Question

The following figure shows how to construct a module of chips that can store 1 MB based on a group of four 256-Kbyte chips. Let’s say this module of chips is packaged as a single 1-MB chip, where the word size is 1 byte. Give a high-level chip diagram of how to construct a 8-MB computer memory using 8 1-MB chips. How many memory addresses lines do you need (In the example, it is 20 lines -MAR)?

Question 8 options:

A	20
B	27
C	23
D	25

Question 9 (3 points)

For a given computer system, the main memory is 256Mbyte (=2^28 Bytes); word size is 4 bytes (=2^2 Bytes); block size is 64 bytes (2^6 bytes); cache size is 64 Kbytes (2^16 Bytes). When direct mapping is used as the mapping function, how many bits is needed to be the tag?

Question 9 options:

A	10
B	16
C	18
D	28

Answer 1

1. 20 bits is the length of memory address.Total size of the memory is 1 MB=1*2¹⁰*2¹⁰

So 20 memory address lines are needed.

Data bus width is 8 bit since the word length is 1 byte.

2. Memory address is 28 bit,since the size of memory is given as 256MB.

It is given that cache size is 64 Kbytes and the block size is 64 bytes.So total number of blocks=64K/64=1K=1024.

So 10 bits are needed to recoganize one block(2¹⁰=1024)

Out of 28 bit,10 bits are needed to identify block,2 bits for word and the remaining 16 bits are for tag.So answer is option B.