1、乘用車物流運輸計劃問題介于此題與運籌學知識相關(guān)，我們初步運用了線性規(guī)劃和遺傳算法對此做出了分析，遺傳 HYPERLINK /lemma/ShowInnerLink.htm?lemmaId=105662 t _blank 算法（Genetic HYPERLINK /lemma/ShowInnerLink.htm?lemmaId=105662 t _blank Algorithm）是一類借鑒 HYPERLINK /lemma/ShowInnerLink.htm?lemmaId=2107883&ss_c=ssc.citiao.link 生物界的進化規(guī)律（ HYPERLINK /lemma/ShowI

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4、自適應(yīng)地調(diào)整搜索方向，不需要確定的規(guī)則。計算程序運用了MATLAB?？紤]到成本問題，影響成本高低的首先是轎運車使用數(shù)量；其次，在轎運車使用數(shù)量相同情況下，1-1型轎運車的使用成本較低，2-2型較高，1-2型略低于前兩者的平均值，但物流公司1-2型轎運車擁有1-2型轎運車使用量不超過1-1型轎運車使用量的20%；再次，在轎運車使用數(shù)量及型號均相同情況下，行駛里程短的成本低，注意因為該物流公司是全國性公司，在各地均會有整車物流業(yè)務(wù)，所以轎運車到達目的地后原地待命，無須放空返回。最后每次卸車成本幾乎可以忽略。盡量先把1-2型的裝滿 然后根據(jù)題目說的盡量不留位置所以都是選留下空位較少的那個方案，裝完1

5、-2然后再來考慮1-1，剩下的裝在1-1，考慮放完的，都是先同樣車型放一起。最后在混放如果有剩余的話?？紤]到第四問中涉及到路徑最優(yōu)問題，我們選用最短路徑算法解決這類問題。 flag = Passenger_num=0;temp Car_carriers(2,1) - (flag(1)*A_2_2(:,1)*Passenger(1,1)+flag(2)*A_2_2(:,2)*Passenger(2,1). +flag(3)*A_2_2(:,3)*Passenger(3,1)+0.1*sum(A_2_2,2);,ib = min(temp);temp_res = A_2_2(ib,:);resul

6、t = ;for i = 1:50 loop_end = round(i/0.2); temp = ; num = 0; for j = 0:100 for k = 0:100 if j+k=loop_end con_1 = 3*temp_res(1)*i+2*restrict(1,1)*j; con_2 = 3*temp_res(2)*i+2*restrict(2,1)*k; if con_1=Passenger_num(1) & con_2=Passenger_num(2) num = num+1; temp(num,:) = i j k; break; end end end end i

7、f size(temp,1)0 result = result;temp(1,:); clear temp; end endres(:,1) = result(:,1);res(:,2) = result(:,2)+result(:,3);final = ;final(1,:) = res(1,:); for i = 2:size(res,1) temp = sum(res(i,:)-sum(res(i-1,:); if (res(i,1)res(i-1,1) & res(i,2)res(i-1,2) | temp=0 continue; end final = final;res(i,:);

8、end? Undefined function or variable Passenger_num. temp_res = 5 4;12 5;result = ;for i = 1:50 loop_end = round(i/0.2); temp = ; num = 0; for j = loop_end:100 con_1 = temp_res(2,1)*i+temp_res(1,1)*j; con_2 = temp_res(2,2)*i+temp_res(1,2)*k; if con_1=Passenger_num(2) & con_2=Passenger_num(3) num = num

9、+1; temp(num,:) = i j; break; end end if size(temp,1)0 result = result;temp(1,:); clear temp; endendres = ;res(:,1) = result(:,1);res(:,2) = result(:,2);final = ;final(1,:) = res(1,:);for i = 2:size(res,1) temp = sum(res(i,:)-sum(res(i-1,:); if (res(i,1)res(i-1,1) & res(i,2)res(i-1,2) | temp=0 conti

10、nue; end final = final;res(i,:);end? Undefined function or variable k. temp_res = 5 1 2;5 12 1;result = ;for i = 1:50 % 剩余多少車 remain_1 = Passenger_num(1)-temp_res(2,1)*i; remain_2 = Passenger_num(2)-temp_res(2,2)*i; remain_3 = Passenger_num(3)-temp_res(2,3)*i; % 先滿足3型車 res_1 = floor(remain_3/restric

11、t(3,1); % 盡可能多放1-1上層只放一種 % % 只放第一種 remain_3_1 = remain_3-restrict(3,1)*res_1; remain_2_1 = remain_2-restrict(2,1)*res_1; res_2 = floor(remain_1/restrict(1,1)/2); remain_1_1 = remain_1-2*restrict(1,1)*res_2; res_3 = 1; temp = find(A_1_2(:,3)=remain_3_1); for j = 1:size(temp,1) temp_res_1 = A_1_2(temp

12、(j),:); remain_2_1 = remain_2_1-temp_res_1(2); remain_1_1 = remain_1_1-temp_res_1(1); flag = remain_1_1=A_1(:,1) & remain_2_1res(i-1,1) & res(i,2)res(i-1,2) | temp=0 continue; end final = final;res(i,:) i;end% 1-1和1-2數(shù)量限制temp = final(:,2)*0.2=final(:,1);final_1.final = final(temp,:);final_1.plan = r

13、esult(final(temp,3),:);final = final_1;? Undefined function or method Passenger_num for input arguments of type double. function final = cc(Passenger_num)% 請輸入運載矩陣% 裝滿的所有情況（加入限制）%1-1 的第一層所有的情況% Passenger_num = 100 68 0;xianzhi;% 把最少的情況求出來if Passenger_num(3) = 0 case_1;elseif Passenger_num(1) = 0 cas

14、e_2;else case_3;endend? function final = cc(Passenger_num) |Error: Function definitions are not permitted in this context. cc? Undefined function or variable cc. % % 車型的矩陣Passenger = 4.61 1.7 1.51;3.615 1.605 1.394;4.63 1.785 1.77;% 轎運車的矩陣Car_carriers = 19 2.7 2.7;24.3 3.5 2.7;% 轎運車最多可以放車型個數(shù)（緊長度限制）r

15、estrict = 4 5;5 6;4 5;% 裝滿的所有情況（加入限制）%1-1 的第一層所有的情況A_1 = ;num = 0;%for i = 0:restrict(1,1) num = num+1; temp_sum = 0; j = 0; while j=restrict(2,1) & temp_sumCar_carriers(1,1) temp = i j; j = j+1; temp_sum = i*Passenger(1,1)+j*Passenger(2,1)+0.1*(i+j-1); end A_1(num,:) = temp; end%1-1 的第二層所有的情況A_1_2

16、= ;num = 0;for i = 0:restrict(1,1) for j = 0:restrict(2,1) num = num+1; k = 0; temp_sum = 0; temp_sum = i*Passenger(1,1)+j*Passenger(2,1)+k*Passenger(3,1)+0.1*(i+j+k-1); while k=restrict(3,1) & temp_sumCar_carriers(1,1) temp = i j k; k = k+1; temp_sum = i*Passenger(1,1)+j*Passenger(2,1)+k*Passenger(

17、3,1)+0.1*(i+j+k-1); end A_1_2(num,:) = temp; end endA_1_2 = unique(A_1_2,rows);% 裝滿的所有情況（加入限制）%1-2 的第一層所有的情況A_2 = ;num = 0;%for i = 0:restrict(1,2) num = num+1; temp_sum = 0; j = 0; while j=restrict(2,2) & temp_sumCar_carriers(2,1) temp = i j; j = j+1; temp_sum = i*Passenger(1,1)+j*Passenger(2,1)+0.

18、1*(i+j-1); end A_2(num,:) = temp; end%1-2 的第二層所有的情況A_2_2 = ;num = 0;for i = 0:restrict(1,2) for j = 0:restrict(2,2) num = num+1; k = 0; temp_sum = i*Passenger(1,1)+j*Passenger(2,1)+k*Passenger(3,1)+0.1*(i+j+k-1); while k=restrict(3,2) & temp_sum % % 車型的矩陣Passenger = 4.61 1.7 1.51;3.615 1.605 1.394;4

19、.63 1.785 1.77;% 轎運車的矩陣Car_carriers = 19 2.7 2.7;24.3 3.5 2.7;% 轎運車最多可以放車型個數(shù)（緊長度限制）restrict = 4 5;5 6;4 5;% 裝滿的所有情況（加入限制）%1-1 的第一層所有的情況A_1 = ;num = 0;%for i = 0:restrict(1,1) num = num+1; temp_sum = 0; j = 0; while j=restrict(2,1) & temp_sumCar_carriers(1,1) temp = i j; j = j+1; temp_sum = i*Passeng

20、er(1,1)+j*Passenger(2,1)+0.1*(i+j-1); end A_1(num,:) = temp; end%1-1 的第二層所有的情況A_1_2 = ;num = 0;for i = 0:restrict(1,1) for j = 0:restrict(2,1) num = num+1; k = 0; temp_sum = 0; temp_sum = i*Passenger(1,1)+j*Passenger(2,1)+k*Passenger(3,1)+0.1*(i+j+k-1); while k=restrict(3,1) & temp_sumCar_carriers(1,1) temp = i j k; k = k+1; temp_sum = i*Passenger(1,1)+j*Passenger(2,1)+k*Passenger(3,1)+0.1*(i+j+k-1); end A_1_2(num,:) = temp; end endA_1_2 = unique(A_1_2,rows);% 裝滿的所有情況（加入限制）%1-2 的第一層所有的情況A_2 = ;num = 0;%for i = 0:restrict(1,2) num = num+1; temp_sum = 0; j