Question

Assume a color display using 8 bits for each of the primary colors (red, green, blue) per pixel and a frame size of 1280 × 1024.

a. what is the minimum size in bytes of the frame buff er to store a frame?
b. how long would it take, at a minimum, for the frame to be sent over a 100 mbit/s network?

Answer 1

a) 3.93216Mb

b) 0.3145728 seg

Step-by-step explanation:

a) Each color use 8 bits equal to one byte, therefore, three colors uses 24 bits or 3 bytes, additionally, the frame has a size of 1280x1024 equal to 1310720 pixels, using this two results the memory size of the frame is 3 Bytes * 1310720 Pixels = 3932160 bytes equivalent to 3.93216Mb.

b) The speed is equal the frame size divided by the time, to make the operation we convert the 100 mbit/s to Mb/s -> 100/8 = 12.5 Mb/s, therefore, speed = 3.93216/12.5 = 0.3145728 seg

Mb -> Megabytes

Answer 2

A.) 1280 x 1024 pixels = 1,310,720 pixel

1,310,720 x 3 = 3,932,160 bytes/frame

B.) 3,932,160 bytes x ( 8bits/byte ) / 100E6bits/second = 0.31 seconds