请懂Lingo的人帮忙解释一下这段代码的意思,min=@sum(zijin:y*a)+@sum(yaoqiu(i):d(i)*0.5*@min(link(i,j):y(i)*k(i,j))); @for(zijin:@bin(y));sets:zijin/1..49/:y,a;yaoqiu/1..49/:d;link(build,require):k;endsetslink(zijin,yaoqiu):
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/05 17:39:07
![请懂Lingo的人帮忙解释一下这段代码的意思,min=@sum(zijin:y*a)+@sum(yaoqiu(i):d(i)*0.5*@min(link(i,j):y(i)*k(i,j))); @for(zijin:@bin(y));sets:zijin/1..49/:y,a;yaoqiu/1..49/:d;link(build,require):k;endsetslink(zijin,yaoqiu):](/uploads/image/z/10397706-42-6.jpg?t=%E8%AF%B7%E6%87%82Lingo%E7%9A%84%E4%BA%BA%E5%B8%AE%E5%BF%99%E8%A7%A3%E9%87%8A%E4%B8%80%E4%B8%8B%E8%BF%99%E6%AE%B5%E4%BB%A3%E7%A0%81%E7%9A%84%E6%84%8F%E6%80%9D%2Cmin%3D%40sum%28zijin%3Ay%2Aa%29%2B%40sum%28yaoqiu%28i%29%3Ad%28i%29%2A0.5%2A%40min%28link%28i%2Cj%29%3Ay%28i%29%2Ak%28i%2Cj%29%29%29%3B+%40for%28zijin%3A%40bin%28y%29%29%3Bsets%3Azijin%2F1..49%2F%3Ay%2Ca%3Byaoqiu%2F1..49%2F%3Ad%3Blink%28build%2Crequire%29%3Ak%3Bendsetslink%28zijin%2Cyaoqiu%29%3A)
请懂Lingo的人帮忙解释一下这段代码的意思,min=@sum(zijin:y*a)+@sum(yaoqiu(i):d(i)*0.5*@min(link(i,j):y(i)*k(i,j))); @for(zijin:@bin(y));sets:zijin/1..49/:y,a;yaoqiu/1..49/:d;link(build,require):k;endsetslink(zijin,yaoqiu):
请懂Lingo的人帮忙解释一下这段代码的意思,
min=@sum(zijin:y*a)+@sum(yaoqiu(i):d(i)*0.5*@min(link(i,j):y(i)*k(i,j))); @for(zijin:@bin(y));
sets:
zijin/1..49/:y,a;
yaoqiu/1..49/:d;
link(build,require):k;
endsets
link(zijin,yaoqiu):
请懂Lingo的人帮忙解释一下这段代码的意思,min=@sum(zijin:y*a)+@sum(yaoqiu(i):d(i)*0.5*@min(link(i,j):y(i)*k(i,j))); @for(zijin:@bin(y));sets:zijin/1..49/:y,a;yaoqiu/1..49/:d;link(build,require):k;endsetslink(zijin,yaoqiu):
第一句是目标函数 求最小值
有两部分 第一部分对zijin集里面的 y*a 求和 第二部分对yaoqiu集求和 求和的是d(i)*0.5*@min(link(i,j):y(i)*k(i,j))
其中@min(link(i,j):y(i)*k(i,j))写的有问题 最好应该是写@min(zijin(j):y(i)*k(i,j)) 是指对所有的y(i)*k(i,j)取最小值
第二句就是所有y为0-1变量