Answer:
True
Explanation:
Both HLOOKUP and VLOOKUP are excel functions used for searching through tables for a specified lookup value to either get an exact or approximate match. The VLOOK and HLOOKUP functions have identical syntax which only differs with the HLOOKUP requiring the row index number to search through the rows while, VLOOKUP requires the column index number to search through the columns. Other than this difference, the other syntax values are the same.
VLOOKUP(lookup_value, table_array, col_index_num, [range_lookup])
HLOOKUP(lookup_value, table_array, row_index_num, [range_lookup])
Suppose a byte-addressable computer using set-associative cache has 2 16 bytes of main memory and a cache size of 32 blocks and each cache block contains 8 bytes. a) If this cache is 2-way set associative, what is the format of a memory address as seen by the cache; that is, what are the size of the tag, set, and offset fields
Answer:
Following are the responses to these question:
Explanation:
The cache size is 2n words whenever the address bit number is n then So, because cache size is 216 words, its number of address bits required for that cache is 16 because the recollection is relational 2, there is 2 type for each set. Its cache has 32 blocks, so overall sets are as follows:
[tex]\text{Total Number of sets raluired}= \frac{\text{Number of blocks}}{Associativity}[/tex]
[tex]=\frac{32}{2}\\\\ =16\\\\= 2^4 \ sets[/tex]
The set bits required also are 4. Therefore.
Every other block has 8 words, 23 words, so the field of the word requires 3 bits.
For both the tag field, the remaining portion bits are essential. The bytes in the tag field are calculated as follows:
Bits number in the field tag =Address Bits Total number-Set bits number number-Number of bits of words
=16-4-3
= 9 bit
The number of bits inside the individual fields is therefore as follows:
Tag field: 8 bits Tag field
Fieldset: 4 bits
Field Word:3 bits