The total number of pages that make up the executable program is shown (no fragm

Question

The total number of pages that make up the executable program is shown (no fragmentation The total size of a frame is equal to 512 bytes The CPU needs to access data that is at the logical address 0 10 10 10 10101 (12 bits) The size of a word in the system is 32 bits total program logical address pages page table 0 19 17 21 18 16 13 20 29 2 6 7 8 9 10 15 bina decimal A. What is the logical page number? B. What is the logical page offset? C. What is the total size in bytes of the program? bytes D. What is the actual physical address in binary (12 bits) E. What is the total size of a page in bytes in this system? bytes 20 points total

Explanation / Answer

Answer

given by

The page frame size is= 512 bytes

Means that sheet offset is 2^9=> 9 bits be for frame

The rational address is= 010101010101 (12 bits)

A)

The size of the statement in the organization is 32 bits = 2^5, this means

The upper 5 bits are logical sheet number= 01010= 10

B)

the rational page offset = 7 bits, since 12-5 = 7, 5 bits are for rational adress

C)

Total size of the plan is = 2^12 bits = 2^7 * 32 bit

in bytes = 2^7 * 2^2 = 512 bytes

or since one statement here is 32 bit, this means = 128 words

E)

entire size = 2^7 = 128

word = 128*32 bits = 2^7*2^2 bytes= 2^9 bytes = 512 bytes.

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