For a bus with width=32Bits, speed=33MHz, what is the theoretical throughput in MiBytes?
a. 32*33*(10^6)/(2^20)
b. 32/8*33* (2^20)/(10^6)
c. 32/8*33*(10^6)/(2^20)
d. 8*33*(10^6)/(2^20)

Respuesta :

Answer:

Option C

Explanation:

Given:

- Clock speed f = 33 MHz

- The width of the bus w = 32 bits

Find:

what is the theoretical throughput in MiBytes?

Solution:

- First step is to convert the width of the bus to bytes as follows:

                                 Bytes = 32 bits * (1 Bytes / 8 bits)

                                 Bytes = 32 / 8

- Second step is to evaluate the time of the cycle:

                                 Time period of clock T = 1 / f

                                 Time period of clock T = 1 s / (33*10^6)  

                                                                     T = (1 / 33*10^6)

- Third step is to formulate the number of byte:

                                 Number of byte = Bytes * T

                                                             = (32 / 8*33*10^6)  

- Fourth step is to convert to Mi bytes:

                                 Mibytes = Number of byte / 2^20

                                 Mibytes = (32 / 8*33*10^6 * 2^20)

- The correct option is C