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,

$\dfrac{460800}{1024}=450\text{ KB}$