Tutorial Work

1

$$\begin{array}{rl}1& \to 3\\ 4& \to 2\\ 5& \to 0\\ 7& \to 1\end{array}$$

title: Correct answer.

2

Virtual address is $16$ bits wide. Physical address is $15$ bits side. Page size is $2^{12}\text{ bytes}=4\text{ kB}$. Page table has $8$ bit wide entries.

1. How many bits correspond to the page number and the offset (within the page)? We have $16$ total bits for the address, the page offset needs $12$ bits, so:

   • $12$ bits for the offset
   • $4$ bits for the page number

2. How many pages fit into the physical memory?

   $$\frac{2^{15}}{2^{12}}=2^3=8\text{ pages}$$

3. Where are virtual pages $3$, $6$ and $11$ located?

   • $3$ is currently in memory at physical page $2$
   • $6$ is currently in memory at physical page $1$
   • $11$ is currently not in memory

4. Modify the content of the page table.

   Index	Valid Bit	Physical Page #
   0	1	5
   1	1	7
   2	1	6
   3	1	2
   4	0	-
   5	0	-
   6	0	-
   7	0	-
   8	0	-
   9	0	-
   10	0	-
   11	1	1
   12	1	0
   13	1	3
   14	0	-
   15	1	4

title: Correct answers.

3

Assume the page table of a system is stored in main memory. Suppose a main memory access takes $200\text{ ns}$. The page fault rate is $1\%$. Accessing a page not in main memory takes $10\text{ ms}$.

Calculate the EAT.

$$\begin{array}{rl}EAT& =\text{hit rate}\times (2\times \text{memory access})+\text{miss rate}\times (\text{out of memory access})\\ & =0.99\times (2\times 200)+0.01\times (10,000,000)\\ & =100,396\text{ ns}\end{array}$$

title: Correct answer.