Y(i+1, j+1, k) = c_i * c_j * sum(sum(X(:,:,k) .* cos(pi*(2*(0:m-1)+1)*i/(2*m)).' .* cos(pi*(2*(0:n-1)+1)*j/(2*n))));

这段代码中的 " .* " 表示矩阵的逐元素乘积，也称为 Hadamard 乘积。如果您在使用 " .* " 时出现错误，可能是因为您在使用它的语境上下文中出现了问题。例如，如果您尝试将一个向量与一个标量相乘，就会出现错误，因为这两个对象的大小不匹配。

为了解决这个问题，您需要检查代码中的变量和矩阵尺寸，确保它们在使用 " .* " 时具有相同的大小。如果问题仍然存在，请提供更具体的信息，以便我可以更好地帮助您。

相关问题

DD=xlsread('residual.xlsx') P=DD(1:621,1)' N=length(P) n=486 F =P(1:n+2) Yt=[0,diff(P,1)] L=diff(P,2) Y=L(1:n) a=length(L)-length(Y) aa=a Ux=sum(Y)/n yt=Y-Ux b=0 for i=1:n b=yt(i)^2/n+b end v=sqrt(b) Y=zscore(Y) f=F(1:n) t=1:n R0=0 for i=1:n R0=Y(i)^2/n+R0 end for k=1:20 R(k)=0 for i=k+1:n R(k)=Y(i)*Y(i-k)/n+R(k) end end x=R/R0 X1=x(1);xx(1,1)=1;X(1,1)=x(1);B(1,1)=x(1); K=0;T=X1 for t=2:n at=Y(t)-T(1)*Y(t-1) K=(at)^2+K end U(1)=K/(n-1) for i =1:19 B(i+1,1)=x(i+1); xx(1,i+1)=x(i); A=toeplitz(xx); XX=A\B XXX=XX(i+1); X(1,i+1)=XXX; K=0;T=XX; for t=i+2:n r=0 for j=1:i+1 r=T(j)*Y(t-j)+r end at= Y(t)-r K=(at)^2+K end U(i+1)=K/(n-i+1) end q=20 S(1,1)=R0; for i = 1:q-1 S(1,i+1)=R(i); end G=toeplitz(S) W=inv(G)*[R(1:q)]' U=20*U for i=1:20 AIC2(i)=n*log(U(i))+2*(i) end q=20 C=0;K=0 for t=q+2:n at=Y(t)+Y(q+1); for i=1:q at=-W(i)*Y(t-i)-W(i)*Y(q-i+1)+at; end at1=Y(t-1); for i=1:q at1=-W(i)*Y(t-i-1)+at1 end C=at*at1+C K=(at)^2+K end p=C/K XT=[L(n-q+1:n+a)] for t=q+1:q+a m(t)=0 for i=1:q m(t)=W(i)*XT(t-i)+m(t) end end m=m(q+1:q+a) for i =1:a m(i)=Yt(n+i+1)+m(i) z1(i)=P(n+i+1)+m(i); end for t=q+1:n r=0 for i=1:q r=W(i)*Y(t-i)+r end at= Y(t)-r end figure for t=q+1:n y(t)=0 for i=1:q y(t)=W(i)*Y(t-i)+y(t) end y(t)=y(t)+at y(t)=Yt(t+1)-y(t) y(t)=P(t+1)-y(t) end D_a=P(n+2:end-1); for i=1:a e6_a(i)=D_a(i)-z1(i) PE6_a(i)= (e6_a(i)/D_a(i))*100 end e6_a PE6_a 1-abs(PE6_a) mae6_a=sum(abs(e6_a)) /6 MAPE6_a=sum(abs(PE6_a))/6 Z(1)=0;Xt=0 for i =1:q Xt(1,i)=Y(n-q+i) end for i =1:q Z(1)=W(i)*Xt(q-i+1)+Z(1) end for l=2:q K(l)=0 for i=1:l-1 K(l)=W(i)*Z(l-i)+K(l) end G(l)=0 for j=l:q G(l)=W(j)*Xt(q+l-j)+G(l) end Z(l)=K(l)+G(l) end for l=q+1:aa K(l)=0 for i=1:q K(l)=W(i)*Z(l-i)+K(l) end Z(l)=K(l) end r=Z*v+Ux r(1)=Yt(n+2)+r(1) z(1)=P(n+2)+r(1) for i=2:aa r(i)=r(i-1)+r(i) z(i)=z(i-1)+r(i) end D=P(n+2:end-1) for i=1:aa e6(i)=D(i)-z(i) PE6(i)= (e6(i)/D(i))*100 end e6 PE6 1-abs(PE6) mae6=sum(abs(e6)) /6 MAPE6=sum(abs(PE6))/6把单步预测的完整代码单独摘出来

以下是该段代码的注释：

% 读取数据
DD = xlsread('residual.xlsx');
P = DD(1:621, 1)';
N = length(P);
n = 486;

% 获取一阶差分、二阶差分、原始序列的部分
F = P(1:n+2);
Yt = [0, diff(P, 1)];
L = diff(P, 2);
Y = L(1:n);

% 计算Ux、v、Y的z-score
Ux = sum(Y) / n;
yt = Y - Ux;
v = sqrt(sum(yt.^2) / n);
Y = zscore(Y);

% 计算R、X、U、AIC2、C、K、m、y、e6、PE6、mae6、MAPE6等
R0 = sum(Y.^2) / n;
R = zeros(1, 20);
for k = 1:20
    for i = k+1:n
        R(k) = R(k) + Y(i) * Y(i-k) / n;
    end
end

X1 = R(1);
xx(1, 1) = 1;
X(1, 1) = X1;
B(1, 1) = X1;
K = 0;
T = X1;
for t = 2:n
    at = Y(t) - T * Y(t-1);
    K = at^2 + K;
end
U(1) = K / (n-1);

for i = 1:19
    B(i+1, 1) = R(i+1);
    xx(1, i+1) = R(i);
    A = toeplitz(xx);
    XX = A \ B;
    XXX = XX(i+1);
    X(1, i+1) = XXX;
    K = 0;
    T = X(1, 1:i+1);
    for t = i+2:n
        r = 0;
        for j = 1:i+1
            r = T(j) * Y(t-j) + r;
        end
        at = Y(t) - r;
        K = at^2 + K;
    end
    U(i+1) = K / (n-i+1);
end

q = 20;
S(1,1) = R0;
for i = 1:q-1
    S(1, i+1) = R(i);
end
G = toeplitz(S);
W = inv(G) * [R(1:q)]';
U = 20 * U;

for i = 1:20
    AIC2(i) = n*log(U(i)) + 2*(i);
end

C = 0;
K = 0;
for t = q+2:n
    at = Y(t) + Y(q+1);
    for i = 1:q
        at = -W(i) * Y(t-i) - W(i) * Y(q-i+1) + at;
    end
    at1 = Y(t-1);
    for i = 1:q
        at1 = -W(i) * Y(t-i-1) + at1;
    end
    C = at * at1 + C;
    K = at^2 + K;
end

p = C / K;
XT = [L(n-q+1:n+a)];
for t = q+1:q+a
    m(t) = 0;
    for i = 1:q
        m(t) = W(i) * XT(t-i) + m(t);
    end
end
m = m(q+1:q+a);

for t = q+1:n
    y(t) = 0;
    for i = 1:q
        y(t) = W(i) * Y(t-i) + y(t);
    end
    y(t) = y(t) + Y(t) - Yt(t+1);
    y(t) = P(t+1) - y(t);
end

D_a = P(n+2:end-1);
for i = 1:a
    e6_a(i) = D_a(i) - (P(n+i+1) + m(i));
    PE6_a(i) = (e6_a(i) / D_a(i)) * 100;
end
mae6_a = sum(abs(e6_a)) / 6;
MAPE6_a = sum(abs(PE6_a)) / 6;

Z(1) = 0;
Xt = 0;
for i = 1:q
    Xt(1, i) = Y(n-q+i);
end
for i = 1:q
    Z(1) = W(i) * Xt(q-i+1) + Z(1);
end
for l = 2:q
    K(l) = 0;
    for i = 1:l-1
        K(l) = W(i) * Z(l-i) + K(l);
    end
    G(l) = 0;
    for j = l:q
        G(l) = W(j) * Xt(q+l-j) + G(l);
    end
    Z(l) = K(l) + G(l);
end
for l = q+1:aa
    K(l) = 0;
    for i = 1:q
        K(l) = W(i) * Z(l-i) + K(l);
    end
    Z(l) = K(l);
end

r = Z * v + Ux;
r(1) = Yt(n+2) + r(1);
z(1) = P(n+2) + r(1);
for i = 2:aa
    r(i) = r(i-1) + r(i);
    z(i) = z(i-1) + r(i);
end

D = P(n+2:end-1);
for i = 1:aa
    e6(i) = D(i) - z(i);
    PE6(i) = (e6(i) / D(i)) * 100;
end
mae6 = sum(abs(e6)) / 6;
MAPE6 = sum(abs(PE6)) / 6;

该代码主要实现了单步预测的过程，其中主要的计算包括：

1. 计算一阶差分、二阶差分和原始序列的部分；
2. 计算Ux、v、Y的z-score；
3. 计算R、X、U、AIC2、C、K、m、y、e6、PE6、mae6、MAPE6等；
4. 实现单步预测过程，计算得到预测值z1和z；
5. 计算e6、PE6、mae6和MAPE6等评价指标。

在matlab 中求解$$\max \sum_{i=1}^n\sum_{j=1}^{10}x_{i,j}$$ $$\text{s.t.}\begin{cases}y_i=10\sum_{j=1}^{10}x_{i,j}, i=1,2,\cdots,n\z_i=\frac{1}{4}\pi h_i^2, h_i=\sum_{j=1}^{10}jx_{i,j}, i=1,2,\cdots,n\d_{i,j}\geq 2.5+r_i+r_j, r_i=\frac{1}{2}\sum_{k=1}^{10}jx_{i,j}, r_j=\frac{1}{2}\sum_{k=1}^{10}jx_{j,k},i,j=1,2,\cdots,n\c_i=10h_i+10, i=1,2,\cdots,n\y_i\leq 500, i=1,2,\cdots,n\z_i\leq 2.8x_{i,1}+5.5x_{i,2}+8.5x_{i,3}+11.9x_{i,4}+14.5x_{i,5}, i=1,2,\cdots,n\x_{i,j}\in{0,1}, i=1,2,\cdots,n, j=1,2,\cdots,10\end{cases}$$

这是一个混合整数规划问题，可以使用MATLAB中的intlinprog函数进行求解。其中，目标函数的系数向量为10的向量，线性不等式约束和等式约束右侧的系数矩阵分别为$n\times 10$和$n\times n$的矩阵，线性等式约束右侧的系数向量为$n$维向量，所有变量的上下界均为0或1。以下是MATLAB代码实现：

% 目标函数系数向量
f = -10 * ones(1, 10 * n);

% 不等式约束系数矩阵和右侧系数向量
A = zeros(4 * n, 10 * n);
b = zeros(4 * n, 1);
for i = 1:n
    % y_i = 10 * sum_j(x_ij)
    A(i, (i-1)*10+1:i*10) = -10 * ones(1, 10);
    b(i) = 0;
    % delta_ij >= 2.5 + r_i + r_j
    for j = 1:n
        if j ~= i
            r_i = 0.5 * (1:10) * x((i-1)*10+1:i*10)';
            r_j = 0.5 * (1:10) * x((j-1)*10+1:j*10)';
            A(n+(i-1)*n+j, (i-1)*10+1:i*10) = 0.5 * (1:10);
            A(n+(i-1)*n+j, (j-1)*10+1:j*10) = 0.5 * (1:10);
            b(n+(i-1)*n+j) = 2.5 + r_i + r_j;
        end
    end
    % y_i <= 500
    A(2*n+i, (i-1)*10+1:i*10) = 10 * ones(1, 10);
    b(2*n+i) = 500;
    % z_i <= 2.8*x_i,1 + 5.5*x_i,2 + 8.5*x_i,3 + 11.9*x_i,4 + 14.5*x_i,5
    A(3*n+i, (i-1)*10+1:i*10) = -1/4*pi*(1:10).^2;
    A(3*n+i, (i-1)*10+1:i*10) = A(3*n+i, (i-1)*10+1:i*10) + ...
                                [2.8 5.5 8.5 11.9 14.5 zeros(1,5) zeros(1,5) zeros(1,5) zeros(1,5) zeros(1,5)];
    b(3*n+i) = 0;    
end

% 等式约束系数矩阵和右侧系数向量
Aeq = zeros(2 * n, 10 * n);
beq = zeros(2 * n, 1);
for i = 1:n
    % z_i = 1/4*pi*h_i^2
    Aeq(i, (i-1)*10+1:i*10) = 0.25 * pi * (1:10).^2;
    beq(i) = 0;
    % c_i = 10*h_i + 10
    Aeq(n+i, (i-1)*10+1:i*10) = 10 * ones(1, 10);
    beq(n+i) = 10;
end

% 变量的上下界
lb = zeros(10 * n, 1);
ub = ones(10 * n, 1);

% 求解混合整数规划问题
[x, fval, exitflag] = intlinprog(f, 1:10*n, A, b, Aeq, beq, lb, ub);

% 输出结果
if exitflag > 0
    fprintf('最大值为：%f\n', -fval);
    for i = 1:n
        fprintf('x_%d = [', i);
        for j = 1:10
            fprintf('%d ', x((i-1)*10+j));
        end
        fprintf(']\n');
    end
else
    fprintf('求解失败！\n');
end