name MAR (binary 16 bits) LOGICAL ADDRESS (1 bits) STBR (binary 16 bits) 1001 0

Question

name MAR (binary 16 bits) LOGICAL ADDRESS (1 bits) STBR (binary 16 bits) 1001 0 0 0 b0001 1 1 binary MAIN frameMEMORY 01000 01001 01010 01011 01100 01101 01110 PAGE memory SEGMENT TABLE TABLE 127 128 129 130 131 132 133 134 135 136 255 256 257 258 259 260 261 262 283 264 10000 10001 GIVEN THE FOLLOWING: process P5 needs to access the instruction in memory FILL IN ALL KNOWN VALUES ABOVE FOR THE FOLLOWING WORD IS 4 BYTES STBR MAR SEGMENT TABLE PAGE TABLE FRAME SIZE IS 512 BYTES TOTA TOTAL NUMBER OF SEGEMENTS IN THE PROGRAM IS 8 all answers in binary) LOGICAL ADDRESS IS GIVEN

Explanation / Answer

Answer

given by

Word is 4 bytes.

Frame extent is 512 bytes = 29 bytes

So, offset field of fundamental address willpower be 9-2 = 7 bits

As the total digit of segments is given as 8 so, the piece number playing field of the logical talk to will be 3 bits.

residual bits i.e 14 -(3 + 7 ) = 4 bits [as total size of commonsense address is 14 bits] will be in the piece of paper number field of the sound address.

logical address is specified as = 10010000000111

The initial three bits are segment number

Therefore, segment number = 100

next 4 bits are page figure

then, page number = 1000

at this time offset = 0000111

Now STBR is nothing but slice table base record .it contains the base address of the part table (which is given as 127(in decimal)).

Therefore STBR will enclose the value 0000000001111111 .

now STBR + segment number = 127 + 4(as segment number is 100) = 131

So in the section table 131th entry will hold the page figure value i.e = 1000 = 8(in decimal)

Now top of page bench address is 255.

So 255 + 8 = 263

So, 263th entry of the folio table will contain the cost of the edge number where the lessons is present i.e = 01111

Now MAR will hold the address of this instruction which is nothing but border number + offset = 0000011110000111.

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