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 buffer 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?

Short answer

a) The minimum size in bytes of the frame buffer to store the frame is 3932160 bytes.

b) The minimum time taken for a frame to be sent over 100Mbit/s is 0.314 s.

Step-by-step solution

Determine the formulas

Write the formula for calculating no. of bits per pixel

$Numberofbitsperpixel=Numberofbitspereachcolor\times Numberofcolors$…(1)

Write the formula for converting bits into bytes.

$Numberofbytes=\frac{Numberofbits}{8}$ …(2)

Write the formula for the size of the frame in MBits

$Sizeofframe\left(Mbits\right)=\frac{Sizeofframe\left(bits\right)}{\left({10}^6\right)}$ …(3)

Write the formula for calculating bytes for the frame buffer to store a frame$Numberofbytestostoreaframe=Bytes\times Framesize$ …(4)

Write the formula for the time taken for a frame to be sent over a network.

$Time=\frac{Sizeoftheframe}{Speedofthenetwork}$ …(5)

Determine the minimum size in bytes of the frame buffer to store the frame

a)

Number of bits per pixel $$\begin{gathered} =8\times 3 \\ =24\raisebox{1ex}{$bits$}\!\left/ \!\raisebox{-1ex}{$pixel$}\right. \end{gathered}$$

Number of bytes per pixel $$\begin{gathered} =\frac{24}{8} \\ =3\raisebox{1ex}{$bytes$}\!\left/ \!\raisebox{-1ex}{$pixel$}\right. \end{gathered}$$

Number of bytes to store a frame in frame buffer $$\begin{gathered} =3\times 1280\times 1024 \\ =3932160bytes \end{gathered}$$

Determine the time taken for the frame to send over a network

b)

size of the frame(in bits)

$$\begin{gathered} =3932160bytes\times 8 \\ =31457280bits \end{gathered}$$

size of the frame (in Mbits)

$$\begin{gathered} =\frac{31457280}{{10}^6} \\ =31.45728Mbits \end{gathered}$$

Time is taken for the frame to be sent over the network

$$\begin{gathered} =\frac{31.45728}{100} \\ =0.3145728seconds \end{gathered}$$