Question

11. Suppose that a 32M × 16 memory built using 512K × 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 11f for low-order interleaving.

Answer 1

Given memory size = 32M * 16

= 32 * 1024 *1024 * 16

= 2⁵ * 2¹⁰ * 2¹⁰ * 16

= 2²⁵ * 16

Chip size = 512K × 8

= 512 * 1024 * 8

= 2⁹ * 2 ¹⁰ * 8

= 2¹⁹ * 8

a) How many RAM chips are necessary?  

Number of RAM Chips = Memory Size/ Chip Size

= 2²⁵ * 16 / 2¹⁹ * 8

= 2 ⁷

  128 RAM chips Needed

b) If we were accessing one full word, how many chips would be involved?

• As given RAM chips holds 8 bits ( 512K × 8)for a word. Main memory holds 16 bits (32M * 16 ).
• So that design needs 16 bit word which means two byes. And chip hold only 1 byte.
• So that 2 chips would be involved to access one word

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

For 512K × 8 chip, bits required are 512K

=> 512 * 1024 = 2⁹ * 2¹⁰ = 2¹⁹

So 2¹⁹ bits required.

And 19 address bits required for each chip.

d) How many banks will this memory have?

The number of banks = main memory addressable / chip addressable items

So that, no.of Banks = 2²⁵ / 2 ¹⁹

= 2⁶ = 64 Banks

Therefore, number of banks equals 64

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

Number of bits required for memory is 32 M = 2⁵ * 2¹⁰ * 2¹⁰

= 2²⁵

Therefore, 25 bits are needed to address main memory.

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

In high order memory interleaving, the address are distributed such that each bank in the memory contains consecutive addresses. In this case, the first 512 K words goes to bank 0.  So, the address 14 can be found in bank 0(000 in decimal)

g) Repeat Exercise 11f for low-order interleaving.

We already found that memory has 25 address bits (A₀-A₂₄)

In Low-order interleaved memory, addresses will be placed in different memory banks.

So that, address 14 is 0xE in hexadecimal and in binary as 25- address format as follows:

00000000 00000000 000001110.

Field	Offset	Bank
Bits	A7-A24	A0-A6
Address 14	00000000 00000000 00	0001110

Therefore, for address 14, the bank is binary 10001110