下面来介绍一下进行数学建模解决露天矿卡车调度问题的思路以及在MATLAB中实现的源码，具体如下：
为了便于问题的研究,对题目中的不确定因素做一些约定和假设：
(1)电铲在一个班次内不改变铲位,也就是说每台电在一个班次内只在一个铲位上工作。这主要是因为电铲的转移不方便并且电铲的转移需要占用时间，影响公司的效益。
(2)矿石漏和铁路倒装场只是卸矿石的不同地方，它们的开采对露天矿的经济效益无影响。同样,卸岩石的岩石漏和岩场的属性也不影响开采公司的经济效益。开采公司的经济效益主要与开采量与运输成本有关。
(3)卸点的品位是指在一个班次内在卸点内所卸总矿石铁的综合含量，并不要求任何部分矿石的铁含量达到品位限制要求。
(4)卡车每次运输都按线重量满载运输，并不考虑因颠簸而使岩石或矿石减少的情况另外，卡车运输始终以28km/h的平均速度行驶,发动和刹车所占用的时间忽略不计。
(5)在同一班次内，每辆卡车所走的路线是不定的，即卡车选择条路线是随机的。
下面来讲解一下建模思路：
1、用线性规划的方法求出从每个铲位到每个卸点所发的车次，从而求解出Go矩阵；
2、从Go矩阵判断所需出动的电铲的台数和铲位分配；
3、依据Go矩阵提供的信息，用线性规划方法求出由每个卸点返回到每个铲位的车次，从而给出了Return矩阵；
4、依据Go矩阵和Return矩阵，根据卡车的充分利用条件求出在一个班次内所需卡车的数量。
下面是MATLAB中的实现过程：
首先在MATLAB的主界面编辑器中写入下列代码：
c=[5.26 5.19 4.21 4 2.95 2.74 2.46 1.9 0.64 1.27 1.9 0.99 1.9 1.13 1.27 2.25 1.48 2.04 3.09 3.51 4.42 3.86 3.72 3.16 2.25 2.81 0.78 1.62 1.27 0.5 5.89 5.61 5.61 4.56 3.51 3.65 2.46 2.46 1.06 0.57 0.64 1.76 1.27 1.83 2.74 2.6 4.21 3.72 5.05 6.1];
A=[0.0154 0.0154 0.0154 0.0154 0.0154 0.0154 0.0154 0.0154 0.0154 0.0154 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -0.0154 -0.0154 -0.0154 -0.0154 -0.0154 -0.0154 -0.0154 -0.0154 -0.0154 -0.0154 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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];
b=[1.2154 -1.2 1.3154 -1.3 1.3154 -1.3 0 0 0 0 0 0 61.68831169 68.18181818 64.93506494 68.18181818 71.42857143 81.16883117 68.18181818 84.41558442 87.66233766 81.16883117 81.16883117 71.42857143 87.66233766 68.18181818 74.67532468 87.66233766 68.18181818 74.67532468 87.66233766 81.16883117 160 160 160 160 160 96 96 96 96 96 96 96 96 96 96]';
Aeq=[];
beq=[];
lb=zeros(1,50);
for i=1 :50
ub(i)=inf;
end
[x,z]=linprog(c,A,b,Aeq,beq,lb,ub)
然后命名保存至自定义路径下，点击运行，结果如下：
执行该程序求出一组解，但是该组解并非整数，所以用手工改动的方法对求出的结果进行优化处理，原则是对所得结果进行向上或向下取整，并在满足限制条件下使目标函数尽可能的大，这样便得到Go矩阵，请大家继续关注！！！