I am quoting a paragraph from the book Operating System Principles by Galvin.Usually, each page-table entry is 4 bytes long, but that size can vary as well. A 32-bit entry can point to one of $2^{32}$ physical page frames. If frame size is 4 kB, then a system with 4-byte entries can address $2^{44}$ bytes (or 16 TB) of physical memory.Now, I know we have $2^{44}$ bytes of memory because we have $2^{32}$ page frames and each frame size is 4 kB, i.e. $2^{12}$ bytes of memory, so physical memory is $2^{32}\cdot2^{12} = 2^{44}$.Please help me to understand the following:How did we get the number of frames as $2^{32}$?If the logical memory space is $2^{32}$ then what should the physical memory space be (considering both the fully used and partially used logical address space concept)?
Operating System Paging concept
operating systems;memory management;virtual memory;paging