Question

A discrete memoryless source emits message symbols $m_1$ and $m_2$ with probabilities $0.8$ and $0.2$ respectively. Using the Huffman source code to convert these symbols into binary codewords, the code efficiency that can be obtained is

• A. $72.2$ %
• B. $50$ %
• C. $100$ %
• D. $66.67$ %

Answer 1

The correct option is A $72.2$ %

$H\left(m\right)=-0.8{\log }_{2}0.8-0.2{\log }_{2}0.2$
$=0.722$ bits/symbols
$\overline{L}=0.8\times 1+0.2\times 1=1$ bits/symbol
$\because$ Two symbols are there, each symbol can be represented with one bit.
$\eta =\frac{H\left(x\right)}{\overline{L}}=0.722\Rightarrow 72.2\%$