4.) [27 pts.] Consider the following parameters for a system with 24-bit address

Question

4.) [27 pts.] Consider the following parameters for a system with 24-bit address / 32-bit data CPU. Page Size TLB Size TLB Organization Cache Size Block Size Set Size 2 KB (2 32-entry (2 2-way set associative 256KB (2 Two 32-bit words 4 blocks per set Assume that the virtual address 0x345678 maps to the physical address 0xBA9E78. Also assume that when the processor tried to read the 32-bit word at the virtual location 0x345678, there was a hit in TLB and a hit in the data cache as wel a.) (6 pts.) Write down the virtual address 0x345678 in binary and show its division into page-offset and virtual page number fields. Further show the division of the virtual page number into set and tag fields based on the TLB parameters provided above Address bits to complete in binary and then indicate the divisions. 2322212019 18 17 1615 14 13 12 1 10 9876 543 2 /BE[3:0 b.) (2 pts.) If the sixteen sets in the TLB are numbered in binary from 0000-1111, which set was chosen? c.) (4 pts.) How many comparisons and of what size (include the valid bit) were performed to see if the translation was available in the TLB? d.) (3 pts.) How sets does the 256KB data cache consist of? e.) (6 pts.) Write down the physical address 0xBA9E78 in binary and mark the word, set, and tag fields based on the data cache parameters provided above Address bits to complete in binary and then indicate the divisions 23 22 21 2019 18 17 16 15 14 13 12 1 10 9 8765 4 32 BE3:0 f.) (1 pts.) Which set did the cache controller choose (index)? g.) (4 pts.) How many comparisons and of what size (include the valid bit) were performed to see if the block was available in the cache?

Explanation / Answer

so we have address:
0x345678 which is a 24 bit address which is equivalent to:
001101000101011001111000

Since the page size is 2KB, Least Significant 11 bits will be considered as Page Offset. so the page offset is 11001111000
and the remaining 13 bits (0011010001010) are virtual pagenumber bits.

Since we have 2 way set associative TLB with 32 entry, we shall have 16 different sets. To identify the set we need 4bit as set bit

So set bits are the 4 LSB of the virtual page number i.e. 1010 and the remaining 9bits are TAG field 001101000.

b. As already said, Set 1010 was choosen

c
2 comparison (set size) of the tag bits were made in the TLB to determine is the page requested is available in the TLB or not. Each comparion consist of the tag bits including a valid bit. So 10 bits of comparison is required.

d
The 256 KB set consist of block size 2*32 bit = 64bit = 8byte
Each set consist of 4 block, so the size of each set is 32byte
Total size is 256KB and set size is 32B, so we shall have (256*1024/32)8192 such sets.

As you probably know according to chegg protocol, we are suppose to solve 4 subparts. Am sorry buddy, for the same reason I cannot answer your remaininh queqtions. :(

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