标签：培训 第一周 sum Len 数模 zeros result end Gen
易拉罐下料问题
问题: 某公司采用一套冲压设备生产一种罐装饮料的易拉罐,这种易拉罐是用镀锡板冲压制成的.易拉罐为圆柱形,包括罐身、上盖和下底,罐身高10 cm,上盖和下底的直径均为5 cm.该公司使用两种不同规格的镀锡板原料:规格1的镀锡板为正方形,边长24 cm;规格2的镀锡板为长方形,长、宽分别为32 cm和28 cm.由于生产设备和生产工艺的限制,对于规格Ⅰ的镀锡板原料,只可以按照图1中的模式1、模式⒉或模式3进行冲压;对于规格2的镀锡板原料只能按照模式4进行冲压.使用模式1模式⒉模式3、模式4进行每次冲压所需要的时间分别为1.5 s,2 s,1 s,3 s.
该工厂每周工作40 h,每周可供使用的规格1规格⒉的镀锡板原料分别为5万张和2万张.目前每只易拉罐的利润为0.10元,原料余料损失为0. 001元/cm3(如果周末有罐身、上盖或下底不能配套组装成易拉罐出售,也看做是原料余料损失).
求解&结果
model :
sets :
row/1..4/ : remain, x, stampingTime, spec1, spec2, body, bottom;
endsets
calc :
pi = @acos(-1); !pi值;
s1 = pi * 5 *10; !罐身面积;
s2 = pi * 5 ^ 2 / 4; !罐底或盖面积;
remain(1) = 24 * 24 - s1 - 10 * s2; !模式1余料损失;
remain(2) = 24 * 24 - 2 * s1 - 4 * s2; !模式2余料损失;
remain(3) = 24 * 24 - 16 * s2; !模式3余料损失;
remain(4) = 32 * 28 - 4 * s1 - 5 * s2; !模式4余料损失;
totalTime = 40 * 3600; !总时间;
endcalc
data :
stampingTime = 1.5, 2, 1, 3; !冲压时间;
spec1 = 1, 1, 1, 0; !规格1;
spec2 = 0, 0, 0, 1; !规格2;
totalSpec1 = 50000; !规格1总数;
totalSpec2 = 20000; !规格2总数;
body = 1, 2, 0, 4; !每个模式的罐身个数;
bottom = 10, 4, 16, 5; !每个模式的罐底或者盖子个数;
enddata
max = 0.1 * count - 0.001 * (@sum(row : remain * x) + s1 * L1 + s2 * L2);
@sum(row : stampingTime * x) <= totalTime;
@sum(row : spec1 * x) <= totalSpec1;
@sum(row : spec2 * x) <= totalSpec2;
count <= @sum(row : body * x);
2 * count <= @sum(row : bottom * x);
L1 = @sum(row : body * x) - count;
L2 = @sum(row : bottom * x) - 2 * count;
@for(row : @gin(x));
@gin(count); @gin(L1); @gin(L2);
end
Global optimal solution found.
Objective value: 4298.014
Objective bound: 4298.014
Variable Value Reduced Cost
COUNT 160250.0 -0.1000000
L1 0.000000 0.1570796
L2 0.000000 0.1963495E-01
X( 1) 0.000000 0.2225708
X( 2) 40125.00 0.1833009
X( 3) 3750.000 0.2618407
X( 4) 20000.00 0.1695067
天然肠衣问题
问题:天然肠衣制作加工是我国的一个传统产业。肠衣经过清洗整理后被分割成长度不等的小段(原料)，进入组装工序。传统的生产方式依靠人工，边丈量原料长度边心算，将原材料按指定根数和总长度组装出成品(捆)。原料按长度分档，通常以o.5米为一档，如: 14-14.4米按14米计算，14.5米-14.9米按14.5米计算，其余的依此类推。
所有模式
%%
clear, close, clc
tic
%%
l = [14 : 0.5 : 22.5, 23.5, 25.5]; % 20种长度
a = [35,29,30,42,28,42,45,49,50,64,52,63,49,35,27,16,12,2,6,1]; %20种原料的根数
cnt1 = floor(sum(a) / 4); % 由总根数求得最多捆数
cnt2 = floor(l * a' / 88.5); % 由总米数求得最多捆数
fprintf('最大捆数为：%5d\n', min(cnt1, cnt2))
n = size(l, 2);
aMin = zeros(1, n); % 初始化各种长度的类型肠衣的最多根数
for i = 1 : n
aMin(i) = min([5, floor(89.5 / l(i)), a(i)]); % 各种长度的类型肠衣的最多根数
end
count = 0;
res = zeros(1, 20);
result = zeros(1, 20);
for i1 = 0 : aMin(1)
result(1) = i1;
Len = result * [l(1 : 1), zeros(1,n - 1)]';
Gen = sum(result(1 : 1));
if Len > 89.5 || Gen > 5
break;
end
for i2 = 0 : aMin(2)
result(2) = i2;
Len = result * [l(1 : 2), zeros(1,n - 2)]';
Gen = sum(result(1 : 2));
if Len > 89.5 || Gen > 5
break;
end
for i3 = 0 : aMin(3)
result(3) = i3;
Len = result * [l(1 : 3), zeros(1,n - 3)]';
Gen = sum(result(1 : 3));
if Len > 89.5 || Gen > 5
break;
end
for i4 = 0 : aMin(4)
result(4) = i4;
Len = result * [l(1 : 4), zeros(1,n - 4)]';
Gen = sum(result(1 : 4));
if Len > 89.5 || Gen > 5
break;
end
for i5 = 0 : aMin(5)
result(5) = i5;
Len = result * [l(1 : 5), zeros(1,n - 5)]';
Gen = sum(result(1 : 5));
if Len > 89.5 || Gen > 5
break;
end
for i6 = 0 : aMin(6)
result(6) = i6;
Len = result * [l(1 : 6), zeros(1,n - 6)]';
Gen = sum(result(1 : 6));
if Len > 89.5 || Gen > 5
break;
end
for i7 = 0 : aMin(7)
result(7) = i7;
Len = result * [l(1 : 7), zeros(1,n - 7)]';
Gen = sum(result(1 : 7));
if Len > 89.5 || Gen > 5
break;
end
for i8 = 0 : aMin(8)
result(8) = i8;
Len = result * [l(1 : 8), zeros(1,n - 8)]';
Gen = sum(result(1 : 8));
if Len > 89.5 || Gen > 5
break;
end
for i9 = 0 : aMin(9)
result(9) = i9;
Len = result * [l(1 : 9), zeros(1,n - 9)]';
Gen = sum(result(1 : 9));
if Len > 89.5 || Gen > 5
break;
end
for i10 = 0 : aMin(10)
result(10) = i10;
Len = result * [l(1 : 10), zeros(1,n - 10)]';
Gen = sum(result(1 : 10));
if Len > 89.5 || Gen > 5
break;
end
for i11 = 0 : aMin(11)
result(11) = i11;
Len = result * [l(1 : 11), zeros(1,n - 11)]';
Gen = sum(result(1 : 11));
if Len > 89.5 || Gen > 5
break;
end
for i12 = 0 : aMin(12)
result(12) = i12;
Len = result * [l(1 : 12), zeros(1,n - 12)]';
Gen = sum(result(1 : 12));
if Len > 89.5 || Gen > 5
break;
end
for i13 = 0 : aMin(13)
result(13) = i13;
Len = result * [l(1 : 13), zeros(1,n - 13)]';
Gen = sum(result(1 : 13));
if Len > 89.5 || Gen > 5
break;
end
for i14 = 0 : aMin(14)
result(14) = i14;
Len = result * [l(1 : 14), zeros(1,n - 14)]';
Gen = sum(result(1 : 14));
if Len > 89.5 || Gen > 5
break;
end
for i15 = 0 : aMin(15)
result(15) = i15;
Len = result * [l(1 : 15), zeros(1,n - 15)]';
Gen = sum(result(1 : 15));
if Len > 89.5 || Gen > 5
break;
end
for i16 = 0 : aMin(16)
result(16) = i16;
Len = result * [l(1 : 16), zeros(1,n - 16)]';
Gen = sum(result(1 : 16));
if Len > 89.5 || Gen > 5
break;
end
for i17 = 0 : aMin(17)
result(17) = i17;
Len = result * [l(1 : 17), zeros(1,n - 17)]';
Gen = sum(result(1 : 17));
if Len > 89.5 || Gen > 5
break;
end
for i18 = 0 : aMin(18)
result(18) = i18;
Len = result * [l(1 : 18), zeros(1,n - 18)]';
Gen = sum(result(1 : 18));
if Len > 89.5 || Gen > 5
break;
end
for i19 = 0 : aMin(19)
result(19) = i19;
Len = result * [l(1 : 19), zeros(1,n - 19)]';
Gen = sum(result(1 : 19));
if Len > 89.5 || Gen > 5
break;
end
for i20 = 0 : aMin(20)
result(20) = i20;
Len = result * [l([1 : 20]), zeros(1,n - 20)]';
Gen = sum(result);
if Len >= 88.5 && Len <= 89.5 && Gen >= 4 && Gen <= 5
count = count + 1;
res(count, :) = result;
if i20 == 1
result(20) = 0;
break;
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
end
%%
toc
%% 文件写入
fid = fopen('changyi.txt', 'w');
for i = 1 : size(res, 1);
for j = 1 : 20
fprintf(fid, "%3d", res(i, j));
end
fprintf(fid, "\n");
end
fclose(fid);
结果
所有符合的模式1.txt
求解最优模式
Model :
Sets :
row/1..2783/ : x, y;
col/1..20/ : l, a;
link(row, col) : b;
Endsets
Data :
l=14,14.5,15,15.5,16,16.5,17,17.5,18,18.5,19,19.5,20,20.5,21,21.5,22,22.5,23.5,25.5;
a=35,29,30,42,28,42,45,49,50,64,52,63,49,35,27,16,12,2,6,1;
b = @file('changyi1.txt');
@text('30.txt') = @writefor(row(i) | x(i) #gt# 0 : @name(x), ' ', x(i), @newline(1));
Enddata
Min = z3;
137 = @sum(row : x); !目标函数1;
3 = x(1) + x(2) + x(3); !目标函数2;
z3 = @sum(row : y); !目标函数3;
@for(col(j) : @sum(row(i) : x(i) * b(i, j)) <= a(j)); !原料约束;
@for(row : x <= 10000 * y);
@for(row : x >= y);
@for(row : @gin(x));
@for(row : @bin(y));
End
Feasible solution found.
Objective value: 14.00000
Objective bound: 8.000000
Infeasibilities: 0.000000
Extended solver steps: 126503
Total solver iterations: 4318795
打印
%%
clear, close, clc
%% 读入数据
load 1.txt
totalModel = X1;
[~, modelRow, ~, modelNum] = textread('10.txt','%s %d %s %d');
x = [modelRow, modelNum];
%%
l = [14 : 0.5 : 22.5, 23.5, 25.5]; % 20种长度
a = [35,29,30,42,28,42,45,49,50,64,52,63,49,35,27,16,12,2,6,1]; %20种原料的根数
totalGen = 0; % 总根数
totalKun = 0; % 总捆数
totalLen = 0; % 总长度
for i = 1 : length(modelRow)
preLen = 0; % 当前长度
preGen = 0; % 当前根数
i1 = modelRow(i);
for k = 1 : length(l)
preLen = preLen + totalModel(i1, k) * l(k);
preGen = preGen + totalModel(i1, k);
end
fprintf("第%02d行，模式%04d为：", i, x(i, 1));
for j = 1 : length(l)
if totalModel(i1, j) > 0
fprintf('(%3.1f米, %1d根 )', l(j), totalModel(i1, j))
end
end
fprintf('长度%3.1f 米，%1d根，%2d捆 \n', preLen, preGen, x(i, 2))
totalLen = totalLen + preLen * x(i, 2);
totalKun = totalKun + x(i, 2);
totalGen = totalGen + preGen * x(i, 2);
end
fprintf('总长度%6.1f， 总根数%6d， 总捆数%5d\n',totalLen, totalGen, totalKun);
结果
如有错误以及可以改进的地方欢迎在下方评论区留言！
标签：培训,第一周,sum,Len,数模,zeros,result,end,Gen 来源： https://blog.csdn.net/zy440458/article/details/118633185
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