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)

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