A virtual memory system uses 32-bit logical addresses and 4 KB page size. If physical memory is 512 MB, how many frames are in physical memory?
$Physical Memory = 512 MB = 512 * 2^20 bytes = 2^29 bytes. Page/Frame size = 4 KB = 4 * 2^10 = 2^12 bytes. Number of frames = 2^29 / 2^12 = 2^17 = 131,072 frames.$
A 2-level paging system has 32-bit virtual addresses, 4 KB page size, and 4-byte PTEs. Outer page table index is 10 bits and inner page table index is 10 bits. If physical memory is 64 MB, what is the size of the Physical Frame Number (PFN) field in bits?
$Physical memory = 64 MB = 2^26 bytes. Frame size = 4 KB = 2^12 bytes. Number of physical frames = 2^26 / 2^12 = 2^14 = 16,384 frames. PFN requires 14 bits.$
Consider a paging system with TLB. TLB lookup time = 20 ns, Main Memory access time = 100 ns. If TLB hit ratio is 0.85, calculate Effective Memory Access Time (EMAT) in nanoseconds.
$EMAT = Hit_rate * (TLB + Memory) + Miss_rate * (TLB + 2 * Memory) = 0.85 * (20 + 100) + 0.15 * (20 + 200) = 0.85 * 120 + 0.15 * 220 = 102 + 33 = 135 ns.$
In a demand paged system, main memory access time is 100 ns. Page fault service time is 8 ms (8,000,000 ns). If the required Effective Access Time is 200 ns, what is the maximum page fault rate p * 10^6? Enter integer value of p * 10^6.
$EAT = (1 - p)*100 + p*8,000,000 = 200 => 100 + 7,999,900 p = 200 => p = 100 / 7,999,900 \approx 1.25 * 10^{-5}. p * 10^6 = 12.5.$
A process references pages in sequence: 1, 2, 3, 4, 1, 2, 5, 1, 2, 3, 4, 5. How many page faults occur using FIFO with 3 frames (initially empty)?
Trace FIFO 3 frames: [1] (f1), [1,2] (f2), [1,2,3] (f3), [2,3,4] (f4), [3,4,1] (f5), [4,1,2] (f6), [1,2,5] (f7), [1,2,5] (hit), [1,2,5] (hit), [2,5,3] (f8), [5,3,4] (f9), [5,3,4] (hit). Total = 9 page faults.