一、背景

为进一步实现了矿山的高效开采，保证采矿作业的安全，降低充填成本是某金矿提高经济效益的重要途径。矿山开采二步骤对采场充填体强度的要求为7d达到0.3 MPa，28d达到0.5 MPa，56d达到0.8 MPa。其中7d、28d、56d的单轴抗压强度分别为：

7d单轴抗压强度：

Y1=-29.32637+0.858937X1+0.204500X2+0.046460X3-0.002600X1X2-0.000620X1X3-0.000784X2X3-0.006344X12-0.000545X22+0.001053X32（Adjusted R2= 0.9844）

28d单轴抗压强度：

Y2=-18.18050+0.665938X1+0.106150X2-0.454540X3-0.002075X1X2+0.005520X1X3+0.000552X2X3-0.005563X12-(3.53492E-17) X22+ 0.004192X32（Adjusted R2= 0.9545）

56d单轴抗压强度：

Y3=-35.55125+1.07144X1+0.252775X2-0.161840X3-0.003400X1X2 + 0.002780X1X3-0.001112X2X3-0.008125X12-0.000440X22+0.002259X32（Adjusted R2=0.9801）

采用多目标函数寻优方法优化料浆配比，以单位体积充填成本最低为优化目标即 min f，各材料充填成本为：矿山中和渣由全尾砂与酸性废水制作而成，加工费为1.2元/t；废石破碎加工价格为17元/t；水泥的价格为380元/t；工业用水的价格为1.6元/t。

现根据料浆质量分数(X1)、废石掺量(X2)和灰砂比(X3) 分别计算出1 m3充填料浆中和渣的质量Ma、废石质量Ms、水泥质量Mj和水的质量Mw，以充填体强度为约束条件，即Y1≥0.3 MPa；Y2≥0.5 MPa；Y3≥0.8 MPa，建立优化模型下式所示：

二、C++代码

#include<iostream>

using namespace std;

void linspace(double start,double end,int n,float* y){
	
	double interval = (end-start)/n;
	
	for(int i = 0;i < n ;i++){
		y[i] = start + interval * i;
	}
}

int main(void){
	int length = 100;
	float ma[length];
	float ms[length];
	float mj[length];
	float mw[length];
	
	float X1,X2,X3,y1,y2,y3,result;
	
	linspace(0.001,2.418,length,ma);
	linspace(0.001,2.67,length,ms);
	linspace(0.001,3,length,mj);
	linspace(0.001,1,length,mw);
	
	long long count = 0;
	int solveCount = 0;
	for(int i = 0;i < length;i++){
		for(int j = 0;j < length;j++){
			for(int k = 0;k < length;k++){
				for(int l = 0;l < length;l++){
					count++;
					if(ma[i]/2.418 + ms[j]/2.67 + mj[k]/3 + mw[l] - 1 < 0.01 && ma[i]/2.418 + ms[j]/2.67 + mj[k]/3 + mw[l] - 1 > -0.01){
//						cout << ma[i] << " " << ms[j] << " " << mj[k] << " " << mw[l] << endl;
						X1 = (ma[i]+ms[j]+mj[k])/(ma[i]+ms[j]+mj[k]+mw[l])*100;
	                    X2 = ms[j]/(ma[i]+ms[j])*100;
	                    X3 = mj[k]/(ma[i]+ms[j])*100;
//	                    cout << X1 << " " << X2 << " " << X3 << endl;
						if(X1 <= 60 && X1 >= 56 && X2 >=30 &&  X2 <= 40 && X3 >= 12.5 && X3 <= 25){
//						if(true){
							y1 = -29.32637+0.858937*X1+0.204500*X2+0.046460*X3-0.002600*X1*X2-0.000620*X1*X3-0.000784*X2*X3-0.006344*X1*X1-0.000545*X2*X2+0.001053*X3*X3;
	//						cout << count << " " << X1 << " " << X2 << " " << X3 << " " << y1 << endl;
							if(y1 >= 0.3){
								y2 = -18.18050+0.665938*X1+0.106150*X2-0.454540*X3-0.002075*X1*X2+0.005520*X1*X3+0.000552*X2*X3-0.005563*X1*X1-(3.53492E-17)*X2*X2+ 0.004192*X3*X3;
								if(y2 >= 0.5){
									y3=-35.55125+1.07144*X1+0.252775*X2-0.161840*X3-0.003400*X1*X2 + 0.002780*X1*X3-0.001112*X2*X3-0.008125*X1*X1-0.000440*X2*X2+0.002259*X3*X3;
									if(y3 >= 0.8){
										result = 1.2*ma[i] + 17*ms[j] + 380*mj[k] + 1.6*mw[l];
										cout << count << " " <<ma[i] << " " << ms[j] << " " << mj[k] << " " << mw[l] << " " << X1 << " " << X2 << " " << X3 << " " <<  result << endl;

									}
								}
							}	
						}    
					}
				}
			}
		}
	}
		return 0;
}

暴力求解:代码采用最简单的暴力求解方法，即将待求值中和渣的质量Ma、废石质量Ms、水泥质量Mj和水的质量Mw采用linspace进行创建等差数列（组合密度取决于变量length），采用循环遍历每一种组合，并判断每一种组合是否满足模型的各个优化条件，若满足则输出。

三、结果

count	Ma	Ms	Mj	Mw	X1	X2	X3	f
19100665	0.46023	0.2679	0.18094	0.64036	58.6713	36.7929	24.85	74.8884
19100666	0.46023	0.2679	0.18094	0.65035	58.2954	36.7929	24.85	74.9043
19110565	0.46023	0.29459	0.15095	0.64036	58.583	39.0278	19.9981	63.9459
19110664	0.46023	0.29459	0.18094	0.63037	59.7498	39.0278	23.9713	75.3261
19110665	0.46023	0.29459	0.18094	0.64036	59.3711	39.0278	23.9713	75.3421
20090665	0.4844	0.24121	0.18094	0.64036	58.6039	33.2424	24.9363	74.4636
20090666	0.4844	0.24121	0.18094	0.65035	58.2279	33.2424	24.9363	74.4796
20100565	0.4844	0.2679	0.15095	0.64036	58.5154	35.6108	20.0651	63.5212
20100664	0.4844	0.2679	0.18094	0.63037	59.685	35.6108	24.0516	74.9014
20100665	0.4844	0.2679	0.18094	0.64036	59.306	35.6108	24.0516	74.9174
20110564	0.4844	0.29459	0.15095	0.63037	59.5997	37.8169	19.3777	63.9589
20110565	0.4844	0.29459	0.15095	0.64036	59.2205	37.8169	19.3777	63.9749
21090565	0.50857	0.24121	0.15095	0.64036	58.4476	32.1708	20.1326	63.0964
21090664	0.50857	0.24121	0.18094	0.63037	59.6199	32.1708	24.1324	74.4766
21090665	0.50857	0.24121	0.18094	0.64036	59.2408	32.1708	24.1324	74.4926
21100564	0.50857	0.2679	0.15095	0.63037	59.5343	34.5023	19.4405	63.5342
21100565	0.50857	0.2679	0.15095	0.64036	59.155	34.5023	19.4405	63.5502
22090564	0.53274	0.24121	0.15095	0.63037	59.4688	31.1661	19.5038	63.1095
22090565	0.53274	0.24121	0.15095	0.64036	59.0892	31.1661	19.5038	63.1254

多目标条件下矿山充填材料配比优化C++实现

一、背景

二、C++代码

三、结果

猜你喜欢