Consider a raster system with resolution $640$ by $480$. What size is frame buffer in bytes for this system to store $12$ bits per pixel?
Resolution of raster system is
$640 \times 480$
Total number of pixels is
$640 \times 480=307200$
Each pixel requires $12$ bits.
So, total memory required is
$307200 \times 12=3686400$ bits
Convert bits into bytes:
$\dfrac{3686400}{8}=460800$ bytes
Now,
$1\text{ KB}=1024\text{ bytes}$
So,
$\dfrac{460800}{1024}=450\text{ KB}$
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