Question

Suppose an image is represented on a display screen by a square array containing 256 columns and 256 rows of pixels. If for each pixel, 3 bytes are required to encode the color and 8 bytes to encode the intensity, how many byte-size memory cells are required to hold the entire picture?

Answer 1

Solution and Explanation:

Given data: number of rows = 256, number of columns = 256, number of bytes that are required for each pixel to encode the color = 3, number of bytes that are required to encode the intensity = 8

calculation of vyte size memory cells is as follows:

the number of pixels = number of rows mulitply with number of columns

= 256 mulitply with 256 = 65536

each pixel require 8 bit, therefore, each pixel require 1 byte

the size of memory must be 65536 bytes.

This is 65536 divide 1024 = 64KB

Answer 2

The byte-size memory cells that are required to hold the entire picture is 64KB.

According to this question,we were given the number of rows as (256).

The number of columns is been given as 256.

The required number of bytes to encode the color  as 3.

To encode the intensity, number of bytes  required is 8.

The number of pixels  =( number of row * number of columns)

$= (256 *256 )= 65536$

The byte-size memory cells that are required to hold the entire picture

$=(65536 *1024 )= 64KB$

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