Consider the following two-dimensional array $D$ in the C programming language, which is stored in row-major order: int D[128][128]; Demand paging is used for allocating memory and each physical page frame holds 512 elements of the array $D$. The Least Recently Used (LRU) page-replacement policy is used by the operating system. A total of 30 physical page frames are allocated to a process which executes the following code snippet:for (int i = 0; i < 128; i++) for (int j = 0; j < 128; j++) D[j][i] *= 10;The number of page faults generated during the execution of this code snippet is ___________.

To determine the total number of page faults:

1. Array Layout and Page Size:

   • The array size is $128 \times 128 = 16384$ elements stored in row-major order.

   • Each row has $128$ elements.

   • Each page frame holds $512$ elements, which corresponds to:

     $$\frac{512}{128} = 4\text{ rows per page}$$

   • Total number of pages required to store array $D$:

     $$\frac{16384}{512} = 32\text{ pages (labeled } P_0, P_1, \dots, P_{31}\text{)}$$

2. Access Pattern:

   • The nested loops access the array column by column: for a fixed outer index i, the inner loop iterates j from $0$ to $127$, accessing $D[j][i]$.

   • For $j = 0, 1, 2, 3$, the elements belong to Page $P_0$.

   • For $j = 4, 5, 6, 7$, the elements belong to Page $P_1$, and so on up to Page $P_{31}$.

   • Thus, for each value of i, the access sequence of pages is:

     $$P_0, P_0, P_0, P_0, P_1, P_1, P_1, P_1, \dots, P_{31}, P_{31}, P_{31}, P_{31}$$

3. Page Replacement Behavior:

   • A total of $30$ frames are allocated using the LRU replacement policy.

   • In each outer iteration (for a given i), $32$ distinct pages are accessed in sequential order ($P_0 \to P_{31}$).

   • Accessing $P_0$ through $P_{29}$ fills all 30 frames ($30$ page faults).

   • Accessing $P_{30}$ replaces the least recently used page $P_0$ (1 page fault).

   • Accessing $P_{31}$ replaces $P_1$ (1 page fault).

   • Hence, each column traversal generates exactly $32$ page faults.

   • Because all $32$ pages are cycled through in each column traversal and only $30$ frames are available, every page is evicted before it can be reused in the next column traversal ($i+1$).

4. Total Page Faults:

   • Outer loop runs $128$ times.

   • Total page faults = $128 \times 32 = 4096$.