使用Optimization Toolbox时如何解决数组索引必须为正整数或逻辑值错误

Let's walk through exactly why your code is throwing that error and how to fix it, plus clean up a few other issues to get your cut-set boundary curve working correctly.

1. Why the Index Error Happens #

The line r(M) = -ropt is the culprit. You're using M (a floating-point number like 0, 0.1, 0.2...) as an index for the array r. MATLAB requires array indices to be positive integers (1, 2, 3...) or logical values—decimals and 0 are invalid here.

2. Step-by-Step Fixes #

Here's how to adjust your code to resolve this and other minor issues:

a. Pre-Initialize the Results Array Correctly #

Instead of redefining r inside the loop (which wipes previous results), create an array that matches the length of your M sequence upfront:

N = 60; t = 3; K = 130;
M_values = 0:0.1:1.4; % Store M sequence in a variable
r = zeros(size(M_values)); % Initialize r to match the length of M_values

b. Use Integer Indices for Assignments #

Loop over the index of M_values instead of the values themselves, so you have a valid integer index for r:

for idx = 1:length(M_values)
    M = M_values(idx); % Get the current M value
    % Rest of your loop code here...
    r(idx) = -ropt; % Use integer index idx to store results
end

c. Fix the Logical Condition #

Your original condition if f || c is backwards. You want to run the optimization only when 1/K ≤ M ≤ t*N/K (approx 0.0077 ≤ M ≤ 1.3846). So use logical AND (&&) instead of OR (||):

lower_bound = 1/K;
upper_bound = t*N/K;
f = M >= lower_bound;
c = M <= upper_bound;
if f && c
    % Run optimization here
end

d. Clean Up the Objective Function Initialization #

Your original r=zeros(size(1:M)); was invalid for M < 1 (since 1:M would be an empty array). We already fixed this by pre-initializing r outside the loop.

3. Full Corrected Code #

N = 60; t = 3; K = 130;
M_values = 0:0.1:1.4;
r = zeros(size(M_values)); % Pre-allocate results array

lower_bound = 1/K;
upper_bound = t*N/K;

for idx = 1:length(M_values)
    M = M_values(idx);
    
    % Check if M is within the valid range
    is_valid = (M >= lower_bound) && (M <= upper_bound);
    
    if is_valid
        % Define the objective function (note: fmincon minimizes, so we keep the negative)
        R2 = @(s) -(s - (s./floor(N./s)).*M);
        LB = 1;
        UB = min(N, K);
        options = optimoptions('fmincon','Algorithm','interior-point');
        
        % Run optimization
        [sopt, ropt] = fmincon(R2, 1, [], [], [], [], LB, UB, [], options);
        r(idx) = -ropt; % Store the positive delivery rate
    else
        r(idx) = NaN; % Mark invalid M values as NaN (won't plot)
    end
end

% Plot the results
plot(M_values, r, 'r-', 'LineWidth', 2);
xlabel('Cache Capacity (M)');
ylabel('Delivery Rate (R)');
grid on; % Optional: add grid for readability

4. What This Fixes #

• The index error is resolved by using integer indices (idx) for the r array.
• Valid M values are correctly filtered with the logical AND condition.
• Results are preserved across loop iterations because r is pre-allocated outside the loop.
• Invalid M values are marked as NaN, so they won't show up in your plot (keeping only the valid cut-set boundary curve).

内容的提问来源于stack exchange，提问作者Fatma Ahmed