Given a virtual memory system with the following specification . The memory is b

Question

Given a virtual memory system with the following specification . The memory is byte addressable Each memory access is to 1-byte word. Each virtual address is 16-bit wide. Each physical address is 12-bit wide. The page size is 32 bytes (25) The TLB is 4-way set associative with 16 total entries. The L1 data-cache is physically addressed and direct mapped, with a 4 byte line size and 16 total sets. If the CPU issues the following virtual address: Ox43d8: What is the value of the Virtual Page Offset or VPO (give your answer in binary)? What is the value of the Physical Page Offset or PPO (give your answer in binary)? What is the value of the TLB index (give your answer in binary)? What is the value of the TLB tag (give your answer in binary)?

Explanation / Answer

1. The value of the virtual offset is 10110 .

See page size is given as 32B = 2^5 .So 5 bits are used for offset. See the last 5 bits in that given 16bit binary number. Remember one thing : for virtual address or physical address doesnt matter ,the offset always remains the same.

2. The value of the TLB index is 110 .

8 = 2^3 so 3 bits are used for TLB index or SET index .So just see the next 3 bits in that 16 bit binary number leaving the 5 bits for offset . It is 110.

3. The value of the tag is 00100011

This one you can do easily. 6 bits were there ,5 bits gone for offset , 3 bits for index...remaining 8 are for tag sp 00100011.

4. This one is also easy , just skip the offset bits , rest is your virtual page numbr( tag + index) = 00100011110

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