Chapter 4, Problem 11E

a)

Explanation of Solution

Required number of RAM chips:

From given scenario, the $\text{16M×16}$ bits can convert into $\text{16}\times {\text{2}}^{\text{20}}{\text{×2}}^{\text{4}}{\text{= 2}}^4\times {\text{2}}^{\text{20}}{\text{×2}}^{\text{4}}=2^{28}\text{ bytes}$ and $\text{512K×8}$ can convert into ${\text{2}}^{\text{19}}{\text{×2}}^3\text{ bytes}$

b)

Explanation of Solution

Number of involved chips:

The numbers of involved chips are “two”. Because, the memory has 16-bit word...

c)

Explanation of Solution

Number of address bits needed for each RAM chip:

From given $\text{512K×8}$ RAM chips, the number of needed address bits is calculated as ${}^{}$

d)

Explanation of Solution

Number of banks in given memory:

The number of banks is calculated using the addressable item in each main memory and the RAM chips.

$\begin{array}{c}\text{Number of banks = }\frac{\text{16M}}{\text{512K}}\\ \text{= }\frac{{\text{2</}}^{}}{}\end{array}$

e)

Explanation of Solution

Number of needed address bits:

The main memory have 16M in addressable item. So, the number of needed bits for all the memory is ${\text{16M = 2}}^{}$

f)

Program Plan Intro

High-order interleaving:

It is the type of memory interleaving which splits the memory into various banks. The higher order of bits is used to access the banks in memory.

g)

Program Plan Intro

Low-order interleaving:

It is the type of memory interleaving which splits the memory into various banks. The low order of bits are used to access the banks in memory.