LOGICAL ADDRESS (14 bits) STBR (binary 16 bits) MAR (binary 16 bits) 1 001000 00
ID: 3700518 • Letter: L
Question
LOGICAL ADDRESS (14 bits) STBR (binary 16 bits) MAR (binary 16 bits) 1 001000 0000111 binary MAIN frame MEMORY 01000 01001 01010 01011 01100 01101 01110 01111 instruction 10000 10001 PAGE addressTABLE memerySEGMENT TABLE address 127 128 129 130 131 132 133 134 135 136 255 256 257 258 259 260 261 262 263 264 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 FRAME SIZE IS 512 BYTES TOTAL TOTAL NUMBER OF SEGEMENTS IN THE PROGRAM IS 8 LOGICAL ADDRESS IS GIVEN STBR MAR SEGMENT TABLE PAGE TABLE all answers in binary)Explanation / Answer
Word is 4 bytes.
Frame size is 512 bytes = 29 bytes
So, offset field of virtual address will be 9-2 = 7 bits
As the total number of segments is given as 8 so, the segment number field of the logical address will be 3 bits.
Remaining bits i.e 14 -(3 + 7 ) = 4 bits [as total size of logical address is 14 bits] will be in the page number field of the logical address.
logical address is given as = 10010000000111
The first three bits are segment number
therefore , segment number = 100
next 4 bits are page number
therefore , page number = 1000
now offset = 0000111
Now STBR is nothing but segment table base register .it contains the base address of the segment table (which is given as 127(in decimal)).
Therefore STBR will contain the value 0000000001111111 .
now STBR + segment number = 127 + 4(as segment number is 100) = 131
So in the segment table 131th entry will contain the page number value i.e = 1000 = 8(in decimal)
Now base of page table address is 255.
So 255 + 8 = 263
So, 263th entry of the page table will contain the value of the frame number where the instruction is present i.e = 01111
Now MAR will contain the address of this instruction which is nothing but frame number + offset = 0000011110000111.
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