Lecture 05
Lecture 05 ADM - Cash and Receivables-1 现金及应收账款
CASH
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Limitations of internal control
• Cost vs. Benefit – Cost of establishing procedure should not exceed expected benefit
• Human element – Fatigue, carelessness, indifference
The role of internal audit is to provide independent assurance that an organisation's risk management, governance and internal control processes are operating effectively. Internal Audit teams duty is to provide an unbiased and objective view. ... In sum, internal auditors help organisations to succeed.
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Principles of internal control (Continued) 5. Independent internal verification
➢ Checking procedures to ensure segregation of duties ➢ Monitoring by supervisors ➢ Rotation of duties ➢ Verification by internal auditor
3. Physical, mechanical and electronic controls
Lecture Five
Ancient Chinese Education最早的学校教育•据古籍文献记载，古代中国早在四千多年前的虞舜时代就已经出现学校教育。
在河南安阳出土的三千多年前的商代甲骨文中，多处出现「学」字，就是当时进行教育活动的证明。
西周时期，学校教育已形成制度。
随后历经春秋战国五百年动荡，官学瓦解，私学兴起，出现了古代最伟大的教育家孔子和百家争鸣的学术繁荣。
Development of Ancient Education•Spring and Autumn Period–Confucius’private school:•3000 disciples;•72 virtuous and talented students•Han Dynasty Education–Great Academy or “Taixue”（太学）Development of ancient education •From Sui and Tang Dynasties to Ming andQing Dynasties•The Guozijian (国子监): Imperial Academy orImperial College–Song Dynasty•Shuyuan(书院): private educationalorganizationDevelopment of Ancient Education •Shang Xiang (上庠: shàng xiáng), was a schoolfounded in the Yu Shun 虞舜era in China. Shun (舜, 2257 BCE–2208 BCE), the Emperor of theKingdom of Yu 虞, or 有虞Youyu, founded twoschools. One was Shang Xiang (shang 上,means up, high), and the other one was XiaXiang 下庠, xia 下means down, low. ShangXiang was a place to educate noble youth.Teachers at Shang Xiang were generally erudite 博学的, elder and noble persons.Development of Ancient Education •Confucius’educational ideas:–“Education should be for all, irrespective of theirsocial status.”（有教无类）–“Six arts”: ritual, music, archery, chariot-riding, writing,and arithmetic.（六艺：礼、乐、射、御、书、数）–“Confused is one who learns without pondering;endangered is one who ponder without learning.”(学而不思则罔，思而不学则殆)–Extrapolation（推论）: if I hold up one corner and astudent cannot come back to me with the other three,I do not go on with the lesson. (举一隅不以三反，则不复也)Taixue （太学）•Taixue (太学) which literally means Greatest Study or Learning was the highest rank of educational establishment in Ancient China between the Han Dynasty and Sui Dynasty. It was replaced by the Guozijian（国子监）. The first nationwide government school system in China was established in 3 CE under Emperor Ping of Han（汉平帝）, with the Taixue located in the capital of Chang'an and local schools established in the provinces and in the main cities of the smaller counties.Taixue （太学）•Taixue taught Confucianism and Chinese literature among other things for the high level civil service, although a civil service system based upon examination rather than recommendation was not introduced until the Sui and not perfected until the Song Dynasty (960–1279).汉代学制结构Guozijian（国子监）•The Guozijian (国子监), the School of the Sons of State sometimes called the Imperial Central School, Imperial Academy or Imperial College was the national central institute of learning in Chinese dynasties after the Sui. It was the highest institute of learning in China's traditional educational system.唐代学制结构Guozijian（国子监）•Guozijian were located in the national capital of each dynasty --Chang'an, Luoyang, Kaifeng, and Nanjing. In Ming there were two capitals; thus there were two Guozijian, one in Nanjing and one in Beijing. The Guozijian, located in the Guozijian Street(or Chengxian street) in the Eater District, Beijing, the imperial college during the Yuan, Ming and Qing dynasties (although most of its buildings were built during the Ming Dynasty) was the last Guozijian in China and is an important national cultural asset.北京国子监Shuyuan (书院：Academies)•The Shūyuàn (书院), usually known in English as Academies or Academies of Classical Learning, were a type of school in ancient China. Unlike national academy and district schools, shuyuan were usually private establishments built away from cities or towns, providing a quiet environment where scholars could engage in studies and contemplation without restrictions and worldly distractions.Shuyuan (书院：Academies)•Shuyuan as a modern term•In the late Qing dynasty, schools teaching Western science and technology were established. Many such schools were called Shuyuan in Chinese. Despite the common name, these shuyuan are quite modern in concept and are quite different from traditional academies of classical learning.Shuyuan (书院：Academies)•Notable Academies•In discussing the shuyuan, it is common to speak of the "Four Great Academies" (四大书院) of ancient China.Usually the "Four Great Academies" refers to the Four Great Academies of the Northern Song. However,sources give a number of different lists, sometimesexpanded to Six or Eight Great Academies. Only oneacademy, the Yuelu Academy, appears in all lists. Each school went up or down the list in different periods.White Deer Grotto Academy had long been anoutstanding academy. As for the impact on the politics of China, Donglin Academy in the Ming Dynasty isespecially notable.The Four Great Academies •Also known as the Four Great Academies of the Northern Song or the Four Northern Song Academies.–Songyang Academy–Yingtianfu Academy–Yuelu Academy–White Deer Grotto Academy•白鹿洞书院为宋代四大书院之首。
Lecture05_WaveletPackets
Usage
Wavelet Transform
Wavelet transform is applied to low pass results (approximations) only:
S
A1
D1
A2
D2
A3
D3
Not optimal
From the point of view of compression, where we want as many small values as possible, the standard wavelet transform may not produce the best result, since it is limited to wavelet bases (the plural of basis) that increase by a power of two with each step. It could be that another combination of bases produce a more desirable representation.
Cost functions: threshold
The threshold cost function counts the number of values in a wavelet packet tree node whose absolute value is greater than a threshold value t.
The cost function is additive in the sense that K([a b]) = K(a) + K(b) for all finite length vectors a and b. K(0) = 0, where 0 denotes the zero vector
英语演讲与辩论课件Lecture Five
An argument can be a fight or a
quarrel, but it can also mean a set of reasons given in support of something in order to persuade. Argument is the process of explaining why a point of view should be accepted. ▼
?convincetheaudienceandjudgesthatyourargumenthasthegreaterstrength?showthatyouranalysisoftheproblemisvalidandthorough?showhowthechangeyouproposeisworthydespitethechallengesofferedbytheopposition?bepersuasivethroughclearreasoninganduseofpersuasivelanguage?discussion讨论?crossexamination盘问?publicspeaking演说?freeargumentation自由辩论?parliamentarystyledebating议会式辩论discussionmaybejustconversationorconversationthatisregulatedwithrulesandprocedureswhichmaybedifferentindifferentcircumstances
which investigate the truth of opinions that challenge and question conventional wisdom. These discussions make us search for greater understanding of the world around us, and think about and look for better solutions to the problems we encounter.
lecture05
EE 338L CMOS Analog Integrated Circuit DesignLecture 5, Single-Stage Amplifiers (2)Calculations of Small Signal Input and Output ImpedancesHow to calculate input and output impedances (or admittances) of an amplifier? In the following sections, we assume that the amplifier is a voltage amplifier, whose input and output are both voltages. But we can easily extend the principles to any other types of amplifiers, such as current amplifiers (input and output are both currents), transimpedance amplifiers (input: current, output: voltage), and transconductance amplifiers (input: voltage, output: current).1. Input impedanceMethod A:i) Apply tst v at the input (* seenote below), draw the smallsignal diagram.ii) Calculate )(tst tst v f i =.iii) The input impedance isgiven by tsttst in i vz =, and the inputadmittance is tsttst in v iy =.Method B:i) Apply tst i at the input, draw thesmall signal diagram. ii) Calculate )(tst tst i f v =.iii) The input impedance isgiven by tsttst in i vz =, and the inputadmittance is tsttst in v iy =.* Note: If the amplifier requires an output termination, we should terminate the output accordingly. The load condition may affect the input impedance.iv tst2. Output impedanceMethod A:i) Set 0=in v , or if the input is a signal current, set 0=in i (** see note below).ii) Apply tst v at the output, draw the small signal diagram. iii) Calculate )(tst tst v f i =.iv) The output impedance is given by tsttst out i vz =, and the output admittanceis tsttst out v iy =.Method B:i) Set 0=in v , or if the input is a current, set 0=in i (** see note below).ii) Apply tst i at the output, draw the small signal diagram.iii) Calculate )(tst tst i f v =.iv) The output impedance is given by tsttst out i vz =, and the output admittanceis tsttst out v iy =** Note: If the amplifier requires some input termination, we should terminate the input accordingly. The input termination may affect the output impedance.Example : Calculate the output impedance of the following circuit, assuming both M1 and M2 work in saturation region. The small signal parameters of M1 and M2 are shown in the following table.Transconductance Bulk transconductance Drain-sourceconductanceM1 g m1 g mb1 g ds1 M2 g m2 g mb2 g ds2Note: As g mb is not 0, we take bulk (or body) effect into consideration. As g ds is not 0, we also consider channel length modulation effect.v tsttstSolution:(1) Set v in =0, and draw the small signal digram.(2) Apply i tst at the output. The small signal diagram is shown in Fig. 1.v intstFig. 1.Note that v gs1=0, and v bs1=0, thus the two voltage controlled current sources inFig. 1 are actually 0 (see Fig. 1). We redraw Fig. 1 as Fig. 2.tstFig. 2(3) For Fig. 2, according to KCL,tst i i i i i =++=2322211Thusv V1112ds tst ds s g i g i v ==Note that, v g2=v b2=0, thus22222222221)()(s m s m s g m gs m v g v g v v g v g i −=−=−== (3)22222222223)()(s mb s mb s b mb bs mb v g v g v v g v g i −=−=−== (4)and)()(222222222s tst ds s d ds ds ds v v g v v g v g i −=−== (5)Eq. (3)+Eq. (4)+Eq. (5), and combine with Eq. (1), we arrive at,tsts mb ds m tst ds s mb s tst ds s m i v g g g v g v g v v g v g i i i =++−=−−+−=++22222222222232221)()( (6)From Eq. (6) we can write,222221ds tsts ds mb m tst g i v g g g v +⎟⎟⎠⎞⎜⎜⎝⎛++=Substitute Eq. (2) into Eq. (7),tst ds ds ds ds mb m ds tst ds tst ds mb m tst i g g g g g g g i g i g g g v ⎟⎟⎠⎞⎜⎜⎝⎛+++=+⎟⎟⎠⎞⎜⎜⎝⎛++=21212221222111 (8) Thus,21212211ds ds ds ds mb m tst tst out g g g g g g i v r +++==(9a)Note that 111ds ds g r =, 221ds ds g r =, and m mb g g η=, Eq. (9a) can be re-written as, 21222122221212212122]1)1([]1)[()1()(ds ds ds m ds ds ds mb m ds ds ds ds m ds ds ds ds mb m tst tstout r r r g r r r g g r r r r g r r r r g g i v r +++=+++=+++=+++==ηη (9b)v V 122122212122)()(ds ds ds mb m ds ds ds ds mb m out r A r r g g r r r r g g r =+≈+++=where2222)(ds mb m r g g A +=Example : 1) Assuming 0=λ, and 0=γ, what is the input impedance of the amplifier? 2) if 0≠λ, and 0≠γ, please repeat 1). Note that V B is a DC bias voltage.Solution:1) As 0=λ, and 0=γ, we have 0=ds g and 0=mb g . Draw the small signal diagram, and apply v tst at the input.g =0iFig. 1From Fig. 1, we have,tst m tst m s g m gs m d tst v g v g v v g v g i i =−−=−−=−=−=)0()(Thus,V DD v inm tst tst in g i v r 1==2) As 0≠λ, and 0≠γ, we have 0≠ds g and 0≠mb g .Draw the small signal diagram as Fig. 2, and apply v tst at the input. According to KCL,)(232221i i i i tst ++−=(1)gs =-g m v tsti 23=g mb v bs =-g i 22=g ds v dsFig. 2From Fig. 2, we can write tst m tst m s g m gs m v g v g v v g v g i −=−=−==)0()(21 (2))()(22tst out ds s d ds ds ds v v g v v g v g i −=−== (3) tst mb tst mb s b mb bs mb v g v g v v g v g i −=−=−==)0()(23 (4)D tst out R i v =(5)Substitute Eqs. (2)-(5) into Eq. (1), we havetst D ds tst ds mb m out ds tst ds mb m tst i R g v g g g v g v g g g i −++=−++=)()( (6a) Simplify Eq. (6) as tst ds mb m tst D ds v g g g i R g )()1(++=+(6b)Thusdsmb m D ds tst tstin g g g R g i v r +++==1Source Followers (Common Drain Amplifiers)(b)Input terminal: gate; output terminal: source.Note that, not all process technologies allow the source and the bulk of NMOS transistors to be connected together. The above schematics only show the concept.Large signal behavior Input/Output Voltage(V)Vout(d)VinVout(b)Input Voltage(V)(V DD=3.3V, W/L=8(6/0.9), R S=5K, I B=120u)When V in<V T, M1 is off, and Vout is 0.When V in>V T, M1 turns on in saturation. As V in increases further, V out follows Vin with a difference of V GS.When V in increases to a certain voltage (exceeding V DD), M1 enters triode region, the output voltage flattens out and clips at V DD.Small signal analysisWe only perform small signal analysis for the schematic above. The small signal diagram is shown in the right side.Note that, 2322211i i i i ++=(1)and SoutR v i =1 (2) )()(21out in m s g m gs m v v g v v g v g i −=−== (3) out ds out ds ds ds v g v g v g i −=−==)0(22(4)out mb out mb s b mb bs mb v g v g v v g v g i −=−=−==)0()(23 (5) Substitute Eqs. (2)-(5) into Eq. (1), and after some simplification, we obtain, out ds mb m in m Soutv g g g v g R v )(++−= (6)ThusSds mb m minout v R g g g g v v A 1+++==(7)Example . For the amplifier shown below (left side), assuming, 0≠λ, and 0≠γ.1) What is the small signal voltage gain inout v v vA =?2) What is the small signal output resistance out r ?outsSmall signal diagramFig. 1, Original schematic and small signal diagram for calculating small signalvoltage gain1) What is the small signal voltage gain inoutv v v A =? Solution:The small signal diagram is drawn in the right side in Fig. 1 above. According to KCL, we have 0232221=++i i i (1a) where )()(21out in m s g m gs m v v g v v g v g i −=−== (1b) out ds out ds s d ds ds ds v g v g v v g v g i −=−=−==)0()(22 (1c)out mb out mb s b mb bs mb v g v g v v g v g i −=−=−==)0()(23 (1d)Combine Eqs. (1a)-(1d), we have dsmb m min out g g g g v v ++= (2)2) What is the small signal output resistance out r ? Solution:Following the steps to calculate output resistance. (i) Set v in = 0.(ii) Apply v tst at the output, and draw the small signal diagram as shown below.Fig. 2, Small signal diagram for calculating output resistance(iii) According to KCL, we have0232221=+++tst i i i i (3) where tst m tst m s g m gs m v g v g v v g v g i −=−=−==)0()(21 (4a) tst ds out ds s d ds ds ds v g v g v v g v g i −=−=−==)0()(22 (4b)tst mb out mb s b mb bs mb v g v g v v g v g i −=−=−==)0()(23 (4c)Substitute Eqs. (4a)-(4c) into Eq. (3)tst ds mb m tst v g g g i i i i )()(232221++=++−= (5)(iii) Thus dsmb m tst tst out g g g i v r ++==1Example . For the amplifier shown below, assuming, 0≠λ, and 0≠γ.1) What is the small signal voltage gain inout v v vA =,2) What is the small signal output resistance out r ?V outFig. 1, Original schematic and small signal diagram for calculating small signalvoltage gain1) What is the small signal voltage gain inoutv v v A =? Solution:The small signal diagram is drawn in the right side in Fig. 1 above. Since there is no body effect , according to KCL, we have 02221=+i i (1a)where )()(21out in m s g m gs m v v g v v g v g i −=−== (1b)out ds out ds s d ds ds ds v g v g v v g v g i −=−=−==)0()(22 (1c)Combine Eqs. (1a)-(1c), we have dsm min out g g g v v += (2)2) What is the small signal output resistance out r ? Solution:Following the steps to calculate output resistance. (i) Set v in = 0.(ii) Apply v tst at the output, and draw the small signal diagram as shown below.i 21=g m v i 22=g v =-g v Fig. 2, Small signal diagram for calculating output resistance(iii) According to KCL, we have02221=++tst i i i (3) where tst m tst m s g m gs m v g v g v v g v g i −=−=−==)()(021 (4a)tst ds out ds s d ds ds ds v g v g v v g v g i −=−=−==)()(022(4b) Substitute Eqs. (4a)-(4b) into Eq. (3)tst ds m tst v g g i i i )()(+=+−=2221 (5)(iii) Thus dsm tst tst out g g i v r +==1Common Gate AmplifiersV DDv inFig. 1, Common-gate stage with direct coupling at inputIn a common-gate amplifier, the input signal is applied to the source terminal, as is shown in Fig.1. It senses the input at the source and generate the output at the drain. The gate is connected to a dc voltage to establish proper operating conditions. Note that the bias current of M1 flows through the input signal source.Large signal behaviorFig. 2 Large signal behaviorWhen V in >V B -V T , M1 is off, and Vout is V DD .As V in decreases, so does Vout, M1 is in saturation until()T B D T in B ox n DD V V R V V V LW C V −=−−−2121)(µ (1)After that, M1 is driven into the triode region.Small signal analysisDraw the small-signal diagram below when the transistor works in saturation, assuming 0=λ and 0≠γ,i 22=g mb v bs =-g i 21=g m v gs =-g m v inFig. 3 Small signal diagram to calculate the voltage gainWe can have,in D mb m D in mb in m out v R g g R v g v g v ⋅⋅+=⋅−−−=)()( (1)Thus,)()(η+⋅⋅=⋅+==1D m D mb m inout v R g R g g v vA (2)Example : What is the input impedance of the common gate amplifier discussed above (assuming 0=λ, and 0≠γ)?Solution : To obtain the impedance seen at the source, we use the following equivalent circuit:i 21=g m v gs =-g m v tsti 22=g mb v bs =-gFig. 4 Small signal diagram to calculate the input impedanceAccording to KCL, we have,)(2221i i i tst +−=(1)From Fig. 1, we have tst m tst m s g m gs m v g v g v v g v g i −=−=−==)()(021 (2)tst mb tst mb s b mb bs mb v g v g v v g v g i −=−=−==)()(022 (3)Substitute Eqs. (2)-(3) into Eq. (1), we havetst mb m tst mb m tst v g g v g g i )()(+=+= (4a) Simplify Eq. (4a), we have tst mb m tst v g g i )(+=(4b)Thus mbm tst tst in g g i v r +==1Example: What is the voltage gain of the following amplifier (assuming 0=λ, and 0=γ)?VFig. 1, Original schematic and small signal diagram for calculating small signalvoltage gainSolution:According to KCL, we have, 1121i i =(1)From Fig. 1, we haves m gs m v g v g i 22221−== (2))(in s m sg m v v g v g i −==11122 (3)Substitute Eqs. (2)-(3) into Eq. (1), we have in m m m s v g g g v 211+=(4)ThusD m m m m in D s m in out v R g g gg v R v g v v A 21212+===(5)。
lecture05哈工大软件学院c++教学课件
© 版权所有 哈尔滨工业大学（威海）软件学院
The C++IBM research Programming Language
© 版权所有 哈尔滨工业大学（威海）软件学院
… digit1 digit2 digit4 digit5
= = = =
number number number number
/ % % %
10000; 10000 / 1000; 10000 % 1000 % 100 / 10; 10000 % 1000 % 100 % 10;
for statement examples Vary control variable from 1 to 100 in increments of 1 for ( int i = 1; i <= 100; i++ ) Vary control variable from 100 to 1 in increments of -1 for ( int i = 100; i >= 1; i-- ) Vary control variable from 7 to 77 in steps of 7 for ( int i = 7; i <= 77; i += 7 )
The C++IBM research Programming Language
C++程序设计
© 版权所有 哈尔滨工业大学（威海）软件学院
The C++IBM research Programming Language
上节课内容回顾
1. if、if…else、while 2. 计数器控制的循环和标记控制的循环 3. 确定ATM系统中类的属性
Lecture five 虚拟语气
Presented by：
Jack Zhu
Subjunctive Mood
Focus 什么是虚拟语气？其最基本 的作用是什么？有那些词可以用 来引导虚拟语气？ 虚拟语气有哪 些特殊的表达形式？如何来辨别 和使用这些特殊形式？
一、什么是虚拟语气？
语气是动词的一种形式，用来表示讲话某件 事情的看法或态度。英语中有三种语气： 陈述语气 (Indicative Mood) 语 气 祈使语气 (Imperative Mood)
D
三、虚拟语气的特殊形式
虚拟语气的特殊形式是这一语法项目的难点是安 徽省专升本考试中的频率较高的考点，其表现形 式我们从以下几个方面进行学习记忆。
1
一些动词在接宾语从句时，要用虚拟语气 suggest, command, order
这些动词有：
propose, recommend, ask
require, request, intend, insist, etc.
decide, suggest, decree, demand, insist, move, order, prefer, propose,recommend, request, require, vote
上述动词所接的虚拟语气基本句型是：主句依据 说话者要求，采用正常的陈述句相对应的时态， 从句用 “should” + “动词原形”，常常省略 “should”。 1. Mother insists that the wallet I picked up in the street ( should) be given back to the owner. 2.He intended that she do it right now.
Lecture Five
Lecture FiveLong Sentence (I)Main Contents Of This Lecture♦I. Two Stages and Six Steps in Translating Long Sentences♦II. Methods Of Translation English Long SentencesI.Two Stages and Six Steps in Translating Long Sentences♦Stage 1♦Comprehension♦Stage 2♦Presentation♦英语长句在翻译时所涉及的基本问题，一是汉英语序上的差异，二是汉英表达方法上的差异。
前者主要表现为定状修饰语在语言转换中究竟取前置式（pre-position）、后置式（post-position）抑或插入式(insertion or parenthesis)。
♦表达方法上的差异涉及的问题比较复杂，主要表现为论理逻辑或叙述逻辑（logic in reasoning and narration）的习惯性与倾向性，因为语言中没有一成不变的表达方法。
所谓论理逻辑或叙述逻辑，主要指：行文层次及主次（包括句子重心，先说原因还是先说结果，先说条件或前提还是先说结论，先说施事者还是先说受事者等等）。
♦表达此外，表达方法还涉及修辞学范畴中的一些修辞格问题（如反复、排比、递进等），翻译中都应加以注意。
♦英语长句之所以很长，一般由三个原因构成：♦A、修饰语多♦B、联合（并列）成分多♦C、结构复杂，层次叠出♦英语长句可以严密细致地表达多重而又密切相关的概念，这种复杂组合的概念在口语语体中毫无例外是以分切、并列、递进、重复等方式化整为零地表达出来的。
此外我们在翻译英语长句时，还应体会到长句的表意特点和交际功能，尽量做到既能从汉英差异出发处理好长句翻译在结构形式上的问题，又要尽力做到不忽视原文的文体特征，保留英语长句所表达的多重致密的思维的特色，不要使译句产生松散和脱节感，妥善处理译句的内在联接问题。
Lecture05-造船生产计划准备工作
编制施工要领的目的
Ì 贯彻落实建造方针，达到预订的综合目 贯彻落实建造方针，
收到预期的经济效益。 标，收到预期的经济效益。 Ì 将各专业和各阶段的施工工艺，技术要 将各专业和各阶段的施工工艺， 求和作业顺序、 求和作业顺序、作业方法等详细地加以 系统的综合和协调。 系统的综合和协调。 Ì 指导生产设计，施工要领的内容要反映 指导生产设计， 到各专业生产设计的图表和技术文件中。 到各专业生产设计的图表和技术文件中。 Ì 指导生产和管理。 指导生产和管理。
生产设计准备
辛国栋 gdxin@
主要内容
Ì 分段划分 Ì 施工要领的编制 Ì 生产设计计划准备
生产设计准备定义
Ì 生产设计准备是在开展生产设计之前， 生产设计准备是在开展生产设计之前，
从全厂性、全船性、 从全厂性、全船性、综合性的角度对设 生产和管理等工作进行技术 技术和 计、生产和管理等工作进行技术和计划 方面的前期沟通和协调。 方面的前期沟通和协调。
r 它主要由船厂生产负荷计划、各阶段负荷计 它主要由船厂生产负荷计划、
划和分阶段负荷计划三部分组成。 划和分阶段负荷计划三部分组成。
船厂生产负荷计划 船厂生产负荷计划 生产
Ì 含义及作用
r
产品配置
预 测 潜在 合 同 客户 主生产计划 ( MPS ) 粗能力计划 ( RCCP ) 可行？ 物料需求计划 ( MRP ) 能力需求计划 ( CRP ) 可行？ 执行物料计划 ( 加工、采购 ) 执行能力计划 (投入/产出控制)
Ì 将1、2年内每个月的各种船舶 曲线的 年内每个月的各种船舶S曲线的 、 年内每个月的各种船舶
负荷值叠加，可以得出 、 年内各月工 负荷值叠加，可以得出1、2年内各月工 厂主要车间的负荷分布累计图。 厂主要车间的负荷分布累计图。
RM lecture 05 handout
11
Zero or Spot rates
A zero rate (or spot rate), for maturity T is the rate of interest earned on an investment that provides a payoff only at time T A chart of spot rates against maturity is called the yield curve or term-structure of interest rates Term structure animation /freecharts/yieldcurve.php
– Eurodollar futures and swaps are used to extend the LIBOR yield curve beyond one year – The overnight indexed swap rate is increasingly being used instead of LIBOR as the risk-free rate
Measuring Interest Rates
The compounding frequency used for an interest rate is the unit of measurement
The difference between semi-annual and annual compounding is analogous to the difference between miles and kilometers E.g. a $100 investment can yield $105 (100 principal + $5 interest) in multiple ways
lecture five
impose a penalty on sth
Reducing the population favorable to economic growth one-child policy has helped the fight against global warming by avoiding 300 million births.
What problems do you think are there in the environment? Do you think people are aware of environmental issues? What do you think the government should do to protect the environment? What should the public do to protect the environment? Do you feel optimistic that environmental problems will be solved in the future?
Beside air pollution, what other kinds of pollutions are there in your city? Which is more important, increasing people’s living standard, or protecting the environment?
New Labour law 新劳动法 Traffic Laws 交通法 Environmental Protection law 环境保 护法 Taxation law 税务法 New plastic-bags standards塑料袋使 用规定 One Child Policy 独生子女政策
Lecture_05
从问题到程序
9
复杂关系，如： 3≤x≤5 时 z 取值 2，否则取值 1。可以用 关系表达式和条件表达式描述：
z = (x >= 3 ? (x <= 5 ? 2 : 1) : 1);
有了关系表达式和条件表达式，足以描述任何可能的复杂 条件。但写出的描述可能比较复杂，不太直观 逻辑运算符专门用于基于简单的逻辑构造复杂的逻辑关系。 用逻辑运算符可直接描述
只能写出平铺直叙的直线型程序。这种程序从头开始执
行，一个个语句执行到最后
不能区分情况，不能用简单方式描述重复的类似动作
例如，要写一个函数，返回两参数中的较大值：
double dmax (double, double) 需要：
比较数据大小（判断） 根据判断结果决定做什么（分情况处理）
函数 dmax 的定义：
double dmax (double x, double y) { return x > y ? x : y; }
例：z = dmax(2, 3);
注意它的计算过程
条件表达式
从问题到程序
8
符号函数 sign对应的C函数： double sign (double x) { return x > 0 ? 1 : (x == 0 ? 0 : -1); }
这里假定了a不为0，处理a为0的情况也不困难（加判断）
连续 if 条件判断
从问题到程序
22
前例中出现了连续的条件判断：
成立时做某些操作 不成立时判断下一条件
这种情况下的建议书写形式（清晰，又不会退格太多）： if (...) { ... ... } else if (...) { ... ... } ... ... else { ... ... }
lecture 5
Character
Definition of “character” (p.24) Character means the features and trait that form the individual nature of a person, and by extension, refers to one who typifies some definite quality.
Reading Questions
√
Mrs. Mallard, the heroine in the short story is A.a round character B. flat character In the short story, Mrs. Mallard’s sister Joseph, her husband friend Richard, and her husband Mr. Mallard are A. main characters B. foils C. antagonists D. protagonists
Reading Questions
What is the best word to describe Mrs. Mallard? A. lucky B. happy C. brave D. wretched (可怜的）
Reading Questions
Kate Chopin focuses on Mrs. Mallard’s presentation of ______ A. appearance B. character C. mental conflicts D. speech
Lecture 5
The Story of an Hour By Tian Ying
华中科技大学费剑平 高级微观经济学课程讲义 Lecture05
Microeconomic Theory: Lecturer. J. Ping FeiLecture 05- 1Lecture 05 Optimization and Firms IV: Cost FunctionThe cost function c(w, y)= wx (w, y) measures the minimum cost of producing a givenlevel of output for some fixed factor prices. As such it summarizes information about the technological choices available to the firms. It turns out that the behavior of the cost function can tell us a lot about the nature of the firm’s technology.If we have the economic data we can recover things like the demand function for factors of production for the firm just by taking derivatives of costs functions with respect to prices.These cost functions in general will have a particular shape, namely they will be concave. This is equivalent to just thinking about the value of information. This is equivalent to saying that the ability to make decisions when we know the realization of a random variable is more valuable than having to make our decision based simply on the distribution from which that random variable is chosen. And that observation is equivalent to saying that these cost functions must be concave in prices regardless of the shape of the underlying technology or isoquant.The objective here is to get you to learn how to describe the technology of a firm. The most common way in intermediate course is production function. In this lecture, we will discuss some economically relevant aspects of a firm’ technology.1. Total cost ,average cost and marginal cost.We will graph cost as a function of output and hold the factor prices constant. It is understood that factor prices are included but they will not be changing. Let’s start with asimple case. Suppose that the only factor of production that is variable is labor and that there is a fixed cost. First, let’s draw a graph for the isoquants. See figure 1.wL+ fixed costKTotal costFCLqFigure 1Figure 2In the short run we will fix the amount of capital. The only way the firm can vary theamount of output it produces is to vary the labor that it hires. Under some reasonableconditions such as constant return to scale (to both variable and fixed factors), the isoquantswill be getting further apart. In other words, each time I have to hire more and more labor toproduce one more unit of output while holding K fixed. If I produce no output at all, I stillhave to pay the cost associated with the fixed amount of capital. As I start producing someoutput, my cost increases. (My total cost in this case is wages times the amount of labor plusthe fixed cost). Initially, my cost increases slowly but later I have to hire more and more laborto produce an additional unit of output, so my cost increases at an increasing rate as shown infigure 2.We can now define two notions: total cost and average cost. Total cost is shown infigure 2. Average cost is total cost divided by the units of output produced, or TC/q. In1Microeconomic Theory: Lecturer. J. Ping FeiLecture 05- 2terms of figure 2, average cost is the slope of the ray from the origin to the total cost curve.This would be the slope of the diagonal lines in Figure 2. When output is close to zero,average cost is really big (the slope is really steep). If my total cost is $10,000 and I produceone unit of output, my average cost is $10,000. If I only produce 1/2 unit of output, my average cost is $20,000, assuming the variable cost of ½ unit and 1 unit are the same. Asoutput increases, the slope decreases until it reaches a minimum. Beyond that minimum,average cost starts to rise again. So, my average cost curve might be expected to have somekind of a U-shape (see figure 3).Marginal cost. Marginal cost is the cost of producing one more unit of output. Marginalcost is the derivative of total cost with respect to output (q). As output increases, the slope ofmarginal cost gets steeper. Since capital is fixed (say capital is the amount of machines Ihave), the only way I can increase my output is by hiring more unit of labor. However, themore people I hire to work with the same number of machines, the less additional output Iwill be getting from each additional unit of labor. Eventually, I will run into some capacitylimitation and my output will be very difficult to increase.MCACTotal cost for the small factoryTotal cost for the large factoryqFigure 3Figure 4The marginal cost curve goes through the minimum of the average cost curve (see figure 3)because the minimum average cost is where the average cost is tangent to total cost (seefigure 2), and the slope of the tangent to total cost is marginal cost.For the intuition, think about bowling scores. You know that if you score less than youraverage, then you lower your average and if you score more than your average, then you raiseyou average. If your average stays the same, it must be because your score on the last gamewas the same as your average.Long run total cost and short run total cost. In the long run I can choose the amount ofthe fixed input, capital in this case. For example, I can choose between a small factory and alarge one. In terms of the graph, the total cost curve of a small factory will have a lower levelof fixed cost than the large factory. However, the total cost curve of the small factory willrise faster than the total cost curve of the larger factory (see figure 4).Which factory would I rather have? It would depend on how much I want to produce. IfI want to produce any amount up to where the short run total cost curves intersect, I wouldrather have the smaller factory because it produces it more cheaply. On the other hand, if Iwant to produce a large amount of output, then I want to have the larger factory because itproduces that level of output at a lower cost. If these two factories are my only alternatives,in the long run my total cost curve would look like the thick line in figure 4.If we have more alternatives in terms of size of factories (each alternative has a differenttotal cost curve), in the long run the cost of any particular level of output is going to be thecost associated with the cheapest means of producing that level of output. The long run totalcost curve will be the lower bound of all the short run total cost curves (see figure 5).2Microeconomic Theory: Lecturer. J. Ping FeiLecture 05- 3SRTCSRTC (1) SRTC LRTCAC (1)AC (3) AC (2)LRACqq11Figure 5Figure 6The LRTC curve is just the lower ‘envelope’ of all the SRTC curves. The LRTC curvedoes not have to start at the origin if there are fixed costs. For any particular set of fixedfactors, short run total cost is the minimum total cost over all of the factors that are variable.Long run total cost for a given level of output is the minimum of short run total cost overfixed factors of production given the level of output. And this is equal to the minimum (overfixed factors) of the minimum (over variable factors) of total cost given q.The question is in what order we are doing the minimization. We turned this into atwo-stage problem. My costs depend not only on the level of output but also on how much ofthe variable factors I hire and how much of the fixed factors I hire. We are defining short runcost as what you get when you minimize over the variable factors. And we are defining longrun cost as what you get when you minimize the short run cost over the fixed factor. What weend up with is a long run total cost obtained by minimizing costs over both fixed and variablefactors. The order of the minimization does not matter.Long run average cost and short run average cost. Figure 6 shows the average costcurves corresponding to the total cost curves in figure 5.short-run total cost=STC=wv xv (w, y, xf ) + wf x fshort-run average cost=SAC=c(w, y, xf ) / yshort-run average variable cost=SAVC=wv xv (w, y, xf ) / yshort-run average fixed cost=SAFC=wf x f / yshort-run marginal cost=SMC=∂c(w, y, x f ) / ∂ylong-run average cost=LAC=c(w, y) / ∂ylong-run marginal cost=LMC=∂c(w, y) / ∂yTo derive the long run and short run average cost curves, suppose we choose a certain level of output, say q1. At output level q1, long run total cost was obtained by choosing the minimum of all ways of producing q1. Output level q1 was produced at lowest cost by factory 1. At q1, the average total costs are the same in the short run and the long run. If factory 1 produces at any level of output different from q1, short run total costs are higher than long run total costs, and hence short run average costs are higher than long run average costs.We can repeat this analysis with factories 2 and 3. So the long run average cost curve is the envelope of the short run average cost curves, in the same way that my long run cost curve is the envelope of the short run total cost curves. Also, long run and short run average costs will be the same for the output level for which the particular factory is the least costly at producing the given level of output.3Microeconomic Theory: Lecturer. J. Ping FeiLecture 05- 4Note that the place where the SRAC curves and the LRAC curve are tangent is notnecessarily the minimum of each SRAC curve. When the LRAC curve is declining thetangency point occurs before the minimum of the SRAC curve and where the LRAC curve isincreasing, the tangency point occurs beyond the minimum of the SRAC curve. The onlypoint where the tangency point occurs at the minimum of a SRAC curve is at the minimum ofthe LRAC curve.This may sound confusing because we are saying that to produce a level of output q1, the best plant to use is plant 1. However, q1 is not the best level of output for that particular plant. Plant 1 could produce a higher level of output at a lower average cost. But if we want toproduce at that higher level of output, plant 1 would not be the best plant to use.We are not interested in the level of output that a plant produces most cheaply for itself.Instead, we are interested in how well that plant produces a particular level of output, relativeto other plants.If I want to produce 100 units of output and plant 1 produces the 100 units of output atthe lowest cost, I do not care if plant 1 can produce 110 units at a lower average cost. What Icare about is which plant produces 100 units in the cheapest way.A baseball analogy: I am interested in putting the best second baseman at second base.The fact that he may be a better third baseman does not interest me if I have a guy who playsthird base still better than he. The fact that you do third base best does not interest me if youhappen to be the best second base player I have and if I have a guy who can play third basebetter than you can.To summarize, what is important to you is not running a plant at its minimum averagecost. What is important to you is deciding what level of output you want to produce anddeciding how to produce that level of output in the cheapest way.How do we decide what level of output I want to produce? Suppose I can sell myoutput at price p per unit of output. If I sell q units, my revenue is pq. My profit is my revenueminus my cost. Graphically, my profit is the vertical difference between the revenue line andthe total cost curve (it does not matter if it is short run total cost or long run total cost) (seefigure 7).$TC Revenue (pq)LRMC LRACNegative profitsPPositive profitsqq*q*Figure 7Figure 8If I want to maximize my profit, I have to choose the point where the curves are thefurthest apart, that is, where the slopes are the same. Since the slope of the revenue line is theprice of output, and the slope of the total cost curve is marginal cost, I maximize my profitwhere price equals marginal cost (where output = q* in figure 7). If I produce less than that,price is greater than marginal cost and I can make more money by producing more. If Iproduce at a point where marginal cost is greater than price, then I can make more money byproducing less.4Microeconomic Theory: Lecturer. J. Ping FeiLecture 05- 5In figure 8, I maximize my profit if I produce at the level of output where the LRMC is equal to the price (at q*).So, I want to produce a level of output q*, and I want to produce it in the cheapest way. The factory associated with the SRAC curve in figure 8 is the factory that produces q* in the cheapest way. In this case, q* is at a point beyond the minimum for that factory.2．Examples:1). The short-run Cobb-Douglas cost functions. min w1x1 + w2 x2 s.t. x2 = k and y = x1a x12−a1−a⇒ SAC = w1( y / k) a + w2k / y1−aSMC=w1( y / k) a / a Proposition 1: Constant returns to scale. If the production function exhibits constant returns to scale, the cost function may be written as c(w, y)= y c(w,1).Proof: let x* be a cheapest way to produce one unit of output at prices w. Notice that yx* is feasible to produce y since the technology is constant returns to scale. Suppose it does not minimize cost; instead let x′ be the cost-minimizing bundle to produce y at w so wx′/y<wyx*/y. Then x′/y can produce 1 since the technology is constant returns to scale.Proposition 2: Marginal costs equal average costs at the point of minimum average costs.Proof: Let y* denote the point of minimum average cost, then to the left of y* averagecosts are declining so that for y≤y* d (c( y) / y) / dy ≤ 0 which implies c′( y) ≤ c( y) / y ∀y ≤ y * . A similar analysis shows that c′( y) ≥ c( y) / y ∀y ≥ y * .Proposition 3:limy→0cv( y) y= cv′ (0)2). CD cost curves1c( y) = Ky a+b1− a −bAC( y) = Ky a+bMC( y) = c′( y) =K1−a −by a+ba+bProposition 4: The long- and short run (average) cost curves are tangent.3. Factor prices and cost functionsProperties of the cost function (1) Non-decreasing in prices. If w′≥w, then c(w′, y) ≥c(w, y). (2) Homogeneous of degree 1 in w. c(tw, y)= t c(w, y) for t>0. (3) Concave in w. c(tw + (1− t)w′, y) ≥ tc(w, y) + (1− t)c(w′, y) ∀0 ≤ t ≤ 1(4) Continuous in w. c(w, y) is continuous as a function of w for w ? 0. (5) Shephard’s lemma. (the derivative property.) Let xi(w, y)be the firm’s conditional factor demand for input i. Then if the cost function is differentiable at (w, y), and wi>0 for i=1, …, n,5Microeconomic Theory: Lecturer. J. Ping FeiLecture 05- 6then xi (w, y) = ∂c(w, y) / ∂wi i = 1,L, n . Proof: Construct g(w)= c(w, y)−wx*, since is thecheapest way to produce y, ∂g(w)/∂wi=0. We can also prove it by the definition of c(w, y)= wx(w, y) or directly by the envelopetheorem. The concavity of the cost function is also a nice geometrical argument.4. The envelope theorem for constrained optimizationdM (a) = ∂L(x, a)da∂ax= x(a )=∂g ∂ax=x(a)−λ∂h ∂aApplications: Shephard’s lemma⇒∂c(w, ∂wiy)=xixi = xi ( w, y )=xi (w,y),marginal cost L = w1x1 + w2 x2 − λ[ f (x1, x2 ) − y] ∂c(w1, w2 , y) / ∂y = λ5. Comparative statics using the cost function1) The cost function is nondecreasing in factor prices⇔ Factor demand x(w,y) is nonnegative. 2) The cost function is homogeneous of degree 1 in w ⇔ x(w,y) is homogeneous of degree 0. 3) The cost function is concave in w has the following implications.a) the cross-price effects are symmetric. b) the own-price effects are nonpositive. c) dwdx≤06。
lecture 05 一致性 假设检验
weighted least squares estimator (WLS). 因此，该式的OLS估计量被称为加权 最小二乘估计量（WLS）
FGLS估计
• 如果h(x)不知道，应该怎么办？
• 下面的方法不是去猜，而是去估计一个函
数。
uˆi2 0 1x1 2 x2 ...k xk vi
• 这种方法的一个好处，预测值不能保证为 正。
)
xi
2
x
2
如果有异方差呢？
• 上面的推导变为：
var(ˆ1)
xi x 2 var(ui ( xi x 2 )2
)
xi
2
x
2
• 看下面这个式子
xi x 2 var(ui ) ( xi x 2 )2
xi x 2uˆi2 ( xi x 2 )2
• 经过严格的证明，还真可以。
多元回归的情况
刻画，其中hi =h(x) 只依赖于可观测特征x • 把原方程的左右两边分别除以 h(x)：
yi h(x)
0
1 h(x)
1
x1 h(x)
...k
xk h(x)
ui h(x)
• 上面方程的残差：
var(ui*) var(
ui ) var(ui ) 2
h(x) h(x)
y* i
* 0
x* 1 i1
趋于零，故ˆ1的概率极限即1。
plimˆ1
1
n1 n1
xi1 x1 ui xi1 x1 2
1.
7
一个更弱的假定
• 要获得估计量的无偏性，我们假定零条件期望 – E(u|x1, x2,…,xk) = 0
• 而要获得估计量的一致性，我们可以使用更弱的假定： 零期望和零相关性假定。
Lecture05-Clipping 计算机图形学ppt课件
else if code AND 0010 then begin // xmax y = y0 + dy * (xmax – x0) / dx; x = xmax
▪ No clipping necessary
2. One endpoint in, one out (line CD)
▪ Clip at intoth endpoints outside
the region:
a. No intersection (lines EF, GH)
Line Clipping
• Strategies for clipping:
a) Check (in inner loop) if each point is inside Works, but slow
if (x >= xmin && x <= xmax && y >= ymin && y <= ymax)
drawPoint(x,y,c);
b) Clip invalid coordinate(s) to boundary Incorrect results
if
(x < xmin) x = xmin;
else if (x > xmax) x = xmax;
if
(y < ymin) y = ymin;
else if (y > ymax) y = ymax;
• Test:
if code0 OR code1 = 0000 accept (draw)
0101
0100
0110
else if code0 AND code1 0000 reject (don’t draw)
翻译 Lecture 5
（三）译成祈使句
145.好好学习，天天向上。 Study hard and make progress every day. 146.学习好，工作好，身体好。 Study hard, work well, and keep fit. 147.按照“三个代表”要求加强和改进党的建 设。 Strengthen and improve the Party building in accordance with the requirements of the “Three Represents”.
（五）译成非谓语结构
151. 巩固和加强农业基础地位，千方百计增加农民 收入。 Consolidating and strengthening the position of agriculture as the foundation of the economy and doing everything possible to increase farmers incomes. 152. 做好加入世贸组织的各项准备，把对外开放提 高到新水平。 Preparing for China’s entry into the WTO and opening up even more to the outside world.
（六）动词+宾语+分词宾补
179.重力使所有物体不能飞离地球。 Gravity keeps all objects from flying away from the earth.
180. 他给我的那一点点温暖至今还使我的心颤 动。（巴金《朋友》） The little warmth that he gave me has been keeping my heart throbbing with gratitude.
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0
Z = 0.12
© 2004 Prentice-Hall, Inc.
Only One Table is Needed
Chap 6-10
Standardizing Example
Z X 51.2 50 0.12
s
10
Normal Distribution
s 10
Standardized Normal Distribution
The Normal Distribution(正态分布) The Standardized Normal Distribution(标准正
态分布) Evaluating the Normality Assumption The Uniform Distribution(均匀分布)© 2004 Prentic源自-Hall, Inc.sZ 1
51.2 X
50
© 2004 Prentice-Hall, Inc.
0.12
Z 0
Z
Chap 6-11
Example:
P47.9 X 52.1 .1664
Z X 47.9 50 .21
s
10
Normal Distribution
s 10
Z X 52.1 50 .21
Varying the Parameters s and , We Obtain
Different Normal Distributions
© 2004 Prentice-Hall, Inc.
Chap 6-6
The Standardized Normal Distribution
When X is normally distributed with a mean and a
X
Mean Median Mode
© 2004 Prentice-Hall, Inc.
Chap 6-4
The Mathematical Model
f X
1
e(1/ 2) X /s 2
2 s
f X : density of random variable X
3.14159; e 2.71828
Chap 6-8
Which Table to Use?
Infinitely Many Normal Distributions Means Infinitely Many Tables to Look Up!
© 2004 Prentice-Hall, Inc.
Chap 6-9
Solution: The Cumulative Standardized Normal Distribution
The Normal Distribution
“Bell Shaped”
f(X)
Symmetrical
Mean, Median and Mode are Equal
Interquartile Range Equals 1.33 s
Random Variable Has Infinite Range
standard deviation s , Z X follows a s
standardized (normalized) normal distribution with a
mean 0 and a standard deviation 1.
f(Z) f(X)
s
X ~ N ,s 2
Z ~ N 0,1
Cumulative Standardized Normal Distribution Table (Portion)
Z .00 .01 .02
0.0 .5000 .5040 .5080
Z 0 sZ 1
.5478
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
Probability 0.3 .6179 .6217 .6255
Chap 6-2
Continuous Probability Distributions
Continuous Random Variable
Values from interval of numbers Absence of gaps
Continuous Probability Distribution
Basic Business Statistics
(9th Edition)
Lecture 5 The Normal Distribution (正态分布) and Other Continuous Distributions
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Chap 6-1
Topics
: population mean
s : population standard deviation
X : value of random variable X
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Chap 6-5
Many Normal Distributions
There are an Infinite Number of Normal Distributions
Distribution of continuous random variable
Most Important Continuous Probability Distribution
The normal distribution
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Chap 6-3
s
10
Standardized Normal Distribution
.0832 s Z 1
.0832
47.9 52.1 X
50
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0.21 0.21
Z 0
Z
Chap 6-12
Example:
P47.9 X 52.1 .1664 (continued)
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Z 0
sZ 1
X
Z
Chap 6-7
Finding Probabilities
Probability is the area under the curve!
f(X)
Pc X d ?
X
cd
© 2004 Prentice-Hall, Inc.