Chapter 1, Problem 1.4E

a.

Explanation of Solution

The number of pixels present in the frame is $\text{1280}\text{×}\text{1024}\text{=}\text{1,310,720}\text{pixles}$.

The number of primary colors is 3.

The minimum size of the frame buffer required to store a frame in terms of bytes can be calculated using the following formula.

$\text{Minimum}\text{size}\text{=}\text{Number}\text{of}\text{pixels}$

b.

Explanation of Solution

The number of pixels present in the frame is $\text{1280}\text{×}\text{1024}\text{=}\text{1,310,720}\text{pixles}$.

The number of primary colors is 3.

The minimum size of the frame buffer required to store a frame in terms of bytes can be calculated using the following formula.

$\text{Minimum}\text{size}\text{=}\text{Number}\text{of}\text{pixels}\text{×}\text{Number}\text{of}\text{colors}$.

Substitute, “1,310,720” for “Number of pixels” and “3” for “Number of colors” is given below:

$\begin{array}{c}\text{Minimum size =}\text{1,310,720}\text{×}\text{3}\\ \text{=}\text{3,932,160}\end{array}$

Therefore, the minimum size of the frame buffer is 3,932,160 bytes/frame