Question

2. Determine how many sets of cache blocks will be there for the following Cache memory size (in bytes) Direct Mapped Blocks Size (in bits) 32 64 218 2-way Set Associative Block Size (in bits) 32 64 A 2A6 [0.5 * 16 = 8] 4-way Set Associative Block Size (in bits) 32 64 SK 64K 256K 3. The physical memory address generated by a CPU is converted into cache memory addressing scheme using the following mapping scheme (Fig. B3 of book). Fill in the following table considering a cache size of 64K [0.5*16=8] Block address Tag Block offset Index Direct Mapped Physical Address size 32 bits 40 bits 32 64 2-way Set Associative Physical Address size 32 bits 40 bits 64 5 I 4-way Set Associative Physical Address size 32 bits 40 bits 6 5 Blocks Size (in bits) # of block offset bits # of index bits # of tag bits 18

question 2 and 3

Answer 1

Aivect mapped! No Ni # of cacahe Blocks. Bleek size e Bore) * Block sizel Bis) 2 32 © CoS = 8K:- 2013 8 * Block -Size = 64 CS SK:- BS 95 c.s= 64k:- 16 l C.S= 64k. NOS BS 15 - = r 2 ® CIS = 256K:- Ble 13 12 @ cis = 256K! 2018 N=cs - 2 2 B.s 25 2-way set. Associative: - SE #of sets. > 2 Blocs i set * Block 8126 = 64 s=N/ * Block- size z 82 Cis: 8. sa er @ CIS 264K: - 2 ® CiS = 256K :- SE 2'03 4-way set. Associative: - * Block size = 64 @ 5:25 * Block size = 32 COS - 8k; S: 26 ② cos a 64ks - 29 & c.s= 256K, anned with CamScanner

3) Direct mapped: ① PA= 32-bit 10 PA-40-bit BS=32 bit 248 2 B => obtset=2 bits 85 = 64 bits = 288=2 oboset = 3 bits CS z 64k 13 CS 216 No. of borracho 23 index = 11 - TAG index offset 1 19 121 32 bits index = 13 bits. :. TAG = 40 - ( 13+3) = 24 bits 24 13 T 3 T TAG index obbset 2-way set ASS:- ① PA= 32-bit offset = 5 bits BiS = 25 #ob blocks en) == 2" #of sets (S) = 2"-210 index = 10 bits TAG - 32 - (10+5) = 17 bits TAG index offset © PA= uobits Block size z 64bit) € 8B22B offset = 3 bits N=9216 213 N 8223 22 > index = 812 bits TAG = 40-(3+12) = 25 bib TAG index offset 1 25 1 12 3 I 4-way set. ASS: - ☺ PA= 40 bits odoser = sbits D.A = 32-bits oftset = 6 bib TAG = 18 bits index = 32-(18+6) = 8 bits TAG index offset 18 18 16 5:2 2 index = 9 bits TAG= 40- (0 5+9)= 26 bits TAG oindex offset 1026 1 9 15 T CSScanned with CamScanner