Q1.
a. A block-set-associative cache consists of a total 64 blocks divided into 4 - block sets. The main memory contains 4096 blocks, each consisting of 128 words. Find: i. How many bits are there in the main memory address? Ii. How many bits are there in
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They used word for a 16-bit quantity, while longword referred to a 32-bit quantity. ... Data structures containing such different sized words refer to them as WORD (16 bits/2 bytes), DWORD (32 bits/4 bytes) and QWORD (64 bits/8 bytes) respectively.
1 byte=8 bit
i word=16 bit
128 words=128*16=2048 bits
2048 bits are there in main memory
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In a computer, a maximum number of bytes which are going to be addressed depends upon the bits in the memory address.
When a user uses a 16-bit machine, then it allow 16 bits for the address.
Similarly, the 32-bit machine allows 17 to 32 bits for th address.
When you know,
1 byte=8 bit
128 words is equal to 128*16=2048 bits
Totally, 2048 bits are there in the main memory, so 2048 bits is the right answer.
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