Chapter02-15[1]
曼昆《经济学原理》Chapter 02
10
Figure 2
The production possibilities frontier
Quantity of Computers Produced
3,000 2,200 2,000 F C The production possibilities frontier shows the combinations of output—in this case, cars and computers— that the economy can possibly produce. The economy can produce any combination on or inside the frontier. Points outside the frontier are not feasible given the economy’s resources.
6
The Economist as a Scientist
• Firms
– Produce goods and services – Use factors of production / inputs
• Households
– Own factors of production – Consume goods and services
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语言学练习Exercise2
Chapter 2 5
Exercises
I. Fill in the blanks. 11. _____ is the smallest linguistic unit. Phoneme 12. According to _____, when there is a choice as to where to place a consonant, it is put into the onset rather than the coda. the maximal onset principle
Chapter 2 7
Exercises
II. Choose the best answer. 15. An aspirated p, an unaspirated p and an unreleased p are _____ of the p phoneme. A. analogues B. tagmemes C. morphemes D. allophones D 16. The opening between the vocal cords is sometimes referred to as _____. A. glottis B. vocal cavity C. pharynx D. uvula A
Chapter 2 14
Exercises
III. Decide whether the following statements are true [T] or false [F]. __ 29. Two sounds are in free variation when they occur in the same environment and do not contrast, namely, the substitution of one for the other does not produce a different word, but merely a different pronunciation. T __ 30. [p] is voiced bilabial stop. F
罗伯特 麦基虚构艺术三部曲
作者介绍
同名作者介绍
罗伯特·麦基(Robert McKee,1941年1月30日-),出生于美国底特律,编剧,作家,导演。他从小热爱 戏剧,早年做过演员,受美国南加利福尼亚大学之邀,开设了“故事”培训班,同时在好莱坞制作电视节目,他 的“故事”培训班已经成为全球最大的故事与写作机构,代表作有《故事》等。
罗伯特·麦基虚构艺术三部曲
读书笔记模板
01 思维导图
03 读书笔记 05 目录分析
目录
02 内容摘要 04 精彩摘录 06 作者介绍
思维导图
本书关键字分析思维导图
读者
银幕
三部曲
银幕
原理
对白
艺术
艺术
人物
故事 人物
场景
罗伯特麦 基
设计
舞台
作家
结构
故事
角色
内容摘要
内容摘要
罗伯特·麦基虚构艺术三部曲包含了《故事:材质、结构、风格和银幕剧作的原理》(2001,新版2014), 《对白:文字、舞台、银幕的言语行为艺术》(2018),直到最新作《人物:文本、舞台、银幕剧作与卡司设计 的艺术》于2022年10月出版,三部曲至此完结,麦基的电影理论宇宙也臻于完备。这个过程几乎横跨了二十余年, 麦基先生从他1981年创办“故事”培训班开始便宣扬讲授的“故事”理论,也在这一过程中不断发展、完善,直 到《人物》为三部曲画上了“完美句号”。译者周铁东先生这样评价:《故事》《对白》和《人物》,虽跨度二 十余年,但作为“虚构艺术三部曲”,三本书相辅相成,相得益彰,堪称一个完美整体。从“三部曲”的副标题 便能窥知,其读者面是层层递进的:《故事》主要是面向专业编剧,即银幕剧作家;《对白》则扩展到了“文本” 和“舞台”;《人物》则更具普适价值,因为它几乎就是一部关于“人性”的哲学、心理学、社会学和文艺学等 领域的具有可读性和娱乐性的科普著作。只要是对“人性”这个永恒命题感兴趣的读者,都能从中找到自己的阅 读趣味。哪怕是无心成为专业写手的业外读者,也能对书中的条分缕析叹为观止并感同身受,并对自己这辈子在 各种文艺作品中已经邂逅和将要涉猎的各种伟大的“虚构人物”有更加直观生动和细致入微的认识。
chapter02汇率决定理论
即:e = e(Y,Y*,P,P* ,i,i*,Eef )
各变量对汇率的影响:
1)其他条件不变,如果本国 Y 增加,通过边际进
口倾向促进进口上升,则外汇需求扩大,外汇汇率
上升,本币汇率下降。外国 Y* 的增加,带来本国
2、一般形式 假设:
①本国投资者选择投资一年期存款,利率为i、i* ②直接标价法,即期汇率为e,远期汇率为f,期
满时的即期汇率为ef
一般形式 ρ = i- i*
3、经济含义
①汇率的远期升贴水取决于两国货币相对利率水 平。
②高利率货币在外汇市场上表现为贴水,低利率 货币在外汇市场上表现为升水。
4、评价:
理论价值
货币分析法将购买力平价这一形成于商品市场上 的汇率决定理论引入到资产市场分析,抓住了汇 率的特殊性质,在一定程度上符合当今世界国际 资本高度流动的客观事实。
该模型引入了诸如货币供应量、国民产出等经济 变量,分析了这些变量对汇率的影响,特别是突 出强调了货币因素在汇率决定中的作用。
27
购买力平价说的局限性 :
购买力平价理论将决定汇率水平的许多实际因素 忽略了,如:国际资本流动、非贸易品、交易成 本等。
忽视了其他经济因素对汇率的影响。 购买力平价并不是一个完整的汇率决定理论。 购买力平价在计量检验中存在很大的技术上的困
难。
28
计量检验中的困难:
采用何种物价指数最为恰当。 在商品分类上,要求不同国家的分类具有一致性
某一时点上的均衡汇率:
e P P*
19
定义
➢ 本国货币的购买力=1/ 本国物价水平(P) ➢ 外国货币的购买力=1/ 外国物价水平(P*)
《最终幻想15》图文攻略全章节流程图文攻略
《最终幻想15》图⽂攻略全章节流程图⽂攻略 C H A P T E R01-启程 主⾓诺克提斯·路希斯·切拉姆,为了出席在邻国举办的⾃⾝的婚礼,⽽准备与三位好友驾着⽗亲的爱车“雷格⾥亚”启程。
不过这段旅途似乎在⼀开始就遭遇了⼀些⼩意外,雷格⾥亚在半道罢⼯了,四⼈只得将车推去附近的“锤头鲨”进⾏修理。
这⾥根据屏幕右侧出现的提⽰,按R2来推车前进吧。
抵达“锤头鲨”后,在这⾥迎接的是⽼板的孙⼥希德妮,虽然她爷爷希德还是个⽼头⼦,不过总算15代的C i d⾥也有⼀个妹⼦了。
希德妮接下了雷格⾥亚的维修⼯作。
看诺克提斯初来乍到,还给予了她⼀份本地区域的地图。
得到地图后,会出现提⽰建议前往餐厅。
【餐厅】是⼀个综合性的场所,在这⾥不但可以收集周边的资讯(包括露营地点,资源采集点灯),还能够享⽤美⾷,或者是承接怪兽讨伐任务,来获得丰厚的奖赏。
不过最低的怪物讨伐也是5级的,还需要升级后再来进⾏。
来到加油站附近跟伊格尼斯对话,触发任务【为⽣活费烦恼的王⼦】 众⼈发现修车已经花掉了⾝上所有的费⽤,必须设法赚取⼀些外界的货币才能继续旅⾏。
%{p a g e-b r e a k|C H A P T E R01-启程(1)|p a g e-b r e a k}% 回头找到希德妮商量,选择“拜托她把维修费打个折”,希德妮会告知爷爷刚才接了⼀份委托,要四⼈去打倒野兽赚钱。
选择这个选项,可以获得10点A P。
关于A P的帮助讯息,游戏中会有详细说明,此处就不赘述了。
之后就是奔赴三处地点分别击败盘踞在区域中的怪物,每次消灭⼀波就会显⽰下⼀波怪物的地点,依次赶赴消灭。
当三波怪物都解决后,希德妮会打开电话,告知她已经完成任务,就可以完成任务获得奖励。
并且希德妮还会请求四⼈在附近找⼀个叫戴夫的赏⾦猎⼈,触发任务【协寻奖⾦猎⼈】 ⾸先前往标⽰位置的⼩屋之中。
发现⼩屋⾥并没有猎⼈的踪迹,但是在⼀个桌柜上有⼀张字条,上前互动,发现字条是⼀张讨伐委托,⽬标是⼀只叫“⾎腥双⾓兽”的怪物。
chapter02课后作业
北大青鸟-徐州中博课后习题(Lesson2)一、选择题1.下列选项中,( )可以打开一个无状态的页面A.window.open("advert.html")B.window.open("advert.html","广告","toolbar=1,scrollbars=0,status=1");C.window.open("advert.html","","scrollbars=1,location=0,resizable=1");D.window.open("advert.html","","toolbars=1,location=1,status=no");2.下列关于Date对象的getMonth()方法返回值的描述,正确的是()A.返回系统时间的当前月B.返回值为1-12C.返回系统时间的当前月+1D.返回值为0-113.setTimeout("adv()",20)表示的意思是( )A.间隔20秒后,adv()函数就会被调用B.间隔20分钟后,adv()函数就会被调用C.间隔20毫秒后,adv()函数就会被调用D.adv()函数被持续调用20次4.下列()可以使窗口显示前一个页面。
A.back()B.forward()C.go(1)D.go(-1)5.某页面中有一个id为mobile的图片,下面()能够正确获取此图片对象。
A.document.getElementsByName("mobile");B.document.getElementById("mobile");C.document.getElementsByTagName("mobile");D.以上选项都可以6.下面()不是document对象的方法A.getElementsByTagName()B.getElementById()C.write()D.reload()7.下面()可以实现刷新当前页面的功能A.reload()B.replace()北大青鸟-徐州中博C.hrefD.referrer二、简答题1.简述prompt(),alert()和confirm()三者的区别,并举例说明。
茶树育种学Chapter-02第二章 引 种
鐘鳴谷應
第二节 引种驯化的原理
一、 引种驯化的遗传学原理
G + E =江P 山千秋俊
P是引种的效果,G是适应性的宽窄,E
昌隆萬代興 是原产地与引种地生态环境的差异。环境
条件的差异是可以度量的,G就成为决定 引种效果P的关键因素。
鐘鳴谷應
第一节 引种驯化的概念和意义
系统发育:长短不同,遗传可塑性有差异。栽培种比 野生种、新育成品种比地方品种、中小叶种比大叶种系 统发育短 。
鐘鳴谷應
第二节 引种驯化的原理
(二)限制茶树引种驯化的主要生态因子 茶树的生存和繁殖,必须有一定的环境条件,
对茶树的生长发育有明显影响和直接同化的因素
江山千秋俊 称生态因子生态因子有气候的、土壤的、生物的,
这种起综合作用的一些生态因素的复合体称生态
昌隆萬代興 环境,生态环境中各个因子的作用不是相等的。 限制茶树引种的主要生态因子有水分、温度、 光照和土壤酸碱度等。
江山千秋俊 州差异要小)引进已在当地引种成功的云南大叶种种子
(与当地品种自然杂交后代)到杭州。 良种中有不少抗寒性较强的品种、与当地抗寒性品种杂
昌隆萬代興 交培育、播种茶籽。
(2)栽培措施 早封园、茶园铺草等;寒冷地区宜选择春季移栽;海拔
宜在1000m以下,避风阳坡。
鐘鳴谷應
第二节 引种驯化的原理
2.降水 降水量要求:年降水量> 700mm;生长季节,月降
水量> 100mm。
江山千秋俊 全国降水规律:东南向西北逐渐减少,夏季降水量
比春、秋、冬三季降水量多。华南茶区、西南茶区和
昌隆萬代興 江南茶区年降水量满足茶树生长,江北茶区应加以注
意。 干旱症状:出苗困难;营养生长缓慢,发芽轮次少;
intermediate accounting 中级会计(wiley)ch02
First level
LO 2 Describe efforts to construct a conceptual framework.
First Level: Basic Objective
OBJECTIVE
“To provide financial information about the reporting entity that is useful to present and potential equity investors, lenders, and other creditors in making decisions in their capacity as capital providers.”
ELEMENTS 1. 2. 3. 4. 5. Assets Liabilities Equity Income Expenses
Second level
Chapter 2-8
OBJECTIVE Provide information about the reporting entity that is useful to present and potential equity investors, lenders, and other creditors in their capacity as capital Providers.
Chapter 2-6
LO 2 Describe efforts to construct a conceptual framework.
Conceptual Framework
Overview of the Conceptual Framework
Three levels:
First Level = Basic objective Second Level = Qualitative characteristics and elements of financial statements Third Level = Recognition, measurement, and disclosure concepts
《运营管理》课后习题答案
Chapter 02 - Competitiveness, Strategy, and Productivity3. (1) (2) (3) (4) (5) (6) (7)Week Output WorkerCost@$12x40OverheadCost @1.5MaterialCost@$6TotalCostMFP(2) ÷ (6)1 30,000 2,880 4,320 2,700 9,900 3.032 33,600 3,360 5,040 2,820 11,220 2.993 32,200 3,360 5,040 2,760 11,160 2.894 35,400 3,840 5,760 2,880 12,480 2.84*refer to solved problem #2Multifactor productivity dropped steadily from a high of 3.03 to about 2.84.4. a. Before: 80 ÷ 5 = 16 carts per worker per hour.After: 84 ÷ 4 = 21 carts per worker per hour.b. Before: ($10 x 5 = $50) + $40 = $90; hence 80 ÷ $90 = .89 carts/$1.After: ($10 x 4 = $40) + $50 = $90; hence 84 ÷ $90 = .93 carts/$1.c. Labor productivity increased by 31.25% ((21-16)/16).Multifactor productivity increased by 4.5% ((.93-.89)/.89).*Machine ProductivityBefore: 80 ÷ 40 = 2 carts/$1.After: 84 ÷ 50 = 1.68 carts/$1.Productivity increased by -16% ((1.68-2)/2)Chapter 03 - Product and Service Design6. Steps for Making Cash Withdrawal from an ATM1. Insert Card: Magnetic Strip Should be Facing Down2. Watch Screen for Instructions3. Select Transaction Options:1) Deposit2) Withdrawal3) Transfer4) Other4. Enter Information:1) PIN Number2) Select a Transaction and Account3) Enter Amount of Transaction5. Deposit/Withdrawal: 1) Deposit —place in an envelope (which you’ll find near or in the ATM) andinsert it into the deposit slot2) Withdrawal —lift the “Withdrawal Door,” being careful to remove all cash6. Remove card and receipt (which serves as the transaction record)8.Chapter 04 - Strategic Capacity Planning for Products and Services2. %80capacityEf f ective outputActual Ef f iciency ==Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4capacityDesign outputActual =n Utilizatiojobs 204.8capacity Ef f ective output Actual Capacity Design ===10. a. Given: 10 hrs. or 600 min. of operating time per day.250 days x 600 min. = 150,000 min. per year operating time.Solutions_Problems_OM_11e_Stevenson3 / 22Total processing time by machineProductABC 1 48,000 64,000 32,000 2 48,000 48,000 36,000 3 30,000 36,000 24,000 460,000 60,000 30,000 Total 186,000208,000122,000machine181.000,150000,122machine 238.1000,150000,208machine224.1000,150000,186≈==≈==≈==C B A N N NYou would have to buy two “A” machines at a total cost of $80,000, or two “B” machines at a total cost of $60,000, or one “C” machine at $80,000.b.Total cost for each type of machine:A (2): 186,000 min ÷ 60 = 3,100 hrs. x $10 = $31,000 + $80,000 = $111,000B (2) : 208,000 ÷ 60 = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133 C(1): 122,000 ÷ 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400Buy 2 Bs —these have the lowest total cost.Chapter 05 - Process Selection and Facility Layout3.Desired output = 4Operating time = 56 minutesunit per minutes 14hourper units 4hourper minutes 65output Desired time Operating CT ===Task # of Following tasksPositional WeightA 4 23B 3 20C 2 18D 3 25E 2 18F 4 29G 3 24H 1 14 I5a. First rule: most followers. Second rule: largest positional weight.Assembly Line Balancing Table (CT = 14)Solutions_Problems_OM_11e_Stevenson5 / 22b. First rule: Largest positional weight.Assembly Line Balancing Table (CT = 14)c. %36.805645stations of no. x CT time Total Ef f iciency ===4. a. l.2. Minimum Ct = 1.3 minutesTask Following tasksa 4b 3c 3d 2e 3f 2g 1h3. percent 54.11)3.1(46.CT x N time)(idle percent Idle ==∑=4. 420 min./day 323.1 ( 323)/1.3 min./OT Output rounds to copiers day CT cycle=== b. 1. inutes m 3.224.6N time Total CT ,6.4 time Total ==== 2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time]Assign f, g, and h to station 2: 2.3 minutes3. 420182.6 copiers /2.3OT Output day CT ===4.420 min./dayMaximum Ct is 4.6. Output 91.30 copiers /4.6 min./day cycle==7.Solutions_Problems_OM_11e_Stevenson7 / 22Chapter 06 - Work Design and Measurement3.Element PR OT NT AF job ST1 .90.46.414 1.15 .4762 .85 1.505 1.280 1.15 1.4723 1.10.83.913 1.15 1.05041.00 1.16 1.160 1.15 1.334Total4.3328. A = 24 + 10 + 14 = 48 minutes per 4 hours.min 125.720.11x70.5ST .min 70.5)95(.6NT 20.24048A =-=====9. a. Element PR OT NT A ST1 1.10 1.19 1.309 1.15 1.5052 1.15 .83 .955 1.15 1.09831.05.56.588 1.15 .676b.01.A 00.2z 034.s 83.x ==== 222(.034)67.12~68.01(.83)zs n observations ax ⎛⎫⎛⎫===⎪ ⎪⎝⎭⎝⎭ c. e = .01 minutes 47 to round ,24.4601.)034(.2e zs n 22=⎪⎭⎫⎝⎛=⎪⎭⎫ ⎝⎛=Chapter 07- Location Planning and Analysis1. Factor Local bank Steel mill Food warehouse Public school1. Convenience forcustomers H L M–H M–H2. Attractiveness ofbuilding H L M M–H3. Nearness to rawmaterials L H L M4. Large amounts ofpower L H L L5. Pollution controls L H L L6. Labor cost andavailability L M L L7. Transportationcosts L M–H M–H M8. Constructioncosts M H M M–HLocation (a) Location (b)4. Factor A B C Weight A B C1. Business Services 9 5 5 2/9 18/9 10/9 10/92. Community Services 7 6 7 1/9 7/9 6/9 7/93. Real Estate Cost 3 8 7 1/9 3/9 8/9 7/94. Construction Costs 5 6 5 2/9 10/9 12/9 10/95. Cost of Living 4 7 8 1/9 4/9 7/9 8/96. Taxes 5 5 5 1/9 5/9 5/9 4/97. Transportation 6 7 8 1/9 6/9 7/9 8/9Total 39 44 45 1.0 53/9 55/9 54/9 Each factor has a weight of 1/7.a. Composite Scores 39 44 45 7 7 7B orC is the best and A is least desirable.b. Business Services and Construction Costs both have a weight of 2/9; the other factors eachhave a weight of 1/9.5 x + 2 x + 2 x = 1 x = 1/9c. Composite ScoresA B C 53/9 55/9 54/9B is the best followed byC and then A.Solutions_Problems_OM_11e_Stevenson9 / 225. Location x yA3 7 B 8 2 C4 6 D4 1E 6 4 Totals 25 20-x = ∑x i = 25 = 5.0-y =∑y i = 20= 4.0n5n5Hence, the center of gravity is at (5,4) and therefore the optimal location.Chapter 08 - Management of Quality1.ChecksheetWork Type FrequencyLube and Oil 12 Brakes 7 Tires 6 Battery 4 Transmission1Total30ParetoLube & Oil BrakesTiresBatteryTrans.2 .The run charts seems to show a pattern of errors possibly linked to break times or the end of the shift. Perhaps workers are becoming fatigued. If so, perhaps two 10 minute breaks in the morning and again in the afternoon instead of one 20 minute break could reduce some errors. Also, errors are occurring during the last few minutes before noon and the end of the shift, and those periods should also b e given management’s attention.4break lunch break3 2 1 0∙ ∙ ∙∙ ∙ ∙ ∙∙∙ ∙ ∙ ∙∙∙∙∙∙∙ ∙∙∙ ∙∙ ∙ ∙∙ ∙ ∙∙∙Solutions_Problems_OM_11e_Stevenson11 / 22Chapter 9 - Quality Control4. Sample Mean Range179.48 2.6 Mean Chart: =X ± A 2-R = 79.96 ± 0.58(1.87) 2 80.14 2.3 = 79.96 ± 1.083 80.14 1.2UCL = 81.04, LCL = 78.884 79.60 1.7 Range Chart: UCL = D 4-R = 2.11(1.87) = 3.95 5 80.02 2.0LCL = D 3-R = 0(1.87) = 0680.381.4[Both charts suggest the process is in control: Neither has any points outside the limits.]6. n = 200 Control Limits = np p p )1(2-±Thus, UCL is .0234 and LCL becomes 0.Since n = 200, the fraction represented by each data point is half the amount shown. E.g., 1 defective = .005, 2 defectives = .01, etc.Sample 10 is too large.7. 857.714110c == Control limits: 409.8857.7c 3c ±=± UCL is 16.266, LCL becomes 0.All values are within the limits.14. Let USL = Upper Specification Limit, LSL = Lower Specification Limit,= Process mean, σ = Process standard deviationFor process H:}{capablenot ,0.193.93.04.1 ,938.min 04.1)32)(.3(1516393.)32)(.3(1.14153<===-=σ-=-=σ-pk C X USL LSL X 0096.)200(1325==p 0138.0096.200)9904(.0096.20096.±=±=For process K:.1}17.1,0.1min{17.1)1)(3(335.3630.1)1)(3(30333===-=σ-=-=σ- C X USL LSL X pk Assuming the minimum acceptable pk C is 1.33, since 1.0 < 1.33, the process is not capable.For process T:33.1}33.1,67.1min{33.1)4.0)(3(5.181.20367.1)4.0)(3(5.165.183===-=σ-=-=σ- C X USL LSL X pk Since 1.33 = 1.33, the process is capable.Chapter 10 - Aggregate Planning and Master Scheduling7. a. No backlogs are allowedPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory TotalSolutions_Problems_OM_11e_Stevenson13 / 22b.Level strategyPeriodForecast Output Regular Overtime Subcontract Output - Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Backlog Total8.Period Forecast Output Regular Overtime Subcontract Output- Forecast Inventory Beginning Ending Average Backlog Costs: Regular Overtime Subcontract Inventory Backlog TotalChapter 11 - MRP and ERP1. a. F: 2G: 1H: 1J: 2 x 2 = 4 L: 1 x 2 = 2 A: 1 x 4 = 4 D: 2 x 4 = 8 J: 1 x 2 = 2 D: 1 x 2 = 2Totals: F = 2; G = 1; H = 1; J = 6; D = 10; L = 2; A = 44. Master ScheduleSolutions_Problems_OM_11e_Stevenson10. Week 1 2 3 4Material 40 80 60 70Week 1 2 3 4Labor hr. 160 320 240 280Mach. hr. 120 240 180 210a. Capacity utilizationWeek 1 2 3 4Labor 53.3% 106.7% 80% 93.3%Machine 60% 120% 90% 105%b. C apacity utilization exceeds 100% for both labor and machine in week 2, and formachine alone in week 4.Production could be shifted to earlier or later weeks in which capacity isunderutilized. Shifting to an earlier week would result in added carrying costs;shifting to later weeks would mean backorder costs.Another option would be to work overtime. Labor cost would increase due toovertime premium, a probable decrease in productivity, and possible increase inaccidents.15 / 22Chapter 12 - Inventory Management2.The following table contains figures on the monthly volume and unit costs for a random sample of 16 items for a list of 2,000 inventory items.a. See table.b. To allocate control efforts.c. It might be important for some reason other than dollar usage, such as cost of astockout, usage highly correlated to an A item, etc.3. D = 1,215 bags/yr. S = $10 H = $75a. bags HDS Q 187510)215,1(22===b. Q/2 = 18/2 = 9 bagsc.orders ordersbags bags Q D 5.67/ 18 215,1== d . S QD H 2/Q TC +=Solutions_Problems_OM_11e_Stevenson17 / 22350,1$675675)10(18215,1)75(218=+=+=e. Assuming that holding cost per bag increases by $9/bag/yearQ ==84)10)(215,1(217 bags71.428,1$71.714714)10(17215,1)84(217=+=+=TCIncrease by [$1,428.71 – $1,350] = $78.714.D = 40/day x 260 days/yr. = 10,400 packagesS = $60 H = $30a. oxes b 20496.2033060)400,10(2H DS 2Q 0====b. S QD H 2Q TC +=82.118,6$82.058,3060,3)60(204400,10)30(2204=+=+=c. Yesd. )60(200400,10)30(2200TC 200+=TC 200 = 3,000 + 3,120 = $6,1206,120 – 6,118.82 (only $1.18 higher than with EOQ, so 200 is acceptable.)7.H = $2/month S = $55D 1 = 100/month (months 1–6)D 2 = 150/month (months 7–12)a. 16.74255)100(2Q :D H DS2Q 010===83.90255)150(2Q :D 02==b. The EOQ model requires this.c. Discount of $10/order is equivalent to S – 10 = $45 (revised ordering cost)1–6 TC74 = $148.32180$)45(150100)2(2150TC 145$)45(100100)2(2100TC *140$)45(50100)2(250TC 15010050=+==+==+=7–12 TC 91 =$181.66195$)45(150150)2(2150TC *5.167$)45(100150)2(2100TC 185$)45(50150)2(250TC 15010050=+==+==+=10. p = 50/ton/day u = 20 tons/day200 days/yr.S = $100 H = $5/ton per yr.a. bags] [10,328 tons 40.5162050505100)000,4(2u p p H DS 2Q 0=-=-=b. ]bags 8.196,6 .approx [ tons 84.309)30(504.516)u p (P Q I max ==-=Average is92.154248.309:2I max =tons [approx. 3,098 bags] c. Run length =days 33.10504.516P Q == d. Runs per year = 8] approx.[ 7.754.516000,4QD == e. Q ' = 258.2TC =S QD H 2I max + TC orig. = $1,549.00 TC rev. = $ 774.50Savings would be $774.50D= 20 tons/day x 200 days/yr. = 4,000 tons/yr.Solutions_Problems_OM_11e_Stevenson19 / 2215. RangeP H Q D = 4,900 seats/yr. 0–999 $5.00 $2.00 495 H = .4P 1,000–3,999 4.95 1.98 497 NF S = $50 4,000–5,999 4.90 1.96 500 NF 6,000+ 4.85 1.94503 NFCompare TC 495 with TC for all lower price breaks:TC 495 = 495 ($2) + 4,900($50) + $5.00(4,900) = $25,4902 495 TC 1,000 = 1,000 ($1.98) + 4,900($50) + $4.95(4,900) = $25,4902 1,000 TC 4,000 = 4,000 ($1.96) + 4,900($50) + $4.90(4,900) = $27,9912 4,000 TC 6,000 = 6,000 ($1.94) + 4,900($50) + $4.85(4,900) = $29,6262 6,000Hence, one would be indifferent between 495 or 1,000 units22. d = 30 gal./day ROP = 170 gal. LT = 4 days,ss = Z σd LT = 50 galRisk = 9% Z = 1.34 Solving, σd LT = 37.31 3% Z = 1.88, ss=1.88 x 37.31 = 70.14 gal.Chapter 13 - JIT and Lean Operations1. N = ?N = DT(1 + X)D = 80 pieces per hourC T = 75 min. = 1.25 hr. = 80(1.25) (1.35)= 3C = 45 45X = .35QuantityTC4. The smallest daily quantity evenly divisible into all four quantities is 3. Therefore, usethree cycles.Product Daily quantity Units per cycleA 21 21/3 = 7B 12 12/3 = 4C 3 3/3 = 1D 15 15/3 = 55.a. Cycle 1 2 3 4A 6 6 5 5B 3 3 3 3C 1 1 1 1D 4 4 5 5E 2 2 2 2 b. Cycle 1 2A 11 11B 6 6C 2 2D 8 8E 4 4c. 4 cycles = lower inventory, more flexibility2 cycles = fewer changeovers7. Net available time = 480 – 75 = 405. Takt time = 405/300 units per day = 1.35 minutes. Chapter 15 - Scheduling6. a. FCFS: A–B–C–DSPT: D–C–B–AEDD: C–B–D–ACR: A–C–D–BFCFS: Job time Flow time Due date DaysJob (days) (days) (days) tardyA 14 14 20 0B 10 24 16 8C 7 31 15 16D 6 37 17 2037 106 44Solutions_Problems_OM_11e_Stevenson21 / 22SPT: Job time Flow time Due date Days Job (days) (days) (days) tardy D 6 6 17 0 C 7 13 15 0 B 10 23 16 7 A 14 37 20 17377924EDD:Job D has the lowest critical ratio therefore it is scheduled next and completed on day 27.b.ardi Flow time Average flow time Number of jobsDays tardy Average job t ness Number of jobs Flow timeAverage number of jobs at the center Makespan==∑=FCFS SPT EDD CR26.50 19.75 21.00 24.75 11.0 6.00 6.00 9.252.86 2.142.272.67c. SPT is superior.9.Thus, the sequence is b-a-g-e-f-d-c.。
Chapter_02
Land
Accounts Receivable
Buildings
Asset Accounts
Supplies
Notes Receivable
Equipment
Prepaid Accounts
Slide 6
C3
LIABILITY ACCOUNTS
Accounts Payable Notes Payable
Page 1
Credit
in ledger.
ACCOUNT No.
PR Debit Credit
2,500
CASH
101
Balance
Date 2009
Description
Dec. 3
Purchased equipment
G1
20,000.00
Slide 18 ########
P1
POSTING JOURNAL ENTRIES
Slide 10
C5
DEBITS AND CREDITS
A T-account represents a ledger account and is a tool used to understand the effects of one or more transactions.
Account Title (Left side) (Right side) Debit Credit
-
+
Debit Credit
+
-
Slide 13
C5
DOUBLE-ENTRY ACCOUNTING
An account balance is the difference between the increases and decreases in an account. Notice the T-Account.
CHAPTER02 出行生成-产生
2-4
TransCAD 交通需求模型
♦ 使用交叉分类计算出行产生量 1. 打开您的出行率表和区域数据库。 2. 选择 Planning-Trip Productions-Cross-Classification 显示交叉分类对话框。
注意: 要想获取有关用 Settings 按钮保存 您的设置的信 息,请参考第十 五章“数据准备 和规划工具”中 的“保存和实用 程序设置”一 节
模型的输入数据来源很广,最佳的情况是调查数据有足够好的质量、时效性和足 够的数量。即使没用调查和收集当地数据,大多基于工作出行的当地数据都可以从以 下来源获得:人口普查交通规划包(CTPP)中的集合数据、公用小型数据样本(PUMS) 中的个体数据、全国个人交通调查数据(NPTS)。最坏情况是没有可用的当地数据、时 间或经费限制而不可能采集新数据,这时可以应用有相似出行特性的区域的有关数 据。另外,有的数据基于全国平均值,比如 NCHRP 187 和 NCHRP 365,“Travel Estimation Techniques for Urban Planning”。
第 2 章 出行生成-产生 2-1
关于出行产生
出行产生的目标是按出行目的估计每个区域产生的或从该区域出发的出行总数。出行 产生是建立出行数量或频率与出行者的、区域的以及交通网络的特性之间的关系。 TransCAD 提供了多种方法进行出行生成分析,从简单交叉分类法到更复杂的统计分析 均有涵盖。
方法
用于出行产生建模的三种主要工具如下:
自变量和数据源
许多不同的因素影响出行产生,这包括有:
因素
例子
个人特性 家庭特性 区域特性 交通网络特性
性别 年龄 个人收入 职业 家庭大小 车辆拥有量 孩子的数量和年龄 家庭收入 土地使用 居民密度 可达性 服务水平
宏观经济学课件chapter02精选全文
可编辑修改精选全文完整版宏观经济学课件chapter02Macroeconomics, 6e (Blanchard/Johnson)Chapter 2: A Tour of the Book2.1 Multiple Choice Questions1) Fill in the blank for the following: GDP is the value of all ________ produced in a given period.A) final and intermediate goods and services produced by the private sector onlyB) final goods and servicesC) final and intermediate goods and services, plus raw materialsD) all of the aboveE) none of the aboveAnswer: BDiff: 12) When using the income approach to measure GDP, the largest share of GDP generally consists ofA) interest income.B) labor income.C) indirect taxes.D) profits.E) capital income.Answer: BDiff: 13) For this question, assume that 1980 is the base year. Given macroeconomic conditions in the United States over the past three decades, we know thatA) nominal GDP is always smaller than real GDP since 1980.B) real GDP and nominal GDP would be equal for the entire period.C) real GDP is larger than nominal GDP from 2002 to 2008.D) real GDP and nominal GDP were equal in 1980.E) none of the aboveAnswer: DDiff: 24) Suppose nominal GDP increased in a given year. Based on this information, we know with certainty thatA) real output has increased.B) the price level (GDP deflator) has increased.C) real output and the price level (GDP deflator) have both increased.D) either real output or the price level (GDP deflator) have increased.E) real output has increased and the price level has decreased.Answer: DDiff: 25) Use the following information to answer this question. If nominal GDP rises from $100 trillion to $120 trillion, while the GDP deflator rises from 2.0 to 2.2, the percentage change in real GDP is approximately equal toA) -10%.B) 10%.C) 20%.D) 9.1%.E) 0%.Answer: DDiff: 26) Hedonic pricing isA) the way that luxury goods are priced in a market economy.B) the tendency for the inflation rate to rise by greater and greater amounts.C) the tendency for nominal GDP to rise when the price level rises.D) the process of translating nominal GDP into real GDP.E) the process of pricing individual characteristics of a good or service.Answer: EDiff: 17) In a given year, suppose a company spends $100 million on intermediate goods and $200 million on wages, with no other expenses. Also assume that its total sales are $800 million. The value added by this company equalsA) $200 million.B) $300 million.C) $500 million.D) $700 million.E) $800 million.Answer: DDiff: 28) A firm's value added equalsA) its revenue minus all of its costs.B) its revenue minus its wages.C) its revenue minus its wages and profit.D) its revenue minus its cost of intermediate goods.E) none of the aboveAnswer: DDiff: 29) Suppose you are provided with the following data for yourcountry for a particular month: 200 million people are working, 20 million are not working but are looking for work, and 40 million are not working and have given up looking for work. The official unemployment rate for that month isA) 7.7%.B) 9.1%.C) 10%.D) 23%.E) 30%.Answer: BDiff: 110) In the United States, someone is classified as unemployed if he or sheA) does not have a job.B) does not have a job, or else has a job but is looking for a different one while continuing to work.C) does not have a job, has recently looked for work, and is collecting unemployment insurance.D) does not have a job, and is collecting unemployment insurance.E) none of the aboveAnswer: EDiff: 111) An individual is said to be a discouraged worker if he or sheA) is working, but prefers not to work.B) is working part time, but would prefer a full time job.C) is working in jobs she/he is not suited for.D) wants to work, and is actively searching for a job.E) wants to work, but has given up searching for a job.Answer: EDiff: 112) Which of the following tends to occur when the unemployment rate increases?A) a reduction in the labor force participation rateB) a reduction in the number of discouraged workersC) an increase in the number of employed workersD) all of the aboveE) none of the aboveAnswer: ADiff: 113) Labor income's share in an advanced country is likely to beA) 70%.B) 45%.C) 29%.D) 10%.E) none of the aboveAnswer: ADiff: 214) The labor force in the United States is defined asA) the total number of individuals who are employed.B) the sum of the total number of individuals who are employed and the officially unemployed.C) the sum of the total number of individuals who are employed, the officially unemployed, and discouraged workers.D) the total number of individuals who are 16 years old and older, but not retired.E) none of the aboveAnswer: BUse the information provided below to answer the following questions.Suppose a country using the United States' system of calculating official unemployment statistics has 100 million people, of whom 50 million are working age. Of these 50 million, 20 million have jobs. Of the remainder: 10 million are actively searching for jobs; 10 million would like jobs but are not searching; and 10 million do not want jobs at all.15) Refer to the information above. The labor force isA) 20 million.B) 40 million.C) 60 million.D) 80 million.E) 100 million.Answer: CDiff: 216) Refer to the information above. The labor force participation rate isA) .2.B) .3.C) .4.D) .6.E) .8.Answer: DDiff: 217) Refer to the information above. The official unemployment rate isA) .1.B) .2.D) .4.E) .66.Answer: CDiff: 218) The GDP deflator provides a measure of which of the following?A) the ratio of GDP to the size of the populationB) the ratio of GDP to the number of workers employedC) the ratio of nominal GDP to real GDPD) the price of a typical consumer's basket of goodsE) real GDP divided by the aggregate price levelAnswer: CDiff: 119) Which of the following calculations will yield the correct measure of real GDP?A) divide nominal GDP by the consumer price indexB) divide the GDP deflator by the consumer price indexC) multiply nominal GDP by the consumer price indexD) multiply nominal GDP by the GDP deflatorE) none of the aboveAnswer: EDiff: 220) The prices for which of the following goods are included in both the GDP deflator and the consumer price index?A) goods bought by householdsB) goods bought by firmsC) good bought by governmentsD) goods bought by foreign households (i.e., exports)E) all of the aboveAnswer: ADiff: 221) Suppose we switch the base year from 2000 to 2008. This change in the base year will causeA) nominal GDP in every year to increase.B) nominal GDP in every year to decrease.C) both nominal and real GDP in every year to decrease.D) real GDP in every year to decrease.E) none of the aboveAnswer: EDiff: 222) Pure inflation occurs whenA) nominal wages rise faster than all prices.B) all prices rise faster than nominal wages.C) all prices and nominal wages rise by the same percentage.D) the GDP deflator and Consumer Price Index rise by the same percentage.E) none of the aboveAnswer: CDiff: 223) One of the reasons macroeconomists have concerns about inflation is that inflation causesA) real GDP to rise.B) nominal GDP to fall.C) wages to rise as fast as prices.D) real GDP to exceed nominal GDP.E) none of the aboveAnswer: EDiff: 124) Changes in GDP in the short run are caused primarily byA) demand factors.B) supply factors.C) technology.D) capital accumulation.E) all of the aboveAnswer: ADiff: 225) Changes in GDP in the medium run are determined primarily byA) demand factors.B) supply factors.C) monetary policy.D) all of the aboveAnswer: BDiff: 226) Changes in GDP in the long run are determined primarily byA) monetary policy.B) fiscal policy.C) demand.D) all of the aboveE) none of the aboveAnswer: EDiff: 227) Which of the following prices will be used when calculating the rate of growth of real GDP between the year's 2005 and 2006 using the chain method?A) prices in the base year (2002)B) prices in 2005C) prices in 2006D) the average of prices in 2005 and 2006E) prices in 2005, 2006, and in 2002 (the base year)Answer: DDiff: 228) Which of the following factors is NOT believed to affect output in the long run?A) technologyB) monetary policyC) the size of the labor forceD) the capital stockAnswer: BDiff: 129) The Okun's law shows the relationship betweenA) inflation and unemployment rate.B) output growth and unemployment.C) inflation and output growth.D) output growth and money supply.Answer: BDiff: 230) The Phillips curve describes the relationship betweenA) output growth and unemployment.B) inflation and output growth.C) output growth and money supply.D) inflation and unemployment .Answer: DDiff: 231) Prices for which of the following are included in the GDP deflator, but not included in the Consumer Price Index?A) firms' purchases of new equipmentB) intermediate goods and servicesC) consumption of goodsD) consumption of servicesAnswer: ADiff: 132) Macroeconomists are concerned about changes in the unemployment rate because changes in the unemployment rate provide information aboutA) the state of the economy.B) the welfare of those who are unemployed.C) none of the aboveD) both A and BAnswer: DDiff: 133) Based on the notation presented in Chapter 2, which of the following expressions represents nominal GDP?A) Y tB) P t Y tC) Y t/P tD) $Y t/P tAnswer: BDiff: 134) Deflation generally occurs when which of the following occurs?A) the consumer price index is greater than the GDP deflatorB) the consumer price index decreasesC) the rate of inflation falls, for example, from 4% to 2%D) nominal GDP does not changeAnswer: BDiff: 135) During the mid-1980s, we observed a significantreduction in oil prices. In the United States, we would expect that this reduction in oil prices would causeA) a larger reduction in the CPI compared to the GDP deflator.B) an equal reduction in the CPI and GDP deflator.C) a larger reduction in the GDP deflator compared to the CPI.D) no change in the CPI and a reduction in the GDP deflator.Answer: ADiff: 236) Suppose nominal GDP in 2009 does not change (compared its previous level in 2008). Given this information, we know with certainty thatA) real GDP increased during 2009.B) the GDP deflator increased during 2009.C) both the GDP deflator and real GDP fell during 2009.D) more information is needed to answer this question.Answer: DDiff: 237) During the late 1990s, Japan experienced reductions in the GDP deflator. Given this information, we know with certainty thatA) real GDP fell during these periods.B) real GDP did not change during these periods.C) the overall price level in Japan decreased during these periods.D) both real GDP and the overall price level decreased during these periods. Answer: CDiff: 238) Hedonic pricing is used toA) convert nominal values to real values.B) calculate the difference between nominal GDP and real GDP.C) measure the rate of change in real GDP.D) obtain chain-weight indexes.E) none of the aboveAnswer: EDiff: 139) GDP in current dollars is equivalent to which of the following?A) real GDPB) GDP in terms of goodsC) GDP in 2000 dollarsD) GDP in constant dollarsE) none of the aboveAnswer: EDiff: 140) Which of the following does NOT represent real GDP?A) GDP in current dollarsB) GDP in terms of goodsC) GDP in base year dollarsD) GDP in constant dollarsAnswer: ADiff: 141) which of the following represents real GDP?A) GDP in constant dollarsB) GDP in terms of goodsC) GDP in base year dollarsD) all of aboveAnswer: ADiff: 142) According to convention, a recession is referred to if an economy goes throughA) at least two consecutive quarters of negative growthB) at least three consecutive quarters of negative growthC) at least four quarters of negative growthD) at least two consecutive months of negative growthAnswer: ADiff: 143) Based on the notation presented in Chapter 2, which of the following expressions represents real GDP?A) Y tB) P t Y tC) Y t/P tD) $Y t/P tAnswer: ADiff: 144) Measures of aggregate output have been published on a regular basis in the United States sinceA) 1947.B) 1933.C) 1917.D) 1946.Answer: ADiff: 145) Which of the following about capital income is NOT correct?A) it refers to a firm's revenue.B) it is also called profit income.C) it goes to the firms.D) it accounts for less than 35% of income in advancedcountries.Answer: ADiff: 146) Which of the following about the Phillips curve is NOT correct?A) It shows the relation between GDP growth and unemployment.B) It has been redefined as a relation between the change in the rate of inflation and the unemployment rate.C) It was first explored by A. W. Phillips.D) The curve is downward sloping.Answer: ADiff: 12.2 Essay Questions1) Explain the three ways GDP can be measured.Answer: GDP can be measured three ways. First, GDP represents the market value of the final goods and services produced in the economy during a given period. This would be obtained by adding C, I, G, and NX. Second, GDP is the sum of the value added by firms. The value added for a firm equals the value of the production (at that stage of the production process) minus the value of the intermediate goods (excluding labor services). The final value of aggregate output can be calculated by either summing the value of all final goods and services OR by summing the value added of all goods and services at each stage of production. And finally, GDP is also the sum of all incomes earned in a given period.2) First, define nominal GDP and real GDP. Second, is it possible for nominal GDP in a year to be less than real GDP in the same year? Explain.Answer: Nominal GDP represents the value of goods and services produced using current prices. Real GDP measures the value of the same goods and services using some base year prices. It is possible for nominal GDP to be less than real GDP in a given year. Given the definitions of the two variables, this will occur if prices in that year are simply less than prices in the base year. If, for example, the base year is 2002, it will generally be the case that nominal GDP will be less than real GDP for those years prior to 2002 given that prices have generally risen in all years. 3) Explain whether it is possible for nominal GDP to increase and real GDP to decrease in the same period.Answer: Nominal GDP can rise because either the price level is rising or the real quantity of goods and services produced has increased. Nominal GDP can increase while real GDP falls if the increase in the aggregate price level is larger (in a proportionate sense) than the drop in real economic activity.4) Explain the difference between the unemployment rate and the participation rate.Answer: The unemployment rate is the percentage of the labor force (those employed and unemployed) that is unemployed. The participation rate is the percentage of the working age population that is in the labor force.5) Explain how the existence of discouraged workers alters the extent to which the official unemployment provides an accurate measure of the use of labor resources.Answer: Discouraged workers are those individuals who have decided to stop searching for employment because they have become "discouraged" about employment opportunities. At some point, these individuals will no longer be considered as part of the labor force. The existence of discouraged workers willcause the official unemployment rate to provide an under-estimate of the underutilization of labor.6) Briefly explain why the reported official unemployment rate in Spain in 1994 may have provided an over-estimate of unemployment in Spain.Answer: The relatively high unemployment rate in Spain is partly the result of a relatively large underground economy. The underground economy is that part of the economy not measured in official statistics. After taking into account those individuals who are "employed" in the underground economy, the unemployment rate in Spain would have been lower (but still relatively high).7) What are the social and economic implications of unemployment? Explain.Answer: Economic implications: signal of economic activity and measure of the utilization of labor. Social implications: the emotional and psychological suffering that occurs as a result of being unemployed.8) Explain what factors cause changes in output in: (1) the short run; (2) medium run; and (3) long run.Answer: In the short run, demand factors primarily cause changes in output. In the medium run, factors such as the technology, amount of capital, and the skill and size of the labor force (supply factors) affect output. And in the long run, the education system, saving rate, and role of government affect economic activity.9) Will the CPI and GDP deflator always move together? Explain.Answer: No they will not. Some of the goods included in the GDP deflator (some investment goods) are not included in theCPI. Some of the goods included in the CPI (foreign goods) are not included in the GDP deflator.10) Explain how inflation can lead to distortions.Answer: First, not all prices and wages adjust automatically when inflation occurs. Second, variations in relative prices (which occur when there is not pure inflation) can lead to uncertainty. Inflation can also lead to distortions if the tax system is not adjusted when inflation occurs (e.g. nominal income tax brackets).11) Explain why economists care about inflation.Answer: Inflation will cause relative prices to change. It will also cause changes in the distribution of income. Inflation will lead to other distortions such as tax distortions and uncertainty.12) Explain Okun's Law.Answer: It shows the relationship between GDP growth and unemployment rate. If output growth is high, unemployment will decrease.13) Explain the Phillips curve.Answer: It shows the negative relationship between inflation rate and unemployment rate. After 1970s, it was redefined as the relationship between the change in the rate of inflation and the unemployment rate.14) Explain why the Phillips curve on average is downward sloping.Answer: When unemployment becomes very low, the economy is likely to overheat and this will lead to upward pressure on inflation.15) Explain why economists care about unemployment.Answer: First, they care about unemployment because of its direct effect on the welfare of the unemployed. Unemployment is often associated with financial and psychological suffering.Second, they care about unemployment because it provides a signal that the economy may not be using some of its resources efficiently.16) Can an economy maintain high output growth, low unemployment, and low inflation at the the same time? Explain.Answer: It would be very hard to achieve the three objectives at the same time. High output growth leads to low unemployment, which is likely to put pressure on inflation.。
计算机科学概论 答案 chapter02
Chapter TwoDATA MANIPULATIONChapter SummaryThis chapter introduces the role of a computer's CPU. It describes the machine cycle and the various operations (or, and, exclusive or, add, shift, etc.) performed by a typical arithmetic/logic unit. The concept of a machine language is presented in terms of the simple yet representative machine described in Appendix C of the text. The chapter also introduces some alternatives to the von Neumann architecture such as multiprocessor machines and artificial neural networks.The optional sections in this chapter present a more thorough discussion of the instructions found in a typical machine language (logical and numerical operations, shifts, jumps, and I/O communication), a short explanation of how a computer communicates with peripheral devices, and alternative machine designs.The machine language in Appendix C involves only direct and immediate addressing. However, indirect addressing is introduced in the last section (Pointers in Machine Language) of Chapter 7 after the pointer concept has been presented in the context of data structures.Comments1. Much of Comment 1 regarding the previous chapter is pertinent here also. The development of skills in the subjects of machine architecture and machine language programming is not required later in the book. Instead, what one needs is an image of the CPU/main memory interface, an understanding of the machine cycle and machine languages, an appreciation of the difference in speeds of mechanical motion compared to CPU activities, and an exposure to the limited repertoire of bit manipulations a CPU can perform.2. To most students at this stage the terms millisecond, microsecond, nanosecond, and picosecond merely refer to extremely short and indistinguishable units of time. In fact, most would probably accept the incorrect statement that activities within a computer are essentially instantaneous. Once a student of mine wrote a recursive routine for evaluating the determinant of a matrix in an interpreted language on a time-sharing system. The student tried to test the program using an 8 by 8 matrix, but kept terminating the program after a minute because "it must be in a loop." This student left with an understanding of microseconds as real units of time that can accumulate into significant periods.3. A subtle point that can add significantly to the complexity of this material is combining notation conversion with instruction encoding. If, for example, all the material in Chapters 1 and 2 is new toa student, the problem, "Using the language of Appendix C, write an instruction for loading register14 with the value 124" can be much more difficult than the same problem stated as, "Using the language of Appendix C, write an instruction for loading register D with the (hexadecimal) value 7C." In general, notation conversion is a subject of minor importance and should not be allowed to cloud the more important concerns.4. If you want your students to develop more than a simple appreciation of machine language programming, you may want to use one of the many simulators that have been developed for the machine in Appendix C. A nice example is included on the Addison-Wesley website at /brookshear or you can find other simulators by searching the Web.5. Here are some short program routines in the machine language presented in Appendix C of the text, followed by their C language equivalents. (These examples are easily converted into Java, C++, and C#.) Each machine language routine starts at address 10. I've found that they make good examples for class presentations or extra homework problems in which I give the students the machine language form and ask them to rewrite it in a high-level language.Address Contents Address Contents Address Contents0D 00 14 20 1B 0F0E 00 15 5A 1C 500F 00 16 30 1D 1210 20 17 0F 1E 3011 5C 18 11 1F 0D12 30 19 0E 20 C013 0E 1A 12 21 00C language equivalent:{int X,Y,Z;X = 92;Y = 90;Z = X + Y;}If the contents of the memory cell at address 1C in the preceding table is changed from 50 to 60 the C equivalent becomes:{float X, Y, Z;X = 1.5;Y = 1.25;Z = X + Y;}Here's another example:Address Contents Address Contents Address Contents0E 00 19 0F 24 200F 00 1A 20 25 0110 20 1B 04 26 5011 02 1C B1 27 0112 30 1D 2C 28 3013 0E 1E 12 29 0F14 20 1F 0E 2A B015 01 20 50 2B 1816 30 21 12 2C C017 0F 22 30 2D 0018 11 23 0EC equivalent:{int X, Y;X = 2; Y =1;while (Y != 4) {X = X + Y; Y = Y + 1;}}6. Here are two C program segments that can be conveniently translated into the machine language of Appendix C.{int X, Limit;X = 0;Limit = 5;do X = X + 1 while (X != Limit);}Program segment in machine language:Address Contents Address Contents Address Contents0E 00 (X) 18 22 (R2 = 1) 22 10 (R0 = Limit)0F 00 (Limit) 19 01 23 0F10 20 (X = 0) 1A 11 (R1 = X) 24 B1 (go to end11 00 1B 0E 25 28 if X == Limit)12 30 1C 50 (X = X+1) 26 B0 (return)13 0E 1D 12 27 1A14 20(Limit = 5)1E 30 28 C0 (halt)15 05 1F 0E 29 0016 30 20 11 (R1 = X)17 0F 21 0E{int X, Y, Difference;X = 33;Y = 34;if (X > Y)Difference = X - Yelse Difference := Y – X}Program segment in machine language:Address Contents Address Contents Address Contents0D 00 (X) 1D 01 2D 16 (Diff = X-Y)0E 00 (Y) 1E 24 (R4=FF) 2E 300F 00 (Diff) 1F FF 2F 0F10 20 (X = 33) 20 96 (R6=not Y) 30 B0 (branch to11 21 21 24 31 3A halt)12 30 22 56 (R6= -Y) 32 90 (R0=not X)13 0D 23 36 33 1414 20 (Y = 34) 24 50 (R0=X-Y) 34 50 (R0 = -X)15 22 25 16 35 0316 30 26 25 (R5=80Hex) 36 50 (R0 = Y-X)17 0E 27 80 37 0218 11 (R1=X) 28 80 (mask low) 38 30 (Diff = Y-X)19 0D 29 50 7 bits) 39 0F1A 12 (R2=Y) 2A B5 (if R0=R5 3A C0 (halt)1B 0E 2B 32 then Y>X 3B 001C 23 (R3=1) 2C 50Answers to Chapter Review Problems1. a. General purpose registers and main memory cells are small data storage cells in a computer.b. General purpose registers are inside the CPU; main memory cells are outside the CPU.(The purpose of this question is to emphasize the distinction between registers and memory cells—a distinction that seems to elude some students, causing confusion when following machine language programs.)2. a. 0010000100000101b. 1010c. 0011001001003. Nine cells with addresses B9, BA, BB, BC, BD, BE, BF, C0, and C1.4. BA5. Program Instruction Memory cellcounter register at 0002 2104 2104 3100 2106 C000 046. To compute x + y - z, each of the values must be retrieved from memory and placed in a register, the sum of x and y must be computed and saved in another register, z must be subtracted from that sum, and the final answer must be stored in memory.A similar process is required to compute (2x) + y. The point of this example is that the multiplication by 2 is accomplished by adding x to x.7. a. Move the contents of register 7 to register E.b. AND the contents of register 0 with the contents of register 8 and place the result in register 0.c. Rotate the contents of register 4 three bits to the right.d. Load register 8 with the value (hexadecimal) 35.e. Compare the contents of registers 3 and 0. If the patterns are equal, jump to the instruction at address AD. Otherwise, continue with the next sequential instruction.8. 16 with 4 bits, 256 with 8 bits9. a. 2766 b. 1766 c. 80F2 d. A403 e. BB3110. The only change that is needed is that the third instruction should be 6056 rather than 5056.11. a. Retrieves from memory cell 3B.b. Is independent of memory cell 3B.c. Changes the contents of memory cell 3B.d. Changes the contents of memory cell 3B.e. Is independent of memory cell 3B.12. a. Place the value 05 in register 4. b. 0513. a. 241B b. 1B3414. a. Load register 0 with the contents of memory cell 04.Store the contents of register 3 in memory cell 45.Halt.b. C0c. 0615. a. 29 b. 0A16. a. 00, 01, 02, 03, 04, 05b. 06, 0717. a. 03 b. 03 c. 0E18. 05. The program is a loop that is terminated when the value in register 0 (initiated at 00) is finally incremented to the value in register 3 (initiated at 05).19. 20 microseconds.20. The point to this problem is that a bit pattern stored in memory is subject to interpretation—it may represent part of the operand of one instruction and the op-code field of another.a. Registers 0, 1, and 2 will contain 32, 20, and 12, respectively.b. 12c. 3221. The machine will alternate between executing the jump instruction at address AF and the jump instruction at address B0.22. It would never halt. The first 2 instructions alter the third instruction to read B000 before it is ever executed. Thus, by the time the machine reaches this instruction, it has been changed to read "Jump to address 00." Consequently, the machine will be trapped in a loop forever (or until it is turned off).23. a. b. c.148D 148D 200034B3 15B3 1145C000 358D B10A34BD22DDB00CC00022CC3288C00024. a. The single instruction B000 stored in locations 00 and 01.b. Address Contents00,01 2100 Initialize02,03 2270 counters.04,05 3109 Set origin06,07 320B and destination.08,09 1000 Now move0A,0B 3000 one cell.0C,0D 2001 Increment0E,0F 5101 addresses.10,11 520212,13 2333 Do it again14,15 4010 if all cells16,17 B31A have not18,19 B004 been moved.1A,1B 2070 Adjust values1C,1D 3071 that are1E,1F 2079 location20,21 3075 dependent.22,23 207B24,25 307726,27 208A28,29 30872A,2B 20742C,2D 30892E,2F 20C030,31 30A432,33 200034,35 20A536,37 B070 Make the big jump!c. Address Contents00,01 2000 Initialize counter.02,03 2100 Initialize origin.04,05 2270 Initialize destination.06,07 2430 Initialize references08,09 1530 to table.0A,0B 310D Get origin0C,0D 1600 value.0E,0F B522 Jump if value must be adjusted.10,11 3213 Place value12,13 3600 in new location.14,15 2301 Increment16,17 5003 R0,18,19 5113 R1, and1A,1B 5223 R2.1C,1D 233C Are we done?1E,1F B370 If so, jump to relocated program.20,21 B00A Else, go back.22,23 2370 Add 70 to24,25 5663 value being26,27 2301 transferred and28,29 5443 update R4 and2A,2B 342D R5 for next2C,2D 1500 location.2E,2F B010 Return (from subroutine).30,31 0305 Table of32,33 0709 locations that34,35 0B0F must be36,37 111F updated for38,39 212B new location.3A,3B 2FFF25.20A1 21A421A2 60016001 30A521A3 C000600126. The machine would place a halt instruction (C000) at memory location 04 and 05 and then halt when this instruction is executed. At this point its program counter will contain the value 06.27. The machine would continue to repeat the instruction at address 06 indefinitely.28. It copies the data from the memory cells at addresses 00, 01, and 02 into the memory cells ataddresses 10, 11, and 12.29. Let R represent the first hexadecimal digit in the operand field;Let XY represent the second and third digits in the operand field;If the pattern in register R is the same as that in register 0,then change the value of the program counter to XY.30. Let the hexadecimal digits in the operand field be represented by R, S, and T;Activate the two's complement addition circuitry with registers S and Tas inputs;Store the result in register R.31. Same as Problem 24 except that the floating-point circuitry is activated.32. a. 04 b. A8 c. FC d. 08 e. F433. a. b. c. d.1066 1034 10A5 10A530BB 21F0 210F 210F8001 8001 80013034 12A6 400121F0A1047001821230A5700230A634. a. 101000 b. 000000 c. 000100 d. 110001 e. 111001 f. 101110g. 010101 h. 111111 i. 010001 j. 101110 k. 010001 l. 00111035. a. AND the byte with 11000011.b. XOR the byte with 11111111.c. XOR the byte with 10000000.d. OR the byte with 10000000.e. OR the byte with 01111111.36. XOR the input string with 10000001.37. First AND the input byte with 10000001, then XOR the result with 10000001.38. a. 11010 b. 00001111 c. 010 d. 001010 e. 1000039. a. 9F b. 86 c. FF d. BB40. a. AB03 b. AB0541. Address Contents00,01 2008 Initialize registers.02,032101220004,05230006,0708,09 148C Get the bit pattern;0A,0B 8541 Extract the least significant bit;0C,0D 7335 Insert it into the result.62120E,0F10,11 B218 Are we done?12,13 A401 If not, rotate registersA30714,1516,17 B00A and go back;18,19 338C If yes, store the result1A,1B C000 and halt.42. The idea is to complement the value at address A1 and then add. Here is one solution:21FF12A1722113A0542334A243. Each character would consist of 8 bits so the rate of 300 bps would translate into approximately37 characters per second. Thus, the printer could just keep up. (In reality, an ASCII characterrequires about 10 bits when transmitted serially because, in addition to a parity bit, start and stopbits are also added to each pattern. As a rule of thumb, 300 bps is considered to be 30 characters persecond.) If the rate were increased to 1200 bps, the printer wouldn't stand a chance.44. The typist would be typing 30 x 5 = 150 characters per minute, or 1 character every 0.4 seconds(= 400,000 microseconds). During this period the machine could execute 20,000,000 instructions.45. The typist would be producing characters at the rate of 2.5 characters per second, whichtranslates to 20 bps (assuming each character consists of 8 bits).46. Address Contents200000,01210102,0304,05 12FE Get printer status06,07 8212 and check the ready flag.08,09 B004 Wait if not ready.0A,0B 35FF Send the data.47. Address Contents00,01 20C1 Initialize registers.210002,03220104,05130B06,0708,09 B312 If done, go to halt.0A,0B 31A0 Store 00 at destination.0C,0D 5332 Change destination0E,0F 330B address,10,11 B008 and go back.12,13 C00048. 14,400 bps is equivalent to 1,800 bytes/sec. So it would take 2960 hours (over 123 days) to fill the20MB drive.49. 14450. Group the 64 values into 32 pairs. Compute the sum of each pair in parallel. Group these sumsinto 16 pairs and compute the sums of these pairs in parallel. etc.51. CISC involves numerous elaborate machine instructions that can be time consuming. RISCinvolves fewer and simpler instructions, each of which is efficiently implemented.52. How about pipelining and parallel processing? Increasing clock speed is another answer.53. In a multiprocessor machine several partial sums can be computed simultaneously.。
Chapter-02
Process State 进程状态
• As a process executes, it changes state 进程执行时,改变状态 – new: The process is being created. 新建:在创建进程 – running: Instructions are being executed. 运行:指令在执行 – waiting: The process is waiting for some event to occur. 等待:进程等待某些事件发生 – ready: The process is waiting to be assigned to a processor. 就绪:进程等待分配处理器 – terminated: The process has finished execution. 终止:进程执行完毕
16
CPU Switch From Process to Process 进程间CPU的切换
17
Context Switch
• When CPU switches to another process, the system must save the state of the old process and load the saved state for the new process. • Context represented in PCB of a process,include -CPU registers -process state -memory-management information • Context-switch time is overhead; the system does no useful work while switching. • Time dependent on hardware support.
软件工程第七版Chapter_02v1过程模型
M odeling analysis design
建模
Cons t ruc t i on code t est
De plo y m e n t d e l iv e ry fe e dbac k
分析 设计
构件 编码 测试
部署 交付 反馈
delivery of
1st increment
12
惯用模型
惯用过程模型提倡有序的软件工程方法 因此导致一些问题… 如果传统过程模型力求实现结构化和有序,那么
对于富于变化的软件世界,这一模型是否合适呢? 如果我们抛弃传统过程模型(以及它们带来的秩
序),取而代之以一些不够结构化的模型,是否 会使如软件工作无法达到协调和一致?
13
瀑布模型
功能和特性需求 开发人员可能对算法的效率、操作系统的兼
容性和人机交互的形式等情况并不确定 2.特点 很少是好用的,可能太慢太大,难以使用。 一般作为被丢弃的系统。
20
特点: 采用循环的 方式逐步加 深系统定义 和实现的深 度,同时降 低风险。 确定一系列 里程碑,确 保利益相关 者都支持可 行的和令人 满意的系统 解决方案。
5
过程模式
一个过程模式
描述了软件工程工作中遇到的过程相关的问题 明确了问题环境 并给出了针对该问题的一种或几种可证明的解决方案
通俗地讲,过程模式提供了一个描述模版 [Amb98]—一种在软件过程的背景下,统一描述 问题解决方案的方法。
6
过程模式描述模板
模式名称:表述该模式在软件过程中的含 义。
(McGraw-Hill, 2009). Slides copyright 2009 by Roger Pressman.
《最终幻想15》全章节流程图文攻略
《最终幻想15》全章节流程图⽂攻略 C H A P T E R01-启程 主⾓诺克提斯·路希斯·切拉姆,为了出席在邻国举办的⾃⾝的婚礼,⽽准备与三位好友驾着⽗亲的爱车“雷格⾥亚”启程。
不过这段旅途似乎在⼀开始就遭遇了⼀些⼩意外,雷格⾥亚在半道罢⼯了,四⼈只得将车推去附近的“锤头鲨”进⾏修理。
这⾥根据屏幕右侧出现的提⽰,按R2来推车前进吧。
抵达“锤头鲨”后,在这⾥迎接的是⽼板的孙⼥希德妮,虽然她爷爷希德还是个⽼头⼦,不过总算15代的C i d⾥也有⼀个妹⼦了。
希德妮接下了雷格⾥亚的维修⼯作。
看诺克提斯初来乍到,还给予了她⼀份本地区域的地图。
得到地图后,会出现提⽰建议前往餐厅。
【餐厅】是⼀个综合性的场所,在这⾥不但可以收集周边的资讯(包括露营地点,资源采集点灯),还能够享⽤美⾷,或者是承接怪兽讨伐任务,来获得丰厚的奖赏。
不过最低的怪物讨伐也是5级的,还需要升级后再来进⾏。
来到加油站附近跟伊格尼斯对话,触发任务【为⽣活费烦恼的王⼦】 众⼈发现修车已经花掉了⾝上所有的费⽤,必须设法赚取⼀些外界的货币才能继续旅⾏。
回头找到希德妮商量,选择“拜托她把维修费打个折”,希德妮会告知爷爷刚才接了⼀份委托,要四⼈去打倒野兽赚钱。
选择这个选项,可以获得10点A P。
关于A P的帮助讯息,游戏中会有详细说明,此处就不赘述了。
之后就是奔赴三处地点分别击败盘踞在区域中的怪物,每次消灭⼀波就会显⽰下⼀波怪物的地点,依次赶赴消灭。
当三波怪物都解决后,希德妮会打开电话,告知她已经完成任务,就可以完成任务获得奖励。
并且希德妮还会请求四⼈在附近找⼀个叫戴夫的赏⾦猎⼈,触发任务【协寻奖⾦猎⼈】 ⾸先前往标⽰位置的⼩屋之中。
发现⼩屋⾥并没有猎⼈的踪迹,但是在⼀个桌柜上有⼀张字条,上前互动,发现字条是⼀张讨伐委托,⽬标是⼀只叫“⾎腥双⾓兽”的怪物。
调查完毕后会有⼀堆饕餮闯⼊。
%{p a g e-b r e a k|C H A P T E R01-启程(1)|p a g e-b r e a k}% 消灭饕餮后离开⼩屋,同伴会提⽰诺克提斯不远处还有⼀间⼩屋。
丧钟为谁而鸣
读书笔记模板
01 思维导图
03 读书笔记 05 作者介绍
目录
02 内容摘要 04 目录分析 06 精彩摘录
思维导图
皮拉尔
本书关键字分析思维导图
罗伯特
丧钟
敌人
美国
西班牙
内容摘要
诺贝尔、普利策双料得主海明威篇幅最大的小说代表作,以美国人参加西班牙人民反法西斯战争为题材,描 写了一九三七年的西班牙内战。故事讲述的是美国青年罗伯特·乔丹在大学里教授西班牙语,对西班牙有深切的 感情。他志愿参加西班牙第二共和国政府军,在敌后进行爆破活动。为配合反攻,他奉命和地方游击队联系,完 成炸桥任务。他争取到游击队队长巴勃罗的妻子皮拉尔和其他队员的拥护,鼓励了已丧失斗志的巴勃罗,并按部 就班地布置好各人的具体任务。在纷飞的战火中,他和皮拉尔收留的被敌人糟蹋过的小姑娘玛丽亚坠入爱河,借 此抹平了玛丽亚心灵的创伤。在这三天中,罗伯特历经爱情与职责的冲突和生与死的考验,人性不断升华。在炸 完桥撤退的时候,自己却被敌人打伤了大腿,独自留下对敌军进行了史诗般的阻击,最终为西班牙人民献出了年 轻的生命。
精彩摘录
这是《丧钟为谁而鸣》的读书笔记模板,可以替换为自己的精彩内容摘录。
谢谢观看
06
CHAPTER 17
05
CHAPTER 16
1
CHAPTER 18
2
CHAPTER 19
3
CHAPTER 20
4
CHAPTER 21
5
CHAPTER 22
CHAPTER 24
CHAPTER 23
CHAPTER 25
01
CHAPTER 26
02
CHAPTER 27
03
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Testing
5 running & testing
Computer running
4 programming Computer solving
int length1, length2, length3; int p,s; length1=3; length2=4; length3=5; p=(length1+ length2+ length3)/2; s=sqrt(p*(p-length1)*(p-length2)(plength3)); printf(“area=%d\n”,s);
8 Copyright © 2012 by The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2.1 Variables: Naming, Declaring, Assigning and Printing Values
Topic Use of standard identifiers such as printf Comment Standard identifiers, such as the function name printf, can be used as variable names. However, their use is not recommended because it leads to confusion. Allowed; however, many programmers use lowercase characters for variable names and uppercase for constant names. Differentiate your identifiers by using different characters rather than different cases
6
Copyright © 2012 by The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2.1 Variables: Naming, Declaring, Assigning and Printing Values
Topics
Naming variables Declaring data types Using assignment statements Displaying variable values
2 Copyright © 2012 by The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Use of blank within an identifier
11 Copyright © 2012 by The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Keywords(32)
auto break case char const continue default do double else enum extern float for goto if int long register return short signed sizeof static struct switch typedef union unsigned void volatile while
The constitution of a program
1.Variable, operator and expression constant, decimal, octal, hexdecimal, variable, identifier, keyword Operator, operand, unary, binary, ternary 2. Data type int, char, float, double 3. Control structure sequence, selection, loop
Concepts
ຫໍສະໝຸດ Variable names must be declared before used “Declare” all your variable names near the beginning of your program Variable names are classified as identifiers first character must be non-digit characters a–z, A–Z, or _ other characters must be non-digit characters a–z, A–Z, _, or digit 0–9 Valid examples apple1 interest_rate xfloat Income one_two Invalid interest_rate% float In come one.two 1apple
10 Copyright © 2012 by The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Table 2.1 Some Constraints on Identifiers
Topic Maximum number of characters in an internal identifier (i.e., identifier within a function) Use of C reserved words, also called keywords, as identifier Comment ANSI C allows a maximum number of 31 characters for names of internal identifiers Not allowed; i.e., do not use these words: auto break case char const continue default do double else enum extern float for goto if int long register return short signed sizeof static struct switch typedef union unsigned void volatile while Not allowed, because an identifier is a token
To calculate the area of 10,000 triangles. Given lengths of three sides Need to make up your own variable names, e.g. lengths: a, b, c For programming in C, the situation is similar choose the variable names, consist of entire words rather than single characters easier to understand your programs if given very descriptive names to each variable
How to solve the problem with computer?
1 problem description Abstraction 2 mathematic modeling Mathematic language 3 algorithm design Computer language
C Programming
a Q & A Approach
by H.H. Tan, T.B. D’Orazio, S.H. Or & Marian M.Y. Choy
Chapter 2
Variables, Arithmetic Expressions and Input/Output
Copyright © 2012 by The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
We may use variable names
Lengths: length1, length2, length3
Much less ambiguous than their algebraic counterparts Expressions look more cumbersome length of expression may span over 1 line Disadvantage which we simply must live with Try your best to make the name descriptive – having mnemonic significance
int month; float expense, income; month = 12; expense = 111.1; income = 100.; printf ("Month=%2d, Expense=$%9.2f\n", month, expense); Result