ELEC365 atibility Mode) Word (Product Activation Failed) UT REFERENCES MAILINGSR
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ELEC365 atibility Mode) Word (Product Activation Failed) UT REFERENCES MAILINGSREV Lacne n VLISS LVLISS [Problem 2] Cache Design rider a processor which uses 15 bit addresses and can address 2415-32K bytes of memory. Suppose t it has one level of cache. The address is split into a t bit tag, an s bit set index, and a b bit block offiset. The cache consists of 512 bytes, with a block size of 32 bytes. [Part A] How many cache lines are there? Part B] What is block field? Part C] What is set? [Part D] What is tag field? [Problem 3] Cache Design Consider a processor which uses a 30 bit addresses and can address 2x301GB bytes of that the processor has one level of direct-mapped cache memory. The address is split into atois ag, an s bit set index, and b bit block offset. The cache consists of 32KB bytes, with a block size of 64 bytes. Suppose [Part A] How many cache lines are there? Part B] What is block field? Part C] What is set? [Part D] What is tag field? ATES)Explanation / Answer
Problem 2
Given Size of cache = 512 bytes
bits required for cache addressing = 15
A. The cache size is 512 bytes and has 32 bytes per lines so number of cache lines are : 512bytes/32bytes i.e 16.
B. Since there are 32 bytes in each block so the block field will contain 5 bits i.e 2^5=32.
C. Since it uses one way associative so each set uses 32 bytes of memory. The entire cache contains 512 bytes of memory. So there are total 512bytes/32bytes=16 bytes sets. So, set field contains 4 bits i.e. 2^4= 16 bytes.
D. So tag field contains the remaining i.e = 32-5-4 =23 bits.
Problem 3:
Given Size of cache = 32 kb
bits required for cache addressing = 30
A. The cache size is 32k bytes and has 64 bytes per lines so number of cache lines are : 32kb/64bytes i.e 512.
B. Since there are 64 bytes in each block so the block field will contain 6 bits i.e 2^6=32.
C. Since it uses one way associative so each set uses 64 bytes of memory. The entire cache contains 32 kb of memory. So there are total 32kb/32bytes=512 bytes sets. So, set field contains 9 bits i.e. 2^9= 512 bytes.
D. So tag field contains the remaining i.e = 32-6-9 =17 bits.
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