A system has a main memory with 16 MB of addressable locations and a 32 KB of di

Question

A system has a main memory with 16 MB of addressable locations and a 32 KB of direct-mapped cache with 8 bytes per block. The minimum addressable unit is 1 byte.

a) How many blocks are in cache?
b) What is the address breakdown for main memory?

---------------------------------

2)

Suppose that a virtual memory system has the following properties: 40-bit virtual byte-address
16 KB pages
32-bit physical address

TLB has 8-entries with fully associative structure
Valid, dirty, and use bits take a total of 3 bits (both TLB and page table have these bits)

Compute the following quantities:

a) Total size of TLB (in bits)

b) Total size of page table (in bits)

---------------

3)

Set-Associative: Parameters:

Main Memory: 1 GB

Block/line size: 64 Byte

Cache: 1 MB 8-way set-associative

Questions:

How many main memory blocks are there

How many cache lines are there?

How many cache lines per set are there?

How many cache sets are there?

How many blocks are map to a single cache set?

What is the address breakdown for cache?

4)

Fully Associative: Parameters:

• • •

Questions:

• • • •

Main Memory: 1 GB Block/line size: 64 Byte Cache: 1 MB fully associative

How many main memory blocks are there How many cache lines are there?
How many blocks map to a single cache line? What is the address breakdown for cache?

---------

5)

Cache design: Direct-Mapped Parameters:

Main memory: 1 GB

Block/line size: 64 Byte

Cache: 1 MB direct-mapped

Questions:

a) How many main memory blocks are there?

b) How many cache lines are there?

c) How many blocks map to a single cache line?

d) What is the address breakdown for cache?

Explanation / Answer

As per Chegg policy I am doing the first question. Please post the rest parts as separate question.

a) How many blocks are in cache?
cache is 8 bytes per block

Total size of cache is 32 KB i.e. 215 bytes per cache

(1 block / 23 bytes) x (215 bytes / cache) = 212 blocks / cache

So we have 212 blocks per cache available and we need 12 bits for indexing

b) What is the address breakdown for main memory?

We have 16 MB of addressable locations i.e.224 bytes. So to address it we need 24 bits.

We have 8 bytes per block i.e. 23 bytes. So we need 3 bits for that.

As solved in previous question we need 12 bits for index.

So memory breakdown is as follows

3 bits as offset, 12 bits for index and 9 bits for tag

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.