Consider a system with 1 K pages and 512 frames and each page is of size 2 KB . How many bits are required to represent the virtual address space memory:
Step-by-step Solution:
\(\text{Virtual address space consists of pages. Given that,}\) \begin{aligned} &\text{Number of pages} = 1K = 2^{10} \\ &\text{Page size} = 2KB = 2^{11} \text{ Bytes} \end{aligned} \(\text{Hence, the virtual address space:} \) \[ 2^{11} \times 2^{10} = 2^{21} \text{ Bytes} \] \(\text{Thus, } 21 \text{ bits are required to represent the virtual address space.} \)