减”栏选择“是”在“正向冲减天数”栏输入5在“反向冲减天数”栏输入5查看主需求计划定义和运行物料需求计划◦定义物料需求计划TA-MRP◦在“选项”中的“计划”栏输入TA-MDS◦选择“计划能力”和“反查”◦保存，点击“生成”、“确定”、“提交”◦记录MRP运行请求编号为4557496 查看物料需求计划生成生产工单◦查找物料需求计划TA-MRP◦在“供应/需求”中查找物料TA-KP和TA-DP◦选中“选定发放”◦保存◦点击“发放”◦查看后记录生产申请号为4557522生成采购计划◦查找物料需求计划TA-MRP◦在“供应/需求”中查找物料TA-PP、TA-TL、TA-XJ、TA-GJ◦选中“选定发放”，在“实施方式”栏选择“采购申请”◦保存◦点击“发放”◦查看后记录采购申请号为4557523创建采购订单◦删除采购员，点击查找◦选择物料TA-PP、TA-TL、TA-XJ、TA-GJ的采购申请，点击“自动”◦点击“发运”◦点击“分配”，在“子库存”栏输入”RIP”◦保存◦点击“审批”、“确定”，状态自动改为“批准”◦记录订单号为5566采购接收入库◦输入采购订单号5566进行查询◦勾选物料TA-PP、TA-TL、TA-XJ、TA-GJ◦保存◦查看后确认子库存RIP中上述物料的现有量与采购订单中的采购量相符查找并释放生产工单◦查找并释放TA-DP的生产工单在“装配件”栏输入TA-DP进行查找在菜单“工具”中点选“更改状态”后选择“释放”保存◦查找并释放TA-KP的生产工单在“装配件”栏输入TA-KP进行查找在菜单“工具”中点选“更改状态”后选择“释放”保存生产子装配件◦在“装配件”栏输入TA-DP◦输入工序序号◦工序步骤从“排队”到“移动”◦在“处理量”栏输入拟生产数量400（与排队量相等） 子装配件完工入库◦在“装配件”栏输入TA-DP◦点击“继续”◦在“子库存”栏输入“RIP”◦在“数量”栏输入子装配件生产的处理量400◦点击“完成”◦查看后确认子库存RIP中TA-DP的现有量为400 生产产成品◦在“装配件”栏输入TA-KP◦输入工序序号◦工序步骤从“排队”到“移动”◦在“处理量”栏输入拟生产数量400（与排队量相等） 产成品完工入库◦在“装配件”栏输入TA-KP◦点击“继续”◦在“子库存”栏输入“FGI”◦在“数量”栏输入产成品生产的处理量400◦点击“完成”◦查看后确认子库存FGI中TA-KP的现有量为400 查看产成品库存实验收获与问题实验收获◦增进了对业务流程的了解◦增进了对业务运作与系统运作间的关联关系的了解◦进一步熟习了ORACLE MRP模块的应用实验中遇到且尚未解决的问题◦预测如何对主需求计划产生影响？◦各种物料的现有量、外购物料的提前期如何录入系统以及如何对主需求计划、物料需求计划产生影响？。
Experimental investigation and numerical simulation for weakening the thermal fluctuations in aT-junctionK.Gao a ,P.Wang b ,T.Lu a ,⇑,T.Song caCollege of Mechanical and Electrical Engineering,Beijing University of Chemical Technology,Beijing 100029,China bSchool of Energy and Power Engineering,Dalian University of Technology,Dalian 116024,China cChina Nuclear Power Technology Research Institute Co.,Ltd,Shenzhen 518124,Chinaa r t i c l e i n f o Article history:Received 25August 2014Received in revised form 17November 2014Accepted 4January 2015Available online 17January 2015Keywords:Experimental investigation Numerical simulation Tee junctionThermal fluctuationa b s t r a c tIn this work,the mixing processes of hot and cold fluids with and without a distributor are predicted by experiments and numerical simulations using large-eddy simulation (LES)on FLUENT platform.Temperatures at different positions of the internal wall and mixing conditions caused by T-junctions at different times are obtained,then the simulated normalized mean and root-mean square (RMS)temperature,velocity vector and temperature contour for the two structures,namely with and without a distributor,are compared.The results show that,compared with the a T-junction without a distributor,the mixing region of hot and cold water in the T-junction with distributor moves to the middle of the pipe,and the inclusion of the distributor reduces the temperature fluctuations of internal wall noticeably and makes the mixing of hot and cold water more efficient.Ó2015Elsevier Ltd.All rights reserved.1.IntroductionTee junction is a familiar structure that is universally used in pipeline systems of power plants,nuclear power plants and chemi-cal plants,it is often applied to mix hot and cold fluid of main and branch pipes.The fluctuations of fluid temperature are transported to the solid walls by heat convection and conduction.This can cause cyclical thermal stresses and subsequent thermal fatigue cracking of the piping (Lee et al.,2009).So far,leakage accidents took place in several light water and sodium cooled reactors due to thermal fati-gue.In 1998,a crack was discovered at a mixing tee in which cold water from a branch pipe flowed into the main pipe in the residual heat removal (RHR)system in a reactor in Civaux,France.Metallur-gical studies concluded that the crack was caused by a high degree of cycle thermal fatigue (Eric Blondet,2002).In 1990,sodium leak-age happened in the French reactor Superphenix (Ricard and Sperandio,1996).It has been established that mixing hot and cold sodium can induce temperature fluctuations and result in thermal fatigue (IAEA,2002).Therefore,it is significant to study how to weaken thermal fatigue of the piping wall to ensure the integrity and safety of the piping system in a nuclear power plant.In the analysis of thermal fatigue,temperature fluctuation is a very important evaluation parameter.A reliable lifetime assess-ment of these components is difficult because usually only thenominal temperature differences between the hot and cold fluids are known,whereas the instantaneous temperatures and heat fluxes at the surface are unknown (Paffumi et al.,2013).Kamaya and Nakamura (2011)used the transient temperature obtained by simulation to assess the distribution of thermal stress and fati-gue when cold fluid flowed into the main pipe from a branch pipe.Numerical simulation of flow in the tee has been carried out Simoneau et al.(2010)to get temperature and its fluctuation curves,and the numerical results were in good agreement with the experimental data.Through the analysis on thermal fatigue stress,it draw the conclusion that the enhanced heat transfer coef-ficient and the temperature difference between hot and cold fluids were primary factors of thermal fatigue failure of tees.Many numerical simulations and experiments have been carried out to evaluate the flow and heat transfer in a mixing tee junction (Metzner and Wilke,2005;Hu and Kazimi,2006;Hosseini et al.,2008;Durve et al.,2010;Frank et al.,2010;Jayaraju et al.,2010;Galpin and Simoneau,2011;Aulery et al.,2012;Cao et al.,2012).Turbulent models such as Reynolds-averaged Navier–Stokes (RANS),Unsteady Reynolds averaged Navier Stokes (URANS),Scale-Adaptive Simulation (SAS),Reynolds stress model (RSM),detached eddy simulations (DES),and LES have all been used in industrial applications.As one of the choices of turbulent model for predicting the mixing flow in tee junctions,the RSM can bemused to describe the momentum conservation of the mixing (Durve et al.,2010;Frank et al.,2010).Turbulent mixing phenomena in a T-junction have been numerically investigated using the k $x/10.1016/j.anucene.2015.01.0010306-4549/Ó2015Elsevier Ltd.All rights reserved.Corresponding author.based baseline Reynolds stress model(BSL RSM)(Frank et al.,2010) for two different cases.Durve et al.(2010)applied the RSM to pre-dict the velocityfield of three non-isothermal parallel jetsflowing in an experiment setup used to simulate theflow occurring at the core outlet region of a fast breeder reactor(FBR),with a Reynolds number of1.5Â104.Theflow in tube of different Reynolds numbers (Re)andflow velocity ratio were studied experimentally with three-dimensional scanning using particle image velocimetry(3D-SPIV) (Brücker,1997).Large-eddy simulation(LES)is an alternative turbulence model with different subgridscale models often employed to predict velocity and temperaturefluctuations.Indeed many numerical studies have shown the capability of LES to model thermalfluctu-ations in turbulent mixing.LES was performed(Lee et al.,2009)to analyze temperaturefluctuation in the tee junction and the simu-lated results were in good agreement with the experimental data. Thermal striping phenomena in the tee junction had been numer-ically investigated using LES(Hu and Kazimi,2006)for two differ-ent mixing cases,and the simulated normalized mean and root-mean square(RMS)was consistent with experimental results. LES in a mixing tee were carried out(Galpin and Simoneau, 2011)in order to evaluate the sensitivity of numerical results to the subgrid scale model by comparing the experimental results, and to investigate the possibility of reducing thefluid computa-tional domain at the inlet.Another simulation that mixing of a hot and a coldfluid stream in a vertical tee junction with an upstream elbow main pipe was carried out with LES(Lu et al., 2013).And the numerical results show that the normalized RMS temperature and velocity decrease with the increases of the elbow curvature ratio and dimensionless distance.In the meantime,many scholars have studied how to weaken the thermalfluctuation.Experiments and simulation were con-ducted(Wu et al.,2003)on a tee junction geometry with a sleeve tube in it.Theflow is divided into three types of jets by theflow velocity ratio in main and branch pipes.Through the analysis of flowfield and velocityfield of various jets types,it indicate that the addition of sleeve tube relieve the thermal shock caused by the coldfluid injection rge-eddy simulation have been used(Lu et al.,2010)to evaluate the thermal striping phe-nomena in tee junctions with periodic porous media,the temper-ature and velocityfield inside the tubes are obtained.The research revealed that the addition of a porous reduces the tem-perature and velocityfluctuations in the mixing tube.As mentioned above,experiments and numerical simulations for both tee junction geometry with a sleeve tube in it(Wu et al., 2003)and for a mixing tee with periodic porous media in it(Lu et al.,2010)have been carried out.The results of previous researches provide a good reference value for this work that anal-yses the role of distributor in weakening the thermalfluctuation of internal piping wall,and this structure has not been studied to date,to the best of our knowledge.In this work,mixing processes have been studied by the experiment and numerically predicted with LES.Then the simulated normalized mean and root-mean square(RMS)temperature,velocity vector and temperature con-tour of the two tees are compared.2.Experiment systemThe Experimentflowchart is presented in Fig.1.The experimen-tal system consists of four main components,a cold water supply line,a hot water supply line,a test section,and a data acquisition unit.The experiment device is shown in Fig.2.Experimentfluid was adjusted to the desired temperature by the heater and chiller, and then was pumped to the test section.After mixing thefluid is returned to the heater for recycling,some of the excessfluid is dis-charged through the overflow pipe.During the mixing of thefluids, the temperature of the mixingfluid is collected and recorded by the thermocouple probe installed on the tube wall.The experiment requires two different structures of the test sec-tion,Fig.3is the T-junction section without the branch liquid dis-tributor and Fig.4is that with the branch liquid distributor.The addition of this structure has two main functions:(1)changing the mixing position of hot and coldfluids:moving the mixing zone to the middle of the tube,and away from the main pipe wall;(2) increasing the intensity of mixing process:adding the fence near the outlet of distributor enhanced the mixed disturbance and the exacerbatedfluid mixing of the inner tube.For the convenience of observing and adjusting the mixing process,the test section is a round pipe made of plexiglass,and other pipes are made of steel. Fig.5is the physical model of the branch liquid distributor.The test conditions in the present experiment are shown in Table1.We collected the instantaneous temperature data of every measurement points by the data collector.The distribution of sam-pling points are shown in Fig.6,there are total eight thermocou-ples in the circumferential direction at each plane.In the T-junction section without the branch liquid distributor,the number of the collected plane is6(x/d m=1,2,3,4,6,8).That is to say there are48thermocouples in the structure without distributor.And in the T-junction section with the branch liquid distributor,the num-ber of the collected plane is5(x/d m=2,3,4,6,8),which means there are40thermocouples in the structure that with the distrib-utor.In both structures,the distance between measuring point the thermocouple probe and the inner wall is30mm.Since the collect-ing frequency of the collector is limited,we use1Hz as the collect-ing frequency after theflowfield is stable,and the total number of collection is800s.Table1shows the specific parameters of the test conditions.NomenclatureT time(s)Pr Prandtl numberLs mixing length of subgrid grid(m)T temperature(K)G acceleration of gravity(m/s2)K von Karman numberCs Smagorinsky numberS ij subgrid strain rate tensorM R momentum ratio of main pipe and branch pipe TÃnormalized mean temperaturesTÃrms normalized RMS temperaturesR d diameter ratioR v velocity ratiox,y,z axial coordinate(m)Greek symbolsqfluid density(kg/m3)b coefficient of thermal expansionl viscosity(Pa s)ltturbulent viscosity(Pa s)k thermal conductivity(w/(m k))C P heat capacity(J/(kg°C))K.Gao et al./Annals of Nuclear Energy78(2015)180–187181182K.Gao et al./Annals of Nuclear Energy78(2015)180–1871\4\11-thermometers 2\5\10-pressure gauge 3\9-flow meter 6-c ooler 7-heater8-overflow 12-test sec tion 13-thermoc ouple data c ollec torFig.1.Experimentflow chart.Fig.5.Physical model of the branch liquid distributor(a)the whole graph(b)theprofile map.Fig.2.Experiment device of thermalfluctuation.Fig.3.Schematic diagram of the T-junction section without the branch liquid distributor.Fig.4.Schematic diagram of the T-junction section with the branch liquid distributor.3.Numerical simulationFig.7is the numerical model based on the experimental section of T junction.The size of the model,boundary conditions are con-sistent with the experiment.In which,hot water enters from the left of main pipe,and cold water enters from the branch pipe,finally the mixingfluidflow out of the right of the main pipe.Dur-ing the calculation,the steady results offlowfield and heat transfer are obtained by Reynolds stress model(RSM)firstly,and then set @q@tþ@q u i@x i¼0ð1Þ@q u i@tþ@q u i u j@x j¼À@ p@x iÀq0bðTÀT0Þgþ@@x jlþltÀÁ@ u i@x jþ@ u j@x i!ð2Þ@q T@tþ@q Tu j@x j¼@@x jkc p@T@x jÀq T00u00j!ð3ÞIn these equations,q,b,l,l t,k and c p represent the density,ther-mal expansion coefficient,molecular viscosity,turbulent viscosity, thermal conductivity and specific heat capacity,respectively.The Smagorinsky–Lilly model is used for the turbulent viscosity,which is described as:lt¼q L2s j S jð4Þj S jTable1Experimental conditions.Main pipe Branch pipeFlow rate (m3/h)Temperature(K)Flow rate(m3/h)Temperature(K)Without distributor0.645304.650.270287.65With distributor0.645304.650.266287.65Fig.6.The distribution of sampling points on the planes.Physical model of T-junction(a)without the branch liquid distributor;(b)with the branch liquid distributor.K.Gao et al./Annals of Nuclear Energy78(2015)180–187183ij ¼12@ u i@x jþ@ u j@x ið7Þwhere k is the Von Karman constant of0.42;d is the distance to the closest wall;C s is the Smagorinsky constant of0.1;V is the volume of the computational cell.4.Results and discussionThe normalized mean and root-mean square temperature are used to describe the time-averaged temperature and temperature fluctuation intensity.The normalized temperature is defined as:ü1NX Ni¼1TÃið8ÞN is the total number of sample times.TÃi¼T iÀT cT hÀT cð9Þwhere T i is the transient temperature,T c is the coldfluid inlet tem-perature and T h is hotfluid inlet temperature.The root-mean square(RMS)of the normalized temperature is defined as:TÃrms¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1X Ni¼1TÃiÀTÃ2rð10Þ184K.Gao et al./Annals of Nuclear Energy78(2015)180–187parison of experimental and numerical resultsAs can be seen from the Fig.8,the numerical normalized mean temperature distributions at the plane x/d m=1and the plane x/ d m=2are in good qualitative agreement and in adequate quantita-tive agreement,and most of them are within the experimental deviation of±20%.Meanwhile,the lifting trends of the data are the same.In the direction of180°,the mean temperatures are both minimal.And with the angle decrease to0°,the temperatures are gradually increased.Quantitative differences between the experi-ment and numerical results are that the normalized mean temper-atures given by LES are larger than the experimental data.That is because we did not add insulation unit on tube wall in the exper-iment,leading the transfer of some heat into the air.And in the process of numerical simulation,we ignored the convective heat transfer between the wall and the air.As shown in Fig.9,although the numerical results and experi-mental results have a little difference at the plane x/dm=2around the location of225°and the plane x/dm=2around the location of 0°and315°,all of them are within the error range that can be accepted.Both the simulations and experimental results give a lar-ger mean temperature in the top half of the main pipe than in the bottom half.This verifies the validity of the LES model for predict-ing the mixing of hot and coldfluids in a tee junction.The normalized RMS temperature on the plane x/d m=1and plane x/d m=2are shown in Fig.10,respectively.Similar to the nor-malized mean temperature,the normalized RMS temperature lines agree very well with the experiment ones.Both of the maximum values appear at the bottom half of the pipe.This indicates that the maximum temperaturefluctuations of main pipe appear on the opposite of the branch pipe inlet in this condition.As shown in Fig.11,the numerical results and the experimental results have the same trend and the numerical data are agreed well with the experimental ones.By comparison with Figs.4and5,dif-ferent from the temperaturefluctuations distribution which with-out the branch liquid distributor,there are two peaks of high fluctuation located at the90°and270°directions along with the tube.This is because the direction is that of the outlet of branch liquid distributor,the coldfluidflowing out from the outlet of branch liquid distributor mixes very fast with the hotfluid,leading to dramatic changes of temperature.In summary,the LES simulation results obtained are generally in good qualitative and quantitative agreement with the experi-mental data for the case of T-junction with/without the branch liquid distributor.Based on this,we analyzed the numerical results further.And the results are reported in the section below.4.2.Numerical results with/without branch liquid distributorThe numerical data were sampled on the inner wall in the plane x/ d m=À1,À0.5,0,0.5,1,2,3,4,6and8.At the same time,the numer-ical data were sampled from points every5mm along the intersec-tional lines of planes of y/d m=0and sections of x/d m=À2,À1,0,1,2, 3,4,5and6,to get the points with the maximum normalized rootK.Gao et al./Annals of Nuclear Energy78(2015)180–187185mean square temperatures in the tee and on the top and bottom walls.Here,the temperature and velocityfields were determined with LES simulations for the case of tee junction with/without branch liquid distributor.The temperature contours and velocity vectors for the T-junction are shown in Figs.12and13,respectively.As can be seen in Fig.12,due to the large branch pipeflow velocity,hot and coldfluid mixing zone is mainly located in both upstream and downstream region of the intersections of the main pipe and the branch pipe.The vigorous mixing offluids in the tube leads to thermalfluctuation on the wall.But in the T-junction with the branch liquid distributor,the mixing region moves to the lower half and downstream region of the main pipe.This indicates that the distributor is advantageous to weaken thermalfluctuations on the wall.The same conclusion can be seen from Fig.13,the dis-tributor weaken thermalfluctuations on the wall of downstream region and the top of the main pipe.186K.Gao et al./Annals of Nuclear Energy78(2015)180–187Fig.14compares the normalized mean temperatures between two tees of different structures.As can be seen,wall temperature changes great in the direction of90°,135°,225°and270°in T-junc-tion with the distributor,because the directions are the distributor outlet directions.This indicates that the coldfluid mixes with hot fluid on the wall afterflows out of the distributor.At the same time,the temperature in the direction of180°also changes dra-matically.That is because the coldfluid moves down in the effects of gravity and buoyancy.As shown in Fig.15,for the tee with distributor,the maximum values of normalized RMS temperature are smaller than that of the tee without distributor in most directions.This indicates that the adding of the distributor can relieve thermalfluctuations on the wall to some extent.And for the T-junction with distributor,tem-perature tends to be stable after the plane of x/d m=6,which indi-cates that twofluids have made a full mixing,while for the initial tee,temperature is still in the dramatic change,and this shows that the improved structure can effectively reduce the mixing length.Fig.16shows the maximum normalized instantaneous temper-aturefluctuations in the tee and on the top and bottom walls in the plane y/d m=0.In the tee,the maximum normalized instantaneous temperaturefluctuations of the case without distributor vary from 0.45to0.8,which means that the hot and coldfluids alternate in this location.However,for the case with distributor,the tempera-turefluctuations in the tee as well as on the top and bottom walls are much smaller than those of the case without distributor.That also implies that the distributor can reduce the temperaturefluctu-ation effectively.The normalized instantaneous temperaturefluctuations cannot describe the relationship between power spectrum density(PSD) and frequency of the temperaturefluctuation.PSD against fre-quency is one of the most important parameter for thermal fatigue analysis,which can directly show how PSD is in a certain fre-quency.The PSDs of the points with maximum temperaturefluctu-ation for the cases with and without distributor against frequency were recorded by fast Fourier transform(FFT)and shown in Fig.17. The temperaturefluctuation of the case without distributor has the highest PSD,at the frequency of0.04Hz,whereas the distributor significantly reduces the PSD of the temperaturefluctuations with the frequency from0.01to0.1Hz.In addition,the PSD of temper-aturefluctuations decreases with the frequency increasing.5.ConclusionsAs thermal stratification can result in thermal fatigue in the pip-ing system of a nuclear power plant,safety and integrity evaluation of the piping system has become an important issue.In this work the temperaturefluctuation has been studied by the experiment and numerically predicted by LES for two types of vertical tee junc-tion:one with distributor in the branch pipe and another without. The numerical results of normalized mean and RMS temperatures for the two structures have been found to be in good qualitative and quantitative agreement with the experimental data,which val-idates the use of LES simulations to evaluate convective mixing in such geometries.At the same time,the simulated normalized mean and root-mean square(RMS)temperature,velocity vector and temperature contour of the two tees are compared.The numerical results show that thefluctuations of temperatures of the tee without the distrib-utor are larger than those of the tee with the distributor,which can be explained by the branch liquid distributor enhancing the mix-ing.Although both tees give the same momentum ratio between the main pipeflow and the branch pipeflow,mixing and convec-tive heat transfer are greatly enhanced by the presence of the branch liquid distributor.These all show that the structure is effec-tive for weakening the thermalfluctuation of tee piping wall when hot and coldfluids mix,and it can make the mixing more sufficient.AcknowledgementsThis work was supported by projects of the National Natural Science Foundation of China(No.51276009),Program for New Century Excellent Talents in University(No.NCET-13-0651),and the National Basic Research Program of China(No.2011CB706900). 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Extensive-Form Games: IntroductionÙIntroduction______________________________________________________________________1 Game Trees______________________________________________________________________1 Information sets___________________________________________________________________6 Games of perfect recall_____________________________________________________________9 Arborescences___________________________________________________________________12 Simultaneity_____________________________________________________________________13 Divide and conquer: subgames______________________________________________________13IntroductionWhen we model a strategic economic situation we want to capture as much of the relevant detail as tractably possible. A game can have a complex temporal and information structure; and this structure could well be very significant to understanding the way the game will be played. These structures are not acknowledged explicitly in the game’s strategic form, so we seek a more inclusive formulation. It would be desirable to include at least the following: 1) the set of players, 2) who moves when and under what circumstances, 3) what actions are available to a player when she is called upon to move, 4) what she knows when she is called upon to move, and 5) what payoff each player receives when the game is played in a particular way. Components 2 and 4are additions to the strategic-form description; component 3 typically involves more specification in the extensive form than in the strategic form.We can incorporate all of these features within an extensive-form description of the game. The foundation of the extensive form is a game tree. First I’ll discuss what a tree is and then describe the additional specifications and interpretations we need to make in order to transform it into a full-fledged extensive-form description of the game. We’ll discuss additional restrictions upon the information structure to reflect a common assumption, viz. perfect recall, which asserts that players don’t forget. We’ll discuss the relation between this traditional, graph-theoretic exposition and a more recent one based on the arborescence. After that we’ll learn how to ease our analysis of complicated games by breaking them up into simpler subgames.Game TreesWe’ll construct our definition of a tree by first defining more primitive graph-theoretic entities. (Iyanaga and Kawada [1980: 234–235]). A graph G=(V,E) consists of a finite set V of vertices and a finite set E of edges of V. An edge e˙E is an unordered pair of distinct elements of V; e.g. the edge e=(v,v’)joins the two vertices v,v’˙V.Ù© 1997 by Jim Ratliff , <jim@>, </gametheory>.Consider an alternating sequence of vertices and edges of a graph G, P={v0,e1,v1,e2,…,e n,v n}. P is a path if 1 each edge e i joins its neighbors v i¥1 and v i and 2 no edge is encountered more than once (i.e.e i≠e j when i≠j).1 In this case we say that v i and v j, for i,j˙{1,…,n}, are connected by P and that P runs from v0to v n.2 (Note that a path may encounter the same vertex more than once.) A path is closed if it begins and ends with the same vertex, i.e. v0=v n. A closed path which never reencounters a vertex, except for the first, is called a circuit. I.e. a closed path P is a circuit if v i≠v j whenever i≠j and {i,j}≠{0,n}. In other words a circuit is a single “loop.” A graph G is connected if for every pair of distinct vertices v,v’˙V, v≠v’, there exists a path in G which connects v and v’.Now we’re ready to define a tree. A tree is a connected graph which contains no circuit.3 In game theory we often refer to the vertices as nodes and, consistent with our dendro-metaphor, to edges as branches.4Because a tree is a connected graph, any node can be reached from any other node by traversing a sequence of branches. Because it has no circuits, a tree is unicursal: for any pair of nodes, there is exactly one path which runs from the first to the second.Example: graphs and treesFigure 1: A graph.The sets of vertices and edges in Figure 1 form a graph. The edges depicted graphically are defined bye1=(v0,v1),e2=(v0,v2),e3=(v1,v2),e4=(v1,v3),e5=(v2,v4),e6=(v3,v4),e7=(v2,v3).The following two alternating sequences of vertices and edges are paths:{v0,e1,v1}, {v0,e2,v2,e5,v4,e6,v3,e7,v2,e3,v1}.Note in the second case that the vertex v2 is encountered twice. The sequence {v0,e1,v2} is not a path. Neither is1Let e=(v,v’) and e’=(v’,v) be edges, where v and v’ are vertices. Because edges are unordered pairs, we have e=e’.2This “runs from… to” terminology is not standard, but I find it useful for expository purposes here.3I.e., no circuit can be constructed from vertices and edges belonging to the graph.4My treatment roughly follows that of Kuhn [1953], which is the hugely influential, seminal paper concerning extensive-form games.{v0,e1,v1,e3,v2,e7,v3,e4,v1,e3,v2},because the edge e3 is encountered twice.The two sequences{v0,e2,v2,e5,v4,e6,v3,e4,v1,e1,v0}, {v0,e1,v1,e3,v2,e7,v3,e6,v4,e5,v2,e2,v0}are closed paths. The first is a circuit; the second is not, because the vertex v2 is encountered twice. Figure 2 shows two connected graphs. The first is a tree; the second is not.Figure 2: Two connected graphs: (a) a tree and (b) a nontree.Now we want to enhance our tree to become a game tree. We henceforth assume that the tree has at least two vertices. We bestow one vertex of the tree with the honor of being the initial node, O. (See Figure 3.) This is where the game begins. It is a decision node.wFigure 3: A game tree.If a noninitial node x is incident with two or more branches, it is also a decision node. Let XÓV be the set of decision nodes. (In Figure 3, X={O,å,∫,∂}.) One of these branches leads to the node; the remaining branches lead away from the node. Since branches are unordered pairs of nodes, how do youknow which branch leads to the node and which lead away? It’s simple: For some noninitial decision node x˙X\{O}, find the unique path which runs from the initial node O to node x. That path will contain exactly one of the branches incident with x; this branch is the one which leads to x. The other branches incident at v are the branches which lead away from x. For example, in Figure 3, there are three edges incident at node ∂: g, U, and D. The unique path running from the initial node to node ∂ is {O,L,å,g,∂}. This path contains only the g edge incident at ∂; therefore g is the branch leading to ∂ and U and D are the edges leading away.We interpret each edge e˙E as an action. Let A be the set of all actions and let f:ÙE§A be the function which assigns an action to each edge. The branches which lead away from a decision node represent the actions available at that node. We require that, for a given decision node x˙X, the function f assign a unique action to each branch leading away from x. (Otherwise it would not be well specified what path would result from a particular action taken at x.) The actions available at any decision node x˙X are given by the set Aªxº=fª{e˙E:Ùe leads away from x}º.5 (So A is a correspondence A:ÙXéA.) In Figure 3, AªOº={L,R}, Aªåº=Aª∫º={g,s}, and Aª∂º={U,D}. More than one edge may correspond to the same action (as long as any such pair of edges does not lead away from the same node). For example, in Figure 3, there are two edges which correspond to the action g. In what follows I will often refer to an edge e˙E by its associated action fªeº.6Any noninitial node with only one incident branch necessarily has no actions available. Such a node is a terminal node. Let ZÓV be the set of terminal nodes. The tree in Figure 3 has five terminal nodes: Z={z1,z2,z3,z4,z5}. The decision and terminal nodes partition the game tree’s nodes: i.e. V=X¨Z and XËZ=õ. The game ends whenever a terminal node is reached. We identify the terminal nodes with outcomes of the game. (Sometimes outcome is used to refer not just to a terminal node but to the unique path from the initial node to that terminal node.)So we have established that the game begins at the initial node and play ends at a terminal node. In what order are the intermediate decisions made? Consider two distinct nodes v,v’˙V, v≠v’. We say that v precedes v’, or that v is a predecessor of v’, if v lies on the unique path which runs from the initial node O to v’. We write vüv’. Equivalently, we say that v’ is a successor of v, and write v’öv. For example, in Figure 3, the unique path running from the initial node O to ∂ is {O,L,å,g,∂}. This path contains å but not contain ∫; therefore å is a predecessor of ∂, but ∫ is not.Predecession, then, defines a binary relation ü on the set of nodes V, viz.ü={(v,v’)˙V2:Ùvüv’}.7,8(1) 5Recall that if f:ÙX§Y is a function and SÓX, then fªSº is the image of S under f, i.e. fªSº={fªxº:Ùx˙S}.6For example, in the previous paragraph I referred to the edge connecting å and ∂ by its action g.7 A binary relation R on a set X can be thought of as a subset of the Cartesian product of X with itself, i.e. RÓX˜X, which includes allthose pairs (x,y)˙X2 such that xRy. I.e. (x,y)˙R⁄xRy.8The succession binary relation ö is the inverse relation of ü; i.e. Åv,v’˙V, vüv’⁄v’öv; it could also be called the dual order (except for a minor problem that ü is irreflexive and therefore not actually providing an ordering). (See later footnote.)It is intuitively obvious—and you can show rigorously—that the predecession relation is irreflexive (Åv˙V, not vüv), asymmetric [vüv’flnotÙÙ(v’üv)], and transitive [(vüv’ and v’üv”) fl vüv”]. This binary relation need not be complete; there may exist a pair of nodes v,v’˙V such that neither vüv’ nor v’üv. For example in Figure 3, å and ∫ are unordered by precedence. A partial listing of the precedence relations in Figure 3 is: Oüz3, åüz2, and åü∂.The initial node is a predecessor of every noninitial node and has no predecessor itself. For every noninitial node x˙V\{O}, let Pªxº be the set of predecessors of x; i.e. Pªxº={v˙V:Ùvüx}. For example in Figure 3, Pª∂º={O,å} and Pªz4º={O,∫}. Although predecession provides only a partial ordering in general on the set V of all nodes, you can show that it totally orders the set Pªxº of predecessors of any noninitial node x˙V\{O}. (I.e. if v,v’˙Pªxº and v≠v’, then either vüv’ or v’üv.9)10We say that v is the immediate predecessor of v’ if there is an action available at v (i.e. a branch incident at v which leads away from the initial node) which leads directly to v’. (This requires that vüv’and that a single edge joins v and v’. You can prove that each noninitial node has a unique immediate predecessor.) Alternatively we could say that v is the immediate predecessor of v’ if v=maxüÙPªv’º, which means that all other predecessors of v’ must also precede v.11 In Figure 3, for example, O is the immediate predecessor of å but not of ∂, and å is the immediate predecessor of ∂.We have thus far identified the locations in the tree at which decisions are made—the decision nodes—and created a structure which relates the various decisions by specifying what actions lead to which other decisions and in what temporal order. But we still haven’t said who makes which decisions. We retain our standard player set I={1,…,n}, but we also allow Nature to be player 0 when required. (We will see later that Nature can be given the job of introducing new information into the game.)So now we assign to each decision node x˙X exactly one player i˙I¨{0}. We specify this assignment by a node-ownership function ô:ÙX§(I¨{0}) where each decision node x˙X is assigned to the player specified by ôªxº. The set of decision nodes assigned to player i˙I¨{0} is X i={x˙X:Ùôªxº=i}. The node-ownership function ô thus induces a player partitionÙ={X0,X1,…,X n} of the set X of decision nodes.12 We say that a node belongs to a particular player or that a player owns a particular node.Every terminal node z˙Z represents some outcome which affects the players. Perhaps it’s success for 9Note that if a node x˙V has only one predecessor, so that Pªxº is a singleton set, then Pªxº is trivially totally ordered by the binary relation ü.10In my experience it is conventional to require that a binary relation R on a set X be reflexive (i.e. Åx˙X, xRx) in order that it impose a preordering upon X. (See Debreu [1959: 7] and Iyanaga and Kawada [1980: 958, 960].) Kreps [1990: 364] explicitly weakens this notion by defining that “a set is totally ordered by a binary relation if any two distinct elements of the set are ordered in one direction or the other.” [emphasis added] This allows the concept of totally ordered to be usefully applied to irreflexive binary relations. Note that for reflexive binary relations the two definitions are equivalent. Kreps [1988: 8] provides alternative terminology for this concept by saying that a binary relation R on a set X is weakly connected if for all x,y˙X, either x=y, xRy, or yRx.11Because Pªv’º is totally ordered by the binary relation ü, we can arrange its elements such that v1üv2üÚüv k. Then v k=maxüÙPªv’º.12 A partition of a set A is a collection of cells A1,…,A n, such that 1 each A iÓA, 2 the collection is exhaustive, i.e. A1¨A2¨Ú¨A n=A,and 3 the cells are mutually exclusive, i.e. when i≠j, A iËA j=õ.one and failure for another; the outcomes may represent profit levels for each firm or lengths of prison sentences for each criminal suspect. Whatever the outcomes are, they are events over which each player has preferences. We assume that each player has von Neumann-Morgenstern preferences, which can be represented by a utility function µi :ÙZ§Â with the expected-utility property, and seeks to maximize her expected utility.13Figure 4 shows an extensive-form game elaborated from the tree in Figure 3. I have indicated the player partition, which assigns players to nodes, by numerical labels and assigned payoff vectors to each terminal node. By convention the upper entry in each payoff vector is the first player’s utility derived from that outcome; the second entry is the second player’s utility. The player partition is given by X 1={O ,∂} and X 2={å,∫}. (Nature does not appear in this game.)Information setsIn the first paragraph I identified five aspects of an economic scenario we’d capture using the extensive form. So far we’ve captured the set of players, who moves when and under what conditions, what actions are available to a player when it’s her turn to move, and what payoffs are received at the end of the game. The remaining challenge is to represent the information which a player knows when called upon to move.What kind of information do I have in mind? We need to be concerned about things which have changed during the course of the game as a result of the actions of the players.14 The key to representing information in the game tree is realizing the connection between nodes and history . At any point in the play of the game the history is a record of who did what when. Remember that game trees are13I’m using µi to represent the utility to player i of an outcome z˙Z because I want to reserve the symbol u i for later use as her utility to a strategy profile. (We will see that an outcome z˙Z could be the result of any of several strategy profiles.)14If information is entering the game in some other fashion, this can be modeled by the addition of Nature as a nonstrategic player who chooses her strategy according to a probability distribution. The revelation of Nature’s moves then communicates the exogenouslygenerated information to the players.Figure 4: An extensive-form game indicating the player partition and payoffs as a function of outcome.unicursal—there’s precisely one path which runs from the initial node to any given node. Therefore, if you know the node you have reached, you also know precisely the history of play that got you there.15,16We can divide extensive-form games into two classes: In games of perfect information, whenever it is your turn to move you know precisely where in the tree you are located. Therefore you know the history of play completely. In games of imperfect information, you might be uncertain about the history of play when you are called on to play. This uncertainty about history corresponds to uncertainty about which of several nodes you might be at.To express this uncertainty we employ the information set. An information set hÓX is a set of decision nodes. In the course of playing the game the players’ actions may cause one of your nodes to be reached. When this occurs, you might not know precisely which of your nodes has been reached. This uncertainty is captured by saying that you only know that a particular information set has been reached; all you know is that you’ve reached one of the nodes x˙h in that information set. It is as if you are sent a message, whenever it is your move, specifying the information set you have reached.There are some obvious restrictions we need to impose on the composition of information sets. First, each decision node must belong to exactly one information set. (If a node belonged to more than one information set, it wouldn’t be well defined what message its owner should receive when that node is reached.) Therefore the set H of all information sets forms a partition of the decision nodes: X=U h˙HÙh and Åh,h’˙H, hËh’=õ. We can also define an information-set membership function h:ÙX§H, which specifies an information set h˙H for each decision node x˙X according to hªxº=h⁄x˙h.Secondly, all of the nodes in an information set must belong to the same player. (I.e. Åh˙H,Åx,x’˙h, ôªxº=ôªx’º.) Otherwise, two or more players may think it is their turn to move.17 (Therefore the information partition H is a refinement of the player partition Ù={X0,X1,…,X n}, because for each information set h˙H there exists a player i˙I¨{0} such that this information set is entirely contained within player i’s set of nodes, i.e. such that hÓX i.18)We can partition the set H of information sets into sets of information sets which belong to each player: For every player i˙I¨{0}, the set H i={h˙H:ÙÅx˙h,ôªxº=i} contains the information sets 15Note that this wouldn’t be the case if we were playing the game represented by the nontree of Figure 1b. If we had reached the bottom node, we wouldn’t know whether we had arrived there via the clockwise or counterclockwise path from the top initial node.16You might object that in the real world there often are many different ways to reach the same result. Such a situation can be modeled within this framework. You would still have a different terminal node for each of these histories; but you would assign the same payoff vector to each of these terminal nodes.17“But what’s wrong with that? Perhaps it’s a simultaneous move game. Then it would be perfectly kosher to have two players believing it’s their turn,” you say. In an extensive-form representation a simultaneous-move game is modeled as a sequential game where the second mover doesn’t get to observe the first mover’s choice at least until after the second mover has played. We’ll discuss simultaneity further later.18One partition is a refinement of a second partition if each cell of the first is a subset of some cell of the second. More formally…. Let ˜={M i}i˙{1,…,m} and ˆ={N i}i˙{1,…,n}be two partitions. M is a refinement of N, denoted ˜üˆ, if ÅM i˙˜, ‰N j˙ˆsuch that M iÓN j. In such a case we say that ˆ is coarser than ˜.owned by player i. Therefore, for all i˙I¨{0}, U h˙HÙh=X i.iIn general, for any information set h˙H, ôªhºfi{ôªxº:Ùx˙h} would refer to the image of the set of nodes h under the node-ownership function ô; i.e. ôªhº would be the set of players who owned one or more nodes in h. However, since all nodes in h are owned by the same player, we see that ôªhº is single valued; i.e. Åh˙H, #ôªhº=1. Therefore we can also consider, with some abuse of notation, ô to be an information-set ownership function ôªhº={i˙I¨{0}:ÙÅx˙h,ôªxº=i}. As with decision nodes, we say that an information set belongs to a particular player or that a player owns a particular information set. Thirdly, every node in an information set must have the same set of available actions. Otherwise the player wouldn’t be sure which actions were truly available to her. I.e. Åh˙H, Åx,x’˙h, Aªxº=Aªx’). It is useful then to refer to the set of actions available at an information set h˙H, viz. the image of h under the correspondence A: Aªhº=U x˙hÙAªxº, because for all nodes in the information set Aªhº is the set of actions available there: Åx˙h, Aªxº=Aªhº.We can summarize these last two restrictions byÅh˙H, Åx,x’˙h, ôªxº=ôªx’º and Aªxº=Aªx’º.(2) An information set may be a singleton—it may consist of only a single node. If every information set of every player is a singleton, we have a game of perfect information; i.e. Åh˙H, #h=1, or equivalently: Åi˙I¨{0}, Åh˙H i, #h=1. If any information set contains more than one node—i.e.‰i˙I¨{0}, ‰h˙H i, such that #h>1—then the game is one of imperfect information. When an information set contains more than one node, we indicate this graphically on the extensive form by connecting all of the nodes in that information set with a dashed line. Singleton information sets require no special ornamentation. For example, the game in Figure 4 has perfect information; therefore H1={{O},{∂}}, H2={{å},{∫}}.(3) Return to the perfect information game of Figure 4. If we change the information structure so that player 1’s first move is unobserved by player 2 by the time player 2 makes his first move, then he doesn’t know at which of his two nodes he is located. We indicate this information imperfection by connecting these two nodes with a dashed line. (See Figure 5a.) Each of player 1’s two nodes belongs to its own information set. In this game then there are three information sets: two belonging to the first player and one belonging to the second. Note that, as required, player 2 has the same two actions, viz. g and s, available at both nodes of his information set; Aªåº=Aª∫º={g,s}. Contrast each player’s collection of information sets in this game, viz.H1={{O},{∂}}, H2={{å,∫}},(3) with those given in (3) for the game in Figure 4.The pseudo–extensive form of Figure 5b has two violations of the restrictions upon information sets. One information set consists of nodes belonging to different players. (Consider the information set h={å,∫}, where ôªåº=2 but ôª∫º=3.) A second information set has different actions available to itsplayer at different nodes. (Consider the information set h’={∂,©}, where Aª∂º={U,D} but Aª©º={L,C,R}.)In summary, then, an extensive-form game Ì is defined by a tree (V,E), a distinguished vertex O , a set of actions A , an action-for-edge assignment function f:ÙE§A , an information partition H of the decision nodes X , a player set I , a node-ownership function ô:ÙX§I¨{0}, and player payoff functions µi :ÙZ§Â for i˙I . From this definition the following entities can be derived: the set X of decision nodes,the set Z of terminal nodes, the set of actions Aªxº available at each decision node x˙X , the player partition Ù, and the precedence relation ü on V . It is further required that, for each information set h˙H ,every node in h shares the same player-owner and shares the same set of available actions: Åh˙H ,Åx,x’˙H , ôªxº=ôªx’º and Aªxº=Aªx’º.Games of perfect recallWe’ve established some restrictions on what nodes we can include in the same information set: We can only draw dotted lines which connect nodes which belong to the same player and at which the same set of actions is available. Now I’ll present two extensive-form fragments which might suggest additional restrictions.In Figure 6a, at node å player 1 can choose Left and turn the move over to player 2 or she can choose Right and immediately move again. If she chooses Right, she reaches node ∫, which is in the same information set as the node å from which she had just come. Note what this particular information set construction implies: When player 1 chose Right at node å, she realized what she was doing. When she then reaches ∫, she doesn’t know whether she’s at ∫ or back at å. Therefore she no longer knows that she chose Right when at å; she has forgotten the action she just took.It is typical in game theory to restrict ourselves to games of perfect recall . In such games a player always know what actions she has taken and never forgets anything she ever knew. How do we(a)(b)Figure 5: A game of (a) imperfect information and (b) implausible information.formalize this restriction?19 We can rule out the game fragment in Figure 6a with the following requirement: If two nodes are in the same information set, then neither can precede the other. (Åh˙H ,Åx,x’˙h , neither xüx’ nor x’üx .)However, that requirement isn’t sufficient. Consider Figure 6b. Nodes ∂ and © are both in the same information set. There are two ways player 1 can reach that information set: She can be at node å and choose stop, or she can start at node ∫ and play Up. Note that å and ∫ are not in the same information set; therefore they represent different information—different histories. If player 1 reaches å she knows that she reached å but never passed through ∫; if she reaches ∫ she knows she reached ∫ but never passed through å. But if she reaches the {©,∂} information set via node å, say, she cannot possibly know that she didn’t pass through ∫. (If she did know this, then she would know for certain that she is at node ∂. But that would violate the information set’s knowledge restriction that she doesn’t know which of the two nodes ∂ and © she is at.) Therefore she has forgotten something she knew earlier; so this would be a violation of perfect recall.The extensive-form fragment of Figure 6b doesn’t violate the prohibition that members of the same information set should not be ordered by precedence; so we need a stronger restriction. The formal characterization of this restriction is extremely arcane and abstruse.20 I’ll try to motivate it the best I can.Consider two nodes ∂ and © which belong to the same information set for player 1. Now consider a player-1 node å which precedes ∂ when player 1 chooses the action a at å. (See Figure 7a.21) When19There are two reasons to be interested in perfect recall. The first is that most if not all games we would be interested in analyzing would satisfy this property. (But for precisely this reason there would be little point in formalizing the restriction.) More important theoretically, however, is that the restriction to games of perfect recall allows a simplification in the analysis of these games: We will see later that in games of perfect recall there is an equivalence between mixed strategies and behavior strategies such that we can choose whichever analytical perspective is more convenient.20See Fudenberg and Tirole [1991: 81], Kreps and Wilson [1982: 867], Kreps [1990: 374–375], and Myerson [1991: 43–44]. The property so defined will prove sufficient for a later demonstration of the equivalence between mixed and behavior strategies. Kuhn [1953: 213] defines a weaker property, which he also calls perfect recall. Although this is a misnomer, his property is necessary and21∂but not necessarily an immediate predecessor.(a)(b)Figure 6: Two game fragments with forgetful players.。
中山大学远程学习方法与技术2012年上半年第二次作业问题 1 得 2 分，满分 2 分一、单选题：（每题 2 分，25 题共50 分）1、以下关于远程教育的看法有误的一项是（）所选答案：D、利用电视、卫星系统、网络实现的教育形式就叫远程教育问题 2 得0 分，满分 2 分2、对于远程教育改变了学生学习的模式的说法有误的一项是（）所选答案：C、问题 3 得 2 分，满分 2 分3、以下哪一项不是远程教育的主要优势？（）所选答案：B、提高全民素质问题 4 得0 分，满分 2 分4、远程教育中最常见的课程教学模式是（）所选答案：A、问题 5 得 2 分，满分 2 分5、远程教育与常规面授教育最本质的区别在于（）所选答案： A 师生处于准永久性分离状态问题 6 得0 分，满分 2 分6、中山大学网络教育学院成立于( )年，2004 年与高等继续教育学院合并，学院全称为高等继续教育学院（网络教育学院），学院办学主要包括远程教育、成人学历教育、继续教育和自学考试。
所选答案：C、问题7 得 2 分，满分 2 分7、中山大学认真贯彻教育部“积极发展、（）、强化服务、提高质量、开拓创新”的远程教育发展方针所选答案： A 、规范管理问题8 得 2 分，满分 2 分8、学生学习资格的获取需要通过哪个环节？（）所选答案： A 、交费注册问题9 得 2 分，满分 2 分9、每学期所要学习的课程最好不要超过多少门？（）所选答案： B 、 5 门问题10 得 2 分，满分 2 分10 、关于网络课件的功能区主要作用是（）所选答案：B、功能区像个文具盒，提供了各种学习小工具，如笔记本，计算器等问题11 得 2 分，满分 2 分11 、小张是一名新生，他非常喜欢登陆学习平台学习，并积极完成平台上的作业和测试题。
有一次他想查询自己的学习成绩，打开平台后却发现自己的成绩显示着“，! ”他不清楚究竟是为什么，然后就仔细的查看教材，马上就弄明白了，原来（）所选答案：D、“！”表示作业中有主观题，暂时不能给定分数问题12 得 2 分，满分 2 分12 、小陈是一名远程教育新生，他想在学习平台上咨询关于期末考试方面的问题，请问他在哪个交流区提问，会得到更快更好的答复？（）所选答案：A、网络教育学院教务问题交流区问题13 得 2 分，满分 2 分13 、小丽想在学习平台上进入自己中心的学习交流区，那么，他可以在学习平台的哪个栏目找到？（）所选答案：B、我的组织问题14 得 2 分，满分 2 分14 、下列做笔记的方法，比较适合听课时用的笔记记录法是（）所选答案：A、5R 笔记法问题15 得 2 分，满分 2 分15 、下列哪句名言说明了掌握学习方法的重要性？（）所选答案：B、授人以鱼不如授人以渔问题16 得 2 分，满分 2 分16 、拟定学习时间表、每月计划表后，我们可以有规律地进行学习，并能够按照自己的计划行事。
Assignment-Form---中山大学经济管理实验教学中心首页
University
ERP系统应用与实践Course:
Instructor: 张宏斌Homework: 第一次作业：MRP模块
梁颖敏、陈丽平、温伟江、刘娜Name:
Class: 08MLE
Date:…20100413…
University
本次作业MRP模块的操作主要包括以下几个部分： 实验设计
实验流程设计
实验流程
实验收获与问题
University
实验设计
某企业生产的一款蛋糕烤盘的产品结构如图所示
款蛋糕烤盘生产的相关信息如下：
◦TA-KP和TA-DP由企业生产
◦TA-PP、TA-TL、TA-XJ、TA-GJ采购，采购价格及采购提前期详见下表：
该款蛋糕烤盘生产的相关信息如下：
◦TA-KP的售价为40元/个，预测4月的销售量为2000件，已接到一张需求量为400件的订单，要求发运时间为4月20
第一周第二周第三周第四周
预测量500 500 500 500
订单量400
University
可用量100 100 200 200
生产计划500 500 500 500
现有量100
该款蛋糕烤盘生产的相关信息如下：
◦TA-DP经由自动喷涂工序（TA-PTA）生产，所需设备为归属于喷涂车间（TA-PTD）的自动喷涂生产线（TA-PTR）◦TA-KP经由装配工序（TA-ZPA）生产，所需设备为归属于装配车间（TA-ZPD）的装配生产线（TA-ZPR）
◦TA-PTR和TA-ZPR的数量均为1，均采用一个白班的班次生产，生产能力均为每天100个
实验流程设计
1.创建物料
2.创建物料清单
3.创建资源
4.创建部门
5.创建工序
6.创建工艺路线
7.创建价目表
8.创建销售订单
9.创建预测
10.定义和装载主需求计划
11.定义和运行物料需求计划
12.生成生产工单
13.生成采购计划
14.创建采购订单
15.采购接收入库
16.查找并释放生产工单
17.生产子装配件
18.子装配件完工入库
19.生产产成品
20.产成品完工入库
实验流程
责任：制造和分销经理
University
组织：M1
创建物料
创建TA-KP
复制自“成品”模板，分配给M1
创建TA-DP
复制自“子装配件”模板，分配给M1
创建TA-PP、TA-TL、TA-XJ、TA-GJ
复制自“采购物料”模板
依据实验设计在“采购”属性的“价目表”栏输入价格，在“总计划”栏输入计划员
分配给M1
将上述物料的“MPS/MRP计划”属性中的“追溯”改为“总装/软反查”
创建物料清单
创建TA-DP的BOM
在“物料”栏输入TA-DP
在“组件”栏输入TA-PP、TA-TL，依据实验设计输入所需数量
创建TA-KP的BOM
在“物料”栏输入TA-KP
在“组件”栏输入TA-DP、TA-XJ、TA-GJ，依据实验设计输入所需数量
创建资源
创建自动喷涂生产线TA-PTR
在“类型”栏选择“设备”
在“单位”栏选择“HR”
在“费用类型”栏选择“WIP移动”
创建装配生产线TA-ZPR
在“类型”栏选择“设备”
在“单位”栏选择“HR”
在“费用类型”栏选择“WIP移动”
创建部门
创建喷涂车间TA-PTD
在“资源”栏选择TA-PTR
取消“24小时可用”，在“数量”栏输入1
选择白班，在“数量”栏输入1
University
创建装配车间TA-ZPD
在“资源”栏选择TA-ZPR
取消“24小时可用”，在“数量”栏输入1
选择白班，在“数量”栏输入1
创建工序
创建自动喷涂工序TAPA
在“资源”栏选择TA-PTR
在“倒数”栏输入每小时产量12.5
在“计划”栏选择“是”
创建装配工序TAZA
在“资源”栏选择TA-ZPR
在“倒数”栏输入每小时产量12.5
在“计划”栏选择“是”
创建工艺路线
创建TA-DP的工艺路线
在“物料”栏选择TA-DP
在“代码”栏选择TAPA
创建TA-KP的工艺路线
在“物料”栏选择TA-KP
在“代码”栏选择TAZA
创建价目表
查找“Corporate”
在“Corporate”的列表行新建一行
在“产品属性”栏选择“项目编号”
在“产品值”栏输入TA-KP
在“价格”栏输入40
创建销售订单
在“客户”栏选择“Business World”
在“行信息”的“订购项目”栏输入TA-KP
在“日期”栏输入4月20日
在“数量”栏输入400
将“发运”中的仓库改为M1
记录订单编号为64031
创建预测
在“预测集”栏输入TA-KPS
在“预测”栏输入TA-KPC
University
在“预测物料”的“物料”栏输入TA-KP
在“详细资料”的“时段”栏选择“天”
在“详细资料”的“日期”栏输入2010/4/10
在“详细资料”的“数量”栏输入100
定义和装载主需求计划
定义主需求计划TA-MDS
在“目标计划”栏输入TA-MDS
在“来源类型”栏选择“特定预测”
在“来源名称” 栏输入TA-KPS
在“改写” 栏选择“所有分录”
在“数量类型”栏选择“原始”
在“冲减”栏选择“是”
在“正向冲减天数”栏输入5
在“反向冲减天数”栏输入5
查看主需求计划
定义和运行物料需求计划
◦定义物料需求计划TA-MRP
◦在“选项”中的“计划”栏输入TA-MDS
◦选择“计划能力”和“反查”
◦保存，点击“生成”、“确定”、“提交”
◦记录MRP运行请求编号为4557496
University
查看物料需求计划
生成生产工单
◦查找物料需求计划TA-MRP
◦在“供应/需求”中查找物料TA-KP和TA-DP
◦选中“选定发放”
◦保存
◦点击“发放”
◦查看后记录生产申请号为4557522
生成采购计划
◦查找物料需求计划TA-MRP
◦在“供应/需求”中查找物料TA-PP、TA-TL、TA-XJ、TA-GJ ◦选中“选定发放”，在“实施方式”栏选择“采购申请”
◦保存
◦点击“发放”
◦查看后记录采购申请号为4557523
创建采购订单
◦删除采购员，点击查找
◦选择物料TA-PP、TA-TL、TA-XJ、TA-GJ的采购申请，点击“自动”
◦点击“发运”
◦点击“分配”，在“子库存”栏输入”RIP”
◦保存
◦点击“审批”、“确定”，状态自动改为“批准”
University
◦记录订单号为5566
采购接收入库
◦输入采购订单号5566进行查询
◦勾选物料TA-PP、TA-TL、TA-XJ、TA-GJ
◦保存
◦查看后确认子库存RIP中上述物料的现有量与采购订单中的采购量相符
查找并释放生产工单
◦查找并释放TA-DP的生产工单
在“装配件”栏输入TA-DP进行查找
在菜单“工具”中点选“更改状态”后选择“释放”
保存
◦查找并释放TA-KP的生产工单
在“装配件”栏输入TA-KP进行查找
在菜单“工具”中点选“更改状态”后选择“释放”
保存
生产子装配件
◦在“装配件”栏输入TA-DP
◦输入工序序号
◦工序步骤从“排队”到“移动”
◦在“处理量”栏输入拟生产数量400（与排队量相等）
子装配件完工入库
◦在“装配件”栏输入TA-DP
◦点击“继续”
◦在“子库存”栏输入“RIP”
◦在“数量”栏输入子装配件生产的处理量400
◦点击“完成”
◦查看后确认子库存RIP中TA-DP的现有量为400
生产产成品
◦在“装配件”栏输入TA-KP
◦输入工序序号
◦工序步骤从“排队”到“移动”
◦在“处理量”栏输入拟生产数量400（与排队量相等）
产成品完工入库
◦在“装配件”栏输入TA-KP
◦点击“继续”
Lingnan MBA Center
Lingnan (University) College, Sun Yat-sen University
◦在“子库存”栏输入“FGI”
◦在“数量”栏输入产成品生产的处理量400
◦点击“完成”
◦查看后确认子库存FGI中TA-KP的现有量为400
查看产成品库存
实验收获与问题
实验收获
◦增进了对业务流程的了解
◦增进了对业务运作与系统运作间的关联关系的了解
◦进一步熟习了ORACLE MRP模块的应用
实验中遇到且尚未解决的问题
◦预测如何对主需求计划产生影响？
◦各种物料的现有量、外购物料的提前期如何录入系统以及如何对主需求计划、物料需求计划产生影响？。
