Question

A 64-bit word computer employs a 128KB cache. The address bus in this system is 32-bits. Determine the number of bits in each field of the memory address register (MAR) as seen by cache in the following organizations (show your calculations): A. Fully associative mapping with line size of 4 words. B. Direct mapping with the line size of 8 words. C. 2-way set associated mapping with the line size of 4 words. D. 8-way set associated mapping with the line size of 2 words.

Answer 1

Answer: provided in the explanation segment

Explanation:

taking a step by step process for this, we will analyze this problem,

we have that the Word size = 8 bytes

Cache size = 128 KB = 128×2¹⁰ bytes = 2¹⁷ bytes

(a). we are asked to cal for fully associative mapping with line size of  4 words.

• fully associative mapping with line size of  4 words = 4 × 8 = 32 bytes
• offset = log₂³² = 5 bit
• tag = 32 - (5+0) = 27 bit
• index bit = 0

(b). Direct mapping with the line size of 8 words:

• here  the line size = 8×8 = 64 bytes
• no of lines = 2¹⁷/2⁶ = 2¹¹ lines
• offset = log ₂(2⁶) = 6 bit
• index = log ₂(2¹¹) = 11 bit
• tag = 32 - (5 +11) = 15 bit

(c). 2-way set associated mapping with the line size of 1 word:

• no of lines = 2¹⁷/2³ = 2¹⁴ lines
• offset = log₂(2³) = 3 bit
• no of sets = 2¹⁴/2 = 2¹³ sets
• index = log₂(2¹³) = 13 bit
• tag = 32 - (3+13) = 16 bit

(d). 8-way set associated mapping with the line size of 2 words:

8-way set associated mapping with the line size of 2 = 2*8 = 16  bytes

• no of lines = 2¹⁷/2⁴ = 2¹³ lines
• no of sets = 2¹³/2³ = 2¹⁰ sets
• offset = log₂¹⁶ = 4 bit
• index = log ₂(2¹⁰) = 10 bit
• tag = 32 - (4+10) = 18 bit

cheers i hope this helps!!!!