Asked in NIMCET Exam 2010

Question

A main memory has an access time of 45 ns. A 5 ns time gap is necessary for the completion of one access to the beginning of the next access. The bandwidth of the memory

(a) 25 MHz                                (b) 20 MHz

(c) 40 MHz                                 (d) 50 MHz

Comments

Memory bandwidth is the rate at which data can be read from or stored into a semiconductor memory by a processor. Memory bandwidth is usually expressed in units of bytes/second, though this can vary for systems with natural data sizes that are not a multiple of the commonly used 8-bit bytes. (wikipedia.org)

Answer 1

Total time required for each memory access is:

45 ns + 5 ns = 50 ns

The bandwidth of the memory can be calculated as the inverse of the time required for each access:

1 / 50 ns = 0.02 GHz

Therefore, the bandwidth of the memory is 0.02 GHz or 20 MHz.

Option B