沿大河漂流问题的分析
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过河问题解题技巧和方法
过河问题是一种经典的数学问题,涉及到如何将一定数量的人或物品从一边河岸运输到另一边河岸。
以下是一些解决这种问题的技巧和方法:
1. 确定问题的基本要素:问题通常包括人物、船只、岸边等元素。
确定这些要素以及它们的数量和限制条件是解决问题的第一步。
2. 确定过河的规则:在运输人或物品时,需要遵守一定的规则。
例如,一艘船只最多只能容纳一定数量的人或重量;在河的任何一边都不能有过多的人或物品;船只必须始终有人或物品在其中等等。
3. 制定计划:根据问题的要素和规则,可以制定一个过河计划。
这个计划应该考虑到每个人或物品的移动方向、时间、船只的位置等细节。
4. 试错法:如果计划不能解决问题,可以采用试错法。
在不违反规则的情况下,尝试不同的策略并观察结果。
如果结果不符合要求,则需要重新制定计划。
5. 简化问题:有时,将问题简化为更小的部分可以更容易地解决问题。
例如,可以考虑只有两个人或一艘船只的情况。
6. 使用图表:图表可以帮助解决过河问题。
可以使用流程图、状态图等来描述问题的不同阶段和状态。
这可以帮助识别可能的问题和解决方法。
7. 分析和优化:当找到解决问题的方法时,可以对其进行分析和优化。
例如,可以比较不同策略的时间和效率,然后选择最佳的方
法。
总之,过河问题是一种有趣而又具有挑战性的数学问题。
通过掌握以上解决方法和技巧,可以更好地解决这种问题并提高数学思维能力。
环形河流中的相遇漂流问题引言本文将讨论环形河流中的相遇漂流问题。
在环形河流中,水流的方向和速度随时间变化,漂流者在漂流过程中可能会相遇。
我们将研究如何计算相遇的可能性,并提供一些简单的策略来解决这个问题。
相遇条件在环形河流中,漂流者相遇的条件可以通过以下两个因素确定:1. 初始位置:漂流者在河流上的初始位置将影响他们是否会相遇。
如果漂流者的初始位置相距较远,他们的相遇可能性较低。
2. 水流速度和方向:河流中的水流速度和方向是另一个影响漂流者相遇的重要因素。
如果水流方向相同且速度相近,漂流者相遇的可能性较大。
相遇计算在环形河流中,计算漂流者相遇的可能性是一个复杂的问题。
相遇的结果取决于漂流者初始位置的随机性,以及水流的速度和方向的变化。
一个简单的计算方法是使用模拟实验。
通过对漂流者的初始位置和水流的速度和方向进行随机抽样,我们可以模拟多组漂流过程,并统计相遇事件的频率。
通过重复模拟实验的过程,我们可以估计漂流者相遇的概率。
策略为了增加漂流者相遇的可能性,我们可以采取以下策略:1. 集中初始位置:如果漂流者的初始位置靠近河流的某个特定位置,他们的相遇可能性会增加。
可以提前确定一个集中初始位置,使漂流者更容易相遇。
2. 相近速度和方向:如果漂流者的初始速度和方向相近,他们的相遇可能性会增加。
可以尽量使漂流者的初始速度和方向接近,以增加相遇的概率。
请注意,这些策略只是一些简单的想法,并不能保证漂流者一定会相遇。
相遇的结果仍取决于环形河流的特定条件和漂流过程的随机性。
结论环形河流中的相遇漂流问题是一个复杂而有趣的话题。
通过模拟实验和采取一些简单的策略,我们可以探索漂流者相遇的可能性。
然而,需要注意的是,这些策略并不能保证相遇一定会发生,因为漂流过程受到多个变量的影响。
希望本文的讨论能够帮助读者更好地理解环形河流中的相遇漂流问题,并为进一步的研究提供一些思路和参考。
参考文献- [引用参考文献1]- [引用参考文献2]。
过河问题解题技巧和方法
过河问题是指在一定条件下,将若干对象从一岸运送到另一岸的问题。
这种问题通常涉及到一些限制条件,如运输工具的数量、容量,运输对象的数量、体积等。
解决过河问题通常需要一些技巧和方法,以下是一些常用的方法: 1. 分析限制条件:首先要了解问题中的限制条件,如运输工具
的数量和容量,运输对象的数量和体积等。
根据这些限制条件,可以确定一些基本的运输策略。
2. 利用递归思想:过河问题通常可以使用递归的思想来解决。
将问题分解为多个子问题,然后解决每个子问题。
通过递归的方式,可以逐步缩小问题的规模,从而得到解决方案。
3. 利用图形法:将问题转化为一个图形,可以更直观地理解问题。
例如,可以绘制一个河岸、船只和对象的图形,然后根据限制条件,将对象依次放入船只进行运输。
4. 利用迭代法:在解决过河问题时,有时需要多次尝试不同的
运输策略,才能找到最优的解决方案。
这时可以使用迭代法,逐步优化运输策略,直到找到最优解。
5. 利用编程工具:计算机编程可以快速地解决过河问题。
一些
编程工具可以实现递归、图形和迭代等方法,帮助解决问题。
同时,编程工具还可以自动化运输策略,使得解决过河问题更加高效。
以上是解决过河问题的一些常用技巧和方法。
在实际问题中,可以根据具体情况选择不同的方法,找到最优的解决方案。
完整版)行程问题流水行船问题本讲将研究流水行船问题。
在江河里航行时,船只除了本身的速度外,还受到流水的推动或顶逆,这就是流水行船问题。
另外,还有一种与之类似的问题是“在风中跑步或行车”的问题,处理方法与流水行船问题一致。
行船问题是一类特殊的行程问题,它的特殊之处在于船只要考虑水流速度的影响。
船速是指在静水中行船,单位时间内所走的路程。
逆水速度是指逆水上行的速度,顺水速度是指顺水下行的速度。
水速是指船只在水中不借助其他外力,只借助水流力量单位时间所漂流的路程,也就是水流速度。
顺水速度等于船速加上水速,逆水速度等于船速减去水速。
顺水行程等于顺水速度乘以顺水时间,逆水行程等于逆水速度乘以逆水时间。
船速等于顺水速度和逆水速度的平均值,水速等于顺水速度和逆水速度的差值的一半。
下面列举几个例子:1.甲、乙之间的水路长234千米,一只船从甲港到乙港需9小时,从乙港返回甲港需13小时,求船速和水速各为每小时多少千米。
2.A、B两港相距560千米,甲船往返两港需要105小时,逆流航行比顺流航行多了35小时,乙船的静水速度是甲船静水速度的2倍,那么乙船往返两港需要多少小时?3.甲河是乙河的支流,甲河水速为每小时3千米,乙河水速为每小时2千米。
一艘船沿甲河顺水航行7小时,行了133千米到达乙河,在乙河中还要逆水航行84千米,问这艘船还要航行几小时?4.一艘轮船在两个港口间航行,水速为每小时6千米,顺水下行需要4小时,返回上行需要7小时,求这两个港口之间的距离。
5.某船从甲地顺流而下,5天到达乙地;该船从乙地返回甲地用了7天,求水从甲地流到乙地用了多少时间?6.一艘小船在XXX,第一次顺流航行33千米,逆流航行11千米,共用11小时;第二次用同样的时间,顺流航行了24千米,逆流航行了14千米,求这艘小船的静水速度和水流速度。
7.一只船在河里航行,顺流而下每小时行18千米,已知这只船下行2小时恰好与上行3小时所行的路程相等,求船速和水速。
行程问题(一)一、考点、热点顺水:行驶速度=静水速度+流水速度逆水:行驶速度=静水速度—流水速度相遇问题:相距距离÷速度和=相遇时间追及问题:相距距离÷速度差=追及时间二、典型例题例1 一只船在静水中每小时行8千米,逆水行4小时航行24千米,求水流速度?例2 一只每小时航行13千米的客船在一条河中航行,这条河的水速为每小时7千米,这只客船顺水航行140千米需要多少小时?例3 甲乙两港间的水路长208千米,一只船从甲港开往乙港,顺水8小时到达,从乙港返回甲港,逆水13小时到达。
求船在静水的速度?例4 甲河是乙河的支流,甲河水流速度为每小时3千米,乙河水流速度为每小时2千米,一艘船沿乙河逆水航行6小时,行了84千米到达甲河,在甲河还要顺水航行133千米,这艘船一共航行多少小时?例5 一艘客船从A港驶往B港顺水下行,每小时航行28千米,到达B港后,又逆水上行回到A港,逆水上行比顺水下行多用2小时,已知水流速度为每小时4千米,求A、B两港相距多少千米?例6 A、B两船分别从上游的甲港和下游的乙港同时相向而行,6小时相遇,然后相并向下游驶去,A船经3小时到达乙港,B船经4小时回到乙港。
已知甲、乙两港间相距936千米,求AB两船的速度及水速各是多少千米?例7 一艘客轮顺水航行60千米需4小时,逆水航行60千米需5小时,现在客轮从上游甲城到下游乙城,已知两城间的水路长75千米。
开船时一旅客从窗口投出一木板,问船到乙城时,木板离乙城还有多少千米?例8 两只木排,甲木排和漂流物同时从A地到B地前行,乙木排也同时从B地向A地前行,甲木排5小时后与漂流物相距75千米,乙木排15小时后与漂流物相遇,两木排的划速相同,AB两地距离多长?三、习题练习1、AB两码头相距360千米,一艘轮船在其间航行,顺流需18小时,逆流需24小时,求水流速度。
2、甲、乙两港相距200千米,有一艘汽艇顺水行完全程需8小时,这条河的水流速度是每小时2.5千米,求逆水行完全程要多少小时?3、一只小船在静水中每小时航行35千米,逆水航行180千米需6小时,顺水航行这段水路需多少小时?4、光明号客船顺水航行200千米要8小时,逆水航行120千米也要8小时,那么在静水中航行200千米需要多少小时?5、一艘客轮每小时行驶27千米,在大河中顺水航行160千米,每小时水速5千米,需要航行多少小时?6、一艘货轮每小时行驶25千米,大河中水速为5千米,要在大河中逆水航行7小时,能行驶多少千米?7、甲乙两地相距270千米,客轮从甲地顺水以每小时27千米的速度航行到乙地要用9小时,这样水速是每小时多少千米?8、一只船顺水行320千米需用8小时,水流每小时15千米,逆水每小时行多少千米?9、惟惟划船,沿河向上游划去,不巧帽子被风刮走了。
四年级奥数流水漂船问题描述流水漂船问题是一道经典的奥数问题,常在四年级的奥数竞赛中出现。
问题的描述如下:在一条宽为 $w$ 米的河上,有一条$v$ 米/秒的水流。
小明和小红分别在两岸等待乘船过河,他们希望选择一条最短的航线来尽快到达对岸。
已知小明和小红两人的行进速度均为 $s$ 米/秒,且可以忽略船在水中的阻力,船只可直线行驶。
请问,在这种情况下,小明和小红应该如何选择航线,才能尽快到达对岸?解决方案要解决这个问题,首先我们需要确定小明和小红的航线。
我们可以利用向量的方法来计算最短航线的长度。
假设小明和小红分别在河岸上的点 $A$ 和点 $B$,对岸的点为 $C$。
在河中间确定一个点 $D$,使得 $CD$ 与河流方向垂直。
如下图所示:我们可以计算向量 $\overrightarrow{AD}$ 和$\overrightarrow{BD}$,得到两者的长度 $AD$ 和 $BD$。
然后,我们可以利用勾股定理,计算出最短航线的长度 $\sqrt{AD^2 +BD^2}$。
算法实现以下是求解流水漂船问题的算法实现步骤:1. 输入河的宽度 $w$,水流速度 $v$ 和行进速度 $s$。
2. 计算航线两端点到中垂线的距离 $AD$ 和 $BD$。
3. 计算最短航线的长度 $\sqrt{AD^2 + BD^2}$。
4. 输出最短航线的长度。
以下是算法的伪代码:function calculateShortestPath(w, v, s):AD = w / 2BD = (w / 2) / (v / s)shortestPath = sqrt(AD^2 + BD^2)return shortestPath示例假设河的宽度为 10 米,水流速度为 2 米/秒,行进速度为 5 米/秒。
根据上述算法,可以计算出最短航线的长度为:calculateShortestPath(10, 2, 5)输出结果为 11.18 米。
流水行船问题船在流水中航行的问题叫做行船问题。
行船问题是行程问题中比较特殊的类型,它除了具备行程问题中路程、速度和时间之间的基本数量关系,同时还涉及到水流的问题,因船在江、河里航行时,除了它本身的前进速度外,还会受到流水的顺推或逆阻。
除了行程问题中路程、速度和时间之间的基本数量关系在这里要反复用到外,行船问题还有几个基本公式要用到。
顺水速度=船速+水速逆水速度=船速-水速如果已知顺水速度和逆水速度,由和差问题的解题方法,我们可以求出船速和水速。
船速=(顺水速度+逆水速度)÷2水速=(顺水速度-逆水速度)÷2例1:船在静水中的速度为每小时13千米,水流的速度为每小时3千米,船从甲港顺流而下到达乙港用了15小时,从乙港返回甲港需要多少小时?【思路导航】根据条件,用船在静水中的速度+水速=顺水速度,知道了顺水速度和顺水时间,可以求出甲乙两港之间的路程。
因为返回时是逆水航行,用船在静水中的速度-水速=逆水速度,再用甲乙两港之间的全长除以逆水速度即可求出乙港返回甲港所需时间。
【思维链接】求乙港返回甲港所需要的时间,实际还是要用甲、乙两港的全程除以返回时的速度,也就是说路程、速度和时间三者关系很重要,只是速度上要注意是顺水速度还是逆水速度。
【举一反三】1、一只船在静水中每小时行12千米,在一段河中逆水航行4小时行了36千米。
这条河水流的速度是多少千米?2、一艘轮船在静水中航行,每小时行15千米,水流的速度为每小时3千米。
这艘轮船顺水航行270千米到达目的地,用了几个小时?如果按原航道返回,需要几小时?例2:一艘小船往返于一段长120千米的航道之间,上行时行了15小时,下行时行了12小时,求船在静水中航行的速度与水速各是多少?【思路导航】求船在静水中航行的速度是求船速,用路程除以上行的时间就是逆行速度,路程除以下行时间就是顺水速度。
顺水速度与逆水速度的和除以2就是船速,顺水速度与逆水速度的差除以2就是水速。
小船过河问题分析与题解【问题概说】(1)船的实际运动是水流的运动和船相对静水的运动的合运动。
(2)三种速度:船相对水的速度为v 船(即船在静水中的速度),水的流速为v 水(即水对地的速度),船的合速度为v (即船对地的速度,船的实际速度,其方向就是船的航向)。
(3)三种情景:①过河时间最短:当船头垂直河岸,渡河时间最短,且渡河时间与水的流速无关。
②过河路径最短:在v 船>v 水的条件下,当船的合速度垂直于河岸时,渡河位移(航程或路径)最小并等于河宽。
在v 船<v 水的条件下,当船头与船的合速度垂直时,渡河位移(航程或路径)最小。
此种情况下,合速度不可能垂直于河岸,无法垂直渡河。
最短航程确定如下:如图所示,以v 水矢量末端为圆心,以v 船矢量的大小为半径画弧,从v 水矢量的始端向圆弧作切线,则合速度沿此切线方向航程最短。
(下图中v 1表船速,v 2表水速)③最小渡河速度:水速和航向一定,船速垂直航向有最小船速。
【典型题例】两河岸平行,河宽d=100m ,水流速度v 1=3m/s ,求:(1)船在静水中的速度是4m/s 时,欲使船渡河时间最短,船应怎样渡河?最短时间是多少?船的位移是多大?(2)船在静水中的速度是6m/s 时,欲使船航行距离最短,船应怎样渡河?渡河时间多长?(3)船在静水中的速度为1.5m/s 时,欲使船渡河距离最短,船应怎样渡河?船的最小航程是多少?[思路分析](1)当船头垂直于河岸时,渡河时间最短:t min =d/v 2=100/4=25s 合速度v=s m v v /543222221=+=+ 船的位移大小s=v t min =125m(2)欲使船航行距离最短,需船头向上游转过一定角度使合速度方向垂直于河岸,设船的开行速度v 2与岸成θ角,则cosθ=216321==v v , 所以θ=600,合速度v=v 2sin600=3s m /3 t=s v d 93100=(3)船在静水中速度小于水流的速度,船头垂直于合速度v 时,渡河位移最小, 设船头与河岸夹角为β,如图所示: cosβ=2135.112==v v 所以β=600 最小位移s min =m d 20060cos 100cos 0==β [答案](1) 船头垂直于河岸时,渡河时间最短:t min =25s ,s =125m ;(2) 船头向上游转过一定角度, 与岸成600角航程最短,t=s 93100; (3) 船头垂直于合速度,船头与河岸夹角600时航程最短,s min =m 200。
流水行船问题一、知识要点当你逆风骑自行车时有什么感觉?是的,逆风时需用很大力气,因为面对的是迎面吹来的风。
当顺风时,借着风力,相对而言用里较少。
在你的生活中是否也遇到过类似的如流水行船问题。
解答这类题的要素有下列几点:水速、流速、划速、距离,解答这类题与和差问题相似。
划速相当于和差问题中的大数,水速相当于小数,顺流速相当于和数,逆流速相当于差速。
划速=(顺流船速+逆流船速)÷2;水速=(顺流船速—逆流船速)÷2;顺流船速=划速+水速;逆流船速=划速—水速;顺流船速=逆流船速+水速×2;逆流船速=逆流船速—水速×2。
二、精讲精练【例题1】一条轮船往返于A、B两地之间,由A地到B地是顺水航行,由B地到A地是逆水航行。
已知船在静水中的速度是每小时20千米,由A地到B地用了6小时,由B地到A 地所用的时间是由A地到B地所用时间的1.5倍,求水流速度。
在这个问题中,不论船是逆水航行,还是顺水航行,其行驶的路程相等,都等于A、B两地之间的路程;而船顺水航行时,其形式的速度为船在静水中的速度加上水流速度,而船在怒水航行时的行驶速度是船在静水中的速度与水流速度的差。
解:设水流速度为每小时x千米,则船由A地到B地行驶的路程为[(20+x)×6]千米,船由B地到A地行驶的路程为[(20—x)×6×1.5]千米。
列方程为(20+x)×6=(20—x)×6×1.5x=4答:水流速度为每小时4千米。
练习1:1、水流速度是每小时15千米。
现在有船顺水而行,8小时行320千米。
若逆水行320千米需几小时?2、水流速度每小时5千米。
现在有一船逆水在120千米的河中航行需6小时,顺水航行需几小时?3、一船从A 地顺流到B 地,航行速度是每小时32千米,水流速度是每小时4千米,212天可以到达。
次船从B 地返回到A 地需多少小时?【例题2】有一船行驶于120千米长的河中,逆行需10小时,顺行要6小时,求船速和水速。
Ⅱ.IntroductionRafting is an extreme sports with thrills, its purpose is to make every travelers to have a memorable experience. In the past few years, this movement is also popular in the world. British long river rafting. Next it is the situation of the British long river rafting.Visitors to the Big Long River (225 miles) can enjoy scenic views and exciting white water rapids. River trips all start at First Launch and exit the river at Final Exit, 225 miles downstream. Passengers take either oar- powered rubber rafts, which travel on average 4 mph or motorized boats, which travel on average 8 mph. The trips range from 6 to 18 nights of camping on the river, start to finish. Currently, X trips travel down the Big Long River each year during a six month period (the rest of the year it is too cold for river trips). There are Y camp sites on the Big Long River, distributed fairly uniformly throughout the river corridor. You must remember that no two sets of campers can occupy the same site at the same time. We should decide a reasonable number of camping area during the drift, as well as the development of suitable drifting to ensure that each traveler can feel the shortage vast field experience, as well as to determine the maximum carryingcapacity of the river vessels.Ⅲ.Problem AnalysisThe issues raised in this article is how to arrange the number of arrangements camping area, and to develop an appropriate drift way to ensure that each traveler can feel the shortage vast field experience, as well as to determine the maximum carrying capacity of the river vessels.Contrary to question one, considering the factors has been limited such as rafting period, rafting the ships not in the same time and place meet during the drifting. It can be abstracted to a simple encounter problems to solve it. Thus we can take advantage of the integer programming model to obtain the optimal number of the camping areas.Contrary to question two, according to the analysis of the question one, in accordance with the distance between camping areas and the number of camping areas, and the speed of the boat. We can divide into travel plans for one stop a day, two stops a day, and a day in three stations. According to the factors of the tourism ,such as weather conditions, fares, travel time, service management and so on. We can take full advantage of the AHP to establish the mathematical model to obtain the travel plans of the heavy weights of decision makers, and to seekappropriate travel plans used to calculate the maximum carryingcapacity.Contrary to question three, according to the problem of two,we can obtained the maximum carrying capacity of the river.And we use the known of knowledge again. We can reasonablyarrange to the different drifting way to meet the requirements ofthe castaways.Ⅳ.Symbol Descriptionis the distance between of the camping areas.tY is the number of the camping areas nearby the Big Longriver.w is the ratio between the first boat from a day to the i days isailing distance and lT is the deadline for the camp.Ⅴ. ModelsBasic Model5.1 Model 1Model one is solving the issue of the maximum number ofcamping trips, we use integer programming methods through thelimited conditions , we can list the constraint equations of themaximum number of the camping trips through the LINGOsoftware, you can get that when t = 6,7,9,11,13 days, and w =30,50,80,110,150 ,we can calculate the maximum ofk1. Through the analysis ,we can list the equation of the maximum number of boat trips.According to the ratio between the distance of the last sailing boat from the first day to the last day and t ∆. And it is nott w .but it can get the end place .we can list the followingequation.1i i w w +> (i=2,3,4……t-1)Due to the departure of the ship in the next day can ’t exceed the departure of the ship on the first day. we can list the following equation:1111(1,2,......1)ji j i i t w W j t =-=+>=-∑∑The last departure ship on the fist day must fall behind from the first departure ship in the next day , we can list the following equation:11i ti W Y ==+∑0i w =According to the above equation, we can use the MATLAB software. We can get the following results. When t = 6,7,9,11,13, w = k1 30,50,80,110,150 ,we can get the maximum of k1. It is in the following table:Table 15.2 Model 2:The best scheduling mode :The daily travel arrangements remain unchanged.According to the analysis of the problem ,we can take it as a cycle of the process,so it is easy to solve the problem. Thus we can come to the six months of the largest number of boat drifting period.(180)1h t k =-⋅And because the t is more bigger , the k1 is more smaller, to make the six months of drifting boat number within themaximum, it should be necessary to reduce the time of tourism.A day for the hair of the first ship speaking, the ship on the first day finally stay in position can through the formula and the:225l⋅=⨯=303044.7151So the first ship will stay in the camp which distance is 44.7 miles one day later, and most of the time on the river is ten hours, if drifters choose rubber raft will not be able to reach the specified camp, so in this situation ,except the last day the first ship should be adopt combination drift mode.Put on the first day of the ship after distance for x l⋅Tox<. Due to the ship's 40xl⋅<, work out the equation,26.8departure time interval can be very short, for ten hours negligible. So in a day of the fifth on the first day of the ship can use rubber raft to drift, and then start ships need particular case is particular analysis. To deal with the first day and the last day of the other time, but also need to adopt the dynamic way drift. If the daily travel arrangement is not the same.And to reduce travel days, if the ship line according to the original arrangement way, there will be a meeting, so need to do some changes can be achieved.Thus we are easy to find, every journey remain unchanged, the bearing capacity of the river be related to time, travel days,and the number of camping on the river, but the biggest capacity for camping, according to the number of every day on the first ship can achieve the longest camp to obtain the number of send ships in one day, and the next day journey all remain unchanged. Therefore, every day the same time the ship on the location of the camp exactly the same, when the first day of the last ship arrived at the last camping ground, at the t-1 day ,the final ship reach the final camping site, so the maximum carrying capacity of river is (1)w k t Y =-= .When t = 6,7,9,11,13 day, Y = 30,50,110,150 separately obtain the maximum carrying capacity of river, the table below :Table 25.3 Model 3:According to the meaning of problems ,we can analysis thatfrom start to finish the entire trip need 6-8 nights. During the trip ,the camping areas must set up more than eighteen. Because the average speed of the two drifting tools are 4 mph and 8mph. The total length of the big river is 225mp.We know that the total time of drifting are 56.25h and 28.125h.Thus We can infer that the tourists pass travel drifting place every day are 1 station , 2 station and 3 station.During the process of drifting , we will take into account weather conditions, fare, travel time, service quality impact on the number of the drifting places.We can get the association graph next:5.3.1 The solving of program layer and criteria layer solvingIn this travel,We can think that it is not suitable for thecontemporary consumer groups , the regardless of too many or too few people. Taking into account today's concept of time are higher than the service concept, Three stops in a day are slightly more popular than one stop in a day. We will design different fares, travel time, quality of service during the implementation of the program. First of all, We will analyze the influence between the program scale layer and fare. We can get the result of the comparison:Table 3According to analysis of the matrix ,We can get the largest eight value of the matrix is three with the help of matlab. We can know that the matrix is consistent matrix. We can treat the column vector ()1,5,2Tas its characteristic vector. Again usingmatlab normalization process to obtain the weight vector ()0.125,0.625,0.251T β=.Secondly, we will study the mutual influence of the programlayer to their travel time. Through the meaning of this problem,we know that different trip tools have different speed. It has a decisive impact on the number of the camping sites. So we can get the travel time t:()225118481i i t c p p ≤⋅⋅⋅+⋅- c is the number of stations on a day; when t=1,the travel tool is the rubber boat; when 0i p =,the travel tool is motor boat. Considering the boat depend on the people, the process of trip time should not be too long or people will be feel tired. But the camping trip will not very short. Thus, we will get the program matrix layer guidelines layer diurnal time. According to compare with each other, we can get the table:Table 4According to analysis of the matrix , We can get the largest eight value of the matrix is 3.0505 with the help of matlab. For the look-up table shows that three matrix RI = 0.58,we know thatmax 0.025,0.0440.11n CI CI CR RI n λ-== ==<-Table 5The positive and negative matrix analysis, can be obtained by matlab the matrix biggest characteristic root for 3, then know this matrix for consistent matrix, take its column vector ()2,1,6TAs its feature vector.Again, the matlab isnormalized the weight vector is obtained()30.22,0.11,0.67T β=5.3.2 The influence of guidelines layer and the target layer Due to the weather changed, we inquire that weather conditions are generally stable in the Mississippi. We only consider that the impact of criteria layer fare, travel time, quality of service to the target layer. Because of its proportion can’t decided, we will be based on each collection of information to compare. We will get the matrix as follows table:Table 6According to analysis of the matrix , We can get the largest eight value of the matrix is 3 with the help of matlab. We can accept inconsistencies of the matrix. We can treat the column vector ()1,7,2Tas its characteristic vector. Again using matlabnormalization process to obtain the weight vector()0.22,0.11,0.67TW =.5.3.3 Seeking the influence of program layer and the target layerFirst, we will get CI and CR. So we get the next formula:()1230.00275CI CI CI CI W =⋅=,,0.00470.1CI CR RI==<Examination shows that the target layer the combination matrix layer of the programs can be used as the basis for decision making. We will get the result:()()131223,,,,TC C C W βββ=⋅1230.1250.750.220.220.310.650.120.110.110.520.170.250.130.670.67C C C ⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪⎪⎪⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎛⎫ ⎪⋅= ⎪ ⎪⎝⎭=So we get the program layer in the share of the impact of the proportion of the target layer of decision-making()()()1:2:30.31:0.52:0.17n n n α==On the basis of this ratio, the proportion of traveling different camping places on a daily is ()()()1,2,3n n n .5.3.4 based on the actual situation analysis solutionAbove is the basic theoretical scheme, in the actual operation than the design more complicated, so in order to make the ships of less as far as possible to meet, but also to prevent the river congestion problems and the same camp stay days of problem arises, the model should consider when day line more camping site ships could not travel on the same day, and because the date line camping site many ships will catch up with the date line camping site less number of ships, this kind of situation will lead to two ship in the same place meet, if the day line stationnumber less in day line station number on the back of the camping site will lead to idle, in order to match the economy and demand synchronization, we need to arrange your time and make the entire river is full day line camp station number of the same ships, only by this way can we avoid meeting and conflicting problem, also makes the camping site no idle phenomenon to appear.Ⅵ.The strengths and weaknesses of the mode:6.1 The advantage of the model(1)In this model, we treat the complex reality into the ideal ofthe pursuit problem, we take advantage of chase ideal obtained optimization program.(2)In this model, we make use of the analytic hierarchy process,we analyze the proportion of each factor, we get the best allocation ratio.(3)In this model, we use the integer programming to solve theallocation problem. It is very easy to understand. We analyzed each problem with the actual solution.(4)In this model, not only we analyze the optimal solution of theidealized. We analyze the problem with actual situation.Finally, we arranged for the travel time list reasonable.6.2The disadvantage of the modelAlthough we combined the ideal model with the actual model each other. But the actual phenomenon is very complex, The results obtained must exit in the assuming range. If it exceeds this rang, we will make other decision.Promotion and optimization of the model:Due to the error caused by the ideal and the reality, During the solution of the model, the error is very difficult to avoid. Especially, when the weather is not very sunny and t he boats’ travel speed are unreasonable. Its will affect the implementation of the program. We can use the computer to simulate and fit the ideal and the reality repeatedly. Making sure the results more flexibility. Finally, we are able to obtain the results for each case.Ⅶ.ReferencesDaganzo, C.F., et al. 1997. Causes and effects of phase transitions in highwaytrafpc. ITS Research Report UCB-ITS-RR-97-8 (December 1997).Halliday, David, Robert Resnick, and Jearl Walker. 1993. Fundamentals ofPhysics. 4th ed. New York: Wiley.Yates, Daniel, David Moore, and George McCabe. 1999. ThePractice of Statistics.New York: W.H. Freeman.. 2000. Voluntary and Involuntary (“Bumped”) Denied Boarding(Jan-Sep2000)./library/stats/blrpt8.htm . Airlines overbooking project. DATE???./~brycw/classes/ date???361/overbook.htm .。