部分作业答案

1、某发电厂有3台300MW 发电机和三条500kV 出线，其中发电机和双绕组主变压器采用单元接线，变压器高压侧的一次接线拟采用一台半断路器接线。

试画出该发电厂的电气主接线图。

解：电气主接线图如下：2、画出两个电源，四条引出线的单母线分段带旁路接线的电气主接线图，并写出线路断路器QF3检修后，恢复线路WL1送电的基本操作步骤。

解：电气主接线图如下：基本操作步骤：(1)恢复WL1供电：合上31QS ,然后合上32QS ，再合上3QF （2）退出旁路母线WP ：先断开1QSP ,然后断开QFP ,再断开1QSP 。

3、图示为什么形式的主接线?说明图中9QS 、10QS 的作用.解：图示为内桥接线9QS 、10QS 的作用：组成跨条，为了减少开环及满足一回进线或出线停运时桥断路器需退出运行。

4、某发电厂厂用电高、低压母线电压等级分别为6kV 和0.38kV 。

高压母线供电的负荷中含有高压电动机，在电动机自启动时，6kV 母线最低电压标幺值为0.83。

高压母线还通过一台额定容量N S 为1MVA ，短路电压%K U 为10的双绕组无激励变压器给低压母线供电。

低压母线上自启动电动机功率为500kW,电动机效率为0.9，功率因数8.0cos =ϕ，启动电流倍数为5,试问：厂用电低压母线和高压母线串接自启动时，低压电动机能否自启动?解：示意图：如图所示，变压器2T 的电抗标幺值为11.0100101.1100%1.1*=×=×=K t U X 低压母线上的合成负荷标幺值为288.09.08.0/500100051cos /11t av t *=××=⋅⋅=⋅=ΣΣϕηP S K S S K X av D 低压母线电压标幺值为：%55%06.60288.0288.011.083.0***1*2*>=×+=⋅+=D D t X X X U U 根据上式可知，低压电动机能自启动。

绪论部分一、填空1、通常认为说出来的话是口语，用笔写下来的是书面语，而从语言学的角度看口语和书面语的差别主要在_________________________方面。

风格2、周秦时代的书面语就是一般所说的文言文，汉魏以后出现了一种跟口语比较接近的书面语，如唐宋时代的语录、宋代的平话、元明清的戏曲和__________________等。

小说3、现代汉语有两种含义，其一是指以北京语音为标准音、以北京话为基础方言、以____________________________________________________为语法规范的普通话。

典范的现代白话文著作4、现代汉语的存在方式包括_______________、_________________、___________________。

口语、书面语、文学语言5、汉语是联合国规定的六种工作语言之一，这六种工作语言分别是__________________、______________、______________、______________、______________、_____________。

汉语、英语、法语、俄语、西班牙语、阿拉伯语6、英语和汉语的句子中都可以有修饰语，但两种语言的修饰语的位置不同，汉语的修饰语包括定语和状语一般在中心语__________________。

前面7、语言的规范化要解决两个问题：一个是__________________；一个是_________________。

规范不明确、规范不普及8、从方言区划分上，上海话属于_____________，南京话属于_____________，广州话属于_____________，南昌话属于________________。

吴方言、北方方言、粤方言、赣方言9、春秋时代，汉民族共同语被称作___________，汉代起被称作___________，明代改称___________，辛亥革命后称为___________，新中国成立后被称作___________。

作业一 一、名词解释 样品 买方样 卖方样 复样 对等样 参考样品 定牌中性 无牌中性 商品品质 商品数量 商品包装 中性包装 指示性标志 警告性标志、选择题1. 在国际贸易中，对于某些品质变化较大而难以规定统一标准的农副产品，其表示品质的方法常用。

( A )A. 良好平均品质B. 看货买卖C. 上好可销品质D. 凭说明书买卖2. 在国际上，对冷冻鱼或冻虾等没有公认规格和等级的商品，其交货时规定品质的方法常用。

( B )良好平均品质 B. 上好可销品质 C. 看货买卖 D. 凭样品买卖 对工业制成品交易，一般在品质条款中灵活制定品质指标，通常使用。

品质公差 B. 品质机动幅度 C. 交货品质与样品大体相等 D. 规定一个约量 买卖6. 对等样品也称之为 ( BC )7. 在国际贸易中最常见的计量办法。

(度不超过。

( C )A. 3%B. 5%C. 10%D. 15% 500 号出版物之规定， 在以信用证支付方式进行散装货物的买卖，若合同中未明确规定机动幅度，其交货数量可有的增减幅度为。

10. 目前，国际贸易中通常使用的度量衡制度有。

( ABCD ) A. 公制 B. 英制 C. 美制 D. 国际单位制 E. 法制11. 在货物买卖合同中，数量机动幅度的选择权可由 ( ABC )A. 船方行使B. 卖方行使C. 买方行使D. 保险公司行使E. 开证银选择12. 数量条款主要涉及 ( ABCD )A. 成交数量确定B. 计量单位确定C. 计量方法确定D. 数量机动幅度的掌握E. 品质公差确定A. 3. A.4. 以实物表示商品品质的方法有 ( AB )A. 看货买卖B. 凭样品买卖C. 凭规格买卖D. 凭等级买卖E. 凭标准买卖5. 在国际贸易中，按样品提供者的不同凭样品成交可分为。

( ABC )A. 凭买方样品买卖B. 凭卖方样品买卖C. 凭对等样品买卖D. 凭图样买卖E. 凭参考样品A. 复样B. 回样C. 确认样D. 卖方样品E. 买方样品A. 毛重B. 净重C. 理论重理D. 法定重量8. 根据国际商会《跟单信用证统一惯例》 500 号出版物之规定，对于“约量”允许其增减幅9. 根据国际商会 《跟单信用证统一惯例》 A. 3% B. 5% C. 10% D. 15%13. 根据包装程度的不同，目前绝大多数商品采用( A )A. 全部包装B. 局部包装C. 软性包装D. 硬性包装14. 直接接触商品并随商品进入零售网点与消费者见面的包装叫( B )A. 运输包装B. 销售包装C. 中性包装D. 定牌15. “唛头”是运输标志中的( A )A. 主要标志B. 目的地标志C. 原产地标志D. 件号标志16. 按照国际贸易惯例，在合同中不作规定时运输标志的提供方一般是( B )A. 开证行B. 卖方C. 买方D. 船方17. 运输包装的主要作用在于( AB )A. 保护商品B. 防止货损货差C. 促进销售D. 宣传商品E. 吸引客户18. 销售包装的标志和说明主要有( ABCD )A. 包装的装潢画面B. 包装上的文字说明C. 包装上的标签D. 条形码E. 唛头19. 按国际标准化组织的建议和推荐，标准运输标志的内容包括( ABCD )A. 收货人的英文缩写字母或简称B. 参考号C. 目的地D. 件数号码E. 条形码20. 包装条款的内容主要包括( ABCDE )A. 包装材料B. 包装方式C. 包装规格D. 包装标志E. 包装费用21. 在卖方同意接受买方提供包装时，合同中包括条款除一般内容外还要订明( ABCD )A. 寄送包装的方法B. 包装送达的日期C. 送交包装迟延的责任D. 运费包括费用的负担E. 包装的技术性能三、判断题1. 国际贷物买卖合同中，品名条款是合同中的次要条款。

过滤作业（书本P357~359）1、为什么粒径小于滤层孔隙尺寸的颗粒会被滤层拦截下来？答：粒径小于滤层孔隙尺寸的颗粒会被滤层拦截下来，不是滤层的机械筛滤作用，而是水中颗粒的迁移作用和颗粒粘附作用。

在过滤时，小颗粒会因重力作用、惯性作用，布朗运动的扩散作用和颗粒表面的水动力作用而使颗粒与滤料层接触，此即为迁移作用；当颗粒与滤料层接触后，则在范德华引力和静电斥力的相互作用下，以及某些化学键和化学吸附力下，被粘附于滤料表面或滤料表面上原先粘附的颗粒上。

2、从滤层中杂质分布规律，分析改善快滤池的几种途径和发展趋势。

答：滤层杂质分布规律：过滤初期，滤料较干净，孔隙率大，空隙流速较小，水流剪力较小，因而粘附作用占优势。

随着时间的延长，滤层中杂质逐渐增多，孔隙率逐渐减小，水流剪力逐渐增大，以至最后粘附颗粒先脱落下来，或则不再有粘附现象，于是，悬浮颗粒便向下层推移，下层滤料截留作用渐次得到发挥。

然而，滤料经过反冲洗后，滤层因膨胀而分层，表面滤料粒径最小，粘附比表面积大，截留悬浮颗粒量最多，而孔隙尺寸又小，因而，过滤一段时间后，表面滤料间孔隙将逐渐被堵塞，使过滤阻力剧增，造成下层滤料截留悬浮颗粒作用远未得到充分发挥时，过滤就停止。

为了改善滤料层中杂质分布状况，提高滤层含污能力，而采取双层、三层或均质滤料。

5、什么叫等速过滤和变速过滤？两者分别在什么情况下形成？分析两种过滤方式的优缺点。

答：等速过滤：当滤池过滤速度保持恒定不变的过滤方式；变速过滤：滤速随过滤时间而逐渐减少的过滤方式。

当滤料粒径，形状滤层级配和厚度以及水温已定时，随着过滤时间的延长，滤层中截留的悬浮物量逐渐增多，滤层孔隙率逐渐减少，在水头损失保持不变的条件下，将引起滤速减少；反之，滤速保持不变时，将引起水头损失的增加，这样就产生了等速过滤和变速过滤。

优缺点：等速过滤各层滤料水头损失的不均匀可能会导致某一深度出现负水头而影响过滤，采用虹吸滤池和无阀滤池的出水口高于滤层的方式可以解决负水头的问题；变速过滤滤后水质较好，过滤初期，滤速较大可使悬浮杂质深入下层滤料，过滤后期滤速减少，可防止悬浮颗粒穿透滤层，普通快滤池和V形滤池即是。

高电压工程第八周作业答案
张灵2012-04-18
4-15绝缘在冲击电压作用下常常是电击穿而不是热击穿，在高频电压下常常是热击穿而不是电击穿，为什么？
【答】冲击电压下，电压作用时间短，热量来不及积聚；高频下由于介损增大，介质发热严重，容易形成热击穿。

4-20为什么油纸组合绝缘的耐电强度比纸和油单一介质时都高？
【答】油纸绝缘的优点主要是优良的电气性能，干纸的耐电强度仅10kV/mm ~ 13kV/mm，纯油的耐电强度也仅是10kV/mm ~ 20kV/mm，二者组合以后，由于油填充了纸中薄弱点的空气隙，纸在油中又起了屏障作用，从而使总体耐电强度提高很多，油纸绝缘工频短时耐电强度可达50kV/mm ~ 120kV/mm。

4-21固体绝缘材料的耐热等级用什么表示，其含义是什么？
【答】国家标准GB 11021-1989“电气绝缘的耐热性评定和分级”将各种电工绝缘材料按其耐热程度划分等级，以确定各级绝缘材料的最高持续工作温度如表4-3 所示。

补充题:
〖批改情况〗本题同学们对于第（1）问都没有问题。

给大家留一个小思考问题：工程上一般采用哪些方法尽可能避免在法兰处的滑闪放电？
第（2）问错了一些同学，原因是对于题意没有正确理解。

原题中是在套管绝缘层中添加5层同心筒形极板，极板并不是绝缘层，不能混淆概念。

所以5层极板将绝缘层分成6层。

HTML4-7章作业1、CSS样式表不可能实现( )功能。

A将格式和结构分离B．一个CSS文件控制多个网页C.控制图片的精确位置D．兼容所有的浏览器2、可用来在一个网页中嵌入显示另一个网页内容的标记符是（）A.<Marquee>B. <Iframe>C.<embed>D.<object>3.我们想要在框架中加入一个叫做list.htm的文件，应该在HTML中如何描述它？A、frame page="list.htm"B、frame target="list.htm"C、frame src="list.htm"D、frame framepage="list.htm"4.我们想要在页面中加入一个层，可以使用哪些HTML标记来描述它？(选择2项)A、<floor>B、<div>C、<span>D、<level>5 .下列哪种CSS样式定义的方式拥有最高的优先级？A、嵌入B、行内C、链接D、导入6 .以下的HTML中，哪个是正确引用外部样式表的方法？A、<style src="mystyle.css">B、<link rel="stylesheet" type="text/css" href="mystyle.css">C、<stylesheet>mystyle.css</stylesheet>7.在HTML文档中，引用外部样式表的正确位置是？A)文档的末尾B)<head>部分C)文档的顶部D)<body>部分8.哪个HTML标签可用来定义内部样式表？A)<style>B)<script>C)<css>9.哪个HTML属性可用来定义内联样式？A)fontB)classC)stylesD)style10.下列哪个选项的CSS语法是正确的？A)body:color=blackB){body:color=black(body}C)body {color: black}D){body;color:black}11.如何在CSS文件中插入注释？A)// this is a commentB)// this is a comment //C)/* this is a comment */D)' this is a comment12 . CSS哪个属性可用来改变背景颜色？A)bgcolor:B)color:C)background-color:13.如何为所有的<h1>元素添加背景颜色？A)h1.all {background-color:#FFFFFF}B)h1 {background-color:#FFFFFF}C)all.h1 {background-color:#FFFFFF}14 . CSS如何改变某个元素的文本颜色？A)text-color:B)color:C)fgcolor:D)text-color=15 .哪个CSS属性可控制文本的尺寸？A)font-sizeB)text-styleC)font-styleD)text-size16 .以下的CSS中，可使所有<p>元素变为粗体的正确语法是?A)<p style="font-size:bold">B)<p style="text-size:bold">C)p {font-weight:bold}D)p {text-size:bold}17 .如何显示没有下划线的超链接？A)a {text-decoration:none}B)a {text-decoration:no underline}C)a {underline:none}D)a {decoration:no underline}18 .如何使文本以大写字母开头？A)text-transform:capitalizeB)无法通过CSS来完成C)text-transform:uppercase19 . css如何改变元素的字体？A)font=B)f:C)font-family:20 .如何改变元素的左边距？A)text-indent:B)margin-left:C)margin:D)indent:21.如何产生带有正方形的项目的列表？A)list-type: squareB)list-style-type: squareC)type: squareD)type: 222 .我们可以在下列哪个HTML元素中放置javascript代码？A)<script>B)<javascript>C)<js>D)<scripting>23.某一站点主页面index.html的代码如下所示，则选项中关于这段代码的说法正确的是（）。

一、在图题1.3所示电路中，已知回路谐振频率0465kHz f =，0100Q =，160N =匝，140N =匝，210N =匝，200pF C =，16k Ωs R =，1k ΩL R =。

试求回路电感L 、有载Q 值和通频带B。

图题1.3i L答案1：002260000000322362223610115864 5.8410(171.2)11010()103.9110160401()10 3.9110(255.7)160161.361e e e L LLL ss s sL e f L uHf CCCQ g sR k g Q g sR g n g sg n g sR k g g g g πωω------∑=⇒===⇒==⨯=Ω=='==⋅=⨯'==⋅⨯=⨯=Ω''∴=++=⨯由并联谐振:折合到线圈两端:5000.70(73.2)4310.8e esR k CQ g f BW kH zQ ω-∑∑=Ω==答案2：答案3：二、在图题1.4所示电路中，0.8μH L =，1220pF C C ==，5pF S C =，10k ΩS R =，20pF L C =，5k ΩL R =，0100Q =。

试求回路在有载情况下的谐振频率0f 、谐振电阻p R （不计S R 和L R ）、LQ 值和通频带B 。

图题1.4C L答案：三、设有一级单调谐回路中频放大器，其通频带4M Hz B =，10uo A =，如果再用一级完全相同的放大器与之级联，这时两级中放总增益和通频带各为多少？若要求级联后的总频带宽度为4M H z ，问每级放大器应如何改变？改变后的总增益是多少？四、 对于图题3.10所示各振荡电路：（1）画出高频交流等效电路，说明振荡器类型；（2）计算振荡频率。

图题3.10E 57ou (a)(b)E 68~125pF五、 试画出图题3.14所示各振荡器的交流等效电路，说明晶体在电路中的作用，并计算反馈系数F 。

液压传动的部分题目及答案 希望对大家有用！！ 第1次部分作业答案1、用恩氏粘度计测得某液压油（ρ=850kg/m 3）200mL 在40℃时流过的时间为t 1=153s ；20℃时200mL 的蒸馏水流过的时间为t 2=51s ，求该液压油在40℃时的恩氏粘度°E,运动粘度v ，和动力粘度u 各为多少？351153210===sst t E sm s m EE v /1083.19/10)31.631.7(262600--⨯=⨯-= sPa sm •⨯=⨯⨯==-22-631069.1/1019.83 850kg/m ρνμ第二次 2：如图1-25所示，一具有一定真空度的容器用一根管子倒置于一液面与大气相通的水槽中，液体在管中上升的高度h=1m ，设液体的密度ρ=103kg/ m 3，试求容器内真空度。

解：由静力学基本方程得gh p p a ρ+=真空度gh p p a ρ=-=m s m m kg 1/8.9/10233⨯⨯ =9800a p3：如图1-25所示，有一直径为d ，质量为m 的活塞浸在液体中，并在力F 的作用下处于静止状态，若液体的密度为ρ，活塞浸入深度为h ，试确定液体在测压管内的上升高度x 。

解：取活塞底面得水平面为等压面（用相对压力）24)(d mgF h x g πρ+=+h gd mg F x -+=ρπ2)(4第3次作业 6、如图1-27所示，一抽吸设备水平放置，其出口与大气相通，细管处截面积A 2=4 A 1，h=1m 。

求开始抽吸时，A 1=24102.3m -⨯，水平管中必须通过的流量q （取a=1，不记损失）。

解：取截面1-1（与细管相交处），2-2（出口处），应用伯努利方程ghw u gh p u gh p +++=++22222211112/2/αραρ (1)121=αα、 0=ghw刚开始抽吸时，根据静力学基本方程有gh p p a ρ+=1 gh p p a ρ-=1=m s m m kg 1/8.9/1010233⨯⨯⨯- =a p 9800-1121222112410u u A A u A u A u p ====21h h =将以上条件代入式（1）得s m s mgh u /57.4/18.9153153221=⨯⨯==min)187(/1046.1102.3/57.4332411<⨯=⨯⨯==--s m m s m A u q7、如图1-28所示，液压泵的流量q=32L/min ，液压泵吸油口距离液面高度h=500mm ，吸油管直径d=20mm ，过滤器的压力降为0.01Mp a ，油液的密度ρ=900kg/m 3，油液的运动粘度v=20×10-6m 2/s 。

《微型计算机原理与接口技术》部分作业及补充题参考答案第7章存储器系统P2176.某SRAM存储芯片，其字位结构为512K×8bit，试问其地址、数据引脚各是多少个？答：∵219＝512K，所以地址引脚需19根；数据引脚需8根（8bit）。

8.现有1024×1bit的存储芯片，若用它组成容量为16K×8bit的存储器。

试求：（1）实现该存储器所需的芯片数量答：(16K×8bit)/(1K×1bit)=128片（2）该存储器所需地址的地址码总位数是多少？其中几位选片？几位用作片内地址？答：∵214 ＝16K，所以地址码总位数为14位。

而1024×1bit存储芯片需要地址10位，因此选片地址为4位，片内地址为10位。

第8章输入/输出系统P2442．接口电路的作用是什么？I/O接口应具备哪些功能？参见教材P2193．什么是端口？端口有几类？参见教材P220 8.1.2 输入输出端口4．I/O端口有哪两种编址方式？PC系列机采用哪种编址方式？答：I/O端口和存储单元统一编址及I/O端口独立编址两种。

PC机采用I/O端口独立编址。

7. 定时/计数器的3个通道在PC系列机中是如何应用的？答：0＃计数器用于系统时钟中断；1＃计数器用于动态存储器刷新定时；2＃计数器用于发声系统音调控制。

10．系统机定时/计数器的通道0定时周期最长是多少？要实现长时间定时,应采取什么措施？如果采用外扩8254定时/计数器实现长时间定时，应采取哪些措施？答：系统机定时/计数器通道0定时周期最长是55ms。

要实现长时间定时，只能使用 INT 1CH 中断的方法，通过对预先设定的中断次数进行计数，达到n倍55ms的定时目的。

采用外扩8254，可以使用三个通道硬件级联的方法实现长时间定时。

补充题：设PC 系统机外扩了一片8254 及相应的实验电路。

(1) 根据由门电路构成的译码电路，分析出该片8254 的四个端口地址。

CHAPTER 9 Virtual Memory Practice Exercises9.1 Under what circumstances do page faults occur? Describe the actions taken by the operating system when a page fault occurs.Answer:A page fault occurs when an access to a page that has not beenbrought into main memory takes place. The operating system verifiesthe memory access, aborting the program if it is invalid. If it is valid, a free frame is located and I/O is requested to read the needed page into the free frame. Upon completion of I/O, the process table and page table are updated and the instruction is restarted.9.2 Assume that you have a page-reference string for a process with m frames (initially all empty). The page-reference string has length p;n distinct page numbers occur in it. Answer these questions for any page-replacement algorithms:a. What is a lower bound on the number of page faults?b. What is an upper bound on the number of page faults?Answer:a. nb. p9.3 Consider the page table shown in Figure 9.30 for a system with 12-bit virtual and physical addresses and with 256-byte pages. The list of freepage frames is D, E, F (that is, D is at the head of the list, E is second, and F is last).Convert the following virtual addresses to their equivalent physical addresses in hexadecimal. All numbers are given in hexadecimal. (A dash for a page frame indicates that the page is not in memory.)• 9EF• 1112930 Chapter 9 Virtual Memory• 700• 0FFAnswer:• 9E F - 0E F• 111 - 211• 700 - D00• 0F F - EFF9.4 Consider the following page-replacement algorithms. Rank these algorithms on a five-point scale from “bad” to “perfect” according to their page-fault rate. Separate those algorithms that suffer from Belady’s anomaly from those that do not.a. LRU replacementb. FIFO replacementc. Optimal replacementd. Second-chance replacementAnswer:Rank Algorithm Suffer from Belady’s anomaly1 Optimal no2 LRU no3 Second-chance yes4 FIFO yes9.5 Discuss the hardware support required to support demand paging. Answer:For every memory-access operation, the page table needs to be consulted to check whether the corresponding page is resident or not and whether the program has read or write privileges for accessing the page. These checks have to be performed in hardware. A TLB could serve as a cache and improve the performance of the lookup operation.9.6 An operating system supports a paged virtual memory, using a central processor with a cycle time of 1 microsecond. It costs an additional 1 microsecond to access a page other than the current one. Pages have 1000 words, and the paging device is a drum that rotates at 3000 revolutions per minute and transfers 1 million words per second. The following statistical measurements were obtained from the system:• 1 percent of all instructions executed accessed a page other than the current page.•Of the instructions that accessed another page, 80 percent accesseda page already in memory.Practice Exercises 31•When a new page was required, the replaced page was modified 50 percent of the time.Calculate the effective instruction time on this system, assuming that the system is running one process only and that the processor is idle during drum transfers.Answer:effective access time = 0.99 × (1 sec + 0.008 × (2 sec)+ 0.002 × (10,000 sec + 1,000 sec)+ 0.001 × (10,000 sec + 1,000 sec)= (0.99 + 0.016 + 22.0 + 11.0) sec= 34.0 sec9.7 Consider the two-dimensional array A:int A[][] = new int[100][100];where A[0][0] is at location 200 in a paged memory system with pages of size 200. A small process that manipulates the matrix resides in page 0 (locations 0 to 199). Thus, every instruction fetch will be from page 0. For three page frames, how many page faults are generated bythe following array-initialization loops, using LRU replacement andassuming that page frame 1 contains the process and the other twoare initially empty?a. for (int j = 0; j < 100; j++)for (int i = 0; i < 100; i++)A[i][j] = 0;b. for (int i = 0; i < 100; i++)for (int j = 0; j < 100; j++)A[i][j] = 0;Answer:a. 5,000b. 509.8 Consider the following page reference string:1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6.How many page faults would occur for the following replacement algorithms, assuming one, two, three, four, five, six, or seven frames? Remember all frames are initially empty, so your first unique pages will all cost one fault each.•LRU replacement• FIFO replacement•Optimal replacement32 Chapter 9 Virtual MemoryAnswer:Number of frames LRU FIFO Optimal1 20 20 202 18 18 153 15 16 114 10 14 85 8 10 76 7 10 77 77 79.9 Suppose that you want to use a paging algorithm that requires a referencebit (such as second-chance replacement or working-set model), butthe hardware does not provide one. Sketch how you could simulate a reference bit even if one were not provided by the hardware, or explain why it is not possible to do so. If it is possible, calculate what the cost would be.Answer:You can use the valid/invalid bit supported in hardware to simulate the reference bit. Initially set the bit to invalid. O n first reference a trap to the operating system is generated. The operating system will set a software bit to 1 and reset the valid/invalid bit to valid.9.10 You have devised a new page-replacement algorithm that you thinkmaybe optimal. In some contorte d test cases, Belady’s anomaly occurs. Is the new algorithm optimal? Explain your answer.Answer:No. An optimal algorithm will not suffer from Belady’s anomaly because —by definition—an optimal algorithm replaces the page that will notbe used for the long est time. Belady’s anomaly occurs when a pagereplacement algorithm evicts a page that will be needed in the immediatefuture. An optimal algorithm would not have selected such a page.9.11 Segmentation is similar to paging but uses variable-sized“pages.”Definetwo segment-replacement algorithms based on FIFO and LRU pagereplacement schemes. Remember that since segments are not the samesize, the segment that is chosen to be replaced may not be big enoughto leave enough consecutive locations for the needed segment. Consider strategies for systems where segments cannot be relocated, and thosefor systems where they can.Answer:a. FIFO. Find the first segment large enough to accommodate the incoming segment. If relocation is not possible and no one segmentis large enough, select a combination of segments whose memoriesare contiguous, which are “closest to the first of the list” andwhich can accommodate the new segment. If relocation is possible, rearrange the memory so that the firstNsegments large enough forthe incoming segment are contiguous in memory. Add any leftover space to the free-space list in both cases.Practice Exercises 33b. LRU. Select the segment that has not been used for the longestperiod of time and that is large enough, adding any leftover spaceto the free space list. If no one segment is large enough, selecta combination of the “oldest” segments that are contiguous inmemory (if relocation is not available) and that are large enough.If relocation is available, rearrange the oldest N segments to be contiguous in memory and replace those with the new segment.9.12 Consider a demand-paged computer system where the degree of multiprogramming is currently fixed at four. The system was recently measured to determine utilization of CPU and the paging disk. The results are one of the following alternatives. For each case, what is happening? Can the degree of multiprogramming be increased to increase the CPU utilization? Is the paging helping?a. CPU utilization 13 percent; disk utilization 97 percentb. CPU utilization 87 percent; disk utilization 3 percentc. CPU utilization 13 percent; disk utilization 3 percentAnswer:a. Thrashing is occurring.b. CPU utilization is sufficiently high to leave things alone, and increase degree of multiprogramming.c. Increase the degree of multiprogramming.9.13 We have an operating system for a machine that uses base and limit registers, but we have modified the machine to provide a page table.Can the page tables be set up to simulate base and limit registers? How can they be, or why can they not be?Answer:The page table can be set up to simulate base and limit registers provided that the memory is allocated in fixed-size segments. In this way, the base of a segment can be entered into the page table and the valid/invalid bit used to indicate that portion of the segment as resident in the memory. There will be some problem with internal fragmentation.9.27.Consider a demand-paging system with the following time-measured utilizations:CPU utilization 20%Paging disk 97.7%Other I/O devices 5%Which (if any) of the following will (probably) improve CPU utilization? Explain your answer.a. Install a faster CPU.b. Install a bigger paging disk.c. Increase the degree of multiprogramming.d. Decrease the degree of multiprogramming.e. Install more main memory.f. Install a faster hard disk or multiple controllers with multiple hard disks.g. Add prepaging to the page fetch algorithms.h. Increase the page size.Answer: The system obviously is spending most of its time paging, indicating over-allocationof memory. If the level of multiprogramming is reduced resident processeswould page fault less frequently and the CPU utilization would improve. Another way toimprove performance would be to get more physical memory or a faster paging drum.a. Get a faster CPU—No.b. Get a bigger paging drum—No.c. Increase the degree of multiprogramming—No.d. Decrease the degree of multiprogramming—Yes.e. Install more main memory—Likely to improve CPU utilization as more pages canremain resident and not require paging to or from the disks.f. Install a faster hard disk, or multiple controllers with multiple hard disks—Also animprovement, for as the disk bottleneck is removed by faster response and morethroughput to the disks, the CPU will get more data more quickly.g. Add prepaging to the page fetch algorithms—Again, the CPU will get more datafaster, so it will be more in use. This is only the case if the paging action is amenableto prefetching (i.e., some of the access is sequential).h. Increase the page size—Increasing the page size will result in fewer page faults ifdata is being accessed sequentially. If data access is more or less random, morepaging action could ensue because fewer pages can be kept in memory and moredata is transferred per page fault. So this change is as likely to decrease utilizationas it is to increase it.10.1、Is disk scheduling, other than FCFS scheduling, useful in asingle-userenvironment? Explain your answer.Answer: In a single-user environment, the I/O queue usually is empty. Requests generally arrive from a single process for one block or for a sequence of consecutive blocks. In these cases, FCFS is an economical method of disk scheduling. But LOOK is nearly as easy to program and will give much better performance when multiple processes are performing concurrent I/O, such as when aWeb browser retrieves data in the background while the operating system is paging and another application is active in the foreground.10.2.Explain why SSTF scheduling tends to favor middle cylindersover theinnermost and outermost cylinders.The center of the disk is the location having the smallest average distance to all other tracks.Thus the disk head tends to move away from the edges of the disk.Here is another way to think of it.The current location of the head divides the cylinders into two groups.If the head is not in the center of the disk and a new request arrives,the new request is more likely to be in the group that includes the center of the disk;thus,the head is more likely to move in that direction.10.11、Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The drive is currently serving a request at cylinder 143, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130Starting from the current head position, what is the total distance (in cylinders) that the disk arm moves to satisfy all the pending requests, for each of the following disk-scheduling algorithms?a. FCFSb. SSTFc. SCANd. LOOKe. C-SCANAnswer:a. The FCFS schedule is 143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130. The total seek distance is 7081.b. The SSTF schedule is 143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774. The total seek distance is 1745.c. The SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86. The total seek distance is 9769.d. The LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86. The total seek distance is 3319.e. The C-SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 86, 130. The total seek distance is 9813.f. (Bonus.) The C-LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 86, 130. The total seek distance is 3363.12CHAPTERFile-SystemImplementationPractice Exercises12.1 Consider a file currently consisting of 100 blocks. Assume that the filecontrol block (and the index block, in the case of indexed allocation)is already in memory. Calculate how many disk I/O operations are required for contiguous, linked, and indexed (single-level) allocation strategies, if, for one block, the following conditions hold. In the contiguous-allocation case, assume that there is no room to grow atthe beginning but there is room to grow at the end. Also assume thatthe block information to be added is stored in memory.a. The block is added at the beginning.b. The block is added in the middle.c. The block is added at the end.d. The block is removed from the beginning.e. The block is removed from the middle.f. The block is removed from the end.Answer:The results are:Contiguous Linked Indexeda. 201 1 1b. 101 52 1c. 1 3 1d. 198 1 0e. 98 52 0f. 0 100 012.2 What problems could occur if a system allowed a file system to be mounted simultaneously at more than one location?Answer:4344 Chapter 12 File-System ImplementationThere would be multiple paths to the same file, which could confuse users or encourage mistakes (deleting a file with one path deletes thefile in all the other paths).12.3 Wh y must the bit map for file allocation be kept on mass storage, ratherthan in main memory?Answer:In case of system crash (memory failure) the free-space list would notbe lost as it would be if the bit map had been stored in main memory.12.4 Consider a system that supports the strategies of contiguous, linked, and indexed allocation. What criteria should be used in deciding which strategy is best utilized for a particular file?Answer:•Contiguous—if file is usually accessed sequentially, if file isrelatively small.•Linked—if file is large and usually accessed sequentially.• Indexed—if file is large and usually accessed randomly.12.5 One problem with contiguous allocation is that the user must preallocate enough space for each file. If the file grows to be larger than thespace allocated for it, special actions must be taken. One solution to this problem is to define a file structure consisting of an initial contiguous area (of a specified size). If this area is filled, the operating system automatically defines an overflow area that is linked to the initial contiguous area. If the overflow area is filled, another overflow areais allocated. Compare this implementation of a file with the standard contiguous and linked implementations.Answer:This method requires more overhead then the standard contiguousallocation. It requires less overheadthan the standard linked allocation. 12.6 How do caches help improve performance? Why do systems not use more or larger caches if they are so useful?Answer:Caches allow components of differing speeds to communicate moreefficiently by storing data from the slower device, temporarily, ina faster device (the cache). Caches are, almost by definition, more expensive than the device they are caching for, so increasing the numberor size of caches would increase system cost.12.7 Why is it advantageous for the user for an operating system todynamically allocate its internal tables? What are the penalties to the operating system for doing so?Answer:Dynamic tables allow more flexibility in system use growth — tablesare never exceeded, avoiding artificial use limits. Unfortunately, kernelstructures and code are more complicated, so there is more potentialfor bugs. The use of one resource can take away more system resources (by growing to accommodate the requests) than with static tables.Practice Exercises 4512.8 Explain how the VFS layer allows an operating system to support multiple types of file systems easily.Answer:VFS introduces a layer of indirection in the file system implementation.In many ways, it is similar to object-oriented programming techniques. System calls can be made generically (independent of file system type). Each file system type provides its function calls and data structuresto the VFS layer. A system call is translated into the proper specific functions for the target file system at the VFS layer. The calling program has no file-system-specific code, and the upper levels of the system callst ructures likewise are file system-independent. The translation at the VFS layer turns these generic calls into file-system-specific operations.。