Question

A 64 bit computer uses a memory of 32KB. Work out the number of bits of
AR, DR, PC, AC, and the HEX code of the address of last memory location.
b) The memory attached to a processor has access time of 400ns. When a cache memory is
introduced in the system, it is found that on average the access time for the computer was
30% greater than the cache access time for a miss ratio of 0.10. Work out the access time of
the cache.
c) Since Hard disks cost per megabyte is much less compared to other memory devices, can
we simplify memory design for a modern computer by building all memory components
(primary , secondary, BIOS, cache) using hard disks? Discuss briefly.

Answer 1

a)

Size of data word in the computer = 64 bits

So the size of the registers that hold data will be 64 bits.

Number of bits in DR (Data register) = 64 bits

Number of bits in AC (Accumulator) = 64 bits

Size of memory = 32 KB = 2^15

So memory address is 15 bits in length

So the size of all the registers that hold memory address will be 15 bits in length

Number of bits in AR (Address Register) = 15 bits

Number of bits in PC (Program Counter) = 15 bits

The last address will be all 1's in binary

111111111111111

Hex of 111111111111111 = 7FFF

b)

Let the access time of the cache be c

Average Access Time = Hit Rate * ( Access time for processor 's memory) + Miss Rate * ( Access time for cache )

1.3c = (0.9 * 400) + (0.1 * (400+c))

c = 333.33 ns

c)

It is true that cost of hard disks is alot less than other forms of memory like cache and RAM. But the speed of hard drives is a lot more compared to caches and RAM. This makes them infeasible for use as for all forms of memory. So hard drives cannot be used because they will signicantly reduce the performance of the computer.