运筹学教程第三章答案

第一篇：运筹学教程第三章答案

大学体验英语综合教程2(第三版)课后翻译答案

Unit1

1. 任何年满18岁的人都有资格投票。(be eligible to, vote)

Answer:Anyone over the age of 18 is eligible to vote.

2. 每学期开学前，这些奖学金的申请表格就会由学校发给每一个学生。(apply for, scholarship) Answer:A form to apply for these scholarships is sent by the university to every student before the start of every semester.

3. 遵照医生的建议，我决定戒烟。(on the advice of)

Answer:On the advice of my doctor, I decided to give up smoking.

4. 公园位于县城的正中央。(be located in)

Answer:The park is located right in the center of town.

5. 这所大学提供了我们所需的所有材料和设备。(facilities)

Answer:The university provides all the materials and facilities we desire.1. 他们花了多年的时间寻找内心的平静，但是收效甚微。(search for)

Answer:They spent many years searching for peace of mind, but with little success.2. 这种新药的成功研制已经使许多疾病的治疗发生了根本性的变革。(revolutionize) Answer:The successful development of the new drug has revolutionized the treatment of many diseases.

3. 由于这个国家的经济不景气，这家公司濒于破产。(on the edge of)

Answer:The company is on the edge of bankruptcy due to the economic depression in the country.

4. 大学毕业后他成为了一名护士。他认为护士这一职业可能很有发展前途。(rewarding) Answer:He became a nurse after college. He thought nursing could be a very rewarding career.

5. 他像往常一样在文件上签了名。(just as)

Answer:He signed his name on the paper just as he has always done it.

Unit2 1. 警察们正忙着填写关于这场事故的各种表格。(fill out)

Answer:The policemen are busy fi lling out forms about the accident.

2. 我想在还车之前把油箱加满。(fill up, fuel tank)

Answer:I want to fi ll up the fuel tank before returning the car.

3. 如果你要投诉，最好遵循正确的程序。(follow the procedure)

Answer:If you want to make a complaint, you’d better follow the correct procedure.

4. 要不是约翰帮忙，我们绝不会这么快就完成实验。(without)

Answer:We couldn’t have fi nished the experiment so soon without John’s help.

5. 暴风雨之后，岸边的人们焦急地搜索湖面以期发现小船的踪迹。(scan for)

Answer:After the storm, the people on the shore anxiously scanned the lake for any sign of the boat.1. 这个国家不大，但是在国际事务中它却发挥着重要作用。(play a role)

Answer:This country is not big, but it plays an important role in international affairs.

2. 正是在我叔叔的帮助下，我得以克服困难，按时完成了任务。(it is ... that)

Answer:It was with the help of my uncle that I overcame the difficulty and completed the assignment in time.3. 毕业时他决定留在北京，而他最好的朋友却选择了去西藏。(while)

Answer:While he decided to stay in Beijing upon graduation, his best friend chose to go to Tibet.

4. 在这次校园英语演讲比赛中，我们班的玛丽获得了第三名。(come in)

Answer:In this Campus English Speaking Contest, Mary from our class came in third.

5. 你应该知道学习弹钢琴需要有时间，有金钱，还要有毅力。(as well as)Answer:You should know that it takes time, money as well as perseverance to learn to play the piano.

Unit3

1. 我发现量入为出地过日子越来越难了。(increasingly)

Answer:I find it increasingly difficult to live within my income.

2. 现代政治家们都试图以电视讲话来影响普通百姓。(reach out)

Answer:Modern politicians try to reach out to ordinary people in their TV speeches.

3. 应该帮助学生对人生采取积极的态度。(adopt)

Answer:Pupils should be helped to adopt a positive attitude to life.

4. 希望全班同学参加这些讨论。(participate in)

Answer:Everyone in the class is expected to participate in these discussions.

5. 如果你犯了罪就必须受到惩罚。(crime)

Answer:If you commit a crime you must expect to be punished.

1.儿子的死让老妇人变得麻木痴呆了。(numb)

Answer:The old woman was numbed by her son’s death.

2.身处所有这些烦恼之中，他依然能保持乐观。(in the midst)

Answer:In the midst of all troubles, he managed to remain cheerful.

3. 虽有困难，他们还是设法坚持试验下去。(carry on)

Answer:They managed to carry on their experiments in spite of the difficulties.

4.我们必须淘汰不合格的申请人。(weed out)

Answer:We have to weed out unqualified applicants.

5. 医生要他减少抽烟。(cut down on)

Answer:The doctor told him to cut down on smoking.

Unit4

1.她在公共汽车站一直等到末班车进站。(come in)

Answer:She waited at the bus stop until the last bus came in.

2.如果我们能帮得上忙，尽管和我们联系。(contact)

Answer:If there is any way we can be of assistance, please do not hesitate to contact us.

3 .他需要多少船务人员才能使他的游艇航行?(crew, yacht)

Answer:How many crew does he need to sail his yacht?

4 .虽然她的新书没有上一本好，但是我还是喜欢它。(not quite as)

Answer:I enjoyed her new book though it’s not quite as good as her last one.

5 .我从未遇到过如此善良的人。(never before)Answer:Never before have I met such a kind person.

1.公共汽车放慢速度并停下，让那位乘客上车。(slow down)

Answer:The bus slowed down and stopped to allow the passenger to get on board.

2.许多车子都驶过去了，可是没有一辆愿意让我们搭便车。(roll by)

Answer:Many vehicles rolled by, but no one offered us a ride.

3 .请勿践踏草地。(get off)

Answer:Please get off the grass.

4.他按妻子的吩咐，一下班就去了市场。(make one’s way)

Answer:He made his way to the marketplace right after work, as his wife had asked him to do.

5 .值得庆幸的是，史蒂夫 (Steve) 从自行车上摔下来时没有摔断骨头。(luckily)Answer:Luckily, Steve didn’t break any bones when he fell off his bike.

Unit5

1. 我们得把感情放在一边，从专业的角度来对待这件事。(from a professional standpoint) Answer:We have to put aside our emotions and take it from a professional standpoint.

2. 这部戏非常精彩，我很快就沉浸于激动人心的剧情之中。(lose oneself in)

Answer:The play was so wonderful that I soon lost myself in the excitement of it.

3 .她没有什么爱好——除非你把看电视也算是一种爱好。(unless)

Answer:She hasn’t got any hobbies — unless you call watching TV a hobby.

4. 他说他是直接从市长本人那里得到这个信息的。(first-hand)

Answer:He said that he had got the information first-hand from the Mayor himself.

5 .既然你不能回答这个问题，我们最好问问别人。(since)

Answer:Since you can’t answer the question, perhaps we’d better ask someone else.

1. 由于公共汽车司机突然刹车，旅客们都不由自主地向前摔倒。(brake, pitch forward) Answer:All the passengers pitched forward because the bus driver braked sharply.

2. 这个协议将冲破对自由贸易设置的障碍。(break through, obstacle)

Answer:This agreement will break through the obstacles to free trade. 3 .我刚放下叫出租车的电话，车就来了。(soon after)

Answer:The taxi arrived soon after I rang for it.

4. 在过去，不管我什么时候到家，我父母总会等我。(no matter)

Answer:My parents always waited up for me no matter what time I got home. 5 .由于没有一方愿意让步，两个公司之间的对话完全破裂了。(break down)

Answer:Talks between the two companies completely broke down, because neither of them wanted to give in.

Unit6

1. 那件工作很难做，不过我想试试看。(have a shot)

Answer:It’s a difficult job, but I’d like to have a shot.

2. 这是一本关于商务实践而非理论的书。(as opposed to)

Answer:This is a book about business practice as opposed to theory.

3. 社会活动从未耽误她的学习。(in the way)

Answer:Social activities never get in the way of her studies.

4. 直到1911年人们才发现第一种维生素。(It is not until ... that)

Answer:It was not until 1911 that the first of the vitamins was identified.

5. 尽管年事已高，爷爷每天还帮忙做家务。(despite)

Answer:Despite advanced years, Grandpa helps do housework every day.1. 我们的婚姻很幸福，但我们也曾经历过许多坎坷。(ups and downs)

Answer:Our marriage is happy but we’ve had many ups and downs.

2. 许多外语教师常常在课堂上自顾自地讲得太多。(be inclined to)

Answer:Many teachers of foreign languages are inclined to talk too much themselves in class.

3. 那位著名演员每天早晨锻炼一小时以保持健康。(work out)

Answer:The famous actor keeps fit by working out for an hour every morning.

4. 与他促膝长谈之后，她的怀疑消失了。(melt away)

Answer:After a long heart-to-heart talk with him, her suspicion melted away.

5. 目前我们没有职位空缺，但我们肯定会记住您的申请。(keep in mind)

Answer:We have no vacancies now, but we’ll certainly keep your application in mind.

Unit7

1. 他并不后悔说过的话，只是觉得他完全可以不用这种方式表达。(could have done) Answer:He did not regret what he had said but felt that he could have expressed it differently.

2. 我们最好等到12月14日。那时大卫已考完试，就能和我们一起去度假了。(will have done)

Answer:We’d better wait till 14 December. David will have had his exam by then, so he’ll be able to go on holiday with us.

3. 他正在做一个新项目，年底前必须完成。(work on)

Answer:He’s working on a new project which has to be finished by the end of the year.

4. 他们让我们使用他们的实验室，作为回报，我们则让他们分享研究成果。(in return) Answer:They are letting us use their lab, and in return, we are sharing with them the results of our research.

5. 诸如打电话、听电话一类的事情占用了这位秘书的大部分时间。(take up)

Answer:Such things as making and answering telephone calls take up most of the secretary’s time.

1. 他们把房子卖了后就到加拿大和女儿住在一起。(go off)

Answer:They sold their house and went off to live in Canada with their daughter.

2. ——你下星期这个时候会在干什么?

——还是像平常一样干活。(will be doing)

Answer:What will you be doing this time next week?I’ll be working as usual.

3. 他既聪明又勤奋，不久就接管了杂志的编辑工作。(editing; before long)

Answer:He was intelligent and hard-working and before long he took over the editing of the magazine.

4. 你话还没说完别人就挂断电话，难道你不恼火吗?(hang up)

Answer:Don’t you hate it when someone hangs up on you before you finish speaking?

5. 让他们感到惊恐的是，他们的房顶着火了。(to one’s horror)

Answer:To their horror, the roof of their house caught fire.

第二篇：网页设计与制作教程-第三次测试答案

1、以下关于FLASH为补间动画分布对象的描述中，正确的是( )。(A) A、用户可以快速将某一帧中的对象分布到各个独立的层中，从而为不同层中的对象创建补间动画 B、每个选中的对象都将被分布到单独的新层中，没有选中的对象也分布到各个独立的层中

C、没有选中的对象将被分布到单独的新层中，选中的对象则保持在原来的位置 D、用户可以快速将全部帧中的对象分布到各个独立的层中，从而为不同层中的对象创建补间动画

2、在FLASH中，执行“视图=〉隐藏边缘”命令的作用是( )。(A) A、隐藏被选择对象的突出显示状态 B、隐藏被选择对象的外框轮廓 C、隐藏被选择对象的填充区域

D、隐藏被选择对象的线条

3、以下关于FLASH逐帧动画和补间动画的说法正确的是( )。(B) A、两种动画模式Flash都必须记录完整的各帧信息 B、前者必须记录各帧的完整记录，而后者不用

C、前者不必记录各帧的完整记录，而后者必须记录完整的各帧记录 D、两种动画模式Flash都不用记录完整的各帧信息

4、在FLASH中，分离操作不会对被分离的对象造成以下后果( )。(D) A、切断元件的实例和元件之间的关系

B、如果分离的是动画元件，则只保留当前帧

C、将位图图像转换为填充对象 D、将位图图像转换为矢量图形

5、在Flash的MP3 压缩对话框中的音质选项中，如果要将电影发布到 Web 站点上，则应选( )。(C)(A) A、中 B、最佳

C、快速 D、无所谓

6、在FLASH中，关于导入PNG格式文件，下面说法错误的是( )。(B) A、导入为电影剪辑，并保留原有层表示将PNG 文件导入为电影剪辑，它所包含的所有帧和层都将出现在电影剪辑元件内

B、导入到当前场景的新层表示将PNG 文档导入到当前Flash 文档单个的新层中，该层将位于所有层的下面

C、如果要将 Fireworks 的 PNG 文件导入为单个的图像，则可以选中作为单个扁平化的位图导入复选框

D、如果选中可单个扁平化的位图导入复选框，所有其他选项都不可用

7、下面关于从浏览器打印电影和从Flash 播放器打印电影的说法错误的是( )。(D) A、从Flash播放器打印电影,可以指定Flash电影中的哪些帧可以被打印 B、从Flash播放器打印电影,可以确定帧的打印区域

C、从Flash播放器打印电影,可以给电影剪辑中的打印帧附加Print 动作

D、从浏览器打印电影肯定比从Flash播放器打印电影效果好

8、在FLASH中，两个关键帧中的图像都是形状则这两个关键帧之间可以创建的补间动画是( )。(C) A、运动补间动画

B、位置补间动画 C、形状补间动画 D、透明度补间动画

9、在FLASH中，对于那些具有复杂颜色效果和包含渐变色的图像，例如照片，最好使用那种方式进行压缩( )。(A) A、JPEG压缩 B、无损压缩 C、有损压缩 D、都不可以

10、以下关于FLASH帧锚记和注释的说法正确的是( )。(D) A、帧锚记和注释的长短都将影响输出电影的大小

B、帧锚记和注释的长短都不影响输出电影的大小

C、帧锚记的长短不会影响输出电影的大小，而注释的长短对输出电影的大小有影响 D、帧锚记的长短会影响输出电影的大小，而注释的长短对输出电影的大小不影响

11、在FLASH中，如果一个对象是以100%的大小显示在工作区中，选择工具箱中的“缩放”工具，并按下Alt键，使用鼠标单击，则对象在工作区中将显示的比例是( )。(A) A、50% B、100% C、200% D、400%

12、在FLASH中，如果要将FreeHand文档中的每一层都转换为Flash 电影中的关键帧应该是FreeHand导入设置对话框的( )。(B) A、层选项中选 Layers(层)选项

B、层选项中选Keyframes(关键帧)选项 C、层选项中选 Flatten(扁平)选项 D、没有这个功能

13、在FLASH中，使用钢笔工具创建路径时，关于调整曲线和直线的说法错误的是( )。(B) A、当用户使用次选工具单击路径时，定位点即可显示

B、使用次选工具调整线段可能会增加路径的定位点

C、在调整曲线路径时，要调整定位点两边的形状，可拖动定位点或拖动正切调整柄 D、拖动定位点或拖动正切调整柄，只能调整一边的形状

14、在Flash的测试模式中，不能以“列表显示对象”命令来显示的是( )。(C)(B) A、帧

B、电影剪辑的实例名 C、目标路径

D、对象类型(形状、电影剪辑或按钮)

15、在FLASH中，将舞台上的对象转换为元件的步骤是( )。(A) A、a.选定舞台上的元素 b.单击"修改〉 转换为元件"，打开转换为元件对话框 c.填写转换为元件对话框，并点击确定

B、a. 单击"修改〉 转换为元件"，打开转换为元件对话框 b.选定舞台上的元素 c.填写转换为元件对话框，并点击确定

C、a.选定舞台上的元素,并将选定元素拖到库面板上 b.单击"修改〉转换为元件"，打开转换为元件对话框 c.填写转换为元件对话框，并点击确定

D、a.单击"修改> 转换为元件"，打开转换为元件对话框 b.选定舞台上的元素,并将选定元素拖到库面板上 c.填写转换为元件对话框，并点击确定

16、当Flash 导出较短小的事件声音(例如按钮单击的声音)时，最适合的压缩选项是( )。(A) A、ADPCM 压缩选项 B、MP3 压缩选项 C、Speech 压缩选项 D、Raw 压缩选项

17、在FLASH中，设置电影属性时，设置电影播放的速度为12fps，那么在电影测试时，时间轴上显示的电影播放速度应该可能是( )。(B) A、大于12fps B、小于等于12fps C、等于12fps D、小于12fps

18、在Internet Explorer 浏览器中，来播放Flash 电影(swf 格式的文件)的技术是( )。(D) A、DLL B、COM C、OLE D、Active X

19、下面对FLASH将舞台上的整个动画移动到其他位置的操作说法错误的是( )。(D) A、首先要取消要移动层的锁定同时把不需要移动的层锁定

B、在移动整个动画到其他位置时，需要使绘图纸标记覆盖所有帧 C、在移动整个动画到其他位置时，对不需要移动的层可以隐藏

D、在移动整个动画到其他位置时，不需要单击时间轴上的编辑多个帧按钮 20、关于Flash影片舞台的最大尺寸，下列说法正确的是( )。(C) A、可以设置到无限大 B、1000px × 1000px C、2880px × 2880px D、4800px × 4800px

21、Flash 影片播放速率最大可以设置到( )。(C) A、99 B、100 C、120 D、150

22、在FLASH中，使用钢笔工具创建路径时，关于调整曲线和直线的说法错误的是( )。(B) A、当用户使用次选工具单击路径时，定位点即可显示

B、使用次选工具调整线段可能会增加路径的定位点

C、在调整曲线路径时，要调整定位点两边的形状，可拖动定位点或拖动正切调整柄 D、拖动定位点或拖动正切调整柄，只能调整一边的形状

23、在Flash中，单选钮的的初始状态是未选中的话，则( )。(B) A、initialState=TRUE B、initialState=FALSE C、initialState=YES

D、initialState=CHOOSED

24、以下关于FLASH按钮元件时间轴的叙述，正确的是( )。(B) A、按钮元件的时间轴与主电影的时间轴是一样的，而且它会通过跳转到不同的帧来响应鼠标指针的移动和动作

B、按钮元件中包含了4帧，分别是Up、Down、Over和Hit帧 C、按钮元件时间轴上的帧可以被赋予帧动作脚本 D、按钮元件的时间轴里只能包含4帧的内容

25、以下关于用FLASH制作形状补间动画，使用形状提示，能获得最佳变形效果的说法中正确的是( )。(B) A、在复杂的变形动画中，不用创建一些中间形状，而仅仅使用开始和结束两个形状 B、确保形状提示的逻辑性

C、如果将形状提示按逆时针方向从形状的右上角位置开始，则变形效果将会更好 D、确保形状提示的逻辑和效果

26、在FLASH中，默认时Flash影片帧频率是( )。(B) A、10 B、12 C、15 D、25

27、下面对创建FLASH蒙板操作的说法错误的是( )。(B) A、通过蒙板的小孔来显示的内容的层在蒙板层的下面

B、对于蒙板上的位图图像、过渡颜色和线条样式等，Flash都将忽略 C、蒙板上的任何填充区域都将是不透明的，非填充区域都将是透明的 D、在蒙板上没有必要创建有过度颜色的对象

28、在Flash中，用以绘制笔直的斜线的绘图工具是( )。(A) A、使用铅笔工具，按住Shift 键拖动鼠标 B、使用铅笔工具，采用伸直绘图模式 C、直线工具 D、钢笔工具

29、在FLASH中，下面关于导入视频说法错误的是( )。(C) A、在导入视频片断时，用户可以将它嵌入到Flash 电影中 B、用户可以将包含嵌入视频的电影发布为Flash 动画 C、一些支持导入的视频文件不可以嵌入到Flash 电影中

D、用户可以让嵌入的视频片断的帧频率同步匹配主电影的帧频率 30、在Flash MX中，要绘制精确的直线或曲线路径，可以使用( )。(B) A、铅笔工具 B、钢笔工具 C、刷子工具

D、油漆桶工具

31、下面关于打印Flash电影说法错误的是( )。(C) A、打印Flash 矢量图形时，可以在任意尺寸上获得清晰的打印效果 B、打印低分辨率的位图图像时，受到像素的影响 C、不可以从浏览器打印Flash电影 D、Flash 播放器的打印功能允许用户打印电影中的目录、联票、单篇、收据、发票或其他文档

32、以下关于使用FLASH元件的优点的叙述，不正确的是( )。(D) A、使用元件可以使发布文件的大小显著地缩减 B、使用元件可以使电影的播放更加流畅 C、使用元件可以使电影的编辑更加简单化 D、使用元件可以使编辑实例

33、在Internet Explorer浏览器中，是通过( )来播放Flash 电影(swf 格式的文件)的。(D) A、DLL库 B、COM程序 C、OLE对象

D、Active X插件

34、不同版本的Flash编辑文件(.FLA)是完全兼容的。(错)

35、在Flash中，分离操作会将被分离的对象由位图图像转换为矢量图形。(对)(错)

36、在Flash中，关键帧是指在动画中定义的更改所在的帧。(对)

37、在Flash中，一般来说逐帧动画文件量比补间动画小。(错)

38、作为发布过程的一部分，Flash 将自动执行某些电影优化操作。(对)

39、在Flash中，引导层是用层名称左侧的辅助线图标表示的。(对) 40、在Flash中，在制作电影时，背景层将位于时间轴的最底层。(对)

41、FLASH的工具箱一般放在屏幕的左边,由四个部分组成,它们分别是工具 查看 颜色 选项。(对)

42、在Flash中，引导层不出现在发布的 SWF 文件中。(对)

43、作为发布过程的一部分，Flash 将自动执行某些电影优化操作。(对)

44、在Flash中，可以在时间轴中排列关键帧，以便编辑动画中事件的顺序。(对)

45、在Flash中，为了在绘画时帮助对齐对象，可以创建引导层。(对)(错)

46、在Flash中，工具箱中选取颜色的工具是( )工具。(吸管)

47、在Flash中，执行“视图=》隐藏边缘”命令的作用是隐藏被选择对象的( )状态。(突出显示)

48、在Flash中，有两种类型的声音事件声音和( )声音。(流式)

49、在Flash中，如果一个对象是以100%的大小显示在工作区中，选择工具箱中的“缩放”工具，按下Alt键，使用鼠标单击，则对象将以( )比例显示在工作区中。(50%) 50、在缺省条件下，Flash影片播放的帧频率是( )帧/秒。(12)

第三篇：运筹学期末试卷及答案

一、判断题(21分)

1、可行解是基本可行解的充要条件是它的正分量所对应的A中列向量线性无关( );

2、如果一个LP问题有最优解，则它的对偶问题也有最优解，且它们的最优解相等( );

3、若线性规划问题有最优解，则一定有唯一的最优解( );

4、若一个原始线性规划问题无界，则它的对偶问题也无界( );

5、设f:RnR1在点xRn处的Hesse矩阵2f(x)存在，若2f(x)0，并且2f(x)正定，则x是(UMP)的严格局部最优解( );

6、若f:RnR1是S上的凸函数，任意实数0则f是S上的凸函数( );

7、设SRn是非空开凸集，f:RnR1二阶连续可导，则f是S上的严格凸函数的充要条件是f的Hesse矩阵2f(x)在 S上是正定的( ).

二、1.将下面的线性规划问题化成标准形(7分)

2，写出下面线性规划的对偶规划(7分)

maxz4x15x26x3

minzx14x23x3

2x13x24x3105x2x8x20123 s.t

x12x25x39x1,x30,x2无约束.2x13x25x32xxx4123 s.t3x1x26x31x10,x30,x2为自由变量.

三、证明题(10分)

设f:RnR1在点xRn处可微.若x是(UMP)的局部最优解，则f(x)0.

四、用对偶单纯形法求解下列线性规划问题(10分)

minz15x124x25x3 6x2x32s.t5x2xx1231

xj0,j1,2,3

五、把线性规划问题(18分)

minZ2x1x2x3 x1x2x36s.tx2x4 记为 (P)

12x1,x2,x30求(1) 用单纯形算法解(p); (2) c2由1变为(3); 由64变为34 

六、用分枝定界法解下述ILP问题(10分)

maxzx1x2

2x1x25s.t4x1x22 x1,x20,且为整数

七、求以下无约束非线性规划问题的最优解(8分)

minf(xx221,2)x1x26x1x1x24x27

八、验证下列非线性规划为凸规划(9分)

minf(x)x2214x29x13x1x211 s.tg1(x)5x17x290gx)2x22x22(12x1x24x270

一、判断题(20分)

1. V ;

2. X;

3.

X;

4. X;

5. X ;

6.

V ; 。

3) b

7. X (

二、1.解：对自由变量x2用x4x5代替;对第一个不等式约束添加松弛变量x6,对第二个不等式约束添加剩余变量x7,再用zz代替原来的目标函数,便得到了标准形式的LP问题 (2分)

minz4x15(x4x5)6x3

(4分)

s.t

2x13(x4x5)4x3105x2(xx)8xx2014536 x2(xx)5xx945371xj0,j1,3,4,5,6,7 (8分)

2.解：这里c(1,4,3)T,b(2,4,1)T,根据定义，其对偶问题是

(2分)

max(21423)

(4分)

s.t

21233134123 56323110,30,2无约束 (7分)

三、证明题(10分)

证：用反证法，若 f(x)0，现令Pf(x)，则有

(2分)

f(x)Pf(x)f(x)f(x)0 (5分)

由定理，必存在0，使当t(0,)时，有

f(xtP)f(x) (8分) T2

成立

但这与假设矛盾.因此必有

f(x)0

(10分)

四、 解：引进非负的剩余变量x40,x50,将不等式约束化为等式约束 6x2x3x42 5x12x2x3x51

x0,j1,,5j将等式两端同乘以(-1)，就直接得到原问题一个基本(不可行)解和对偶问题的一个可行解(检验数向量0)其对应的单纯形标如下

1r161r2r13r04r13r221r1r243r0r22zx4x5152450005620z150051102x21011x511162034081106311133 (6分)

1573170022225111x210444415131x3012222 (8分) z

1117此时，b0,故原问题的最优解为x(0,,)T，其最优值为。

422 (10分)

五、解：(1)在约束条件中加入松弛变量x4,x5得

minz2x1x2x3

x1x2x3x46 s.tx12x2x5

4 它的初始表

x1,,5j (2分)

z211000x4x511211060014r2r1rz2r1

1zx1x5031201210131111016 (5分) 100)其,最优值为z012。

此时检验数向量0,故最优解为x(6,0,T (6分)

(2) x1是非基变量11(c1c1)1 (8分)

zx1x5011112012111101101r231r1r231rzr23

zx2004/37/31/346/310012/31/32/31/31/31/38/310/36x1

03 (10分)

,此时检验数向量0,故最优解为x(8/3,10/3,T0)其最优值为z046。 (12分) 3T(3) 原问题的最优解为x(6,0,0),所对应的可行基B=A110 B1, 11

10A5=,

1110331ccb6  故 bBb z1501147 (16分)

从而新问题对应的单纯形表为

z x1x503120610131111013 7T,其0最优值为z06。 由于b0，故最优解为x(3,0 (18分)

六、解：用图解法解求ILP问题的松弛问题的最优解为(,)T,最优值为z0 (2分)

它的最优解不符合整数的要求,可任选一个变量,如选择x17[]1, (4分) 6786323。 67进行分枝.由于6引进两个约束x11和x12生成两个子问题

maxzx1x2 maxzx1x2

s.t

2x1x254xx212x11x1,x20,且整数

(p1)

和

2x1x254xx212s.t (p2) (6分)

x21x1,x20,且整数ILP问题(p1)的松弛LP问题的最优解x1(1,2)T,最优值z3。(p2)的松弛LP问题的最优解

x2(2,1)T,最优值z3。

(8分)

由于33,故ILP问题的最优解x1(1,2)T,x2(2,1)T,最优值z3。

(10分)

2x1x26

七、解：目标函数的梯度向量为 f(x)，

x2x412 (2分)

令f(x)0，求得f的驻点

x(8/T3。

(4分)

21，2fx的

一、二阶顺序主子式分别为 f的Hesse矩阵为fx122 20

，

211230 (6分)

对xRn,2fx为正定矩阵，因而f是Rn上的凸函数。故 (8分) x(8/3,2T/为它的整体最优解。3

八、解：

f的Hesse矩阵为

232fx38，

(2分)

2fx的

一、二阶顺序主子式本别为

20 ，

233870 ， 因而2fx为正定矩阵，f是严格凸函数. (4分)

4-1而g2x= ，它也是一个正定矩阵，因而g2x也是严格凸函数， -142 (7分)

其它的不等是约束为线性的。由定理知，该非线性规划是一个凸规划。

(9分)

第四篇：运筹学附录D,附录E答案

3附录D 判断题答案

线性规划 1.× 不一定有最优解 2.√ 3.×

不一定 4.√ 5.√ 6.× 是非线性规划模型，但可以转化为线性规划模型 7.√ 8.√ 9.×

不一定是可行基，基本可行解对应的基是可行基 10.√ 11.√ 12.√ 13.√ 14.×

原问题可能具有无界解 15.√ 16.√ 17.√ 18.√

19. √

应为|B|≠0 20.×

存在为零的基变量时,最优解是退化的;或者存在非基变量的检验数为零时，线性规划具有多重最优解 线性规划的对偶理论 21. ×

当原问题是max时。 22.√ 23.× 不一定 24.√ 25.× 对偶问题也可能无界 26.(1)× 应为CX*≥Y*b (2)√ (3)√ (4)√ (5)√ (6)√ 27.√ 28.× 应为对偶问题不可行 29.× 应为最优值相等 30.× 不一定 31.× 影子价格是单位资源对目标函数的贡献 32.× 用单纯形法计算;或原问题不可行对偶问题可行时用对偶单纯形法计算 33.× 原问题无可行解 34.× 求解原问题 35.× 应为 maxibiirbi|ir0brmin|ir0 iir36.√

37.√ 38.× 不一定 39.√ 40.× 同时变化时最优解可能发生变化 整数规划 41.× 取整后不一定是原问题的最优解 42.× 称为混和整数规划 43.√ 44.√ 45.√ 46.√

n47.×

可行解数小于等于2 48.√ 49.× 应是axijj1njbi-Myi

50.√

目标规划 51.× 正负偏差变量全部非负 52.√ 53.√ 54.× 至少一个等于零 55.√ 56.× 应为minZd 57.√ 58.× 一定有满意解 59.√ 60.√

运输与指派问题 61.× 唯一 62.× 变量应为6个 63.× 一定有最优解 64.√ 65. √

66.×有可能变量组中其它变量构成闭回路 67.√ 68.× 有mn个约束 69.√ 70.× r(A)=m+n-1 71.√ 72.√ 73.× 应为存在整数最优解，但最优解不一定是整数 74.× 效率应非负。正确的方法是用一个大M减去效率矩阵每一个元素 75.× 变化后与原问题的目标函数不是一个倍数关系或相差一个常数关系 76.√ 77.√ 78.× 纯整数规划 79.√ 80.× 参看第75题 网络模型 81.× 取图G的边和G的所有点组成的树 82.√ 83.× 没有限制 84.× 容量之和为割量 85.× 最小割量等于最大流量 86.√ 87.√ 88.× 最大流量唯一 89.× 可以通过多条路线 90.× 单位时间内最大通过能力 91.√ 92.√ 93.× 不超过最小割量 94.× 等于发点流出的合流或流入收点的合流 95.× 是求最短路的一种算法 96.× 直到有n-1条边 97.√ 98.× 满足流量 f >0 99.× 最大流量与最大流是两个概念 100.× 遍历每一个点。 网络计划 101.× 等于关键工序时间之和 102.√ 103.√ 104.× 不允许 105.√ 106.√ 107.√ 108.√ 109.√ 110.× 不一定 111.× 是用箭条表示工序 112.√ 113.√ 114.× 最短路线 115.√ 116.√ 117.√ 118.√ 119. × 等于(a+4m+b)/6 120. × 等于(应急成本-正常成本)÷(正常时间-应急时间) 动态规划 121. × 不是一种算法 122. × 变量数作为阶段数，资源限量为状态变量 123. × 不一定 124. √ 125. × 各阶段所有决策组成的集合才是决策集 126. √ 127. √ 128. √ 129. × 到第n阶段的最优指标值 130. √ 排队论 131. √ 132. × 等待时间=逗留时间-服务时间。 132. √ 134.√ 135. × 单队多服务台比多队多服务台效率要高 136. √ 137. √

138. × 当t，系统有n个顾客的概率趋于一个常数时为平稳状态 139. × 不一定 140.√ 存储论 141.√ 142.× 不小于 143.√ 144.× 此结论只适合不允许缺货情形 145.√ 146.× 对模型2和4成立，对模型1和模型3不成立 147.√ 148.√ 149. × 等于Q 150. × 是单位时间内总期望成本最低 决策论 151.√ 152.√ 153. ×不一定 154.× 不一定 155.√ 156. × 依过去的信息由决策者估计的概率 157. × 不同 158.√ 159.√ 160.√ 对策论 161. × 股东的总盈利与总损失不相等，不是零和现象 162. √ 163. √ 164. × 不一定。当所有元素大于零时成立 165. × 不一定 166. √ 167. × 不一定 168. × 不一定 169. √ 170. × 不一定

附录E 选择题答案

1.C 2.B 3.A 4.D 5.A 6.C,D 7.B,D 8.A,C,E 9.B,E 10.B,C,E 11.D 12.B 13.A,C,D 14.A,B 15.A,D 16.B,C 17.D 18.C 19.C 20.D 21.A 22.D 23.A,B,C,D 24.B,D 25.D 26.B 27.D,E 28.A,C,D,E 29.A,B,C 30.C 31.A,D 32.A,D,E 33.A,B 34.B,C,D,E 35.A,B,C,D 36.B,D,E 37.A,D 38.A,B,C 39. B,C,D,E 40.A,B,E 41.B,D 42.C 43.C 44.A,B 45.D 46.C 47.A,C,E 48.A 49.C 50.B 51.A 52.D 53.B,E 54.A,B,E 55.B,C,D 56.B 57.B,C 58.B,D,E 59.D 60.A 61.B,C,E 62.D 63.A,B,C,D 64.B 65.D 66.C 67.B 68.A,B,E 69.A 70.B 71.B,C,D 72.B 73.D 74.A 75.C 76.A,B,D,E 77.A 78.D 79.C 80.B,D 81.C,D 82.A,E 83.D 84.B 85.C

第五篇：数据、模型与决策(运筹学)课后习题和案例答案003

CHAPTER 3 THE ART OF MODELING WITH SPREADSHEETS Review Questions 3.1-1 The long-term loan has a lower interest rate. 3.1-2 The short-term loan is more flexible. They can borrow the money only in the years they need it. 3.1-3 End with as large a cash balance as possible at the end of the ten years after paying off all the loans. 3.2-1 Visualize whe

3.2-2 First, it can help clarify what formula should be entered for an output cell. Second, hand calculations help to verify the spreadsheet model. 3.2-3 Sketch a layout of the spreadsheet. 3.2-4 Try numbers in the changing cells for which you know what the values of the output cells should be. 3.2-5 Relative references are based upon the position relative to the cell containing the formula. Absolute references refer to a specific cell address. 3.3-1 Enter the data first. 3.3-2 Numbers should be entered separately from formulas in data cells. 3.3-3 With range names, the formulas and Solver dialogue box contain descriptive range names rather than obscure cell references. Use a range name that corresponds exactly to the label on the spreadsheet. 3.3-4 Borders, shading, and colors can be used to distinguish data cells, changing cells, output cells, and target cells on a spreadsheet. 3.3-5 Three. One for the left-hand-side, one for the inequality sign, and one for the right-hand-side. 3.4-1 Try different values for the changing cells for which you can predict the correct result in the output cells and see if they calculate as expected. 3.4-2 Control-~ on a PC (command-~ on a Mac). 3.4-3 The auditing tools can be used to trace dependents or precedents for a given cell.

3-1 Problems 3.1 A1234567891011121314151617181920BCDEFGHIJKLLT RateST RateSavings InterestStart BalanceMinimum CashYear200320042005200620072008200920102011201220137%10%3%10.5CashFlow-8-2-4363-47-210LTLoan7.50STLoan0.002.517.275.510.5702.59000(all cash figures in millions of dollars)LTInterest-0.53-0.53-0.53-0.53-0.53-0.53-0.53-0.53-0.53-0.53STInterest0.00-0.25-0.73-0.55-0.060-0.26000LTPaybackSTPayback0.00-2.51-7.27-5.51-0.570-2.59000SavingsInterest0.0150.0150.0150.0150.0150.0709260.0150.1242340.0522110.338027MinimumBalance0.500.500.500.500.500.500.500.500.500.500.503.2

a. The COO will need to know how many of each product to produce. Thus, the decisions are how many end tables, how many coffee tables, and how many dining room tables to produce. The objective is to maximize total profit.

-7.50Balance0.500.500.500.500.502.360.504.141.7411.273.58>=>=>=>=>=>=>=>=>=>=>= b. Pine wood used

Labor used

c. = (3 end tables)(8 pounds/end table) + (3 dining room tables)(80 pounds/dining room table) = 264 pounds = (3 end tables)(1 hour/end table) + (3 dining room tables)(4 hours/dining room table) = 15 hours

Coffee TablesDining Room Tables

End TablesUnit ProfitResource Used per unit ProducedPine WoodLaborEnd TablesUnits ProducedCoffee TablesDining Room TablesTotal Used<=<=AvailableTotal Profit BCDEFG d. A123456789End TablesUnit Profit$50Pine WoodLaborCoffee TablesDining Room Tables$100$220Total Used3000<=200<=Available3000200Total Profit$10,600Resource Used per unit Produced81580124Coffee TablesDining Room Tables4030End TablesUnits Produced0 3-2 3.3 a. Top management will need to know how much to produce in each quarter. Thus, the decisions are the production levels in quarters 1, 2, 3, and 4. The objective is to maximize the net profit. b. Ending inventory(Q1)

Ending inventory(Q2)

Profit from sales(Q1) Profit from sales(Q2) Inventory Cost(Q1) Inventory Cost(Q2) c.

Inventory Holding CostGross Profit from SalesStartingInventoryQuarter 1Quarter 2Quarter 3Quarter 4MaximumProduction<=<=<=<=Demand/EndingSalesInventory>=>=>=>=InventoryCostGross Profitfrom Sales

= Starting Inventory(Q1) + Production(Q1) – Sales(Q1) = 1,000 + 5,000 – 3,000 = 3,000

= Starting Inventory(Q2) + Production(Q2) – Sales(Q2) = 3,000 + 5,000 – 4,000 = 4,000

= Sales(Q1) * ($20) = (3,000)($20) = $60,000 = Sales(Q2) * ($20) = (4,000)($20) = $80,000

= Ending Inventory(Q1) * ($8) = (3,000)($8) = $24,000 = Ending Inventory(Q2) * ($8) = (4,000)($8) = $32,000

ProductionNet Profit M d. A12345678910BCDEFGHIJKLInventory Holding CostGross Profit from SalesStartingInventory1,0000$8$20Production2,0004,000MaximumProduction<=6,000<=6,000Demand/EndingSalesInventory3,0000>=04,0000>=0InventoryCost$0$0Totals$0Net ProfitGross Profitfrom Sales$60,000$80,000$140,000$140,000Quarter 1Quarter 2 e. A123456789101112BCDEFGHIJKLMInventory Holding CostGross Profit from SalesStartingInventory1,0001,0003,0001,000$8$20Production3,0006,0006,0006,000MaximumProduction6,0006,0006,0006,000Demand/EndingSalesInventory3,0001,0004,0003,0008,0001,0007,0000InventoryCost$8,000$24,000$8,000$0Totals$40,000Net ProfitGross Profitfrom Sales$60,000$80,000$160,000$140,000$440,000$400,000Quarter 1Quarter 2Quarter 3Quarter 4<=<=<=<=>=>=>=>=0000 3.4 a. Fairwinds needs to know how much to participate in each of the three projects, and what their ending balances will be. The decisions to be made are how much to participate in each of the three projects. The objective is to maximize the ending balance at the end of the 6 years.

3-3 b. Ending Balance(Y1)

Ending Balance(Y2)

c. Starting Cash= Starting Balance + Project A + Project C + Other Projects

= 10 + (100%)(–4) + (50%)(–10) + 6 = 7 (in $millions)

= Starting Balance + Project A + Project C + Other Projects = 7 + (100%)(–6) + (50%)(–7) + 6 = 3.5 (in $millions)

Year123456ParticipationCash Flow (at full participation, $million)Project AProject BProject CTotalCash FlowFrom ABCOtherProjectsEndingBalance>=>=>=>=>=>=MinimumBalance<=100%<=100%<=100%

EFGHI d. A123456789101112131415BCDStarting Cash10all cash numbers are in $millionsTotalCash FlowFrom ABC00OtherProjects66EndingBalance1622MinimumBalance>=1>=1Year12Cash Flow (at full participation, $million)Project AProject BProject C-4-8-10-6-8-7Participation0%<=100%0%<=100%0%<=100% EFGHI e. A123456789101112131415BCDStarting Cash10all cash numbers are in $millionsTotalCash FlowFrom ABC-10.75-8.125-8.125-0.5-344OtherProjects666666EndingBalance5.253.12516.59.559.5MinimumBalance111111Year123456ParticipationCash Flow (at full participation, $million)Project AProject BProject C-4-8-10-6-8-7-6-4-724-4-5030-3004418.75%<=100%0%<=100%100%<=100%>=>=>=>=>=>= 3-4 3.5 a. Web Mercantile needs to know each month how many square feet to lease and for how long. The decisions therefore are for each month how many square feet to lease for one month, for two months, for three months, etc. The objective is to minimize the overall leasing cost. b. Total Cost = (30,000 square feet)($190 per square foot) + (20,000 square feet)($100 per square foot) = $7.7 million. c.

Month of Lease:1Length of Lease:1Month 1Month 2Month 3Month 4Month 5Cost of Lease(per sq. ft.)Total Cost

1213Month Covered by Lease?112222333451234123414251TotalLeased(sq. ft.)>=>=>=>=>=SpaceRequired(sq. ft.)Lease (sq. ft.) BCDEFG d.

A12345678910Month Covered by Lease?TotalMonth of Lease:112LeasedLength of Lease:121(sq. ft.)Month 11130,000>=Month 21120,000>=Cost of Lease$65(per sq. ft.)$100$65SpaceRequired(sq. ft.)30,00020,000Lease (sq. ft.)10,00020,000ABCDEFGHI0JKLM33N41Total Cost$2,650,000 O42P51QTotalLeased(sq. ft.)30,00030,00040,00030,00050,000RSSpaceRequired(sq. ft.)30,00020,00040,00010,00050,000 e. 12345678910111213Month of Lease:1Length of Lease:1Month 11Month 2Month 3Month 4Month 51211131111411111511111$190Month Covered by Lease?2222312341111111111113211111111$1001$65>=>=>=>=>=Cost of Lease$65$100$135$160(per sq. ft.)Lease (sq. ft.)0000$65$100$135$160$65$100$135$653.6

a. Larry needs to know how many employees should work each possible shift. Therefore, the decision variables are the number of employees that work each shift. The objective is to minimize the total cost of the employees.

30,000000010,000000020,000Total Cost$7,650,0003-5 b. Working 8am-noon: 3 FT morning + 3 PT = 6 Working noon-4pm: 3 FT morning + 2 FT afternoon + 3 PT = 8 Working 4pm-8pm: 2 FT afternoon + 4 FT evening + 3 PT = 9 Working 8pm-midning: 4 FT evening + 3 PT = 7 Total cost per day = (3+2+4 FT)(8 hours)($14/hour) + (12 PT)(4 hours)($12/hour) = $1,584. c. Full Time8am-4pmCost per ShiftShift Covers Time of Day? (1=yes, 0=no)8am-noonnoon-4pm4pm-8pm8pm-midnightWorkers per ShiftTotalWorking>=>=>=>=TotalNeededFull Timenoon-8pmFull Time4pm-midnightPart Time8am-noonPart Timenoon-4pmPart Time4pm-8pmPart Time8pm-midnight

TotalTime of DayFull Time8am-noonnoon-4pm4pm-8pm8pm-midnightTimes TotalPart Time>=>=>=>=TotalCost

CDEFGHIJK d. A12345678910111213141516171819BFull Time8am-4pmCost per Shift$1128am-noonnoon-4pm4pm-8pm8pm-midnightWorkers per Shift11Full Timenoon-8pm$112Full Time4pm-midnight$112Part Time8am-noon$48Part Timenoon-4pm$48Part Time4pm-8pm$48Part Time8pm-midnight$48TotalWorking68126>=>=>=>=TotalNeeded68126112Shift Covers Time of Day? (1=yes, 0=no)1111162Times TotalPart Time448022410TotalCost$1,72843.7

a. Al will need to know how much to invest in each possible investment each year. Thus, the decisions are how much to invest in investment A in year 1, 2, 3, and 4; how much to invest in B in year 1, 2, and 3; how much to invest in C in year 2; and how much to invest in D in year 5. The objective is to accumulate the maximum amount of money by the beginning of year 6. TotalTime of DayFull Time8am-noon4noon-4pm64pm-8pm88pm-midnight6>=>=>=>=3-6 b. Ending Cash (Y1) = $60,000 (Starting Balance) – $20,000 (A in Y1) = $40,000 Ending Cash (Y2) = $40,000 (Starting Balance) – $20,000 (B in Y2)

– $20,000 (C in Y2) = $0 Ending Cash (Y3) = $0 (Starting Balance) + $20,000(1.4) (for investment A) = $28,000 Ending Cash (Y4) = $28,000 (Starting Balance) Ending Cash (Y5) = $28,000 (Starting Balance) + $20,000(1.7) (investment B)

= $62,000 Ending Cash (Y6) = $62,000 (Starting Balance) + $20,000(1.9) (investment C)

= $100,000 c. Beginning BalanceMinimum BalanceInvestmentAYear1Year 1Year 2Year 3Year 4Year 5Year 6Dollars InvestedA2A3A4B1B2B3C2D5EndingBalance>=>=>=>=>=>=MinimumBalance

BCDEFGHIJK d. A12345678910Beginning Balance$60,000Minimum Balance$0InvestmentYearYear 1Year 2Year 3A1-11.4$0A2-1-1$0$0$0A3B1-1B2-1-1$0$0B3C2-1EndingBalance$0$0$84,000MinimumBalance>=$0>=$0>=$0Dollars Invested$60,000 JKLM e. A12345678910111213BCDEFGHIBeginning Balance$60,000Minimum Balance$0InvestmentYearYear 1Year 2Year 3Year 4Year 5Year 6A1-11.41.41.41.4$0$84,000$0$0$0A2-1-1-11.71.71.7$01.9$0-11.3$117,600A3A4B1-1B2-1-1B3C2-1D5EndingBalance$0$0$0$0$0$152,880MinimumBalance$0$0$0$0$0$0>=>=>=>=>=>=Dollars Invested$60,000 3-7 3.8 In the poor formulation, the data are not separated from the formula—they are buried inside the equations in column C. In contrast, the spreadsheet in Figure 3.6 separates all of the data in their own cells, and then the formulas for hours used and total profit refer to these data cells.

In the poor formulation, no range names are used. The spreadsheet in Figure 3.6 uses range names for UnitProfit, HoursUsed, TotalProfit, etc.

The poor formulation uses no borders, shading, or colors to distinguish between cell types. The spreadsheet in Figure 3.6 uses borders and shading to distinguish the data cells, changing cells, and target cell.

The poor formulation does not show the entire model on the spreadsheet. There is no indication of the constraints on the spreadsheet (they are only displayed in the Solver dialogue box). Furthermore, the right-hand-sides of the constraints are not on the spreadsheet, but buried in the Solver dialogue box. The spreadsheet in Figure 3.6 shows all of the constraints of the model in three adjacent cells on the spreadsheet. Cell F16 has –0.47 for LT Interest, rather than –LTRate*LTLoan.

Cell G14 for the 2006 ST Interest uses the LT Loan amount rather than the ST Loan amount.

Cell H21 for LT Payback refers to the 2006 ST Loan rather than the LT Loan to determine the payback amount. Cell G21 for the 2013 ST Interest uses LTRate instead of STRate.

Cell H21 for LT Payback in 2013 as –6.649 instead of –LTLoan.

Cell I15 for ST Payback in 2007 has –LTLoan instead of –E14 (LT Loan for 2006). 3.9 3.10 Case 3.1 a. PFS needs to know how many units of each of the four bonds to purchase, how much to invest in the money market, and their ending balance in the money market fund each year after paying the pensions. The decisions are how many units of each bond to purchase, as well as the initial investment in 2003 in the money market. The objective is to minimize the overall initial investment necessary in 2003 in order to meet the pension payments through 2012.

3-8 b. Payment received from Bond 1 (2004) = (10 thousand units) ($1,000 face value) + (10,000 units) ($1,000 face value) (0.04 coupon rate) = $10.4 million Payment received from Bond 1 (2005) = $0

Payment received from Bond 2 (2004) = (10 thousand units) ($1,000 face value) (0.02 coupon) = $0.2 million Payment received from Bond 2 (2005) = (10 thousand units) ($1,000 face value) (0.02 coupon) = $0.2 million

Balance in money market fund (2003) = $28 million (initial investment)

– $8 million (pension payment) = $20 million

Balance in money market fund (2004) = $20 million (starting balance)

+ $10.4 million (payment from Bond 1)

+ $0.2 million (payment from Bond 2)

– $12 million (pension payment)

+ $1 million (money market interest) = $19.6 million Balance in money market fund (2005) = $19.6 million (starting balance)

+ $0.2 million (payment from Bond 2)

– $13 million (pension payment)

+ $0.98 million (money market interest) = $7.78 million

c. PFS will need to track the flow of cash from bond investments, the initial investment, the required pension payments, interest from the money market, and the money market balance. The decisions are the number of units to purchase of each bond. Data for the problem include the yearly cash flows from the bonds (per unit purchased), the money market rate, and the minimum required balance in the money market fund at the end of each year. A sketch of a spreadsheet model might appear as follows.

Money Market RateMinimum Required BalanceRequiredPensionFlowMoneyMarketInterestMoneyMarketBalance>=>=>=>=>=>=>=>=>=>=0000000000

Bond Cash Flows (per unit)Bond 1Bond 2Bond 3Bond 42003200420052006200720082009201020112012Units PurchasedBondFlowInitialInvestment

3-9 d. The bond cash flows (per unit) are calculated in B7:E9. For example, one unit of Bond 1 costs $0.98 in 2003, and returns the face value ($1) plus the coupon rate ($0.04) in 2004. The total cash flow from bonds is then calculated in column F. The Initial Investment (G7) is both a decision variable and the target cell. It includes all money invested on January 1, 2003 (including enough to pay for the bonds and pension payment in 2003, as well as any initial investment in the money market).

If just years 2003 through 2005 are considered, then 23.44 thousand units of Bond 1 should be purchased at a cost of $22.97 million, along with an initial $8 million investment in the money market fund on January 1, 2003. A1234567891011121314BCDEFGHIJKLMoney Market RateMinimum Required BalanceRequiredPensionFlow-8-12-135%0MoneyMarketInterest0.000.62MoneyMarketBalance0.0012.380.00200320042005Units Purchased(thousands)Cost of BondsBond Cash Flows (per unit)Bond 1Bond 2Bond 3Bond 4-0.98-0.92-0.75-0.801.040.020.030.020.0323.44000BondFlow-22.9724.380.00InitialInvestment30.97>=0>=0>=0all cash figures in $millions0.980.920.750.8 IJ 456789F56789 BondFlow=SUMPRODUCT(B7:E7,UnitsPurchased)=SUMPRODUCT(B8:E8,UnitsPurchased)=SUMPRODUCT(B9:E9,UnitsPurchased)

MoneyMarketBalance=SUM(F7:I7)=MoneyMarketRate*J7=J7+SUM(F8:I8)=MoneyMarketRate*J8=J8+SUM(F9:I9)MoneyMarketInterest

Range NameBondFlowInitialInvestmentMinimumBalanceMinimumRequiredBalanceMoneyMarketBalanceMoneyMarketInterestMoneyMarketRatePensionFlowUnitsPurchasedCellsF7:F9G7L7:L9I2J7:J9I7:I9I1H7:H9B11:E11 3-10 e. Expanded to consider all years through 2012, the spreadsheet is as shown below. PFS should purchase 44.27 thousand units of Bond 1, 51.36 thousand units of Bond 3, and 43.55 thousand units of Bond 4 (at a cost of $116.74 million), and invest an additional $8 million in the money market on January 1, 2003. A12345678910111213141516171819BCDEFGHIJKLMoney Market RateMinimum Required BalanceRequiredPensionFlow-8-12-13-14-16-17-20-21-22-245%0MoneyMarketInterest0.001.771.270.700.001.780.940.001.14MoneyMarketBalance0.0035.3425.4213.990.0035.6718.760.0022.860.002003200420052006200720082009201020112012Units Purchased(thousands)Bond Cash Flows (per unit)Bond 1Bond 2Bond 3Bond 4-0.98-0.92-0.75-0.801.040.020.030.020.031.020.030.031.000.030.030.031.0344.27051.3643.55BondInitialFlowInvestment-116.74124.7447.341.311.311.3152.671.311.3144.860.00>=>=>=>=>=>=>=>=>=>=0000000000all cash figures in $millions IJ F5678910111213141516

BondFlow=SUMPRODUCT(B7:E7,UnitsPurchased)=SUMPRODUCT(B8:E8,UnitsPurchased)=SUMPRODUCT(B9:E9,UnitsPurchased)=SUMPRODUCT(B10:E10,UnitsPurchased)=SUMPRODUCT(B11:E11,UnitsPurchased)=SUMPRODUCT(B12:E12,UnitsPurchased)=SUMPRODUCT(B13:E13,UnitsPurchased)=SUMPRODUCT(B14:E14,UnitsPurchased)=SUMPRODUCT(B15:E15,UnitsPurchased)=SUMPRODUCT(B16:E16,UnitsPurchased)

45678910111213141516MoneyMarketInterest=MoneyMarketRate*J7=MoneyMarketRate*J8=MoneyMarketRate*J9=MoneyMarketRate*J10=MoneyMarketRate*J11=MoneyMarketRate*J12=MoneyMarketRate*J13=MoneyMarketRate*J14=MoneyMarketRate*J15MoneyMarketBalance=SUM(F7:I7)=J7+SUM(F8:I8)=J8+SUM(F9:I9)=J9+SUM(F10:I10)=J10+SUM(F11:I11)=J11+SUM(F12:I12)=J12+SUM(F13:I13)=J13+SUM(F14:I14)=J14+SUM(F15:I15)=J15+SUM(F16:I16)

Range NameBondFlowInitialInvestmentMinimumBalanceMinimumRequiredBalanceMoneyMarketBalanceMoneyMarketInterestMoneyMarketRatePensionFlowUnitsPurchasedCellsF7:F16G7L7:L16I2J7:J16I7:I16I1H7:H16B18:E18 3-11