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露天矿生产的车辆安排(CMCM2003)B摘要本文要解决的问题主要围绕减少实际中的经济成本而展开的。在经济运作中,减少成本是作为露天矿生产带来利润最大化的有效手段之一。 而合理安排有效路线和车次,成为了解决减少成本问题的关键。 鉴于铲点到卸点线路的复杂性,我们把问题分成两个层次加以解决。首先我们采用了整体规划的算法,建立了数学模型以求得最小运输量。其基本思想是提取重要的约束性条件,对于总产量达最小的目标函数进行约束,运用lingo程序求出其最优解,最后得出最小运输量为85628.62吨,且第5、6、7个铲点没有使用。对于层次二,通过解决层次一所得出具体流量计算卡车在各个路线上一个班次最多可以运行的次数及各路线上需要的卡车数,从而得出所有路线要出动13辆卡车.问题重述此题类似与产地与销地的运输的整数规划问题。 10个矿位,5个卸点,运输矿石和岩石两种产品。此题的重点在于限制条件的提取,由于题中所给条件较多如:每个铲位至多能安置一台电铲,电铲的平均装车时间为5分钟。卸货地有卸矿石的矿石漏、2个铁路倒装场和卸岩石的岩石漏、岩场等。每个卸点都有各自的产量要求,岩石卸点的品味限制都为29.5%1%电铲和卸点都不能同时为两辆及两辆以上卡车服务。卡车每次都是满载运输,且在一个班次中不存在卡车等待情况,卡车只在开始时点火一次。卡车的平均卸车时间为3分钟。一个班次为8小时等等。总之,限制条件比较隐晦,需要从题中认真挖掘。本文需要解决的问题是如何设计一个班次使得总运量(吨公里)最小,同时出动最少的卡车,从而运输成本最小。一个班次的生产计划应该包含以下内容:出动几台电铲,分别在哪些铲位上;出动几辆卡车,分别在哪些路线上各运输多少次。因此问题可转化为:1、如何在满足所有限制条件的条件下,使得一个班次内所有卡车的总运量最小。2在总运量最小的情况下,如何设计卡车路线,使得出动的卡车数最少,从而使总成本最低。模型假设.卡车每次都是满载,且不出现堵车情况。.发动机点火时需要消耗相当多的电瓶能量,故每个班次每台车只在开时点火一次。.不存在卡车等待时间.在铲位或卸点处因两条路线(及以上)造成的冲突时,只要平均时间能完成任务即可,不进行排时讨论符号约定Xij:从i号铲位到j号卸点的石料运量,单位:车•次;Cij:从i号铲位到j号卸点的距离,单位:公里;Tij:在i号铲位到j号卸点路线上运行一个周期平均所需时间,单位:分;Aij:从i号铲位到j号卸点最多能同时运行的卡车数,单位:辆;Bij:从i号铲位到j号卸点,一辆车一个班次中最多可以运行次数,单位:次;pi:i号铲位的矿石铁含量qj:j号卸点任务需求cki:i号铲位的铁矿石储量,单位:万吨;cyi:i号铲位的岩石储量,单位:万吨;fi:描述第i号铲位是否使用的0-1变量,不使用的铲位为0,使用的为1;chej:从i号铲位到j号卸点所需要的车的最小辆数(实数)数值上等于xj/Bj模型建立与求解问题一:总运量最小运输路线的设计目标函数:由题意易得目标函数为10 5min 154xjcij1j1(1)道路能力约束:一个电铲(卸点)不能同时为两辆卡车服务,一条路线上最多能同时运行的卡车数是有限制的。卡车从 i号铲位到j号卸点运行一个周期平均所需时间为 (分钟)。由于装车时间5分钟大于卸车时间3分钟,所以这条路线上在卡车不等待条件下最多能同时运行的卡车数为: ;其中最后开始发车的一辆卡车一个班次中在这条路线上最多可以运行的次数为 (其他卡车这里可能比此数多1次)这里是开始装车时最后一辆车的延时时间。个班次中这条固定路线上最多可能运行的总车次大约为:(2)电铲能力约束:一台电铲不能同时为两辆卡车服务,所以一台电铲在一个班次中的最大可能产量为8X60/5X154(吨)。(3)卸点能力约束:卸点的最大吞吐量为每小时 60/3=20车次,于是一个卸点在一个班次中的最大可能产量为8X20X154(吨)。(4)铲位储量约束:铲位的矿石和岩石产量都不能超过相应的储藏量。(5)产量任务约束:各卸点的产量不小于该卸点的任务要求。(6)铁含量约束:各矿石卸点的平均品位要求都在指定的范围内。(7)电铲数量约束:电铲数量约束无法用普通不等式表达,可以引入 10个0-1变量来标志各个铲位是否有产量。(8)整数约束:当把问题作为整数规划模型时,流量 Xij除以154为非负整(9)卡车数量约束:不超过20辆

得到的一种模型为10 5(0)min 154XjCj(0)i1j1s.t.Xj Aj Bij,i1,...,10,j1,...,55Xs.t.Xj Aj Bij,i1,...,10,j1,...,55Xjj1fi860/5,i1,,1010xiji1820,j1, ,5154x154xi1xi2xi5 cki10000154 xi3xi4 cy10000,i1,,10154 xi3xi4 cy10000,i1,,1010xiji1qj/154,j1,,510154 x10154 xij pi30.5 0i15 ,j1,2,5 (6)154 xij pi28.5 0j1xij为整数(8)xij(8)xiji」B20求解模型由于限制条件较多,需要用快速解法利用LINGO程序对以上模型求解sets:cai/1..10/:p,cy,ck,f;xie/1..5/:q;link(cai,xie):a,b,c,t,x,che;endsetsdata:v=28;p=30282932313332313331;q=1.21.31.31.91.3;c=5.26001.90005.89000.64004.42005.19000.99005.61001.76003.86004.21001.90005.61001.27003.72004.00001.13004.56001.83003.16002.95001.27003.51002.74002.25002.74002.25003.65002.60002.81002.46001.48002.46004.21000.78001.90002.04002.46003.72001.62000.64003.09001.06005.05001.27001.27003.51000.57006.10000.5000;cy=1.251.101.351.051.151.351.051.151.351.25;ck=0.951.051.001.051.101.251.051.301.351.25;enddata@for(link:t=120*c/v+8;a=@floor(t/5);b=@floor((485-5*a)/t));min=@sum(link:x*154*c);@for(link:x<=a*b);@for(cai(i):@sum(xie(j):x(i,j))<=f(i)*96);@for(xie(j):@sum(cai(i):x(i,j))<=160);@for(cai(i):x(i,1)+x(i,2)+x(i,5)<=ck(i)*10000/154);@for(cai(i):x(i,3)+x(i,4)<=cy(i)*10000/154);@for(xie(j):@sum(cai(i):x(i,j))>=q(j)*10000/154);@sum(cai(i):x(i,1)*(p(i)-30.5))<=0;@sum(cai(i):x(i,2)*(p(i)-30.5))<=0;@sum(cai(i):x(i,5)*(p(i)-30.5))<=0;@sum(cai(i):x(i,1)*(p(i)-28.5))>=0;@sum(cai(i):x(i,2)*(p(i)-28.5))>=0;@sum(link:x/b)<=20;@sum(cai:f)<=7;@for(link:@gin(x));@for(cai:@bin(f));@for(link:che=x/b);end运行结果最小运输量为85628.62吨且第5、6、7个铲点没有使用问题二:在总运量最小的情况下,如何设计路线,使得卡车的使用辆最小由问题一模型求解结果的以下数据各个路线上的最佳运输车次:xj铲位1铲位2铲位3铲位4铲位5铲位6铲位7铲位8铲位9铲位10矿石漏135411倒装场I4243岩场7015岩石漏8143倒装场n13270各路线上需要的卡车数(实数):chej铲位1铲位2铲位3铲位4铲位5铲位6铲位7铲位8铲位9铲位10矿石漏0.8671.8620.314倒装场I1.0771.162岩场1.8920.326岩石漏1.8411.229倒装场n0.6840.11.489所有路线所需卡车数(实数)的和为12.843o即要最少出动13辆卡车易得:有7辆卡车在一个班次内在固定路线上运输,它们的分布如下铲位1铲位2铲位3铲位4铲位5铲位6铲位7铲位8铲位9铲位10矿石漏1倒装场I11岩场1岩石漏11倒装场II1另外,有6量车需要联合运输来满足总运输量最小。联合运输因尽量以 V型路线为主,避免Z型路线,过程比较复杂,需要用到组合数等理论,在此就暂不求解。模型评价本文根据题中所给条件,建立了整数规划模型,由于题中的限制条件较多,运用LINGO软件快速求解,减少了运算量,方便快捷。优点:.此模型应运面较广,使用类似的产销,集中产品的生产等线性规划问题都可适用。、用LINGO软件求解,方便快捷缺点:. 模型建立不太完善。、没有求出联合派车的具体路线。模型改进和应用此模型在现实生活中比较常见,类似的生产规划问题都能改进此模型,快速求出较优生产计划。附录LINGO程序运行结果:604Globaloptimalsolutionfoundatiteration:

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