数学建模写作训练题目0905
2009全国大学生数学建模竞赛 题目汇总
青光眼 白内障 外伤 白内障(双眼) 视网膜疾病 白内障(双眼) 视网膜疾病 视网膜疾病 白内障 视网膜疾病 视网膜疾病 白内障(双眼) 白内障(双眼) 外伤 青光眼 白内障(双眼) 白内障(双眼) 白内障(双眼) 白内障(双眼) 视网膜疾病 白内障(双眼) 外伤 白内障(双眼) 白内障 青光眼 视网膜疾病 外伤 白内障 白内障(双眼) 白内障(双眼) 视网膜疾病 视网膜疾病 白内障 白内障(双眼) 视网膜疾病 视网膜疾病 青光眼 白内障 外伤 白内障 青光眼 视网膜疾病 白内障 白内障(双眼)
【附录】 2008-07-13 到 2008-09-11 的病人信息
序号 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 类型 外伤 视网膜疾病 白内障 视网膜疾病 青光眼 视网膜疾病 白内障(双眼) 视网膜疾病 白内障(双眼) 白内障 视网膜疾病 白内障(双眼) 白内障(双眼) 青光眼 视网膜疾病 视网膜疾病 视网膜疾病 白内障 青光眼 白内障(双眼) 视网膜疾病 青光眼 白内障 白内障 白内障(双眼) 视网膜疾病 门诊时间 2008-7-13 2008-7-13 2008-7-13 2008-7-13 2008-7-13 2008-7-13 2008-7-13 2008-7-14 2008-7-14 2008-7-14 2008-7-14 2008-7-14 2008-7-14 2008-7-14 2008-7-14 2008-7-14 2008-7-15 2008-7-15 2008-7-15 2008-7-15 2008-7-15 2008-7-15 2008-7-15 2008-7-15 2008-7-15 2008-7-15 入院时间 2008-7-14 2008-7-25 2008-7-25 2008-7-25 2008-7-25 2008-7-26 2008-7-26 2008-7-26 2008-7-26 2008-7-26 2008-7-26 2008-7-26 2008-7-26 2008-7-27 2008-7-27 2008-7-27 2008-7-27 2008-7-27 2008-7-27 2008-7-27 2008-7-27 2008-7-27 2008-7-27 2008-7-27 2008-7-28 2008-7-28 第一次手术 时间 2008-7-15 2008-7-27 2008-7-28 2008-7-27 2008-7-27 2008-7-29 2008-7-28 2008-7-29 2008-7-28 2008-7-28 2008-7-29 2008-7-28 2008-7-28 2008-7-29 2008-7-29 2008-7-29 2008-7-29 2008-7-28 2008-7-29 2008-7-28 2008-7-29 2008-7-29 2008-7-28 2008-7-28 2008-8-4 2008-7-31 第二次手术 时间 / / / / / / 2008-7-30 / 2008-7-30 / / 2008-7-30 2008-7-30 / / / / / / 2008-7-30 / / / / 2008-8-6 / 出院时间 2008-7-19 2008-8-8 2008-7-31 2008-8-4 2008-8-5 2008-8-11 2008-8-2 2008-8-6 2008-8-1 2008-7-30 2008-8-8 2008-8-2 2008-8-2 2008-8-4 2008-8-9 2008-8-6 2008-8-12 2008-7-30 2008-8-5 2008-8-2 2008-8-9 2008-8-6 2008-7-30 2008-7-30 2008-8-9 2008-8-10
自-2009年全国大学生数学建模大赛D题优秀论文
会议筹备优化模型摘要能否成功举办一届全国性的大型会议，取决于会前的筹备工作是否到位。
本文为某会议筹备组,从经济、方便、满意度等方面，通过数学建模的方法制定了一个预订宾馆客房、租借会议室和租用客车的合理方案。
首先,通过对往届与会情况和本届住房信息有关数据的定量分析，预测到本届与会人数的均值是662人，波动范围在6４0至６７９之间。
拟预订各类客房47５间。
其次，为便于管理、节省费用，所选宾馆应兼顾客房价位合适，宾馆数量少,距离近,租借的会议室集中等要素。
为此，依据附件4，借助EXCEL计算，得出7号宾馆为１0个宾馆的中心。
然后,运用LＩNGO软件对选择宾馆和分配客房的０-１规划模型求解,得出分别在1、2、6、7、８号宾馆所预订的各类客房。
最后,建立租借会议室和客车的整数规划模型,求解结果为：某天上下午的会议,均在7、８号宾馆预订容纳人数分别为200、1４0、１40、１60、1３0、１30人的6个会议室;租用４5座客车2辆、３３座客车2辆,客车在半天内须分别接关键词：均值综合满意度EXCEL0-1规划ＬINGＯ软件１.问题的提出1.1基本情况某一会议服务公司负责承办某专业领域的一届全国性会议。
本着经济、方便和代表满意的原则,从备选10家宾馆中的地理位置、客房结构、会议室的规模(费用)等因素出发，同时,依据会议代表回执中的相关信息，初步确定代表总人数并预定宾馆和客房;会议期间在某一天上下午各安排６个分组会议，需合理分配和租借会议室;为保证代表按时参会,租用客车接送代表是必需的(现有4５座、36座、３3座三种类型的客车,租金分别是半天80０元、７00元和6０0元）。
1.2相关信息(见附录)附件1 10家备选宾馆的有关数据。
附件2 本届会议的代表回执中有关住房要求的信息（单位:人)。
附件3以往几届会议代表回执和与会情况。
附件4 宾馆平面分布图。
1.3需要解决的问题1.预测本届会议参会人数,确定需要预定的各类客房的总量；2．选择宾馆,预定客房；3．预订会议室以及制定租车方案和绘制行车路线。
【全国大学生数学建模竞赛获奖优秀论文作品学习借鉴】全国大学生数学建模竞赛2009年赛题B题
2009高教社杯全国大学生数学建模竞赛题目（请先阅读“全国大学生数学建模竞赛论文格式规范”）B题眼科病床的合理安排医院就医排队是大家都非常熟悉的现象，它以这样或那样的形式出现在我们面前，例如，患者到门诊就诊、到收费处划价、到药房取药、到注射室打针、等待住院等，往往需要排队等待接受某种服务。
我们考虑某医院眼科病床的合理安排的数学建模问题。
该医院眼科门诊每天开放，住院部共有病床79张。
该医院眼科手术主要分四大类：白内障、视网膜疾病、青光眼和外伤。
附录中给出了2008年7月13日至2008年9月11日这段时间里各类病人的情况。
白内障手术较简单，而且没有急症。
目前该院是每周一、三做白内障手术，此类病人的术前准备时间只需1、2天。
做两只眼的病人比做一只眼的要多一些，大约占到60%。
如果要做双眼是周一先做一只，周三再做另一只。
外伤疾病通常属于急症，病床有空时立即安排住院，住院后第二天便会安排手术。
其他眼科疾病比较复杂，有各种不同情况，但大致住院以后2-3天内就可以接受手术，主要是术后的观察时间较长。
这类疾病手术时间可根据需要安排，一般不安排在周一、周三。
由于急症数量较少，建模时这些眼科疾病可不考虑急症。
该医院眼科手术条件比较充分，在考虑病床安排时可不考虑手术条件的限制，但考虑到手术医生的安排问题，通常情况下白内障手术与其他眼科手术（急症除外）不安排在同一天做。
当前该住院部对全体非急症病人是按照FCFS（First come, First serve）规则安排住院，但等待住院病人队列却越来越长，医院方面希望你们能通过数学建模来帮助解决该住院部的病床合理安排问题，以提高对医院资源的有效利用。
问题一：试分析确定合理的评价指标体系，用以评价该问题的病床安排模型的优劣。
问题二：试就该住院部当前的情况，建立合理的病床安排模型，以根据已知的第二天拟出院病人数来确定第二天应该安排哪些病人住院。
并对你们的模型利用问题一中的指标体系作出评价。
2009年数学建模集训题目
2009年数学建模集训题目第1题A题纯净水安全监控问题日趋加剧的水污染，已对人类的生存安全构成重大威胁，成为人类健康、经济和社会可持续发展的重大障碍。
据世界权威机构调查，在发展中国家，各类疾病有8%是由于饮用了不卫生的水而传播的，每年因饮用不卫生水至少造成全球2000万人死亡，因此，水污染被称作"世界头号杀手"。
我国政府对纯净水安全问题十分重视，已将纯净水安全作为一项重要的公共管理目标，采取了一系列措施，强化纯净水安全的监管，并取得了初步成效。
但纯净水安全问题的总体形势仍不容乐观，依然存在一系列隐忧，近年来食品安全方面的恶性、突发性事件屡屡发生。
2007年07月12日，南通一纯净水厂发生造假事件。
2008年3月底，贵阳市发生数百人感染甲肝事件，经卫生部中国疾控中心专家组核查，确认“竹源牌”桶装水是造成疫情爆发的主因。
2009年03月25日，某大学B区学生饮用了“清清”牌桶装纯净水后，百余学生先后出现集体腹泻事件。
2009年2月26日，湖南师范某寝室在长沙爱高普纯净水有限公司订购的桶装纯净水中出现了黑色虫子事件。
生物性和化学性污染对纯净水安全的影响愈来愈严重。
本问题主要考虑纯净水的以下危害因素: （按照危害的严重性依次给出）“电导率”: 是纯净水的特征性指标，反映的是纯净水的纯净程度，以及生产工艺控制的好坏，“电导率”根本达不到国家卫生标准要求，与自来水无异，根本不能算做纯净水。
菌落总数: 是指纯净水检样经过处理，在一定条件下培养后所取1ml（g）检样中所含菌落的总数。
它可以作为判定纯净水被污染程度的指标之一。
大肠菌群：反映纯净水加工过程中对大便污染程度的一个指标。
数值越高证明污染越严重。
霉菌：食物霉变后产生，直接引起中毒，或产生致癌物质，毒害人体。
纯净水的安全危机的爆发，往往是日常的监控机制和管理长期存在漏洞的反映。
完整、有效的纯净水安全风险分析监测预控，为政府及有关部门实施控制措施提供决策依据和技术支持，可以有效提高纯净水安全监管效率和管理水平，及时化解可能出现的安全危机。
2005《数学建模》考试题(开卷)
《数学建模》试卷首页“数学建模”考试说明：1．本课程考试为开卷考试，按规定时限交卷。
且开卷考试要求独立完成。
雷同卷一律作废。
2．交卷时间4月8日中午12：00——13：00。
3．认真填写试卷首页各项内容，不能空白。
必需填写内容：1．姓名：2．专业：3．英语水平（考试过级及分数）：4．计算机能力（等级考试级别及使用软件情况）：5．是否希望参加建模培训：6．是否参加省竞赛及同组队员姓名、专业：7．联系电话：8．住址（宿舍）：数学建模组教师联系电话：陈东彦：88592800，82670161 李冬梅：89916997，86642300 王树忠：86823965 田广悦：863982862005哈尔滨理工大学《数学建模》考试题（开卷）1．讨价还价中的数学。
在当前市场经济条件下，在商店，尤其是私营个体商店中的商品，所标价格a 与其实际价值b 之间，存在着相当大的差距。
对购物的消费者来说，希望这个差距越小越好，即希望比值λ接近于1，而商家则希望 1λ>。
这样，就存在两个问题：第一，商家应如何根据商品的实际价值（或保本价）b 来确定其价格a 才较为合理？第二，购物者根据商品定价，应如何与商家"讨价还价"？第一个问题，国家关于零售商品定价有相关规定，但在个体商家实际定价中，常用"黄金数"方法，即按实际价b 定出的价格a ，使:0.618b a ≈。
虽然商品价值b 位于商品价格a 的黄金分割点上，考虑到消费者讨价还价，应该说，这样定价还是较为合理的。
对消费者来说，如何"讨价还价"才算合理呢？一种常见的方法是"对半还价法"：消费者第一次减去定价的一半，商家第一次讨价则加上二者差价的一半；消费者第二次还价要减去二者差价的一半；如此等等。
直至达到双方都能接受的价格为止。
有人以为，这样讨价还价的结果其理想的最终价格将是原定价的黄金分割点。
2005研究生数学建模竞赛优秀论文A题1418-A题
2005研究⽣数学建模竞赛优秀论⽂A题1418-A题⾼速公路⾏车时间估计及最优路径选择问题1 问题复述I⾏车时间的估计对于旅⾏者来说⾮常重要。
因此，有些美国⾼速公路安装了传感器。
⽐如在圣安东尼奥（San Antonio）市在所有的双向六车道的⾼速路上都安装了传感器。
但是车辆往往会不停的变换车道，为了简化问题我们可以忽略换道的影响，⽽只考虑⼀个车道的交通问题（如下图所⽰（参见原题），正⽅形代表传感器）。
1.传感器可以每天24⼩时探测每个车辆的速度。
每辆车的速度信息每20秒刷新⼀次记录。
下表是⼀组真实数据（由于交通数据⾮常巨⼤，因此只记录了每2分钟间隔中最后20秒的数据，单位：英⾥/⼩时）。
请分析⾼速公路上的路况特点（如：拥塞及其疏导。
⼀般来说时速⾼于50英⾥/⼩时认为不存在拥塞问题。
）如果车辆在时间t经过传感器，那么经过多久它通过第5个传感器？请设计⼀种算法来估计车辆的运⾏时间，并证明算法的合理性和精确性。
如果路况信息每20秒（⽽不是每2分钟）刷新⼀次，那么这对你们的估计算法有影响吗？在上⾯问题条件的基础上，如果传感器不仅能探测车辆速度，⽽且能探测单位时间的交通流量（如下表，流量的单位是：车辆数/每20秒）。
这些信息是否有助于算法的合理性和精确性的提⾼？如果是，请重新设计你的算法。
II第⼀张图（参见原题）是美国德克萨斯州圣安东尼奥市的地图。
第⼆张图（参见原题）反映了圣安东尼奥市的路况信息。
旅⾏者可以在车辆内置的导引系统中输⼊当前位置和⽬的地，系统会帮助选择路径并估计需要的时间。
不幸的是，由于每⼀路段（两个⼗字路⼝之间的道路）的路况是随机的，系统不能很好地确定最优（最快）路线和可靠的时间估计。
你能在问题1的基础上改进这个系统吗？1.假设每段路的运⾏时间是互相独⽴的随机变量，请为系统设计⼀种算法来确定最优路线及时间估计。
请明确你的“最优”是如何定义的。
2.每⼀段路的⾏车时间依赖于其出发时间，并且⾏车时间之间具有相关性。
数学建模题目及答案-数学建模100题
资料范本本资料为word版本，可以直接编辑和打印，感谢您的下载数学建模题目及答案-数学建模100题地点：__________________时间：__________________说明：本资料适用于约定双方经过谈判，协商而共同承认，共同遵守的责任与义务，仅供参考，文档可直接下载或修改，不需要的部分可直接删除，使用时请详细阅读内容09级数模试题1. 把四只脚的连线呈长方形的椅子往不平的地面上一放，通常只有三只脚着地，放不稳，然后稍微挪动几次，就可以使四只脚同时着地，放稳了。
试作合理的假设并建立数学模型说明这个现象。
（15分）解：对于此题，如果不用任何假设很难证明，结果很可能是否定的。
因此对这个问题我们假设：（1）地面为连续曲面（2）长方形桌的四条腿长度相同（3）相对于地面的弯曲程度而言，方桌的腿是足够长的（4）方桌的腿只要有一点接触地面就算着地。
那么，总可以让桌子的三条腿是同时接触到地面。
现在，我们来证明：如果上述假设条件成立，那么答案是肯定的。
以长方桌的中心为坐标原点作直角坐标系如图所示，方桌的四条腿分别在A、B、C、D 处，A、B,C、D的初始位置在与x轴平行，再假设有一条在x轴上的线ab,则ab也与A、B，C、D平行。
当方桌绕中心0旋转时，对角线 ab与x轴的夹角记为。
容易看出，当四条腿尚未全部着地时，腿到地面的距离是不确定的。
为消除这一不确定性，令为A、B离地距离之和，为C、D离地距离之和，它们的值由唯一确定。
由假设（1），,均为的连续函数。
又由假设（3），三条腿总能同时着地，故=0必成立（）。
不妨设,g（若也为0，则初始时刻已四条腿着地，不必再旋转），于是问题归结为：已知,均为的连续函数，,且对任意有，求证存在某一，使。
证明：当θ=π时，AB与CD互换位置，故，。
作，显然，也是的连续函数，而，由连续函数的取零值定理，存在，，使得，即。
又由于，故必有，证毕。
2.学校共1000名学生，235人住在A宿舍，333人住在B宿舍，432人住在C宿舍。
2005数学建模课程大作业题目
2005数学建模课程大作业题目论文要求：A、B题中任选一道。
第十六周完成。
论文做封面一张，写上题目、组别、作者、学号、系别、班级、日期论文一律用A4纸打印出来，正文用5号宋体字。
要包含有：标题、摘要、问题的重述、合理假设、问题的分析、模型的建立及求解、结果分析、模型的检验、模型的评价与推广、参考文献、计算机程序等。
交卷时间：12月30日（第16周）周五第一讲。
A题：基金使用计划某校基金会有一笔数额为Ｍ元的基金，打算将其存入银行或购买国库券。
当前银行存款及各期国库券的利率见表1.1。
假设国库券每年至少发行一次，改造时间不定。
取款政策参考银行政策。
表1.1校基金会计划在n年内每年用部分本息奖励优秀师生，要求每年的奖金额大致相同，且在n年末仍保留原基金数额。
校基金会希望获得最佳的基金使用计划，以提高每年的奖金金额。
请你帮助校基金会在如下情况下设计使用方案，并对M=5000万元，n=10年及n=12年给出具体结果：1.只存款不购国库券；2.可存款也可购国库券;3.学校在基金到位后的第三年要塗百年校庆，基金会希望这一年的奖金比其他年度多r=20%B题医疗保障基金额度的分配某集团下设四个子公司：子公司A、子公司B、子公司C和子公司D。
各子公司财务分别独立核算。
每个子公司都实施了对雇员的医疗保障计划，由各子公司自行承担雇员的全部医疗费用。
过去的统计数据表明，每个子公司的雇员人数以及每一年龄段的雇员比例，在各年度都保持相对稳定。
四个子公司各年度的医疗费用支出见表1。
为进一步规范各个子公司的医疗保障计划，集团董事会规定，在2003年底,各个子公司均需以银行活期存款的方式，设立医疗保障基金，基金专门用于支付2004年度雇员的医疗费用。
并规定每个子公司的医疗保障基金只能用于支付本子公司雇员。
已知2004年银行活期存款利率为1%。
董事会综合考虑了各种因素，确定本集团设立的2004年度医疗保障基金的总额度为80万元，这一额度在四个子公司之间分配。
2009年全国大学生数学建模大赛C D题优秀论文
2009高教社杯全国大学生数学建模竞赛题目C题卫星和飞船的跟踪测控卫星和飞船在国民经济和国防建设中有着重要的作用，对它们的发射和运行过程进行测控是航天系统的一个重要组成部分，理想的状况是对卫星和飞船（特别是载人飞船）进行全程跟踪测控。
测控设备只能观测到所在点切平面以上的空域，且在与地平面夹角3度的范围内测控效果不好，实际上每个测控站的测控范围只考虑与地平面夹角3度以上的空域。
在一个卫星或飞船的发射与运行过程中，往往有多个测控站联合完成测控任务，如神州七号飞船发射和运行过程中测控站的分布如下图所示：图片来源/jrzg/2008-09/24/content_1104882.htm请利用模型分析卫星或飞船的测控情况，具体问题如下：1. 在所有测控站都与卫星或飞船的运行轨道共面的情况下至少应该建立多少个测控站才能对其进行全程跟踪测控？2.如果一个卫星或飞船的运行轨道与地球赤道平面有固定的夹角，且在离地面高度为H 的球面S上运行。
考虑到地球自转时该卫星或飞船在运行过程中相继两圈的经度有一些差异，问至少应该建立多少个测控站才能对该卫星或飞船可能飞行的区域全部覆盖以达到全程跟踪测控的目的？3. 收集我国一个卫星或飞船的运行资料和发射时测控站点的分布信息，分析这些测控站点对该卫星所能测控的范围。
2009高教社杯全国大学生数学建模竞赛C题评阅要点[说明]本要点仅供参考，各赛区评阅组应根据对题目的理解及学生的解答，自主地进行评阅。
1．考虑最简单圆形轨道和一般的椭圆轨道假设卫星测控站分布在与卫星轨道共面的地球表面，且卫星的运行轨道为圆。
利用几何关系给出全部覆盖需要的测控站点数与卫星高度的关系。
如卫星高度100 200 300 343 400 500观测站数24 16 12 12 11 10 当卫星的运行轨道为椭圆，卫星运行轨道的一个焦点在地球中心，利用几何关系给出每个测控站的覆盖范围。
然后利用数值方法对测控站点进行优化，给出一些具体结果（数量和位置）。
2009年数学建模集训小题目解答
ii）拟合的 Lingo 程序
model: sets: var/1..201/:xi,yi; endsets data: xi=@file(mdata2.txt); yi=@file(mdata2.txt); enddata min=@sum(var:@exp((-(xi-mu)^2)/(2*sigma^2))/(@sqrt(2*3.14 159)*sigma)-yi)^2; @free(mu); end
记年度为阶段变量 k = 1,2,3,4,5 。令 xk 表示第 k 年初完好机器数， uk 表示第 k 年安排
生产 A 种产品的机器数，则 xk − uk 为第 k 年安排生产 B 种产品的机器数，且 0 ≤ uk ≤ xk 。
则第 k + 1年初完好的机器数 xk+1 = (1− 0.2)uk + (1− 0.1)(xk − uk ) = 0.9xk − 0.1uk
x=linprog(f,a,b,aeq,beq,lb,ub);
v=-x(end)
（2）用 Lingo 编程，必须把数据通过纯文本文件或 Excel 文件传递到 Lingo 程序中。 1）mat 数据转化到 txt 文件数据，后由 lingo 读取。在 matlab 下将 mat 数据转到 txt 数 据的程序如下：
令 vk (xk , uk ) 表示第 k 年的纯收入， f k (xk ) 表示第 k 年初往后各年的最大利润之和，这里
vk (xk ,uk ) = 5uk + 4(xk − uk ) = 4xk + uk
显然
f6 (x6 ) = 0
则
fk (xk )
=
max
0≤uk ≤xk
{v
2005年A题全国数学建模优秀论文3
2005
年全国大学生数学建模竞赛全国一等奖
华南理工大学：李宁、董泽彦、林泽彬，指导教师：陶志穗
∑ Hale Waihona Puke xrij =k =1
n
ki
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∑ (x
k =1
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ki
− xi ) 2
∑ (x
k =1
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(i = 1⋯ n; j = 1⋯ p )
kj
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Step3：求相关系数矩阵 R 的特征值并排序 λ1 ≥ λ2 ≥ ⋯ ≥ λ p ，再求出 R 的特征值的相
分为除 x1 之外的三项指标的综合。 以各个主成分的方差贡献率为权重可得到水质的最终 综合评价： 表 2 综合评价表 第一主成分 第二主成分 第三主成分 -1.4696 -0.68454 -0.91026 -1.7625 -0.94467 0.032861 -0.33083 -0.31954 -1.2501 2.0022 1.8388 -0.30098 -1.4067 -0.02552 0.53935 0.17222 0.14241 -1.0413 -1.2741 0.10172 0.43579 3.7118 0.6009 -1.3863 -2.2066 -1.0968 -0.0973 0.33637 0.27026 -1.1778 -2.0179 -0.26008 0.33508 -0.46208 -0.72673 -0.06801 0.44676 1.8036 2.1496 0.9091 1.4033 -0.13198 5.447 -2.5132 1.3095 0.36553 0.90899 -0.90294 -1.1922 0.044725 0.6528 综合得分 综合排名 -1.2028 14 -1.3049 15 -0.44818 10 1.5801 3 -0.84008 13 -0.01085 7 12 -0.74634 2.3162 2 -1.6381 17 0.09879 6 -1.3068 16 -0.43753 9 0.91285 4 0.81843 5 3.2738 1 -0.39204 8 -0.67255 11
2005-2009全国大学生数学数学建模竞赛参考答案
2003-2009全国大学生数学建模竞赛试题及参考答案2010-7-192005A题: 长江水质的评价和预测 (2)2005 A题评阅要点 (4)2005B题： DVD在线租赁 (6)2005 B题评阅要点 (8)2006A题:出版社的资源配置 (10)2006A题评阅要点 (10)2006B题: 艾滋病疗法的评价及疗效的预测 (13)2006 B题评阅要点 (14)2007A题：中国人口增长预测 (17)2007 A题评阅要点 (18)2007 B题：乘公交，看奥运 (21)2007 B题评阅要点 (22)2008A题数码相机定位 (24)2008 A题评阅要点 (27)2008B题高等教育学费标准探讨 (27)2008B题评阅要点 (29)2009 A题制动器试验台的控制方法分析 (30)2009 A题评阅要点 (32)2009B题眼科病床的合理安排 (35)2009 B题评阅要点 (36)2005A题: 长江水质的评价和预测水是人类赖以生存的资源，保护水资源就是保护我们自己，对于我国大江大河水资源的保护和治理应是重中之重。
专家们呼吁：“以人为本，建设文明和谐社会，改善人与自然的环境，减少污染。
”长江是我国第一、世界第三大河流，长江水质的污染程度日趋严重，已引起了相关政府部门和专家们的高度重视。
2004年10月，由全国政协与中国发展研究院联合组成“保护长江万里行”考察团，从长江上游宜宾到下游上海，对沿线21个重点城市做了实地考察，揭示了一幅长江污染的真实画面，其污染程度让人触目惊心。
为此，专家们提出“若不及时拯救，长江生态10年内将濒临崩溃”（附件１），并发出了“拿什么拯救癌变长江”的呼唤（附件2）。
附件3给出了长江沿线17个观测站（地区）近两年多主要水质指标的检测数据，以及干流上７个观测站近一年多的基本数据（站点距离、水流量和水流速）。
通常认为一个观测站（地区）的水质污染主要来自于本地区的排污和上游的污水。
2009年美国数学建模题目 数学
PROBLEM A: Designing a Traffic CircleMany cities and communities have traffic circles—from large ones with many lanes in the circle (such as at the Arc de Triomphe in Paris and the Victory Monument in Bangkok) to small ones with one or two lanes in the circle. Some of these traffic circles position a stop sign or a yield sign on every incoming road that gives priority to traffic already in the circle; some position a yield sign in the circle at each incoming road to give priority to incoming traffic; and some position a traffic light on each incoming road (with no right turn allowed on a red light). Other designs may also be possible.The goal of this problem is to use a model to determine how best to control traffic flow in, around, and out of a circle. State clearly the objective(s) you use in your model for making the optimal choice as well as the factors that affect this choice. Include a Technical Summary of not more than two double-spaced pages that explains to a Traffic Engineer how to use your modelto help choose the appropriate flow-control method for any specific traffic circle. That is, summarize the conditions under which each type of traffic-control method should be used. When traffic lights are recommended, explain a method for determining how many seconds each light should remain green (which may vary according to the time of day and other factors). Illustrate how your model works with specific examples.PROBLEM B: Energy and the Cell PhoneThis question involves the “energy” consequenc es of the cell phone revolution. Cell phone usage is mushrooming, and many people are using cell phones and giving up their landline telephones. What is the consequence of this in terms of electricity use? Every cell phone comes with a battery and a recharger.Requirement 1Consider the current US, a country of about 300 million people. Estimate from available data the number H ofhouseholds, with m members each, that in the past were serviced by landlines. Now, suppose that all the landlines are replaced by cell phones; that is, each of the m members of the household has a cell phone. Model the consequences of this change for electricity utilization in the current US, both during the transition and during the steady state. The analysis should take into account the need for charging the batteries of the cell phones, as well as the fact that cell phones do not last as long as landline phones (for example, the cell phones get lost and break).Requirement 2Consider a second “Pseudo US”—a country of about 300 million people with about the same economic status as the current US. However, this emerging country has neither landlines nor cell phones. What is the optimal way of providing phone service to this country from an energy perspective? Of course, cell phones have many social consequences and uses that landline phones do not allow.A discussion of the broad and hidden consequences of having only landlines, only cell phones, or a mixture of thetwo is welcomed.Requirement 3Cell phones periodically need to be recharged. However, many people always keep their recharger plugged in. Additionally, many people charge their phones every night, whether they need to be recharged or not. Model the energy costs of this wasteful practice for a Pseudo US based upon your answer to Requirement 2. Assume that the Pseudo US supplies electricity from oil. Interpret your results in terms of barrels of oil.Requirement 4Estimates vary on the amount of energy that is used by various recharger types (TV, DVR, computer peripherals, and so forth) when left plugged in but not charging the device. Use accurate data to model the energy wasted by the current US in terms of barrels of oil per day.Requirement 5Now consider population and economic growth over the next 50 years. How might a typical Pseudo US grow? For each 10 years for the next 50 years, predict the energy needs for providing phone service based upon your analysis in the first three requirements. Again, assume electricity is provided from oil. Interpret your predictions in term of barrels of oil.。
2009年全国大学生数学建模竞赛选手选拔试题与答案
2009年全国大学生数学建模竞赛选拔试题时量：180分钟满分：200分系别：专业：学号：姓名：一、数学模型部分(共90分)1、简述数学建模论文的基本结构。
答：应该主要包含论文标题，摘要，问题重述，问题分析，模型建立，模型求解，模型验证，模型分析与改进，模型评价，参考文献等内容。
2、简述数学建模论文摘要的要求及其应包含的主要内容。
答：摘要要用独立的一页来写，字数为300字左右。
应该主要包含建模的思想，建模的方法，建立的模型，模型的求解，模型的改进，模型的评价，以及主要创新点和亮点。
3、试建立桌子在四条腿脚呈长方形时的数学模型，以说明桌子能否在地面上放稳的问题。
4、请把1~9共9个数字填入3乘以3的正方形格子，使3个行中每个行的数字总和为15，3个列中每个列的数字总和也15，两条对角线数字总和也15。
(1)中间格的数字应该为多少？并证明之。
(2) 用推理或建立模型方法求出其它数字(说明求解过程)，最终结果请填入右图。
解:(1) 把第2行,第2列,两对角线所有数字相加,1,2,3,4,5,6,7,8,9数字各出现1次,而中间数字记为x 多出现了3次,列出方程1543)987654321(⨯=+++++++++x 解方程得 x=5, 中间格x 22为5(2) 数字1不能填对角,否则相应一个对角为9而1对应行,列总和为14,而14=6+8仅有一种排法 由对称性有右图填法 ( 2分) 把余下数分3个一组,按总和为15分为第一组(3,4,8)预放入第1行,第2组(2,6,7) 预放入第3行 ( 2分) 调整次序不难得出右图最终结果 (2)别一法:利用上图列出方程⎪⎪⎪⎩⎪⎪⎪⎨⎧=+=+=+=++=++610141515k c n b m a k n m c b a 解空间是1维, 取k 为自由变量(k=2,3,4,,6,7,8),取k=2时其它变量全为整数。
5、 设一个鞋店平均每天卖出鞋100双，批发一次差旅费为每次200元，每双鞋每存储一天的费用为0.01元。
2005 B M Giving Queueing the Booth
Giving Queueing the BoothMCM Team851February7th,20051Team8512 Contents1Introduction31.1Assumptions (3)2Model62.1r<κ,λ,µ (7)2.2κ<r<λ<µ (7)2.3κ<λ<r<µ (10)2.3.1Section1 (10)2.3.2Section2 (11)2.3.3Section3 (12)2.4κ<λ<µ<r (13)2.4.1Section1 (14)2.4.2Section2 (14)2.4.3Section3 (14)2.4.4Section4 (15)2.4.5Section5 (15)2.4.6Section6 (16)2.5mn <p (17)2.5.1λ<r<µ (17)2.5.2λ<µ<r (18)3Strengths and Weaknesses of our Model193.1Strengths (19)3.2Weaknesses (19)3.3Possible Improvements (20)4Results214.1Estimation of Parameters (21)4.2Results from Program (22)4.3When m=n (26)5Conclusions285.1n=m (28)Team8513 1IntroductionDelays at tolling systems are ubiquitous in major road networks throughout the western world.In fact traffic has created problems for urban centers of civilization throughout the ages.“Vehicular traffic,except for chariots and official vehicles,was prohibited from entering Rome during the hours of daylight”for a period in thefirst century[3,Page3].“At least one(Roman) emperor was forced to issue a proclamation threatening the death penalty to those whose chariots and carts blocked the way”of traffic[1,Section1.2] Most toll booth systems involve a major highway with a certain number of lanes,which suddenly increases to a larger number of lanes.Each of these lanes then passes through a toll booth where a tolling system charges the motorist the required toll.Traditionally this has involved the exchange of actual currency,which delayed traffic,although modern electronic payment systems are eliminating this feature(see[15]).The roadway then squeezes back to the initial number of lanes at the end of the toll area.Increasing the number of booths so that the ratio of booths to initial lanes is much larger than1will result in delays at the actual booth being virtually eliminated, however as the largeflow of traffic is squeezed back down to the initial number of lanes results in major congestion,especially at“rush hour”.If the number of toll booths equals the initial number of lanes,then there is no squeezing of traffic,however large tailback occur at the toll booths themselves.The challenge(faced by traffic authorities throughout the western world)is to find an intermediate value that minimizes congestion.Our aim was that the delay time for a rush hour trip through our system in comparison with a trip through with no traffic be minimized1.1AssumptionsWe made a number of assumptions in order to simplify our system and allow us to construct our model.•We assume all vehicles entering our system are identical.We ignore differences between cars,vans,trucks,lorries,etc.•We assume that all vehicles move at the same speed.We assume traf-fic is slowed in the toll plaza relative to the previous section of road, perhaps by imposing a lower speed limit speed limit.Of course in this scenario we now have to assume no-one breaks the speed limitTeam8514•We assume that the vehicles are evenly distributed throughout the lanes,i.e.there is no preference for any lane.•We allow vehicles to be in one of two states,either moving at afixed speed or stopped.•If the number of vehicles arriving at a particular section(namely plaza, toll booths or squeeze point)of our system is greater than the maximum capacity of our system,then that capacity of vehicles pass through, and the remainder form a“virtual queue”based on a“first-come,first-served”basis.(It should be noted that in this report plaza will refer to the wide section before the toll booth and the whole area to be modelled referred to as the system).If vehicles arrive at a section anda queue has already formed,those vehicles join the back of the queue.This is essentially,in the nomenclature of queueing theory,a blocked customers delayed system with parallel servers,see[10,Chap.3]and [7,Pg.38]respectively•All toll booths are identical.In reality almost all modern toll roads em-ploy an electronic system that allows motorists to pass through without stopping,e.g.[15].we are also ignoring the presence of specialist lanes,nes devoted exclusively to heavy goods vehicles.•Our toll plazas are of large enough length that congestion at the squeeze point does not back up as far as the toll booths,which would delay passage of vehicles through the booths and similarly congestion at the toll booths does not affect traffic entering the plaza.•We divide the lanes into discrete intervals of constant length.This length is the length of the vehicle plus the length between successive vehicles,which we assume to be constant.We then model vehicles to be points moving between successive intervals.Since we have assumed constant vehicle speed this means that there is a characteristic time,θ, associated with the passage of vehicles from one interval to the next.•We assume all vehicles arriving at an empty toll booth pass through in the same time,τ.We ignore all possible technical malfunctions, customer delay,etc.For ease of modelling we assumeτis an integer multiple ofθ,τ=pθ,p>1Team8515•We assume that if a vehicle arrives at the toll booths and there is a free booth the vehicle will pass through that booth.•We ignore all external factors other than the incoming traffic for the toll plaza.For example during the transition from domestic currencies to the single European currency,most European toll booths experienced delays,see,e.g.[14].We ignore these such factors.•We assumed the trafficflow would have the general shape shown in Figure1.1.This is a justified assumption for a radial route in anFigure1:The general trend for trafficflowurban centre,see[1,Chap.6].The accepted time span for the peak in Figure is of the order2hours and we will assume it is exactly2hours when considering explicit examples in the results section.•In order to measure congestion we decided to measure the average delay time of the vehicles who arrive in the system during the period of congestion.In other words we decided to measure the extra time spent waiting during congestion compared to the time it would take to pass through the system if there was absolutely no other traffic present, and average this over all vehicles entering the system while there is congestion.We defined an optimal system to be one which minimizes this average delay time.Team8516 2ModelWe began by considering a function h(t)that gives the instantaneous rate of traffic incoming on our system.While this continuous approach may seem to contradict our assumptions that give a discrete nature to our approach,this issue will be addressed shortly.We then multiplied our function by1+γwhereγis a random real number between−0.01and0.01.This gives us a function f(t)which contains an inherent randomness which we believe is an integral part of trafficflow. We now considerκ,the maximum rate at which cars can leave our system. By the definition ofθthis will beκ=n θwhere n is the number of lanes after the squeeze point(and before the toll plaza).Analogously we will have the maximum rate at which vehicles can pass through the toll booth to beλ=m τwhere m is the number of toll booths.Since we are assuming the same speed and lane-interval length in all parts of our system we will have a maximumrate in the plaza ofµ=m θ.If we denote by r the peak of our function h(t)then the behavior of our system can be characterized by the relative values of r,κ,λandµ.We note that for any traffic approaching our system the max rate that will ever approach the squeeze point isλsince the toll booths will delay any traffic flowing at a greater rate.This means that in order for congestion to occur at the squeeze point in our system we requireκ<λEquivalentlyn θ<mτ⇔mn>τθTeam8517mn>p(1) Condition(1)will be crucial to the analysis of our model.It is a necessary but not sufficient condition for congestion to occur at the squeeze point. 2.1r<κ,λ,µIn this case theflow of traffic into our system is less than the maximum capacity of each individual section of the system.Consequently there is no congestion in our system and so corresponds to light traffic.We neednot concern ourselves with such behavior since varying mn has no effect oncongestion.2.2κ<r<λ<µIt is clear thatτ=pθ>θand soλ=mτ≤mθ=µin general.Suppose congestion occurs at the squeeze point but not at the toll booth.This means condition(1)must hold.If this congestion is to occur we also requireκ<r≤λ.This behavior is shown in Figure2.We nowΜΛΚFigure2:The relationship between r,κ,λandµproceed as follows.We define∆ρ(t)=f(t) for f(t)<λλfor f(t)≥λTeam8518 The second condition may seem redundant since r<λ,but it should be noted that r is the peak of our function h and not of our function f.It is entirely possible that our“noise”termγwould cause f(t)to exceedλand thus∆ρto exceedλ.However∆ρrefers to a discrete arrival rate at the squeeze point,and we cannot have this exceedingλso we apply the above careful conditional definition.Based on this we calculate the number of vehicles arriving at the squeeze point at time t as∆c(t)=θ∆ρ(t)We are now ready to apply our model.We set up the following difference equations for the function c(t)which refers to the number of vehicles trying to go through the squeeze point.It includes those who will get through at time t.It can be thought of as the number of people in the(virtual)queue. In order to simplify matters we take t=0when h(t)=κ,which is valid since we have assumed h(t)is monotonic in the region before the peak and so by the discussion earlier in this section there will be no congestion for t<0. Our initial condition isc(0)=∆c(0)This simply says that the number of vehicles trying to pass through the squeeze point at time0is those that arrive at time0,since there is no congestion before this,and consequently no(virtual)queue.This now givesc(θ)=c(0)−n+∆c(θ)since we have c(0)−n remaining in the(virtual)queue from time0and∆c(θ) arrive at timeθ.Repeating this givesc(iθ)=c((i−1)θ)−n+∆c(iθ)where i=1,2,...Since our function f peaks at some value less thanλand then falls belowκthis means c(t)will eventually begin to fall and at some point become≤0.At this point we define c(t)to be zero and our model is finished since there is now no more congestionc(Nθ)≤0,c(Nθ):=0This means that our model gives the following difference equation with spec-ified initial and terminating conditionsc(0)=∆c(0)c(iθ)=c((i−1)θ)−n+∆c(iθ)(2)c(Nθ)≤0,c(Nθ):=0Team8519ΚThis can be solved recursively.The above approach highlights the discrete nature of our approach alluded to earlier.In effect we are incrementing time in discrete intervals ofθand allowing each vehicle to either move forward one space increment or to remain in situ.During each time interval of sizeθ,c((i−1)θ)−n vehicles are delayed by amountθ,which gives usT delay=Ni=0(c(iθ)−n)θfor the total delay time.The total number of vehicles arriving during con-gestion is,clearly,given byV total=Ni=0∆c(iθ)Hence our“optimization parameter”ist delay=T delayV total=Ni=0(c(iθ)−n)Ni=0∆c(iθ)(3)where the overbar represents mean.It would now seem to simply be a computing problem to vary the ratio m to minimize the mean delay time as given by(13).However this is not quite the case.Suppose we are faced with the challenge of designing an optimal toll plaza for an already existing highway.This means n,τ,θ, f and r are determined by factors outside our control(pre-existing values,Team85110 efficiency of booths,speed of cars in plaza,incoming traffic and incoming traffic respectively).This means that m is our only free parameter,which influencesλ.The crucial point is thatλonly affects whether or not this particular version of our model is valid.To generalize,what our assumptions say is that if the rate of influx of traffic is too low to result in congestion at the toll booths,then the congestion at the squeeze point is independent of m, the number of toll booths.While this is clearly not a perfect model of reality, it is not as far removed from the truth as one might think.Intuitively,there should be a correlation between these two quantities.However the overall number of vehicles“squeezing in”and the number of lanes into which they squeeze are constants so the dependence may not be that strong.If this is the case then our model is a good approximation,especially when we consider that in later sections,the model is much better.2.3κ<λ<r<µTake the case where,congestion occurs at both the squeeze point and the toll booths.Again condition(1)must hold as well asκ<λ<r<µ.We now define temporarily∆ρ(t)=f(t) for f(t)<λλfor f(t)≥λwhich is exactly the same as the previous section,and∆β(t)=f(t) forλ<f(t)<µµfor f(t)≥µrefereing to a discrete arrival at the toll booths this time.To model this in a clear and concise way,the model was broken into three separate sections.2.3.1Section1This starts at the point where f(t)first intersects the lineκand is set to be t=0.Before this point,f(t)is always belowκso no congestion can occur.Above this point however,congestion occurs at the squeeze point. The conditions in this section are the same as the beginning of the model inTeam85111ΛΚ123ΛΚ1232.2,so it is reasonable to take the difference equation for congestion at the squeeze point to bec(0)=∆c(0)(4)c(iθ)=c((i−1)θ)−n+∆c(iθ)with the terminating conditions removed for the moment.Section1ends, however as f(t)increases to the point where it intersectsλ.This is labelled t=t o.2.3.2Section2The second section intuitively begins at the point where congestion begins to occur at the toll booths.A difference equation is constructed here much in the same way as for the squeeze point in2.2.The number of vehicles at time t arriving at the booths is given in time intervals ofτthough∆b(t)=τ∆β(t)Team85112 We now define b(t)to be the number of vehicles trying to go through the toll booths or the number in the(virtual)queue.There will be no congestion for f(t)<λso we can assume the initial condition:b(t0)=∆b(t0)The general difference equation then becomes:b(t0+iτ)=b(t0+(i−1)τ)−m+∆b(t0+iτ)again with i=1,2,....The function f peaks at a value r<µand then decreases belowλ,meaning that at some value of t,b(t)≤0.After this point,there will be no more congestion at the toll booths,and so is the terminal condition for b(t)and is defined as0to avoid negative numbers:b(t0+Mτ)≤0,b(t0+Mτ):=0The difference equation with initial and terminal conditions for the toll booth isb(t0)=∆b(t0)b(t o+iτ)=b(t0+(i−1)τ)−m+∆b(t o+iτ)(5)b(t o+Mτ)≤0,b(t o+Mτ):=0The effect of the congestion at the toll booths on the squeeze point is that the rate of influx of vehicles is constantly atλ.This gives a general difference equation for the section ofc(t0+iθ)=c(t0+(i−1)θ)−n+λθ2.3.3Section3The third and last section begins where section2ends,at t=t0+Mτ.Here, there is no more queuing at the toll booths,but queuing continues at the squeeze point.(As can be seen from the graph above,there is a sharp drop at the start of section2until f[t]is obtained again.Though this seems counter-intuitive, it was noticed,by chance,while sitting over coffee during a break that some-one who joined a long queue at the end of a busy period,still had no oneTeam 85113behind when he reached the counter!)The general difference equation for the squeeze point reverts back to that of 2.2,since f [t ]is now below λagain.The terminating condition can also be taken from the model in 2.2givingc (Nθ)≤0,c (Nθ):=0(6)We can now redefine c (t )properly over all three sections ∆ρ(t )= f (t ) for t <t o ,t o +Mτ<t λfor t o <t <t o +MτThe total delay time is then given by:T delay =[Ni =0(c (iθ)−n )]θ+[M i =0(b (t o +iτ)−m )]τand with the total number of cars delayed given byV total =Ni =0 f (iθ)we get our “optimization parameter”to bet delay =T delay V total =[ N i =0(c (iθ)−n )]θ+[ M i =0(c (t o +iτ)−m )]τ N i =0∆c (iθ)(7)2.4κ<λ<µ<rThis is perhaps the most complicated part of our model,with more traffic than the plaza can handle.The intuitive thinking behind this is that,in the event of unusually large traffic,for example a sports event or concert,the maximum rate for the plaza itself is too low for the rate of traffic.In effect,a queue forms before the plaza itself.The model for this is simply an extra tier onto the model for 2.3.Much the same as 2.3is an extra tier on top of 2.2.Here,the graph of f [x ]is split first into three sections as in part B.Team 85114ΚΛΜ1234562.4.1Section 1This is the same as section 1in 2.3(also the same as the beginning of 2.2)and as such,the start point is set to t =0and end point set to t =t o ,the rest of the difference equation also follows from 2.3c (0)=∆c (0)c (iθ)=c ((i −1)θ)−n +∆c (iθ)(8)2.4.2Section 2In order to simplify the model,section 2was taken to be the same as in part B,i.e.from the end point of section 1to the terminating point of the queuing at the toll booth.As a result of this,the behavior of the queuing at the squeeze point is exactly the same as in part B:c (t 0+iθ)=c (t 0+(i −1)θ)−n +λθThe queuing at the booth and plaza are entirely contained in section 2but will be broken down into further sections later.2.4.3Section 3This is again the last section in terms of chronological order,but labelled in this way,is exactly the same as 2.3.3,and as such the difference equation,by the same arguments isTeam85115c(iθ)=c((i−1)θ)−n+∆c(iθ)(9)c(Nθ)≤0,c(Nθ):=02.4.4Section4Sections4to6are now analogous to2.3withµinstead ofλandλinstead ofκ.There is also,however a time shift,with t=0going to t=t o.Hence the equations for the queuing at the toll booth becomeb(t0)=∆b(t0)(10)b(t o+iτ)=b(t0+(i−1)τ)−m+∆b(t o+iτ) These apply until f(t)>µ,when the congestion at the plaza kicks in. This time is labelled t=t1and is the starting point for section5.2.4.5Section5The queuing at the plaza is described by similar difference equations to the queuing at the squeeze point in that it depends on the characteristic time θ.However,because there are m lanes instead of n,m vehicles get through to the plaza in each time intervalθ.There is also a time scale based on the congestion at the plaza beginning at t=t1.Thefinished equation reads∆α(t)= f(t)∆a(t)=θ∆α(t)(11)c(t1)=∆a(t1)a(t1+iθ)=a(t1+(i−1)θ)−m+∆a(t1+iθ)a(t1+Qθ)≤0,a(t1+Qθ):=0The traffic now arriving at the booth is at a constant rate ofµ.This has the same effect as on the squeeze point in both2.3.2and2.4.2.The congestion is modelled in the following way until there is no more congestion at the plazaTeam85116b(t o+iτ)=b(t0+(i−1)τ)−m+µτThere is one major difference though between the model for2.3and the whole of2.4.2here.Because the queuing at the plaza is measured in time intervals ofθand the queuing at the booth in intervals ofτ=pθ,the congestion at the plaza may not end on an integer number ofτfor the booth.Intuitively,ifθandτare out of sync,there will be a slight delay in the effect onτ.As a result,the next integer value of t1+iτafter the termination of a(t)is taken as the end of2.4.2for the boothb(t1+Dθ)s.t.D= Qθτ2.4.6Section6Here,we have no more congestion at the plaza,and we aren’t considering the queuing at the squeeze point(already covered as section2),so the only concern is the toll booth itself.Hence this section starts at t1+Dτ.It then follows the basic difference equation for the booth,with the appropriate time scaling,terminating in the usual way.b(t o+Mτ)≤0,b(t o+Mτ):=0Now the three different functions for the arriving traffic at each section need to be defined properly∆ρ(t)=f(t) for t<t o,t o+Mτ<t λfor t o<t<t o+Mτ∆β(t)=f(t) for t<t1,t1+Dτ<tµfor t1<t<t1+Dτ∆α(t)= f(t)(12)Team85117 The summations of these functions over their respective time spanT delay=[Ni=0(c(iθ)−n)]θ+[Mi=0(b(t o+iτ)−m)]τ+[Qi=0(a(t1+iθ)−m)]θTotal number of cars delayedV total=Ni=0f(iθ)“optimization parameter”t delay=T delayV total=T delay=[Ni=0(c(iθ)−n)]θ+[Mi=0(b(t o+iτ)−m)]τ+[Qi=0(a(t1+iθ)−m)]θNi=0∆c(iθ)(13)2.5mn<pHere are the last remaining variations of the model,where the condition 1doesn’t hold,i.e.λ<κ.In this scenario,maximum rate of vehicles that leave the booth will never be enough to cause congestion at the squeeze point.The case where m=n is a special case of this scenario.2.5.1λ<r<µIn this situation,traffic congestion occurs at the toll booths,but nowhere else.It is very similar to the model for2.2but with a number of differences. Instead of c(t)we use b(t)because the congestion is at the toll booths and not the squeeze point.θalso changes toτevery where andκandλgo toλandµrespectively.This gives the difference equationsb(0)=∆b(0)b(iτ)=b((i−1)τ)−m+∆b(iτ)b(Nτ)≤0,b(Nτ):=0(14)Team85118ΜΛTotal delay time for systemT delay=[Ni=0(b(iτ)−m)]τThe total number of vehicles arriving during congestionV total=Ni=0∆b(iτ)Our“optimization parameter”ist delay=T delayV total=[Ni=0(b(iτ)−m)]τNi=0∆b(iτ)(15)where the overbar represents mean.2.5.2λ<µ<rIn this section,the peak rate of traffic influx is greater than plaza can deal with as well and so congestion builds up there as well as at the booths.This is analogous to2.3with a few minor adjustments again.c and b go to b and a respectivelyθandτswap andκandλgo toλandµrespectively.Thefinal average comes out ast delay=T delayV total=[Ni=0(b(iτ)−m)]τ+[Mi=0(a(t o+iθ)−m)]θNi=0∆b(iτ)(16)Team85119ΜΛ3Strengths and Weaknesses of our Model 3.1Strengths•The model we have chosen is easily programmed and the program we wrote(in Mathematica5.0)is included as an Appendix.•None of the inputs for our model are specific to certain types of toll booths.Parameters such as h(t),τ,etc can be easily measured by a team of statistical consultants.•If our data is as good an approximation of reality as we think it is,then the behavior of our model should be stable for any realistic variations.•Our model contains a random nature ensured by the presence ofγ, which makes it more realistic of real-life.•Our model will in theory give an optimal ratio of m to n3.2Weaknesses•While we have a random element to our data,repeated studies and standard probability or queueing theory have established the need to consider a quasi-random Poisson arrival model for this situation(see, for example,[6,Section3.5])•Almost all of the assumptions we have made for the nature of the movement of vehicles through our system are highly unrealistic.ForTeam85120 example,vehicles will not be homogenous,will not move in discrete lengths over discrete time intervals,will not arrive in continuous time.•Even if our discrete model is valid,there is absolutely no justificationfor assuming p=τθis an integer if we are considering a general tollplaza.•If1person experiences a1hour delay and99people experience no wait, then,by our definition of optimal,this is better than100people waiting for half a minute.This is not an ideal scenario,and while this example is highly extreme,it does indicate that we may not be optimizing what we think we are.3.3Possible Improvements•Clearly from above,a poisson arrival would be the obvious generaliza-tion of our model•Another added feature that would bring our model closer to reality would be to correlate the likelihood of a vehicle going through a booth to the distance from the lane the vehicle is on to the nearest free toll-booth•While there any many more possibilities,these seem to be the two key aspects of reality that we are ignoring.Team 851214ResultsIn order to obtain real data to test our model,we chose the Westlink M50toll plaza in Dublin,Ireland.Some of our team members had some limited personal experience of this particulartoll plaza.4.1Estimation of ParametersAs already mentioned in the assumptions,we restrict our attention to the two busiest hours,and approximate this as the parabola shown.TimeTrafficFlowOf course the change of flow will never be as uniform as this,so adding randomness to the model gives the following distribution for our rate of flow.On the Westlink M50,2lanes fan out into 5for the toll plaza,so in ournotation,m =5and n =2.From before,we require p <m nin order for congestion to occur at the constriction after the toll booths.Therefore,since mn=2.5,we must take p =1or 2so as to be able to apply our model.We have τ=p Θ,i.e.p relates how long one spends passing through the constriction to the amount of time to go through the toll booth,and so p =2is the more realistic value to take (intuitively,the time spent driving through the booth will be longer,since,for example,the car must stop to pay the toll).The length a particular vehicle could range anywhere,from 3.6meters for a Toyota Yaris for instance (see [12]),up to 25meters for large trucks(seeTeam 85122TimeTrafficFlow[13]).Θ=lv ,where l is the average vehicle length and v the average speedof a vehicle at the constriction,so taking Θ=4seconds or 1900hours is ing statistics relating to the number of vehicles which use the M50,we approximated the maximum traffic flow during the busy two hour period to be,on average,4000vehicles per hour.4.2Results from ProgramThe Mathematica programs in the appendix calculate the average waiting time t delay (in minutes)for a vehicle travelling through the toll system.Wefixed n =2,Θ=1900,and p =2,and then varied m and r ,the peak traffic flow,to get values for the waiting time.m took values from 2to 14,while r =2000,3000,4000,5000,6000.We obtained the data shown in the following tables:Team85123r m t delay20002 2.97507312.15334 1.58025 1.563336 1.587277 1.585218 1.558299 1.5582910 1.5582911 1.5582912 1.5582913 1.5582914 1.55829r m t delay3000274.78013 3.02861418.1931513.90786 2.26458718.2488818.2488918.24881018.24881118.24881218.24881318.24881418.2488Team85124r m t delay40002220.114342.35994 2.96445562.3328632.2738713.54568 3.01919937.74191037.74191137.74191237.74191337.74191437.7419r m t delay50002430.1953116.371429.7735110.105686.7116752.9446829.3646913.602110 3.70615110.02424541258.31831358.31831458.3183Team85125r m t delay60002706.5073217.076474.61985196.886111.3557113.582874.6693948.4941028.1131113.938712 4.46381130.2686461479.3435We now hadfive sets of fourteen data points(m,t delay).We had reckoned r to be4000on average,and so we applied the following weightings to the values for r,based on how likely that level of trafficflow would occur:r weighting20000.0130000.240000.5850000.260000.01We felt that the distributions offlows would be normal with mean4000, and that those weights chosen would approximate this:In each of thefive sets of data points corresponding to different values of r,the t delay was multiplied by the the respective weight.Finally,we summed the t delay components of corresponding points in allfive data sets to give one set of fourteen points.When plotted(see below),we get a graph of m versus the mean weighted delay time:Clearly,for our model of the Westlink M50with n=2roads,m=4and m=8minimize the delay time.The time difference is so small,and will fluctuate with the randomness in the model,so either number of roads will minimize delays.。
数学建模试卷2009(答案)
华中科技大学《数学建模》考试卷（半开卷）2009～2010学年度第一学期 成绩学号 专业 班级 姓名一、选择题 （每题2分，共10分）（1）建模预测天气。
在影响天气的诸多因素及相互关系中，既有已知的又有许多未知的非确定的信息。
这类模型属于（ b ）。
a.白箱模型b.灰箱模型c.黑箱模型（2）在城镇供水系统模型中，水箱的尺寸是（ c ）。
a.常量b.变量c.参数 （3）在整理数据时，需处理和分析观测和实验数据中的误差，异常点来源于（ c ）。
a.随机误差b.系统误差c.过失误差（4）需对一类动物建立身长与体重关系的模型。
在对模型的参数进行估计时，如已有30组数据，且参数估计精度要求较高，应采用（ b ）估计参数。
a.图解法 b.统计法 c.机理分析法（5）在求解模型时，为了简化方程有时会舍弃高价小量（如一阶近似、二阶近似等），由此带来一定的误差，此误差是（ a ）。
a.截断误差b.假设误差c.舍入误差二、填空题 （每空1分，共10分）（1）已知函数 ()()22y 1a sin x a cos x ωωω=++，当a 很小时，一阶近似为（ ()y 1a sin x ω=+ ），当ω很小时，二阶近似为（ 22y 1a x a ωω=++ ），而当x 很小时，一阶近似为( 22y 1a x a ωω=++ ），二阶近似为（ 222421y 1a x a a x 2ωωω=++-）。
（2）学校共有3个系， 甲系103人，乙系63人，丙系34人。
学生会共设有20个成员， 按Hamilton 方法分配名额为（10 ，6 ，4 ），按Q 值法分配名额为（ 11 ，6 ，3 ）。
（3）使用三次样条函数()()3n 123j j 23301j 1x x xxS x x 2!3!3!βαααα-+=-=++++∑进行插值，在两节点间()3S x 是（ 3 ）次多项式，在每个节点上()3S x 既连续又（ 光滑 ）。
2005mcm、icm
2005 年美国大学生数学建模竞赛MCM、ICM 试题2005 MCM A: Flood PlanningLake Murray in central South Carolina is formed by a large earthen dam, which was completed in 1930 for power production. Model the flooding downstream in the event there is a catastrophic earthquake that breaches the dam.Two particular questions:Rawls Creek is a year-round stream that flows into the Saluda River a short distance downriver from the dam. How much flooding will occur in Rawls Creek from a dam failure, and how far back will it extend?Could the flood be so massive downstream that water would reach up to the S.C. State Capitol Building, which is on a hill overlooking the Congaree River?2005 MCM B: TollboothsHeavily-traveled toll roads such as the Garden State Parkway, Interstate 95, and so forth, are multi-lane divided highways that are interrupted at intervals by toll plazas. Because collecting tolls is usually unpopular, it is desirable to minimize motorist annoyance by limiting the amount of traffic disruption caused by the toll p lazas. Commonly, a much larger number of tollbooths is provided than the number of travel lanes entering the toll plaza. Upon entering the toll plaza, the flow of vehicles fans out to the larger number of tollbooths, and when leaving the toll plaza, the flow of vehicles is required to squeeze back down to a number of travel lanes equal to the number of travel lanes before the toll plaza. Consequently, when traffic is heavy, congestion increases upon departure from the toll plaza. When traffic is very heavy, congestion also builds at the entry to the toll plaza because of the time required for each vehicle to pay the toll.Make a model to help you determine the optimal number of tollbooths to deploy in a barrier-toll plaza. Explicitly consider the scenario where there is exactly one tollbooth per incoming travel lane. Under what conditions is this more or less effective than the current practice? Note that the definition of “optimal” is up to you to determine.2005 ICM: Nonrenewable ResourcesSelect a vital nonrenewable or exhaustible resource (water, mineral, energy, food, etc.) for which your team can find appropriate world-wide historic data on its endowment, discovery, annual consumption, and price.The modeling tasks are:1、Using the endowment, discoveries, and consumption data, model the depletionor degradation of the commodity over a long horizon using resource modeling principles.2、Adjust the model to account for future economic, demographic, political andenvironmental factors. Be sure to reveal the details of your model, provide visualizations of the model’s output, and explain limitations of the model.3、Create a fair, practical “harvesting/management” policy that may includeeconomic incentives or disincentives, which sustain the usage over a long period of time while avoiding severe disruption of consumption, degradation or rapid exhaustion of the resource.4、Develop a “security” policy that protects the resource against theft, misuse,disruption, and unnecessary degradation or destruction of the reso urce. Other issues that may need to be addressed are political and security management alternatives associated with these policies.5、Develop policies to control any short- or long-term “environmental effects” ofthe harvesting. Be sure to include issues such as pollutants, increased susceptibility to natural disasters, waste handling and storage, and other factors you deem appropriate.6、Compare this resource with any other alternatives for its purpose. What newscience or technologies could be developed to mitigate the use and potential exhaustion of this resource? Develop a research policy to advance these new areas.。
全国数学建模大赛试题——出租车模型及数据(C)
2005年全国部分高校研究生数学建模竞赛C题城市交通管理中的出租车规划最近几年，出租车经常成为居民、新闻媒体议论的话题。
某城市居民普遍反映出租车价格偏高，而另一方面，出租车司机却抱怨劳动强度大，收入相对来说偏低，甚至发生出租车司机罢运的情况，这反映出租车市场管理存在一定问题，整个出租车行业不景气，长此以往将影响社会稳定，值得关注。
我国城市在未来一段时间内，规模会不断扩大，人口会不断增长，人民生活水平将不断提高，对出租车的需求也会不断变化。
如何配合城市发展的战略目标，最大限度地满足人民群众的出行需要，减少环境污染和资源消耗，协调各阶层的利益关系，是值得深入研究的。
（附录中给出了某城市的相关数据）。
(1)考虑以上因素，结合该城市经济发展和自身特点，类比国内外城市情况，预测该城市居民出行强度和出行总量，同时进一步给出该城市当前与今后若干年乘坐出租车人口的预测模型。
(2)给出该城市出租车最佳数量预测模型。
(3)按油价调价前后（3.87元/升与4.30元/升），分别讨论是否存在能够使得市民与出租车司机双方都满意的价格调整方案。
若存在，给出最优方案。
(4)本题给出的数据的采集是否合理，如有不合理之处，请你给出更合理且实际可行的数据采集方案。
(5)请你们站在市公用事业管理部门的立场上考虑出租车规划问题，并将你们的研究成果写成一篇短文，向市公用事业管理部门概括介绍你们的方案。
附录11、2004年某城市的城市规模和道路情况如下：（1）城市现辖6区，2004年城市建成区面积181.77平方公里，人口185.15万。
（2）道路总长度998公里，道路铺装面积928万平方米，道路广场面积1371.45万平方米，道路网密度7.71公里/平方公里，人均道路长度0.7米，人均道路面积6.16平方米。
（3）城市总体规划人口城市总体规划人口规模（单位：万人）通过对出行特征的分析，把出行特征相近的人口划归为一类，常住人口和暂住人口称为第一类人口，短期及当日进出人口称为第二类人口。