source address to a destination address, 1 byte at a time. st = 0×5FC0, src = ®x1A10, length = ex3080 ache has 16 byte blocks, 2-way set associative, and 256 rows. 'irtual Memory: 64KB (16 bits) Page Size: 4K hysical Memory: 16MB (24 bits) "AGE TABLE: VirtPage#: PhysPage#: Ox0 ---> 0x200 Ox1 ---> 0×31F Ox2 ---> 0x5B2 Ox3 ---> 0x33C Ox4 ---> 0×4FF Øx5 ---> 0x5C5 Ox6 ---> 0x662 Ox7 ---> 0x3A7 Ox8 ---> 0x808 Ox9 ---> 0X9D0 OxA ---> 0×44A f the virtual address is Ø×53AC what is the corresponding physical address?

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Question Statement at the bottom, could you explain how the physical address is found?

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Given the following C instruction, memcpy( 0×5FC0, ex1A10, Øx3080), we copy length b
a source address to a destination address, 1 byte at a time.
dst = 0×5FC0, src =
Øx1A10, length - 0x3080
Cache has 16 byte blocks, 2-way set associative, and 256 rows.
Virtual Memory: 64KB (16 bits) Page Size: 4K
Physical Memory: 16MB (24 bits)
PAGE TABLE: VirtPage#: PhysPage#:
---> 0x200
Ox0
Ox1 ---> 0X31F
Ox2
---> 0x5B2
Ox3
---> Ох33С
Ox4 ---> 0×4FF
Ox5 ---> 0x5C5
Ox6
---> 0x662
Ox7
---> ОхЗА7
Ox8 ---> 0x808
Өx9 ---> Өх9D0
OxA ---> 0×44A
If the virtual address is O×53AC what is the corresponding physical address?
Transcribed Image Text:Given the following C instruction, memcpy( 0×5FC0, ex1A10, Øx3080), we copy length b a source address to a destination address, 1 byte at a time. dst = 0×5FC0, src = Øx1A10, length - 0x3080 Cache has 16 byte blocks, 2-way set associative, and 256 rows. Virtual Memory: 64KB (16 bits) Page Size: 4K Physical Memory: 16MB (24 bits) PAGE TABLE: VirtPage#: PhysPage#: ---> 0x200 Ox0 Ox1 ---> 0X31F Ox2 ---> 0x5B2 Ox3 ---> Ох33С Ox4 ---> 0×4FF Ox5 ---> 0x5C5 Ox6 ---> 0x662 Ox7 ---> ОхЗА7 Ox8 ---> 0x808 Өx9 ---> Өх9D0 OxA ---> 0×44A If the virtual address is O×53AC what is the corresponding physical address?
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