a陶子數(shù)學(xué)模型食品安全抽檢問題論文(編輯修改稿)
2025-07-04 13:31 本頁面
【文章內(nèi)容簡介】 0 0 0 + 0 0 0 w=x(:,1)/sum(x(:,1))w = （=，權(quán)重向量為w=（，））=()/(31)=RI==,表明A通過了一致性驗證。 B1=[1 3 5。1/3 1 3。1/5 1/3 1]B1= [x,y]=eig(B1)x = + + y = 0 0 0 + 0 0 0 w=x(:,1)/sum(x(:,1))w = （1=，權(quán)重向量w1=（，）） B2=[1 3 7。1/3 1 5。1/7 1/5 1]B2 = [x,y]=eig(B2)x = + + y = 0 0 0 + 0 0 0 w2=x(:,1)/sum(x(:,1))w2 = （2=，權(quán)重向量w2=（，））附件2： f=*(9/44*+2/11*+3/22*+15/88*+13/88*+7/44*)/((9/44*^2+2/11*^2+3/22*^2+15/88*^2+13/88*^2+7/44*^2)+440*()^2)f = f=*(9/44*+2/11*+3/22*+15/88*+13/88*+7/44*)/((9/44*^2+2/11*^2+3/22*^2+15/88*^2+13/88*^2+7/44*^2)+440*()^2)f = f=*(9/44*+2/11*+3/22*+15/88*+13/88*+7/44*)/((9/44*^2+2/11*^2+3/22*^2+15/88*^2+13/88*^2+7/44*^2)+440*()^2)f = f=*(9/44*+2/11*+3/22*+15/88*+13/88*+7/44*)/((9/44*^2+2/11*^2+3/22*^2+15/88*^2+13/88*^2+7/44*^2)+440*()^2)f = f=*(9/44*+2/11*+3/22*+15/88*+13/88*+7/44*)/((9/44*^2+2/11*^2+3/22*^2+15/88*^2+13/88*^2+7/44*^2)+440*()^2)f = f=*(9/44*+2/11*+3/22*+15/88*+13/88*+7/44*)/((9/44*^2+2/11*^2+3/22*^2+15/88*^2+13/88*^2+7/44*^2)+440*()^2)f = 90，80，60，75，65，70，，，=n=8+7+5+7+6+6=39 假設(shè)t=,p=60,Pmax=6000，Tmax=176, =11附件3：c=[39]。 A=[39。39。1]。 b=[。100。11]。 Aeq=[]。 beq=[]。 vlb=[0。0。0]。 vub=[]。[x,fval]=linprog(c,A,b,Aeq,beq,vlb,vub)x = fval = w=[1 1 3 3 5 5 6 7 1 9 9。1 1 3 3 5 5 6 7 1 9 9。1/3 1/3 1 1 5/3 5/3 2 7/3 1/3 3 3。1/3 1/3 1 1 5/3 5/3 2 7/3 1/3 3 3。1/5 1/5 3/5 3/5 1 1 6/5 7/5 1/5 9/5 9/5。1/5 1/5 3/5 3/5 1 1 6/5 7/5 1/5 9/5 9/5。1/6 1/6 1/2 1/2 5/6 5/6 1 7/6 1/6 3/2 3/2。1/7 1/7 3/7 3/7 5/7 5/7 6/7 1 1/7 9/7 9/7。1 1 3 3 5 5 6 7 1 9 9。1/9 1/9 1/3 1/3 5/9 5/9 2/3 7/9 1/9 1 1。1/9 1/9 1/3 1/3 5/9 5/9 2/3 7/9 1/9 1 1。]w = [x,y]=eig(w)x = Columns 1 through 7 + + + + + + + + + + Columns 8 through 11 + + + + + + + + + +
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