Question

5. (14 points) Suppose that a 2M x 16 main memory is built using 256K × 8 RAM chips and memory is word-addressable. a) How many RAM chips are necessary? b) If we were accessing one full word, how many chips would be involved? c) How many address bits are needed for each RAM chip? d) How many banks will this memory have? e) How many address bits are needed for all of memory? f) If high-order interleaving is used, where would address 14 (which is E in hex) be located? g) Repeat Exercise f) for low-order interleaving.

Answer 1

a. 16 RAM chips

b. two chips

c. 18 address bits

d. 8 banks

e. 21 bits

f. bank 0 (0000)

g. 1110

Step-by-step explanation:

a. How many RAM chips are necessary?

To find the RAM chips required, we need to divide the memory size by chip size

Memory size
`=2M*16=2^(21)*2^(4) =2^(25)`

Chip size
`=256K*8=2^(18)*2^(3) =2^(21)`

`(2^(25) )/(2^(21) ) =2^(4) =16`

Hence 16 RAM chips are needed

b. To access one full word, how many chips are involved?

since each chip holds one byte (8 bits) for each addressable item, 2 chips per 16-bit word are necessarily required.

c. How many address bits are needed for each RAM chip?

`256K=2^(18)`

18 address bits for each chip are required.

d. How many banks will this memory have?

The number of banks depends on the number of addressable items in each of the main memory and the RAM chips

`Banks=(main memory)/(chips)`

`(2M)/(256K) =(2^(21) )/(2^(18) ) =2^(3) =8`

Hence this memory would have 8 Banks.

e. How many address bits are needed for all of memory?

`2M=2^(21)`

21 address bits are needed

Out of these 21 address bits, 18 bits go to each bit ( see part c) and remaining 3 bits are for bank selection (see part d)

f. If high-order interleaving is used, where would address 14 (which is E in hex) be located?

If high order interleaving is used, then the first 256K words will be found in bank 0. so address 14 will be in bank 0.

g. Repeat Exercise f) for low-order interleaving.

in part d we found that we have 8–way interleaving.

The memory has 21 bit addressing and the address lines are labeled A0 - A20.

Address bits A0 - A2 illustrates the bank and remaining address bits A3 -A20 corresponds to the offset within the bank.

The bank number is binary 1110