CSCI 223 Computer Org. & Assembly Language Homework #7 DUE: 5/1/2018 (Tuesday),
ID: 3720982 • Letter: C
Question
CSCI 223 Computer Org. & Assembly Language Homework #7 DUE: 5/1/2018 (Tuesday), 9:30 AM Submission: Submit the hard copy of your homework in class (after printing if electronically written) Simulate a small virtual memory system with TLB and L1 D-cache with following assumptions .The memory is byte addressable .Memory accesses are to 1-byte words (not 4-byte words) Virtual addresses are 14 bits wide (n-14) Physical addresses are 12 bits wide (m - 12) The page size is 64 bytes (P 64) .The TLB is four-way set associative with 16 total entries The L1 D-cache is physically addressed and direct-mapped, with a 4-byte line size and 16 total sets Figure below shows the format of the virtual and physical addresses. Since each page is 26-64 bytes, the low-order 6 bits of the virtual and physical addresses serve as the VPO and PPO respectively. The high-order 8 bits of the virtual address serve as the VPN. The high-order 6 bits of the physical address serve as the PPN 13 1210 9 87 65 43 21 0 address Virtual page number) (Virtual page offset) 10 9 8 76 543 20 (Physical page number) (Physical page offset) Figure below shows a snapshot of our little memory system including the TLB (a), a portion of the page table (b), and the L1 cache (c). Above the figures of the TLB and cache, we have also shown how the bits of the virtual and physical addresses are partitioned by the hardware as it accesses these devices. 13 12 11 10 9 76 5 3 2 0 Set Tag PPN Valid Tag PPN Valid Tag PPN Valid Tag PPN Vali (a) TLB: Four sets, 16 entries, four-way set ssociative 07 03 2D 04 0 03 07 03 OD 0A 34 VPN PPN Valid VPN PPN Valid 08 13 01 02 03 OA OB oC 05 06 07 OE (b) Page table: Only the first 16 PTEs are shownExplanation / Answer
CPU generates load instruction 0x03d7
Therefore virtual address = 00 0011 1101 0111
STEP 1: SEARCH IN THE TLB:
The VPN part of the virtual address is = 00 0011 11
TLBI is = 11 and TLBT = 00 0011
Therefore we will search in set number 3 and match the TAG as 03.
In set number 3 there is TAG as 03 so this is a TLB HIT.
The corresponding PPN = 0D = 0000 1101 and as valid bit is 1 that means the page is present in the memory.
The corresponding Physical address(PA) is = 0000 1101 + VPO = 00 1101 01 0111
STEP 2 : SEARCH THE CACHE:
CO = 11
CI = 0101
CT = 001101
So index = CI = 0101 = 5
TAG = CT = 001101 = 0D
So, we seach the index 5 with the TAG 0D and it is a CACHE HIT.
Therefore we access Blk0 = 36 , Blk1 = 72 , Blk2 = F0 , Blk3 = 1D