Question

A system has a Translation Lookaside Buffer (TLB) that has a reach of 1 MB. TLB
reach is defined as the total amount of physical memory that can be accessed
through the TLB entries. The paging system uses pages of size 4 KB. The virtual
address space is 64 GB and physical address space is 1 GB. If each TLB entry stores
a 4-bit process id, page number, frame number, and a 2-bit control field, then the
size of the TLB (in bytes) is ___________. (answer in integer)
Note: \(1\mathrm{K}=2^{10}\), \(1\mathrm{M}=2^{20}\), \(1\mathrm{G}=2^{30}\)

Hint:

First use TLB reach and page size to get the number of TLB entries. Then use virtual address space and physical address space (divided by page/frame size) to get the widths of the page number and frame number fields. Add process id + page number + frame number + control bits to get one entry's size in bytes, then multiply by the number of entries.

Answer 1

Correct Answer: 1536

This question can be solved by breaking the TLB into 'how many entries' and 'how big is each entry,' then multiplying the two.

Finding the entry count: The TLB reach tells us how much physical memory the TLB can cover in one shot. Since reach = entries \(\times\) page size, we get:

\[entries = \frac{1\,MB}{4\,KB} = \frac{2^{20}}{2^{12}} = 2^8 = 256 \text{ entries}\]

Finding the width of each field in an entry:

• Process id field: given directly as 4 bits.
• Page number field: must be wide enough to represent every possible virtual page. With a 64 GB virtual space and 4 KB pages, there are \(2^{36}/2^{12} = 2^{24}\) distinct pages, so this field needs 24 bits.
• Frame number field: must be wide enough to represent every possible physical frame. With a 1 GB physical space and 4 KB frames, there are \(2^{30}/2^{12} = 2^{18}\) distinct frames, so this field needs 18 bits.
• Control field: given directly as 2 bits.

Adding up one entry's width:

\[4 + 24 + 18 + 2 = 48 \text{ bits} = 6 \text{ bytes per entry}\]

Scaling up to the whole TLB: multiply the per-entry size by the number of entries computed from the reach:

\[256 \times 6 \text{ bytes} = 1536 \text{ bytes}\]

So the TLB occupies a total of \(1536\) bytes.