9. Virtual Memory

Question 1

What main problems are addressed by virtual memory?

Answer 1

Capacity (we wish for the physical memory to be smaller than the address space)

Safety (we want to prevent user programs from accessing the memory used by the OS and want to make sure that user programs cannot interfere with each other)

Question 2

Give a brief summary of the concept of virtual memory.

Answer 2

Each program thinks it owns the entire memory –> the virtual address space.

Virtual addresses are translated to physical addresses by the OS and hardware.

Question 3

Swap space is located in which memory?

Answer 3

Secondary.

Question 4

What is a page?

Answer 4

The equivalent to a block of virtual memory. Page sizes tend to be 4-16 KB.
Mapping is done through a page table.

Question 5

Does a cache or physical memory have a higher miss rate?

Answer 5

Cache.

Question 6

A virtual address is divided up into what parts in a page table?

Answer 6

Virtual page number (equivalent of tag) and page offset.

Question 7

Does virtual memory use write-through or write-back?

Answer 7

Write-back

Question 8

How does the computer know where the page table for a program is stored?

Answer 8

The start of the table is pointed to by the page table register.

Question 9

Why is no tag required for a page table?

Answer 9

Each page has its own mapping.

Question 10

What acts like the index in a page table?

Answer 10

The full block address (virtual page number)

Question 11

Describe how a virtual address is translated into a physical address.

Answer 11

The first 20 bits of the virtual page number is the index of the page table. The 18-bit address at that entry is the physical page number.

The last 12 bits are the offset, and are kept the same.

The physical page number is added to the front of the offset to get the physical address.

Question 12

What is the size of the page table given a 32-bit virtual address space, 4 KB physical pages and 1 GB of main memory?

Answer 12

4 KB = 4000 B, so 12 bits are required for offset.
Then, we have 20 bits for virtual page number.

1 GB = 1,000,000,000 B, so 30 bits are needed for the physical address. We therefore have 18 bits for virtual page number.

There are 2²⁰ entries in the table with 18 bits required for each.

Thus, 18*2²⁰ bits = 2.25 MB

Question 13

What are the ARM special bits in a page table?

Answer 13

Access (set when the page is accessed, and reset periodically)
Modified (set if data needs to be written back to disk)
Residence (set if page table is in memory)

Question 14

What bits control the permissions of a particular entry in a page table?

Answer 14

W (read & write)
R (read only)
X (execute only)

Question 15

What is the only manager that can edit a page table?

Answer 15

Operating System.

Question 16

Why is a sole page table too slow to be useful?

Answer 16

There are two memory accesses required - one to look up the address and one to load/store.

Question 17

What is a TLB?

Answer 17

A TLB (Translation Lookaside Buffer) is a cache of page table entries.

The tag is the virtual page number, and physical page number is the entry.

Question 18

What are the 5 TLB status bits?

Answer 18

V (Valid)
D (Dirty)
R (Read)
W (Read & Write)
X (Execute)