GATE CS 1995 Q2.7 — Memory Management and Virtual Memory

Since each page contains $100$ records, the page number for any given address $A$ can be determined as $\lfloor A / 100 \rfloor$.

Let's map the given address sequence to its corresponding page numbers:

• $0100 \rightarrow \lfloor 100/100 \rfloor = 1$
• $0200 \rightarrow \lfloor 200/100 \rfloor = 2$
• $0430 \rightarrow \lfloor 430/100 \rfloor = 4$
• $0499 \rightarrow \lfloor 499/100 \rfloor = 4$
• $0510 \rightarrow \lfloor 510/100 \rfloor = 5$
• $0530 \rightarrow \lfloor 530/100 \rfloor = 5$
• $0560 \rightarrow \lfloor 560/100 \rfloor = 5$
• $0120 \rightarrow \lfloor 120/100 \rfloor = 1$
• $0220 \rightarrow \lfloor 220/100 \rfloor = 2$
• $0240 \rightarrow \lfloor 240/100 \rfloor = 2$
• $0260 \rightarrow \lfloor 260/100 \rfloor = 2$
• $0320 \rightarrow \lfloor 320/100 \rfloor = 3$
• $0370 \rightarrow \lfloor 370/100 \rfloor = 3$

The resulting page reference string is:
$1, 2, 4, 4, 5, 5, 5, 1, 2, 2, 2, 3, 3$

With only $1$ free main memory frame, a page fault occurs every time a referenced page is different from the page currently residing in the frame:

1. Reference $1$ $\rightarrow$ Fault (Frame: $[1]$)
2. Reference $2$ $\rightarrow$ Fault (Frame: $[2]$)
3. Reference $4$ $\rightarrow$ Fault (Frame: $[4]$)
4. Reference $4$ $\rightarrow$ Hit (Frame: $[4]$)
5. Reference $5$ $\rightarrow$ Fault (Frame: $[5]$)
6. Reference $5$ $\rightarrow$ Hit (Frame: $[5]$)
7. Reference $5$ $\rightarrow$ Hit (Frame: $[5]$)
8. Reference $1$ $\rightarrow$ Fault (Frame: $[1]$)
9. Reference $2$ $\rightarrow$ Fault (Frame: $[2]$)
10. Reference $2$ $\rightarrow$ Hit (Frame: $[2]$)
11. Reference $2$ $\rightarrow$ Hit (Frame: $[2]$)
12. Reference $3$ $\rightarrow$ Fault (Frame: $[3]$)
13. Reference $3$ $\rightarrow$ Hit (Frame: $[3]$)

Counting the page faults, we get a total of $7$ page faults.