Question

Suppose that a 1M x 64 main memory is built using 256K × 16 RAM chips and memory is word-addressable.

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 6f for low-order interleaving.

Please explain with steps

Answer 1

e.

We first calculate the number of 256K * 16 modules required

= (Total memory) / (Size of 1 module)

= (1M * 64) / (256K * 16)

= (2²⁶) / (2¹⁸ * 2⁴)

= 2⁴ = 16

Thus, 16 modules are required, labelled from module-0 to module-15.

Now, each module has 256K addresses

= 2⁸ * 2¹⁰ addresses

= 2¹⁸ addresses

Thus, 18 bits are required for addressing a particular chip.

Also, to select one of the 16 chips, we require further log₂16 = 4 address lines.

Thus, total address bits required

= 18 + 4

= 22 address bits

f.

In high-order interleaving, the highest 4 bits of the physical address will determine the module-number, and (22 - 4 =) 18 low bits will determine the offset within the selected module-number.

Now, address 14 = 0xE

Writing the address in 22-bit binary, we have

0xE = 00 0000 0000 0000 0000 1110

The highest 4 (bolded) bits are 0000, equivalent to module-number 0.

The next 18 bits determine the offset = 0xE = decimal 14

Thus, in high order interleaving, 0xE will lie in module-number 0, with an offset of 0xE (decimal 14).

g.

In low order interleaving, the address mapping will be as follows :

• 0 = module 0, offset 0
• 1 = module 1, offset 0
• 2 = module 2, offset 0
• ..
• ..
• 14 = module 14, offset 0
• 15 = module 15, offset 0
• 16 = module 0, offset 1
• 17 = module 1, offset 1
• ..
• ..
• 31 = module 15, offset 1
• 32 = module 0, offset 2

and so on ..

The general formulae are given by :

• Module number :
  • (Address) mod 16
• Offset :
  • 
    Address/16
    

Thus, for address 0xE = decimal 14,

• Module number :
  • 14 mod 16 = 14
• Offset :
  • 
    | 14/16
    = 0

Thus, in low order interleaving, 0xE will lie in module-number 14, with an offset of 0x0 (decimal 0).