Question

1. For a 512 k × 32 memory system, how many unique address locations are there? Give the exact number.

2. For a 512 k × 32 memory system, what is the capacity in bits?

3. For a 512 k × 32 memory system, how wide does the incoming address bus need to be in order to access every unique address location?

Answer 1

1. 512 k × 32 = 2⁹ * 2¹⁰ * 2⁵

= 2²⁴

For a 512 k × 32 memory system, how many unique address locations are 2²⁴

2. 512 k × 32 memory system, the capacity is 24 bits

3. 512 k × 32 = 2²⁴

= 2⁴ * 2²⁰  

= 16MB

The incoming address bus need to be 16MB in order to access every unique address location.