Question

A discrete memoryless channel with three inputs and outputs has transition probabilities as shown. The parameter $\alpha \in [0.25,1]$. Find the value of $\alpha$ that maximizes channel capacity (rounded to two decimals).

Hint:

For cyclic symmetric channels, capacity is maximized when output entropy is minimized, which occurs at boundary values of the probability parameter.

Answer 1

Correct Answer: 1

The channel is symmetric in a cyclic sense: - Input $a$ outputs $a$ with probability $1-\alpha$ and transitions to $b,c$ with probability $\alpha$. - Similarly, $b$ and $c$ have the same pattern shifted. Thus this is a weakly symmetric channel, and its capacity is: \[ C = \log_2(3) - H(\alpha,\, \alpha,\, 1-2\alpha), \] where the output distribution for each input is: \[ \{1-\alpha,\ \alpha,\ \alpha\}. \] The entropy term is: \[ H = -(1-\alpha)\log_2(1-\alpha) - 2\alpha\log_2\alpha. \] Capacity increases as the distribution becomes more skewed (lower entropy). We check the boundary of the valid region $\alpha \in [0.25,1]$. - At $\alpha=0.25$: \[ \{0.75,\ 0.25,\ 0.25\} \Rightarrow \text{smaller entropy}. \] - At $\alpha=1$: \[ \{0,1,1\} \quad\text{(invalid distribution)}. \] - For $\alpha \in (0.25,1)$, entropy increases; capacity decreases. Thus capacity is maximized at the minimum allowable $\alpha$: \[ \boxed{1.00} \] Rounded to two decimals: \[ \boxed{1.00} \]