I need a step by step solutiona nd explainationg for A and B.! please The total

Question

I need a step by step solutiona nd explainationg for A and B.! please

The total number of pages that make up the executable program is shown (no internal fragmentation). The total size of a frame is equal to512 bytes The CPU needs to access data that is at the logical address 0 0 1 1 0 1 0 1 0 1 1 1 (12 bits) The size of a word in the system is 32 bits. What is the logical page number? What is the logical page offset? What is the total size in bvtes of the program? bytes Wha is the total size of a page in bvtes in this system? bytes The page of the data that was needed was put into the physical frame as shown above. Fill in the correct index of the page table with the exact data that it holds.

I know The total size of a page = the total frame size

total size in bytes of the program = # of pages * total size of a frame.

I need help with logical page number and logical page offset.

Explanation / Answer

Given data:

The total size of a frame = 512bytes

The logical address= 0 0 1 1 0 1 0 1 0 1 1 1 (12 bits).

The size of a word in the system = 32 bits.

The total size of a frame = 512bytes

                                       = 128 *32bits

So, the number of words = 128words           

Hence the page contains 128words.

The logical address= 0 0 1 1 0 1 0 1 0 1 1 1 (12 bits).

High order 5 bits are denotes the page number =2^5 =32bits.

Therefore, the logical page number = 32bits

In logical address, the lower 7 bits are page offset = 2^7 =128bits

Therefore, the logical page offset = 128bits.

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