homework 1
清华大学随机过程答案1
3. 质点在直线上做随机运动，即在 t = 1, 2, 3, · · · 时质点可以在 x 轴上往右或往左做一个 单位距离的随机游动。若往右移动一个单位距离的概率为 p，往左移动一个单位距离的 概率为 q，即 P {ξ (i) = +1} = p，P {ξ (i) = −1} = q，p + q = 1，且各次游动是相互统 ∑n 计独立的。经过 n 次游走，质点所处的位置为 ηn = η (n) = ξi。
参考答案：
(1) V = a 时，一条样本轨道为典型的正弦曲线。 2
(2) ξ (0) = 0，fξ(0)(x) = δ(x)；ξ (π/2ω) = V ，其概率密度同 V 一样。
(π) ξ
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=
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解法二：因各次游走是相互统计独立的，则 E [η (n)] = E[ξi] = (p − q)n。
i=1
清华大学电子工程系版权所有
3
(2) 假设 n1 < n2,
Rηη (n1, n2) = E [η (n1) η (n2)] = E {η (n1) [η (n1) + η (n2) − η (n1)]} = E[η (n1)]2 + E [η (n1)] E [η (n2) − η (n1)] = {E [η (n1)]}2 + V ar [η (n1)] + (p − q)2n1 (n2 − n1) = (p − q)2n21 + n1V ar [ξi] + (p − q)2n1 (n2 − n1) = (p − q)2n1n2 + n1[1 − (p − q)2]
上外全新版大学英语综合教程讲义-book1-unit6
Book one Unit 6 Animal IntelligenceText AThe first time:The first period:I. Preview:1. Let the students be familiar with some important words and phrases in Text A, which will enable them to better understand the text: controversy, encounter, convince, suspicious, blank, relieve, figure out, assess, inaccessible, release, deceive and wipe out.2. Ask the students to search for some information about animal intelligence through Internet or the Encyclopedia while keeping the question in mind that whether animals do have intelligence and what kind of presentation they may show.3. Read Text A in general and paragraph 5&6 in Text B.II. Check the homework left last time when the fifth unit was finished.III. Introductory remarks (Ts’ words):In unit 5, we’ve talked about romance. Of course, it’s about the romance of us human beings. But, do we believe that animals will also fall in love? In this unit, Text B, we have examples showing that animals do have certain affections to the one of the same kind but the opposite sex. As we’ve read paragraphs 5&6 in Text B, the point has been proved in someway.Well, from the stories, we see that animals really have their own way of thinking. But, what animals really think? Do they have intelligence as we do?IV. Check the preview work through discussion:1. What’s the meaning of intelligence? And, what about animal intelligence?→Intelligence refers to the power of perceiving, learning, understanding, and knowing. It also means mental ability.As for animal intelligence, it involves such factors as the ability to learn, to solve new problems, or to create novel solutions to familiar problems and reasoning. The last one means that some animals have the ability to benefit from the experiences of others. That’s what we can find in the third story in Text A, that Towan is clever to do the similar trick as Melati.2. Ask Ss to tell some interesting stories concerning animals’ presentation of intelligence. Here are also two samples.→Story one: My parents once raised a dog named Doudou. One day, my father left his coat behind in the scallion field. Doudou stayed beside the coat until my father returned to the scallion field and looked for his coat. Then, he went home happily together with my father.Story two: Once in Yunnan Province, a boa saved the son of his master from a river.Ts may list some other animal behaviors to arouse the Ss’ interests:E.g.: Bees communicate a food source to other members of the hive by means of various “dancing”.Some other species can communicate by sight, taste or odor, electrical impulse (fish), or touch, and many animals have more than one system.Conclusion: From all above, we find that animals do display intelligence in some areas. Let’s read the text and find more examples of animal intelligence.V. Global reading tasks.The second period: Detailed reading Tasks:Part 1 (paragraphs 1-2)I. Let the students reread the first two paragraphs, and think about the questions:1.What is the traditional way to examine the intelligence of animals?→experiments designed to teach human signs.PS:For example, in one early study, the intelligence level of animals was measured in terms of the human capacities required to perform parallel acts. Such a method led to gross overestimation of intelligence of the animal being studied. (Ts may take this as part of the Background Information.)2.What is the author’s idea?→to observe the animal’s performance for their own purposesII. Difficult Sentences:1. Wondering whether there might be better ways to explore animal intelligence than experiments... not when scientists ask them to. (Paragraph 1)------ I want to know if there might be better ways to find out more about animal intelligence than those experiments which are designed to teach human language to animals. Then I realized one thing that seems to be very clear now. If animals can think, they will probably do their best thinking for themselves. Animals will not think very hard when scientists ask them to do so.2. Most do not study animal’s intelligence, but they encounter it, and the lack of it, every day. (Paragraph 2)----- Most of the zookeepers do not research on animal intelligence, but they meet animal intelligence unexpectedly every day. They also meet the situation that animals show little intelligence everyday.3. The stories they tell us reveal what I’m convinced is a new window on animal intelligence: the kind of mental feats animal perform when dealing with capacity and the dominant species on the planet—humans. (Paragraph 2)----- The stories they tell us show something that I believe is a new aspect of animal intelligence. They tell us about the marvelous thinking that animals do when they deal with the situation of being caught and human, the most powerful species on the earth.III. Important Words and expressions:controversy over/about something encounter problems/difficulties/oppositions/resistanceconvince somebody of/to do something do one’s best thinkingserve one’s own proposePart 2 (paragraphs 3-6)I. Reread this part and do the exercises:1. True or false questions:a. Colo’s example indicates that some animals know quite well how to negotiate with the keepers. _T__b. Colo is a good bargainer, and this case shows that animals may really have intelligence. T_c. Mile’s experiment to teach Chantek to share things proves to be a great success. F_→Mile’s experiment to teach Chantek to share things proves to be a great failure.2. Ask the students to fill in the blanks with the information mentioned in this part, esp. the animals’ action. Then figure out the author’s purpose.a. Colo: Broke the key chainAuthor’s purpose:Animals know how to negotiate with people.b. Chantek: 1) Expanded the money supply by breaking plastic chips in two;2) Found pieces of tin foil and tried to make copies of metal chipsAuthor’s purpose: Animals can even handle the moneyIII. Difficult Sentence:Arriving on the scene, Jendry offered Colo some peanuts, only to be met with a blank stare. (Paragraph 3) ----- When she got to the place where Colo was, Jendry gave Colo some peanuts, but here good will was not well accepted and she got a blank stare from Colo.IV. Important Words and Expressions:a blank stare maintain eye contact raised the stakes careful bargainerspend on treats trade one thing for another catch up with somebody/somethingonly to do why (not ) do be suspicious of/aboutHomework: 1. Do the exercises of structure on pages 181&182. (only to do, why do/not do)2. Preview the left part of the text.The second time:Part 3 (paragraphs 7-10)I. Check the homework and go on to the new stories.II. Five minutes are given to the students to finish reading this part; two questions will be put forward.1. What did Orky do in this section?→Allowed somebody to stand on his head to reach up and release the baby2. What can you get from this example?→Animals can assess the situation and make a right decision.III. Difficult sentence:This is true, but I do not think it goes far enough. (Paragraph 7) ----- What behaviorist say is right, but I do not think their explanation helps very much.IV. Important Words and Expressions:in one’s interest make judgments emergency care go wrong assess based onthrive at throw up size up keep steady slide intoPart 4 (Paragraphs 11-12)I. Ask the students to read through this part quickly , then to answer the following questions.1.How did Melati and Towan play the trick on Shewman ?→Melati: Hid an orange in her other hand. Towan:Hid an orange underneath his foot.2.What did the author want to tell us？→Animal intelligence can be seen in their attempts to deceiveII. Let the students do some true or false questions according to Part 3 and Part 4.1.Behaviorists say that animals cooperate with human beings for their own benefit. ( T )ule believed that Orky, a killer whale, was the most intelligent animal she had ever seen. ( F )→Laule believed that Orky, a killer whale, was the most intelligent animal she had ever worked with.3.Some animals’intelligence can be seen in their attempts to deceive. ( T )4.Orky allowed somebody to stand on his head to reach up and release the baby because he had been trained to do so.( F )→Orky let somebody stand on his head to reach up and release the baby, but he had not been trained to do so.5.Animals can learn from each other in playing some tricks. Towan is a good example. ( T )III. Difficult sentence:Instead of moving away to get it, Melati looked Shewman in the eye and held out her hand. (Paragraph 11)------ Melati looked directly and steadily at Shewman without moving away to get it.Part 5 (Paragraph 13)I. While reading this paragraph, think about the function of the last paragraph. How do you understand it?→It concludes the whole passage--- animal intelligence not only exist but also plays a very important role in their survival. II. Difficult sentences:1. If life is about survival of a species---- and intelligence is meant to serve that survival--- then we can’t compare with pea-brained sea turtles, which were here long before us and survived the disaster that wiped out the dinosaurs.----- If the purpose of life is to survive and intelligence is for survival, then our intelligence can’t be compared with that of the pea-brained turtle. They were on this earth long before we human beings and they even survived the survived the disaster that destroyed thedinosaurs.2. … even if their horizons are more limited than ours.---- even though they can’t assess the world in the same way as we do because of more limited knowledge and experience.III. Important Words and Expressions in the two parts above:deceive somebody into doing something look somebody in the eye hold outinaccessible (get/have access to somebody/something ) move off hold one’s gaze steadilygive in be meant to do something wipe outPost-reading tasks:Ts may give a general conclusion about the whole text and answer the questions raised by the students.Homework: 1. Review the important words we’ve learned.2. Finish all the exercises in the book.The third time:I. Check the homework and do some exercises:a. Words Dictation:intelligence controversy encounter convince dominant suspicious blank negotiate maintain relieve undertake thrive emergency release evidence deceive survival disaster figure out wipe outb. Paragraph dictation: The following paragraph will be read three times; Ss should listen to the paragraph carefully for the first two times. For the third time, Ss should write down sentence by sentence of what they heard. Then check it.Why don’t birds get lost on their long flights from one place to another? Scientists have been puzzled over this question for many years. Now they fill in the blank.Not long ago, experiments showed that birds rely on the sun to guide them during daylight hours. But what about birds that fly by night? Tests with artificial stars have proved that certain night flying birds are able to follow the stars in their long-distance flights. What do they do when the stars are hidden by clouds? Apparently, they find their ways by such landmarks as mountain ranges, coast lines and river courses. But when it’s too dark to see these birds circle helplessly, unable to get their ways.II. Deal with the exercises of Text A. (partly & selectively)III. Writing skills:The passages in this unit are expositions. The writing style is not strange to us since we have come to learn about it in Unit Two. The purpose of an exposition is to make something clear to the reader, and its focus is to explain, define and interpret. It is widely used in our daily life. Students’ attention can be directed to some writing techniques of an exposition.1. Using examples to support to a topic. For example, in this article, the author employs the stories of gorilla, three orangutans, and a killer whale to support his point.2. Using transitional devices to make a coherent writing.a. Headings and subheadings provide natural transitions between paragraphs and sections.For example, in this text we have three subheadings for three parts of the main body of the passage. They are: “Let’s Make a Deal”, “Tale of a Whale”, “Primate Shell Game”. The subheading directs readers’attention to another demonstration of animal intelligence.b. Transitional sentences.c. Conjunctions.The sixth periodText B Pre-reading tasksI.Before explaining text B, we will do some listening work together. Listen to the following passage carefully, then, answer the following questions.When lionesses give birth to their young, they usually have three or four cubs to a litter, and stay with them for about two years to protect them from danger when they are tiny, and to teach them the fine art of hunting. They have a very closely-knit family life. But when a lioness has her young she is usually in a very nasty protective disposition. But she is very patient and accommodating towards her cubs. A unique thing about lionesses is that they will nurse cubs from another litter besides their own. They never care less whose cubs they are.II. Ask two Ss to give a report about myquestion left last class. Do animals love each likehuman beings?(hint: some animals do love each other deeplyand faithfully. What’s more, their wooingmethods are even more fantastic. Their devotedaffection is touching to some degree. ) Onepicture about some animals’wooing methodswill be presented to help the Ss have a deepunderstanding of animals’ love.Global-reading tasks (15 minutes )I. Ss are given 10 minutes to finish scanning thetext, to count how many examples arementioned in the text. ( 5: Ado, a male parrot,Timmy, Coyotes, Tibby )II.Then ask the Ss to tell the organization of thetext.Part one: ( 1-2 ) Some animals do displayvarious feelings under different circumstances.Part two: ( 3-14 ) Concrete examples areoffered to prove that animals may fall in love.Part three: ( 15 ) Though rejectionremains, the fact that animals experienced, areexperiencing, will experience joy, anger, sorrow,and happiness everywhere. ( the seven humanemotions: joy, anger, sorrow, fear, love, hate, desire) ( 15 minutes )Starling:椋鸟blue tit:蓝山雀quack: 呱呱叫Detailed-reading tasksPart one: Group discussion1. After scan the fist part in two minutes, Ss have a discussion on the following topics:Topic one: what’s true love? Reference: Two persons must love, trust one another. They should help, care, and cherish each other in any case. If one is ill, disabled, or is dying, another is willing to accompany and take good care of her or him wholeheartedly, not to desert him or her. If necessary, one may dedicate his or her life to his or her lover.Topic two: How do you think animals nurse their babies affectionately?Reference: to a large extent, it should be thought as natural instincts.Homework: Ask Ss to read paragraph3, 5, 6, 7, 10, 12, and 13 carefully, and describe those animals’ behavior mentioned in three—four adjective phrases respectively.Ado in para3: silent, depressed, and alone Athan: indifferent, glad, happyTimmy: reluctant, excited, sad, hurt Coyote: affectionate, gentle, happyTibby: get lost, strange, nostalgicThe fourth timeI First check the answers of the homework left last class. Then go on with what we stopped last time.II. Structure:a. what about-----used to make a suggestion e.g.: What about dinner at my place next week?b. how about----used to make a suggestion about what to do e.g.: No, I’m busy on Monday. How about Tuesday at seven?How about doing somethingIII. Useful expressionsBe cautious about doing sth 干----时很谨慎prefer to do/doing 更喜欢prefer A to B 比起B,更喜欢AAdequate parental care 足够的亲本照顾 a sense of loss 失落感IV. ensure ( US insure ) assure guaranteeAssure: tell sb positively or confidently 向-----保证，cause sb to be sure or feel certain about ~ sb of sth 使确信E.g.: I ~ you they will be perfectly safe with us. Assurance n..保险Ensure: make sure guarantee 确保，保证 e.g.: Please ~ that all the lights are switched off at night.Guarantee: promise (usually in writing) in a transaction will be fulfilled e.g.: the watch comes with a year’s ~.Part twoI. Ask the Ss to underline Ado’s, Athan’s，Timmy’s, coyotes’, and Tibby’s behavior respectively in the text. Then tell us theimplied meaning.Ado behavior: He stood…body, he hung his head…became vacant. didn’t have heart…sharply. Pulled himself together …another mate.Implied meaning: fell sorrowful and a sense of loss.Parrot: behavior: ignored a fine-feathered young female but thought an older female in extremely poor condition was the love of his life. The two birds…eventually produced young.Implied meaning: Instinct may…they will love.Timmy: behavior: declined to…at once.Implied meaning: Many species…of their species.Coyotes: behavior: Observations indicate that…sexually active. Pairs can be observed…curled up to sleep.Implied meaning: There is evidence… each other.Tibby：behavior Tibby made a habit…village. She tried to…his house. Acted strangely, even trying to follow him indoors.Implied meaning: An animal raised… when it grows up.II. Important language points :Lay ~ sth aside: put sth aside lay-off: dismissal of a worker, esp for a short time. Layout: arrangementVacant:not filled or occupied a ~ post blank: a ~ stare, look, a ~ mind 茫然的心情Compare A to B 把A 比成B compare A with B 把A和B向比较Shed/cast light on sth: make sth clear or easy to understandIII. Let the students translate the following phrases into Chinese:Put human emotions in animals 赋予动物人类的情感Cross the bridge of reality 逾越现实的鸿沟Whatever distinctions may be made between the love of two people and the love of two animals, the essence frequently seems the same. 无论人类的男欢女爱与动物雌雄相悦之间有多少取别，两者的本质常常是一致的。
Homework(1)
HomeworkI.Put the following into English.1.下公共汽车________________________________2.上火车____________________________3. 到达学校（三种）____________________________________________________________4. 到达英国（三种）____________________________________________________________5. 到家（三种）_________________________________________________________________6. 复习功课（二种）______________________________________________________________7. 做眼保健操______________________________8. 做早操____________________________ 9. 上学前__________________________________10. 出墙报___________________________ 11. 帮助他人_______________________________12. 感到快乐_________________________ 13. 老人福利院_____________________________14. 帮助他人_________________________ 15. 玩球类游戏_____________________________16. 在这一天_________________________ 17. 结束（两种）___________________________18. 在这一天__________________________19. 很高兴与我们在一起_________________________________________________________20. 帮助他们洗衣服_____________________________________________________________21. 在食堂吃午饭_______________________________________________________________22. 在音乐教室唱歌_____________________________________________________________23. 在电脑房打游戏_____________________________________________________________24. 在操场上踢足球_____________________________________________________________25. 总是_____________ 通常____________经常__________有时候____________plete the sentences according to the given Chinese.(参考TBP.50)1. can (否定形式) 1. ___________2. _________ 2. must （否定形式）1. _________2. ________3. need （否定形式）1. _____________2._____________4. may （否定形式）_____________5. 学生们不应该在课堂上吃或喝。
卫星无线蜂窝网络-homework1
1、How a cellular phone call is made？Cellular telephone, sometimes called mobile telephone, is a type of short-wave analog or digital telecommunication in which a subscriber has a wireless connection from a mobile phone to a relatively nearby transmitter. The transmitter’s span of coverage is called a cell. As the cellular telephone user moves from one cell or area of coverage to another, the telephone is effectively passed on to the local cell transmitter.A cell phone is - put simply - a kind of two-way radio that acts in the same way as a telephone. However, naturally a cellular phone is much more complex than a simple walkietalkie, as cell phones permit two people to speak at the same time, making calls and receiving calls, three-way calls, call holding, voice mail, text messaging, etc.However, similar to walkie talkies, all mobile phone calls are entirely unprotected and can be intercepted by other devices. Cell phones aren't at all as "secure" as wired telephones. They have the same security level as a radio - which, after all, they are.When you place a cellular phone call, you dial the number and press the send button. A number of steps then follow:●Your cell phone scans for the nearest base station in order to provide it with thestrongest signal and, in turn, the best possible connection. It checks 21 different control channels to determine the strongest available signal.●Your cell phone then selects the strongest signal for its use.●An origination message (a very short message of about ¼ second in length) isthen sent by the cellular phone, which includes its MIN (Mobile Identification Number, that is, your cellular phone number), as well as the ESN (Electronic Serial Number), and the number that has been dialed.●Once the cellular service provider verifies that you are among its customers -based on the sent-out MIN and ESN - the base station sends a channel assignment message to the cellular phone (another ¼ of a second in length), telling the phone where the conversation will be.●The cell phone then tunes into that assigned channel and the call begins.All of this has happened by the time you hear the ringing or busy signal on the other end of the phone.2、Why the cellular pattern is hexagon?In a cellular System a land area is divided into regular shaped cells, which can be hexagonal, square, circular or some other irregular shapes, although hexagonal cells are conventional. This is because there are some criteria for the cell shape, which are ●Geometric shape●Area without overlap●Area of the cellAnd the eligible shapes for these criteria are Square, circle, equilateral triangle&hexagon.The Geometric shape & Area without overlap is satisfied by a hexagon,square, equilateral triangle as they can be fitted in a manner where there is no area of overlap. The circle on the other hand would overlap (which implies interference of signals) or leave gaps (which means loss of coverage in those areas) when not overlapping.When the area factor is considered a circle has the highest area however it does not satisfy the second criteria of overlap. Therefore we have to consider a shape which fits correctly and also has maximum area.For this purpose we shall compare the area of the remaining shapes to the area of circle to see which has the maximum area.The area of an equilateral triangle to a circle approx = 17.77%The area of a square to a circle approx = 63.7%The area of a hexagon to a circle approx = 83%Which means hexagon has the highest coverage area after a circle from the lot.Thus of the lot hexagon satisfies all the conditions which is why the shape of a cell is hexagonal in cellular network.More than likely because the antennas used have a radiation pattern of 60 degrees. Thus 6 are required for full 360 degree coverage.If you look at the top of a cell phone tower, you will see the vertical arrays that constitute the antennas, usually with the form of long vertical rectangles or boxes, sometimes arranged in parallel groups. A more detailed inspection would reveal that the vertical arrays are separated 120ºfrom each other. This means that each array is pointing to the center of its hexagonal cell. Hence, each cell is covered by six arrays from six separate towers, one at the apex of each of its six angles.Each provider can use a fixed set of frequencies, which forces them to assign a limited number of frequencies to each adjacent cell; the frequencies start repeating on the non-adjacent cells. Similarly, each antenna array can serve a limited number of subscribers at any given time. All this means that when traffic increases within a cell (more users), the only way of coping with the extra traffic is to reduce the area of the cell so it serves less subscribers. This is done by adding more towers to form smaller cells, hence the constant demand for cell tower sites. When a cell is fractioned in smaller cells, the antenna arrays that serve the new cells are lowered and their power is reduced so their emitted energy does not go beyond the adjacent cells to avoid interfering with those other cells, farther cells that use the same sets of frequencies.Having the cell phone towers closer to each other means that the signal, although purposely weakened, can now reach inside elevators and concrete buildings. This also means that cell phones need to transmit in their high power setting much less time while in use, which allows for smaller handsets whose batteries last longer. Hint,Farthest for interference and largest area of cellFewest number of cellClose to circular pattern , applying omni-direction antenna。
homework (1)-教育文档
要练说，先练胆。说话胆小是幼儿语言发展的障碍。不少幼儿当众说话时显得胆怯：有的结巴重复，面红耳赤；有的声音极低，自讲自听；有的低头不语，扯衣服，扭身子。总之，说话时外部表现不自然。我抓住练胆这个关键，面向全体，偏向差生。一是和幼儿建立和谐的语言交流关系。每当和幼儿讲话时，我总是笑脸相迎，声音亲切，动作亲昵，消除幼儿畏惧心理，让他能主动的、无拘无束地和我交谈。二是注重培养幼儿敢于当众说话的习惯。或在课堂教学中，改变过去老师讲学生听的传统的教学模式，取消了先举手后发言的约束，多采取自由讨论和谈话的形式，给每个幼儿较多的当众说话的机会，培养幼儿爱说话敢说话的兴趣，对一些说话有困难的幼儿，我总是认真地耐心地听，热情地帮助和鼓励他把话说完、说好，增强其说话的勇气和把话说好的信心。三是要提明确的说话要求，在说话训练中不断提高，我要求每个幼儿在说话时要仪态大方，口齿清楚，声音响亮，学会用眼神。对说得好的幼儿，即使是某一方面，我都抓住教育，提出表扬，并要其他幼儿模仿。长期坚持，不断训练，幼儿说话胆量也在不断提高。1. Oral work:
语文课本中的文章都是精选的比较优秀的文章,还有不少名家名篇。如果有选择循序渐进地让学生背诵一些优秀篇目、精彩段落,对提高学生的水平会大有裨益。现在,不少语文教师在分析课文时,把文章解体的支离破碎,总在文章的技巧方面下功夫。结果教师费劲,学生头疼。分析完之后,学生收效甚微,没过几天便忘的一干二净。造成这种事倍功半的尴尬局面的关键就是对文章读的不熟。常言道“书读百遍,其义自见”,如果有目的、有计划地引导学生反复阅读课文,或细读、默读、跳读,或听读、范读、轮读、分角色朗读,学生便可以在读中自然领悟文章的思想内容和写作技巧,可以在读中自然加强语感,增强语言的感受力。久而久之,这种思想内容、写作技巧和语感就会自然渗透到学生的语言意识之中,就会在写作中自觉不自觉地加以运用、创造和发展。2. Written work:
人教PEP版五年级英语上册《Unit1》精品教案教学设计小学优秀公开课
人教PEP版五年级英语上册第一单元教案Unit1What’s he like?教学设计一、内容PEP小学五年级英语上册unit1What’s he like?二、课程标准能听、说、读、写本单元的四会单词；会运用本单元主要句；初步认识陈述句和问句；有兴趣听英语，说英语，背歌谣，唱歌曲，讲故事，做游戏等。
三、内容分析本单元重点学习人物体貌特征和个性语言，重点学习的句型是：Who’s your art teacher?What’s he like?Is she…?Yes,she is/No,she isn’t.这单元主要描写的人物体貌特征，所以我们可以让孩子对所认识的老师进行描述。
四、学情分析五年级学生多，他们聪明活泼、勤奋好学，这些学生英语基础较好，他们对英语感到好奇，而且都有着十分浓厚的兴趣。
因此，我确立的学习目标如下：五、学习目标1、能力目标：（1）能够简单描述自己教师的体貌特征及性格特点，如：We have a new English teacher.He’s tall and strong.He’s very funny.（2）能够询问并介绍学校里教师的情况，如：Who’s your English teacher?Mr Ding.He’s from China.What’s he like?He’s tall and strong.（3）能够听懂一些描述人物特征的简单对话，完成学生用书中的Let’stry部分。
（4）能够听懂、会唱歌曲“My New Teacher”.能够灵活替换歌词中的科目及表示人物特征的形容词。
2、知识目标（1）理解A、B部分Read and write中的会话，并根据提示填充句子或回答问题。
（2）听、说、读、写A、B部分Let’s learn和Read and write中的四会单词和句子。
（3）理解Let’s start、Let’s find out、Let’s sing、Let’s chant和Pair work 等部分的内容。
新9A第一单元Unit1Know yourself教案
Task4 Practice
1. Work in pairs and drill the dialogue.
2. Ask some Ss to act the dialogue out
3.To use adjectives to describe someone’s characteristics.
4: To improve Ss’ ability of reading comprehension and use adjectives to describe someone’s characteristics freely.
课题
Unit 1Comic strip and welcome to the unit
主备人
课型
新授
课时
1
总课时
6
教学设计
二次备课
教学目标
Students can talk about personalities.
Students can master some useful expressions.
(1) Is Wu Wei quiet?
(2) What did Su Ning do before he started to work for the sales department in a big company?
(3)What is Liu Hao’s job at present?
(4) What is Fang Yuan like?
1.Introduce your best friend with different adj .
Homework 1
Homework for March.9一、英译汉1.The person who knows “how” will always have a job. The person who knows “why” will always be his boss.2.The girl in red sitting in the tree loves the boy in blue who is standing under the tree.3.Very smart Romeo who is from a noble family really loves especially gentle Juliet who can speak French fluently.4.In New York, the young without money can realize in this way the dreams in their hearts.二、汉译英1. 我们的村庄在两山之间的峡谷里。
2.我们班里有五个从新疆来的女学生。
3.那个戴着墨镜的漂亮女孩是杨紫。
4.在你的一生中，帮助过你的朋友才是你应该感谢的朋友。
5.莎士比亚时代使用英语的人数是五百万，而今天达2亿6千万。
6.我周一六点去上学。
7.她在公交站等公交车。
8.我吃了晚饭后出去散步。
9.我今天去上班，看到一个小孩子给老人让座。
10.从中国来的新学生在美国的一所大学里努力学习英语。
keys:一、英译汉1.一个知道怎么做的人将来总会找到工作。
一个知道为什么这样做的人将来总是成为他的老板。
2. 一个坐在树上的穿着红衣服的女孩爱上了一个站在树下的穿着蓝色衣服的男孩。
3. 来自贵族家庭的非常聪明的罗密欧的确爱着极其温柔的能讲一口流利的法语的朱丽叶。
4. 在纽约，没有钱的年轻人以这种方式能实现他们心中的梦想。
二、汉译英1. Our village is in a valley between two hills.2. There are five girl students from Xinjiang Province in our class.3.The beautiful girl wearing sunglasses is Yang Zi.4.In your life , a friend that/who has helped you is a friend that/who you should thank.5. The number of speakers of English in Shakespeare's timewas five million .Today it is 260 million.6. I go to school at six on Monday.7. She waited for the bus at the bus stop.8. I took a walk after dinner.9. Today I saw a kid offer his seat to an elderly man on my way to work.10. The new students from China study English hard in a university in America.。
homework1
Homework Set1Xiuli SunDue on:November2/3,2016October19,20161.Let kids denote the number of children ever born to a woman,and let educdenote years of education for the woman.A simple model relating fertility to years of education iskids=β0+β1educ+uwhere u is the unobserved error.(i)What kinds of factors are contained in u?Are these likely to be correlatedwith level of education?(ii)Will a simple regression analysis uncover the ceteris paribus effect of edu-cation on fertility?Explain.Note:Cateris paribus means other(relevant)factors being equal”.2.The following table contains the ACT scores and the GP A(grade point average)for eight college students.Grade point average is based on a four-point scale and has been rounded to one digit after the decimal.Student GPA ACT1 2.8212 3.4243 3.0264 3.5275 3.6296 3.0257 2.7258 3.730(i)Estimate the relationship between GP A and ACT using OLS;that is,obtainthe intercept and slope estimates in the equationˆGP A=ˆβ0+ˆβ1ACTComment on the direction of the relationship.Does the intercept have a useful interpretation here?Explain.How much higher is the GP A predicted to be if the ACT score is increased byfive points?(ii)Compute thefitted values and residuals for each observation,and verify that the residuals(approximately)sum to zero.(iii)What is the predicted value of GP A when ACT=20?(iv)How much of the variation in GP A for these eight students is explained by ACT?Explain.ing data from1988for houses sold in Andover,Massachusetts,from Kieland McClain(1995),the following equation relates housing price(price)to the distance from a recently built garbage incinerator(dist):ˆlog(price)=9.40+0.312log(dist)n=135R2=0.162(i)Interpret the coefficient on log(dist).Is the sign of this estimate what youexpect it to be?(ii)Do you think simple regression provides an unbiased estimator of the ceteris paribus elasticity of price with respect to dist?(Think about the citys decision on where to put the incinerator.)(iii)What other factors about a house affect its price?Might these be correlated with distance from the incinerator?4.Consider the savings functionsav=β0+β1inc+u,u=√·ewhere e is a random variable with E(e)=0and V ar(e)=σ2e .Assume that eis independent of inc.(i)Show that E(u|inc)=0,so that the key zero conditional mean assumption is satisfied.[Hint:If e is independent of inc,then E(e|inc)=E(e).](ii)Show that V ar (u |inc )=σ2e inc ,so that the homoskedasticity Assumptionis violated.In particular,the variance of sav increases with inc .[Hint:V ar (e |inc )=V ar (e ),if e and inc are independent.](iii)Provide a discussion that supports the assumption that the variance of savings increases with family income.5.Consider the standard simple regression model y =β0+β1x +u under the classical OLS assumptions.The usual OLS estimators ˆβ0and ˆβ1are unbiased for their respective population parameters.Let ˜β1be the estimator of β1obtained by assuming the intercept is zero.(i)Find E (˜β1)in terms of the x i ,β0,and β1.Verify that ˜β1is unbiased for β1when the population intercept (β0)is zero.Are there other cases where ˜β1is unbiased?(ii)Find the variance of ˜β1.(Hint:The variance does not depend on β0.)(iii)Show that V ar (˜β1)≤V ar (ˆβ1)).[Hint:For any sample of data,∑n i =1x 2i ≥∑n i =1(x i −¯x)2,with strict inequality unless ¯x =0.](iv)Comment on the tradeoffbetween bias and variance when choosing between ˆβ1and ˜β1.6.Let ˆβ0and ˆβ1be the OLS intercept and slope estimators,respectively,and let ¯u be the sample average of the errors (not the residuals!).(i)Show that ˆβ1can be written as ˆβ1=β1+∑n i =1w i u i where w i =d i /SST x and d i =x i −¯x .(ii)Use part (i),along with ∑n i =1w i =0,to show that ˆβ1and ¯uare uncorrelated.[Hint:You are being asked to show that E [(ˆβ1−β1)·¯u ]=0.](iii)Show that ˆβ0can be written as ˆβ0=β0+¯u −(ˆβ1−β1)¯x .(iv)Use parts (ii)and (iii)to show that V ar (ˆβ0)=σ2/n +σ2(¯x )2/SST x .e the data in SLEEP75.RAW from Biddle and Hamermesh (1990)to studywhether there is a tradeoffbetween the time spent sleeping per week and the time spent in paid work.We could use either variable as the dependent variable.For concreteness,estimate the modelsleep =β0+β1totwrk +uwhere sleep is minutes spent sleeping at night per week and totwrk is total minutes worked during the week.(i)Report your results in equation form along with the number of observationsand R2.What does the intercept in this equation mean?(ii)If totwrk increases by2hours,by how much is sleep estimated to fall?Do youfind this to be a large effect?8.We used the data in MEAP93.RAW.Now we want to explore the relationshipbetween the math pass rate(math10)and spending per student(expend).(i)Do you think each additional dollar spent has the same effect on the passrate,or does a diminishing effect seem more appropriate?Explain.(ii)In the population modelmath10=β0+β1log(expend)+uargue thatβ1/10is the percentage point change in math10given a10%increase in expend.(iii)Use the data in MEAP93.RAW to estimate the model from part(ii).Report the estimated equation in the usual way,including the sample size and R-squared.(iv)How big is the estimated spending effect?Namely,if spending increases by10%,what is the estimated percentage point increase in math10?(v)One might worry that regression analysis can producefitted values for math10that are greater than100.Why is this not much of a worry in this data set?e the data in WAGE2.RAW to estimate a simple regression explaining monthlysalary(wage)in terms of IQ score(IQ).(i)Find the average salary and average IQ in the sample.What is the samplestandard deviation of IQ?(IQ scores are standardized so that the average in the population is100with a standard deviation equal to15.)(ii)Estimate a simple regression model where a one-point increase in IQ changes wage by a constant dollar e this model tofind the predicted increase in wage for an increase in IQ of15points.Does IQ explain most of the variation in wage?(iii)Now,estimate a model where each one-point increase in IQ has the same percentage effect on wage.If IQ increases by15points,what is the approximate percentage increase in predicted wage?10.Textbook:Problems:2.8,2.9,2.10;Computer Exercises:2.2,2.3,2.5,2.6。
新概念英语第一册lesson83-84课件
I had lunch an hour ago. She went to London last year.
一般现在时& 一般过去时 & 现在完成时 & 现在进行时
1. 一般过去时：表示在过去的同某个时间做过的事情或 存在的状态。I did my homework last night.
2. 现在完成时：动作发生在过去，但对现在造成了影响
leave come go arrive 可以代替一般将来时 We’re tomorrow Tom and Ileaving are going to have a
I’m coming soon. holiday. She’s going My father is arriving
We’re going to leave tomorrow. = We’re leaving tomorrow. 当动词是leave, go, come, arrive等时，可 以用现在进行时表将来。 I’m coming. 他今天晚上就到 He’s arriving tonight.
现在完成时vs一般过去时时态用法动词形式时间状语标志词例句现在完已经发生的事对现在的影响havehas动词的过去分词alreadyjust用于肯定ivealreadyhadlunchhaveyouhadsometea
洗澡 在楼上 快准备好了 请抽烟 来杯威士忌 在7点钟 吃晚餐 去饭店
烤牛肉和土豆
Lesson 81 短语 1.Have a bath
否定式: 主语 + 助动词have/has + not + 动词的过去分词.
英语小贴士
already & yet
1. already “已经”，用于肯定、 2. yet 用于疑问、否定句的句尾，含有“仍然”， “还没有”之意。 Have you done it yet? ---- Yes, I have already done it.
新视野商务英语1-5
NEW INSIGHTS into BUSINESS
Unit 1 COMPANY STRUCTURES
Objectives
When the learners finish learning this unit, they should be able to providing information on some companies structures collecting information on some companies structures. a basic understanding of a company structure as well as the responsibility corresponded to every position.
workforce
organization chart
managing director middle management
board of directors Senior management
Lead-in
Listen to seven people talking about their work and decided which department each one works for?
Unit 2 RECRUITMENT
Objectives
When the learners finish learning this unit, they should be able to providing information on recruitment. collecting information on recruitment. learn to read the advertisement. practise writing CVs and letters of application.
外研社版小学四年级上册英语教案_含教学反思1
Module 1 （单元备课）Theme: Numbers.Functions: Talking about numbers 13-29.Target Language : Numbers 13-29.Twenty and one is twenty-one.Vocabulary: one , purple , pink, white, orange, point, winner, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen, twentyChant: Ten big lions at the zoo.Teaching Contents:Unit 1: Here’s a red one.This is a blue one. Ten eleven twelveUnit 2 : I’ve got twenty-two points.I’ve got twenty-six points.I’ve got a bear.I’ve got an elephant. A lion. A cat , a tiger, a dog, a monkey , a birdI’m the winner.Teaching Time: Four periods.课题Module1 Unit1…seventeen, eighteen, nineteen, twenty!课型新授课教学目标知识目标：掌握新授单词： seventeen, eighteen, nineteen, twenty…认知目标：能正确认读数字13—20，总结出这些单词的构成规律。
能力目标：熟记颜色类单词，会用这些单词描述事物。
情感目标：体验自主学习取得成功的乐趣。
教学重点难点重点：让学生根据规律掌握数字13-19，并识记数字20. 难点:熟记颜色类单词，会用这些单词描述事物。
homework-1
Find the word or phrase from the list below that best matches the description in the following questions.1.1.1 Computer used to run large problems and usually accessed via a network ==> 3) servers1.1.2 10^15 or 2^50 bytes (i.e., 10 to the 15 or 2 to the 50) ==> 7) petabyte1.1.3 Computer composed of hundreds to thousands of processors and terabytesof memory ==> 5) supercomputers1.1.4 Today's science fiction application that probably will be available in nearfuture ==> 1) virtual worlds1.1.5 A kind of memory called random access memory==> 12) RAM1.1.6 Part of a computer called central processor unit ==> 13) CPU1.1.7 Thousands of processors forming a large cluster ==> 8) datacenters1.1.8 A microprocessor containing several processors in the same chip==> 10) multi-core processors1.1.9 Desktop computer without screen or keyboard usually accessed via a network==> 4) low-end servers1.1.10 Currently the largest class of computer (i.e., there are moreof these types of computers than any others) that runs one applicationor one set of related applications ==> 9) embedded computers1.1.11 Special language used to describe hardware components ==> 11) VHDL1.1.12 Personal computer delivering good performance to single users at lowcost ==> 2) desktop computers1.1.13 Program that translates statements in high-level language to assemblylanguage==> 15) compiler1.1.14 Program that translates symbolic instructions to binary instructions==> 21) assembler1.1.15 High-level language for business data processing ==> 25) cobol1.1.16 Binary language that the processor can understand==> 19) machine language1.1.17 Commands that the processors understand ==> 17) instruction1.1.18 High-level language for scientific computation ==> 26) fortran1.1.19 Symbolic representation of machine instructions ==> 18) assembly language1.1.20 Interface between user's program and hardware providing a variety ofservices and supervision functions ==> 14) operating system1.1.21 Software/programs developed by the users ==> 24) application software1.1.22 Binary digit (value 0 or 1) ==> 16) bit1.1.23 Software layer between the application software and the hardware thatincludes the operating system and the compilers ==> 23) system software1.1.24 High-level language used to write application and system software==> 20) C1.1.25 Portable language composed of words and algebraic expressions thatmust be translated into assembly language before run in a computer ==> 22) high-level language1.1.26 10^12 or 2^40 bytes==> 6) terabyteExercise 1.21.2.1 For a color display using 8 bits for each of the primary colors (red, green, blue) per pixel, what should be the minimum size in bytes of the frame buffer to store a frame?8 bits × 3 colors = 24 bits/pixel => 4 bytes/pixel.1280 × 800 pixels = 1,024,000 pixels.1,024,000 pixels × 4 bytes/pixel = 4,096,000 bytes (approxitly 4 Mbytes).1.2. 2 How many frames could it store, assuming the memory contains no other information?2 GB = 2000 MbytesNumber of frames = 2000 Mbytes/4 Mbytes = 500 frames1.2.3 If a 256 Kbytes file is sent through the Ethernet connection, how long it would take?1 gigabit network ==> 1 gigabit/per second = 125 Mbytes/second.File size: 256 Kbytes = 0.256 Mbytes.Time= Mbytes/network speed: 0.256/125 = 2.048 ms.Exercise 1.31.3.1 Which processor has the highest performance expressed in instructions per second?P2 has the highest performanceP1 (Performance = instructions/sec) = 2 × 10^9/1.5 = 1.33 × 10^9P2 (Performance = instructions/sec) = 1.5 × 10^9/1.0 = 1.5 × 10^9P3 (Performance = instructions/sec) = 3 × 10^9/2.5 = 1.2 × 10^91.3.2 If the processors each execute a program in 10 seconds, find the number of cycles and the number of instructions.Number of cycles = time × clock rateCycles (P1) = 10 × (2 × 10^9) = 20 × 10^9 sCycles (P2) = 10 × (1.5 × 10^9) = 15 × 10^9 sCycles (P3) = 10 × (3 × 10^9) = 30 × 10^9 sTime = (Number of instructions × CPI)/clock rate; Number of instructions = Number of cycles/CPI Instructions (P1) = 20 × 10^9/1.5 = 13.33 × 10^9Instructions (P2) = 15 × 10^9/1 = 15 × 10^9Instructions (P3) = 30 × 10^9/2.5 = 12 × 10^91.3.3 We are trying to reduce the time by 30% but this leads to an increase of 20% in the CPI. What clock rate should we have to get this time reduction?Time new = Time old × 0.7 = 7 sCPI new = CPI old × 1.2 => CPI (P1) = 1.8, CPI (P2) = 1.2, CPI (P3) = 3Time = Number of instructions × CPI/clock rateTime (P1) = 13.33 × 10^9 × 1.8/7 = 3.42 GHzTime (P2) = 15 × 10^9 × 1.2/7 = 2.57 GHzTime (P3) = 12 × 10^9 × 3/7 = 5.14 GHz1.3.4 Find the IPC (instructions per cycle) for each processor.IPC new = 1/CPI old = Number of instructions/ (time × clock rate)IPC (P1) = (20 x 10^9)/(7 x 3) = 1.42IPC (P2) = (30 x 10^9)/(10 x 2.5) = 2IPC (P3) = (90 x 10^9)/(9 x 4) = 3.331.3.5 Find the clock rate for P2 that reduces its execution time to that of P1.Time new/Time old = 7/10 = 0.7.Clock rate new = Clock rate old/0.7 = 1.5 GHz/0.7 = 2.14 GHz1.3.6 Find the number of instructions for P2 that reduces its execution time to that of P3.Time new/Time old = 9/10 = 0.9.Instructions new = Instructions old × 0.9 = (30 × 10^9)× 0.9 = 27 × 10^9Exercise 1.41.4.1 Given a program with 106 instructions divided into classes as follows: 10% class A, 20% class B, 50% class C and 20% class D, which implementation is faster?P2 has a faster implementation timeClass A: 10^5 instr.Class B: 2 × 10^5 instr.Class C: 5 × 10^5 instr.Class D: 2 × 10^5 instr.Time = Number of instructions × CPI/clock rateP1: Time class A = 0.66 × 10^−4Time class B = 2.66 × 10^−4Time class C = 10 × 10^−4Time class D = 5.33 × 10^−4Total time P1 = 18.65 × 10^−4P2: Time class A = 10^−4Time class B = 2 × 10^−4Time class C = 5 × 10^−4Time class D = 3 × 10^−4Total time P2 = 11 × 10^−41.4.2 What is the global (i.e., average) CPI for each implementation?CPI = time × clock rate/ number of instructionsCPI (P1) = (18.65 × 10^−4) × (1.5 × 10^9)/10^6= 2.79CPI (P2) = (11 × 10^−4) × (2 × 10^9)/10^6 = 2.21.4.3 Find the clock cycles required in both cases.Clock cycles (P1) = (10^5 × 1) + ((2 × 10^5) × 2) + ((5 × 10^5) × 3) + ((2 × 10^5) × 4) = 28 × 10^5Clock cycles (P2) = (10^5 × 2) + ((2 × 10^5) × 2) + ((5 × 10^5) × 2) + ((2 × 10^5) × 3) = 22 × 10^51.4.4 Assuming that arithmetic instructions take 1 cycle, load and store 5 cycles and branch 2 cycles, what is the execution time of the program in a 2 GHz processor?(500 × 1) + (50 × 5) + (100 × 5) + (50 × 2) × (0.5 × 10^–9) = 675 ns1.4.5 Find the CPI for the program.CPI = time × clock rate/ Number of instructionsCPI = ((675 × 10^–9) × (2 × 10^9))/700 = 1.921.4.6 If the number of load instructions can be reduced by one-half, what is the speed-up and the CPI?Time = ((500 × 1) + (50 × 5) + (50 × 5) + (50 × 2)) × (0.5 × 10^–9) = 550 nsSpeed-up = 675 ns/550 ns = 1.22CPI = ((550 × 10^–9) × (2 × 10^9))/700 = 1.57Exercise 1.51.5.1 Assume that peak performance is defined as the fastest rate that a computer can execute any instruction sequence. What are the peak performances of P1 and P2 expressed in instructions per second?1.5.2 If the number of instructions executed in a certain program is divided equally among the classes of instructions except for class A, which occurs twice as often as each of the others. Which computer is faster? How much faster is it?1.5.3 If the number of instructions executed in a certain program is divided equally among the classes of instructions except for class E, which occurs twice as often as each of the others? Which computer is faster? How much faster is it?1.5.4 Assuming that computes take 1 cycle, loads and store instructions take 10 cycles, and branches take 3 cycles, find the execution time of the program on a 3 GHz MIPS processor.1.5.6 Assuming that computes take 1 cycle, loads and store instructions take 2 cycles, and branches take 3 cycles, what is the speed-up of a program if the number of compute instruction can be reduced byone-half?Exercise 1.61.6.1 For the same program, two different compilers are used. The table above shows the execution time of the compiled program. Find the average CPI for the program given that the processor has a clock cycle time of 1 nS.1.6.2 Assume the average CPI found in 1.6.1, but that the compiled program runs on two difference processors. If the execution times on the two processors are the same, how much faster is the clock of the processor running compiler A’s code versus the clock of the processor running compiler B’s code?1.6.3 A new compiler is developed that uses only 600 million instructions and has an average CPI of 1.1. What is the speed-up of using this new compiler versus using Compiler A or B on the original processorof 1.6.1?1.6.4 Assume that peak performance is defined as the fastest rate that a computer can execute any instruction sequence. What is the peak performance of P1 and P2 expressed in instructions per second?1.6.5 If the number of instructions executed in a certain program is divided equally among the classes of instructions in Problem2.36.4 except for class A, which occurs twice as often as each of the others, how much faster is P2 than P1?1.6.6 At what frequency does P2 have the same performance as P1 for the instruction mix given in 1.6.5? Exercise 1.71.7.1 What is the geometric mean of the ratios between consecutive generations for both clock rate and power? (The geometric mean is described in Section 1.7.)Geometric mean clock rate ratio = (1.28 × 1.56 × 2.64 × 3.03 × 10.00 × 1.80 × 0.74)1/7 = 2.15Geometric mean power ratio = (1.24 × 1.20 × 2.06 × 2.88 × 2.59 × 1.37 × 0.92)1/7 = 1.621.7.2 What is the largest relative change in clock rate and power between generations?Largest clock rate ratio = 2000 MHz/200 MHz = 10Largest power ratio = 29.1 W/10.1 W = 2.881.7.3 How much larger is the clock rate and power of the last generation with respect to the first generation?Clock rate: 2.667 × 10^9/12.5 × 106 = 212.8Power: 95 W/3.3 W = 28.781.7.4 Find the average capacitive loads, assuming a negligible static power consumption.Capacitive = Power/Voltage^2 × clock rate80286: C = 0.0105 × 10^−680386: C = 0.01025 × 10^−680486: C = 0.00784 × 10^−6Pentium: C = 0.00612 × 10^−6Pentium Pro: C = 0.0133 × 10^−6Pentium 4 Willamette: C = 0.0122 × 10^−6Pentium 4 Prescott: C = 0.00183 × 10^−6Core 2: C = 0.0294 × 10^−61.7.5 Find the largest relative change in voltage between generations.Pentium Pro/Pentium 4 Willamette = 3.3/1.75 = 1.781.7.6 Find the geon1etric mean of the voltage ratios in the generations since the Pentium.Pentium Pro / Pentium => 3.3/5 = 0.66Pentium 4 Willamette/ Pentium Pro => 1.75/3.3 = 0.53Pentium 4 Prescott / Pentium 4 Willamette => 1.25/1.75 = 0.71Core 2 / Pentium 4 Prescott => 1.1/1.25 = 0.88Geometric mean = 0.68Exercise 1.81.8.1 How much has the capacitive load been reduced between versions if the dynamic power has been reduced by 10%?Power1 = V^2× clock rate × Capacitive; Power2 = 0.9 Power1C2/C1 = 0.9 × 5^2 × 0.5 × 10^9/3.3^2 × 1 × 10^9 = 1.031.8.2 By how much has the dynan1ic power been reduced if the capacitive load does not change?Power2/Power1 = V22 × clock rate2/V12 × clock rate1Power2/Power1 = 0.87 => Reduction of 13%1.8.3 Assuming that the capacitive load of version 2 is 80% the capacitive load of version 1, find the voltage for version 2 if the dynamic power of version 2 is reduced by 40% from version 1.Power2 = V22 × 1 × 10^9 × 0.8 × C1 = 0.6 × Power1Power1 = 52 × 0.5 × 10^9 × C1V22 × 1 × 10^9 × 0.8 × C1 = 0.6 × 52 × 0.5 × 10^9 × C1V2 = ((0.6 × 52 × 0.5 × 10^9)/(1 × 10^9 × 0.8))1/2 = 3.06 V1.8.4 By what factor does the dynamic power scales?Power new = 1 × C old × V2old/(2−1/4)2× clock rate × 21/2 = Power oldAccording to this the power scales by 1.1.8.5 Find the scaling of the capacitance per unit area.1/2−1/2 = 21/21.8.6 Using data from Exercise 1.7, find the voltage and clock rate of the Core 2 processor for the next process generation.Voltage = 1.1 × 1/2−1/4 = 0.92 VClock rate = 2.667 × 21/2 = 3.771 GHzExercise 1.91.9.1 Find the percentage of the total dissipated power comprised by static power.1.9.2 If the total dissipated power is reduced by 10% while maintaining the static to total power rate of problem 1.9.1, how much should the voltage be reduced to maintain the same leakage current?1.9.3 Determine the ratio of static power to dynamic power for each technology.1.9.4 Determine the static power for each version at 0.8 V, assuming a static to dynamic power ratio of 0.6.Power st/Power dyn = 0.6 => Power st = 0.6 × Power dyna)Power st = 0.6 × 35 W = 21 Wb)Power st = 0.6 × 30 W = 18 W1.9.5 Determine the static power and dynamic power dissipation assuming the rates obtained in problem 1.9.1.L lk = Voltage / Power sta) I lk = 21/0.8 = 26.25 Ab) I lk = 18/0.8 = 22.5 AExercise 1.101.10.1 The table above shows the number of instructions required per processor to complete a program on a multiprocessor with 1, 2, 4, or 8 processors. What is the total number of instructions executed per processor? What is the aggregate number of instructions executed across all processors?Instructions per processor = (Arith Inst x Arith Cpi)+( load Inst x load Cpi)+( branch Inst x branch Cpi)program on I, 2, 4, and 8 processors. Assume that each processor has a 2 GHz clock frequency.1.10.3 If the CPI of arithmetic instructions was doubled, what would the impact be on the execution time of the program on 1,2,4, or 8 processors?1.10.4 Assuming a 3 GHz clock frequency, what is the execution time of the program using 1, 2, 4, or 8 cores.1.10.5 Assuming that the power consumption of a processor core can be described by the following equation:Power = (5.0n1A)/(MHz)* Voltage^2Where the operation voltage of the processor is described by the following equation:Voltage = 1/5*Frequency+0.4With the frequency measured in GHz. So, at 5 GHz, the voltage would be 1.4V. Find the power consumption ofthe program executing on 1, 2, 4, and 8 cores assuming that each core is operating at a 3GHz clock frequency. Likewise, find the power consumption of the program executing on I, 2, 4, or 8 cores assuming that each core is operating at 500 MHz.Exercise 1.111.11.1 Find the yield.Wafer area = π × (d/2)2a. Wafer area = π × 7.52 = 176.7 cm2b. Wafer area = π × 12.52 = 490.9 cm2Die area = wafer area/dies per wafera. Die area = 176.7/90 = 1.96 cm2b. Die area = 490.9/140 = 3.51 cm2Yield = 1/ (1 + (defect per area × die area)/2)2a. Yield = 0.97b. Yield = 0.921.11.2 Find the cost per die.Cost per die = cost per wafer/ (dies per wafer × yield)a. Cost per die = 0.12b. Cost per die = 0.161.11.3 If the number of dies per wafer is increased by 100/0 and the defects per area unit increases by 150/0, find the die area and yield.a. Dies per wafer = 1.1 × 90 = 99Defects per area = 1.15 × 0.018 = 0.021 defects/cm2Die area = wafer area/Dies per wafer = 176.7/99 = 1.78 cm2Yield = 0.97b. Dies per wafer = 1.1 × 140 = 154Defects per area = 1.15 × 0.024 = 0.028 defects/cm2Die area = wafer area/Dies per wafer = 490.9/154 = 3.19 cm2Yield = 0.931.11.4 Find the defects per area unit for each technology given a die area of 200 mm2Yield = 1/(1 + (defect per area × die area)/2)2Defect per area = (2/die area) (y−1/2− 1)Replacing values for T1 and T2 we getT1: defects per area = 0.00085 defects/mm2 = 0.085 defects/cm2T2: defects per area = 0.00060 defects/mm2 = 0.060 defects/cm2T3: defects per area = 0.00043 defects/mm2 = 0.043 defects/cm2T4: defects per area = 0.00026 defects/mm2 = 0.026 defects/cm2Exercise 1.121.12.1 Find the CPI if the clock cycle time is 0.333 ns.CPI = clock rate × CPU time/instr. countClock rate = 1/cycle time = 3 GHza. CPI (pearl) = 3 × 10^9 × 500/2118 × 10^9 = 0.7b. CPI (mcf) = 3 × 10^9 × 1200/336 × 10^9 = 10.71.12.2 Find the SPEC ratio.SPECratio = ref. time/execution time.a. SPECratio (pearl) = 9770/500 = 19.54b. SPECratio (mcf) = 9120/1200 = 7.61.12.3 For these two benchmarks, find the geometric mean.(19.54 × 7.6)1/2 = 12.191.12.4 Find the increase in CPU time if the number of instruction of the benchmark is increased by 100/0 without affecting the CPI.CPU time = number of instructions × CPI/clock rateIf CPI and clock rate do not change, then the CPU time will increase equal to the number ofinstructions, which would be by 10%.1.12.5 Find the increase in CPU time if the number of instruction of the benchmark is increased by 10% and the CPI is increased by 5%.CPU time (before) = number of instructions × CPI/clock rateCPU time (after) = 1.1 × number of instructions × 1.05 × CPI/clock rateCPU times (after)/CPU time (before) = 1.1 × 1.05 = 1.155.CPU time is increased by 15.5%1.12.6 Find the change in the SPECratio for the change described in 1.12.5.SPECratio = reference time/CPU timeSPECratio (after)/SPECratio (before) = CPU time (before)/CPU time (after) = 1/1.1555 = 0.86The SPECratio is decreased by 14%.Exercise 1.131.13.1 Find the new CPI.CPI = (CPU × clock rate)/Number of instr.a. CPI = 450 × 4 × 10^9/ (0.85 × 2118 × 10^9) = 0.99b. CPI = 1150 × 4 × 10^9/ (0.85 × 336 × 10^9) = 16.101.13.2 In general, these CPI values are larger than those obtained in previous exercises for the same benchmarks. This is due mainly to the clock rate used in both cases, 3 GHz and 4 GHz. Determine whether the increase in the CPI is similar to that of the clock rate. If they are dissimilar, why?Clock rate ratio = 4 GHz/3 GHz = 1.33.a. CPI at 4 GHz = 0.99, CPI at 3 GHz = 0.7, ratio = 1.41b. CPI at 4 GHz = 16.1, CPI at 3 GHz = 10.7, ratio = 1.50They are different because the CPU time has been reduced by a lower percentage, even though the number of instructions has been reduced by 15%,1.13.3 How much has the CPU time been reduced?a. 450/500 = 0.90CPU time reduction = 10%b. 1150/1200 = 0.958CPU time reduction = 4.2%1.13.4 If the execution time is reduced by an additional 10% without affecting the CPI and with a clock rate of 4 GHz, determine the number of instructions.Number of instructions = CPU × clock rate/CPI.a. Number of instructions = 820 × 0.9 × 4 × 10^9/0.96 = 3075 × 10^9b. Number of instructions = 580 × 0.9 × 4 × 10^9/2.94 = 710 × 10^91.13.5 Determine the clock rate required to give a further 10% reduction in CPU time while maintaining the nun1ber of instructions and CPI unchanged.Clock rate = Number of instructions × CPI/CPU time.Clock rate new = Number of instructions × CPI/0.9 × CPU time = 1/0.9 clock rate old = 3.33 GHz.1.13.6 Determine the clock rate if the CPI is reduced by 15% and the CPU time by 20% while the number of instructions is unchanged.Clock rate = Number of instructions × CPI/CPU time.Clock rate new = Number of instructions × 0.85 × CPI/0.80 CPU time = 0.85/0.80 clock rate old = 3.18 GHz. Exercise 1.141.14.1 One usual fallacy is to consider the computer with the largest clock rate as having the large perfonl1ance. Check if this is true for PI and P2.Number of instructions = 106Tcpu (P1) = 106 × 1.25/4 × 10^9 = 0.315 × 10–3 sTcpu (P2) = 106 × 0.75/3 × 10^9 = 0.25 × 10–3 sClock rate (P1) > clock rate (P2), but performance (P1) < performance (P2)1.14.2 Another fallacy is to consider that the processor executing the largest number of instruction will need a larger CPU time. Considering that processor PI is executing a sequence of 106 instructions and that the CPI of processors PI and P2 do not change, detenl1ine the number of instructions that P2 can execute in the same time that PI needs to execute 106 instructions.P1: 106 instructions, Tcpu (P1) = 0.315 × 10–3 sP2: Tcpu (P2) = N × 0.75/3 × 109N = 1.26 × 1061.14.3 A common fallacy is to use MIPS (millions of instructions per second) to con1pare the performance of two different processors, and consider that the processor with the largest MIPS has the largest perforn1ance. Check if this is true for PI and P2.MIPS = Clock rate × 10−6 /CPIMIPS (P1) = 4 × 10^9 × 10–6/1.25 = 3200MIPS (P2) = 3 × 10^9 × 10–6/0.75 = 4000MIPS (P1) < MIPS (P2), performance (P1) < performance (P2) in this caseAnother common performance figure is MFLOPS (millions of floating-point operations per second), defined asMFLOPS = Number of FP operations/(execution time x 10^6)1.14.4 Find the MFLOPS figures for the programs.a. FP op = 106 × 0.4 = 4 × 105, clock cylesfp = CPI × Number of FP instr. = 4 × 105Tfp = 4 × 105 × 0.33 × 10–9 = 1.32 × 10–4 then MFLOPS = 3.03 × 103b. FP op = 3 × 106 × 0.4 = 1.2 × 106, clock cylesfp = CPI × Number of FP instr. = 0.70 × 1.2 × 106Tfp = 0.84 × 106 × 0.33 × 10–9 = 2.77 × 10–4 then MFLOPS = 4.33 × 1031.14.5 Find the MIPS figures for the programs.5 CPU clock cycles = FP cycles + CPI (L/S) × Number of instr. (L/S) + CPI (Branch) ×Number of instr. (Branch)a. 5 × 105 L/S instr., 4 × 105 FP instr. and 105 Branch instr.CPU clock cycles = 4 × 105 + 0.75 × 5 × 105 + 1.5 × 105 = 9.25 × 105Tcpu = 9.25 × 105 × 0.33 × 10–9 = 3.05 × 10–4MIPS = 106/ (3.05 × 10–4 × 106) = 3.2 × 103b. 1.2 × 106 L/S instr., 1.2 × 106 FP instr. and 0.6 × 106 Branch instr.CPU clock cycles = 0.84 × 106 + 1.25 × 1.2 × 106 + 1.25 × 0.6 × 106 = 3.09 × 106Tcpu = 3.09 × 106 × 0.33 × 10–9 = 1.01 × 10–3MIPS = 3 × 106/ (1.01 × 10–3 × 106) = 2.97 × 1031.14.6 Find the performance for the programs and con1pare with MIPS and MFLOPS.a. Performance = 1/Tcpu = 3.2 × 103b. Performance = 1/Tcpu = 9.9 × 102The second program has the higher performance, but the first program has the higher MIPSfigure.Exercise 1.151.15.1 By how much is the total time reduced if the time for FP operations is reduced by 20%?a. T fp = 35 × 0.8 = 28 s; T p1 = 28 + 85 + 50 + 30 = 193 s; Reduction of 3.5%b. T fp = 50 × 0.8 = 40 s; T p4 = 40 + 80 + 50 + 30 = 200 s; Reduction of 4.7%1.15.3 Can the total time be reduced by 20% by reducing only the time for branch instructions?a. T p1 = 200 × 0.8 = 160 s; T fp + T int + T l/s = 170 s. No the total time cannot be reduced by 20%b. T p4 = 210 × 0.8 = 168 s; T fp + T int + T l/s = 180 s. No the total time cannot be reduced by 20% Assume that each processor has a 2 GHz clock rate.1.15.4 By how much must we improve the CPI of FP instructions if we want the program to run two times faster?Clock cyles = CPI fp × Number of FP instr. + CPI int × Number of INT instr. + CPI l/s × Number of L/Sinstr. + CPIbranch × Number of branch instr.Tcpu = clock cycles/clock rate = clock cycles/2 × 109a. 1 processor=> clock cycles = 8192; T cpu = 4.096 sb. 8 processors=> clock cycles = 1024; T cpu = 0.512 sTo half the number of clock cycles by improving the CPI of FP instructions:CPI improved fp × Number of FP instr. + CPI int × Number of INT instr. + CPI l/s × Number of L/S instr. + CPI branch × Number of branch instr. = clock cycles/2CPI improved fp= (clock cycles/2 − (CPI int × Number of INT instr. + CPI l/s × Number of L/S instr. +CPI branch × Number of branch instr.))/Number of FP instr.a. 1 processor: CPI improved fp = (4096 – 7632)/560 => not possibleb. 8 processors: CPI improved fp = (512 – 944)/80 => not possible1.15.5 By how much must we improve the CPI of L/S instructions if we want the program to run two times faster?Using clock cycle from 1.15.4:To half the number of clock cycles improving the CPI of L/S instructions:CPI fp × Number of FP instr. + CP Iint × Number of INT instr. + CP Iimproved l/s × Number of L/S instr. +CPI branch × Number of branch instr. = clock cycles/2CPI improved l/s = (clock cycles/2 − (CPIfp × Number of FP instr. + CPI int × Number of INT instr. +CPI branch × Number of branch instr.))/Number of L/S instr.a. 1 processor: CPI improved l/s = (4096 – 3072)/1280 = 0.8b. 8 processors: CPI improved l/s = (512 – 384)/160 = 0.81.15.6 By how much is the execution time of the program improved if the CPI of INT and FP instruction is reduced by 40% and the CPI of L/S and branch is reduced by 30%?Clock cyles = CPIfp× Number of FP instr. + CPI int× Number of INT instr. + CPI l/s× Number of L/S instr. +CPI branch × Number of branch instr.T cpu = clock cycles/clock rate = clock cycles/2 × 109CPI int = 0.6 × 1 = 0.6; CPI fp = 0.6 × 1 = 0.6; CPI l/s = 0.7 × 4 = 2.8; CPI branch = 0.7 × 2 = 1.4a. 1 processor => T cpu (before improved) = 4.096 s; T cpu (after improved) = 2.739 sb. 8 processors =>T cpu (before improved) = 0.512 s; T cpu (after improved) = 0.342 s。
Homework 1
• 2. If people mean anything at all by the expression "untimely death", they must believe that some deaths run on a better schedule than others. Death in old age is rarely called untimely--a long life is thought to be a full one. But with the passing of a young person, one assumes that the best years lay ahead and the measure of that life was still to be taken.
• History denies this, of course. Among prominent summer deaths, one recalls those of Marilyn Monroe and James Deans, whose lives seemed equally brief and complete. Writers cannot bear the fact that poet John Keats died at 26, and only half playfully judge their own lives as failures when they pass that year. The idea that the life cut short is unfulfilled is illogical because lives are measured by the impressions they leave on the world and by their intensity and virtue.
六年级下册_unit_1_how_tall_are_you？
Homework 1. 单词三会 10.1并背诵 2. 线上：爱乐奇Review 习题
class 13 两部分题包
3. pep课文 1.1 并背诵 4. 家长签字
How heavy is she?
3.are bigger Your than mine feet(连词成句)
Your feet are bigger than mine.
4.翻译：But I think you are heavier than me.
但是我觉得你比我重。
选择适当的词填空
1. I am 150 ___c_m____ tall.（am, cm）
2. My friend is shorter ___th_a_n____ me.
（than, then）
3. ___H__o_w____ tall are you? （How, What）
4． I’m 12. Lily is 11. I’mol_d_e_r_____
than her. (shorter, older)
词汇教学
bigger
smaller
Language Points
1.That’s the tallest dinosaur in this hall. the + 形容词最高级 表示“最……的”
2.How tall is it ? 它有多高？
3. It’s taller than both of us together. 它比我们俩加起来还高。 taller than “比……高” than 用于比较级之后 both of us 我们俩
I am taller than you.
homework1
X transaction ,buys(X ,C ) buys(X ,K )[s 40%,c 100%]
X transaction ,buys(X ,K ) buys(X ,C )[s 40%,c 40%]
X transaction ,buys(X ,E ) buys(X ,K )[s 80%,c 100%]
age %fat age %fat
0 0.0490 0.7632 0.7723
0 0.5389 0.8158 1.0000
0.1053 0 0.8158 0.6052
(e) Pearson’s product moment coefficient:
rA ,B
i 1
(ai A )(bi B )
Null {} K:5
E:4
C:1
O:4
C:1
通过创建条件模式基挖掘 FP 树： 项 C O E 条件模式基 {{K,E,O:1},{K:1}} {{K,E:4}} {{K:4}} 条件 FP 树 <K:2> <K:4，E:4> <K:4> 产生的频繁模式 {C,K:2} {O,K:4}、{O,E:4}、{O,K,E:4} {K,E:4}
age %fat 23 9.5 23 26.5 27 7.8 27 17.8 39 31.4 41 25.9 47 27.4 49 27.2 50 31.2 52 34.6 54 42.5 54 28.8 56 33.4 57 30.2 58 34.1 58 32.9 60 41.2 61 35.7
故所有的频繁项集为 L={{C},{E},{K},{O},{CK},{EK},{EO},{KO},{EKO}}。 (b) FP-growth 算法：数据库的第一次扫描与 Apriori 算法相同，它导出频繁 1 项集的 集合，并得到它们的支持度计数。频繁项的集合按支持度计数的递减序排序。结果 集记为 L。这样，L={{K:5},{E:4},{O:4},{C:2}}。FP-tree 如下图所示：
高级英语1高级英语第一册第二课PPT
I. Background Information
4. 300-Year-Old Camphor Tree Uprooted By The Blast
5. A Building Brought To Knees By The Blast
I. Background Information
6. A Boy Who Received Radiation Burns On His Whole Body
3) The rather arresting spectacle of little old Japan adrift amid beige concrete skyscrapers is the very symbol of the incessant struggle between the kimono and the miniskirt.
arrival in Hiroshima Question: what was his attitude toward
Hiroshima?
III . Text Analysis
2. the plots the author looks for his destination.
3.the mayor’s speech and two schools of thought.
IV.Special Difficulties
2)The cab driver’s door popped open at the very sight of a traveler.
--As soon as the taxi driver saw a traveler, he immediately opened the door.
牛津译林版八年级英语上册Unit1 全单元课件
12.告诉她有关你自己的一切 tell her everything about yourself
13. 愿意帮助
be ready to help
14.有困难
have problems
15.相信他的话
believe what he says
16.青少年杂志
Teenagers magazine
17.什么人适合做朋友？
6. have problems (with sth. )（在…方面）有问题
He __h_a_s_s_o_m_e__p_ro_b_l_e_m_s__w_i_th____his English. (在英语学习上有些问题)
7. believe what he/she says
相信他/她说的话
=__h_i_s/_h_e_r _w_o_r_d_s___
2. My friend is honeset so bI.bDeolieyovue talk to him/her when
what he / she says.
you are happy?
3. My friend keeps acsecrec.t CsoanIytoeullthelilmhim/ h/ ehrer everything about myself. everything about yourself ?
_b_u_y__so_m__e_m__o_r_e (再买一些). 4. My best friend often _m_a_k_e_s__m_e_h_a_p_p_y_ (让我开心). 5. He always helps me when I _h_a_v_e_p_r_o_b_le_m__s (有困难).
● be honest
仁爱英语说课稿1unit3_topic2_sectionb
•classical music
Beethoven
serious It is _______.
People usually enjoy it at a concert _______ or in a theater _______.
Part Ⅳ The analysis of the learning methods
1.Autonomous ,cooperation and inquiry. 2.Reading ,observation and practice.
Part V. The analysis of the teaching procedu
Read 1a and finish 1b.
pop Wen Wei used to enjoy ___ music, but now folk he likes ____ music. He doesn’t like classical ______ music at all because he thinks it’s serious classical music too______. Jane’s favorite is not _______ pop ______, but she doesn’t mind it. She likes ____ music best.
3) Moral goals:
△To widen the Ss’ musical knowledge and help them appreciate music. △To help the students in different levels to learn English. △To cultivate the Ss learning interests and good hobbies in English.