Two hosts are connected via a packet switch with 107 bits per second links. Each link has a propagation delay of 20 microseconds. The switch begins forwarding a packet 35 microseconds after it receives the same. If 10000 bits of data are to be transmitted between the two hosts using a packet size of 5000 bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is
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A computer has 32-bit instructions and 12-bit addresses. There are 250 two address instructions. How many one address instructions can be formulated?
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5 ANSWERS
Priya Shukla
Priya Shukla
Answered Oct 12, 2016
32 bit instruction and 12 bit two address instruction so total used bit for address are 12+12=24 remaining 32–24=8 bits for opcode . 2^8 combination that means 256 combinations can be made
this you can understand as if we are having 6 bit instruction in which 3 bits are for address so remaining 3 will be for opcode and we can make 2^3 combination as follows
000
001
010
011
100
101
110
110
so accordingly we have total 2^8 combinations according to question that means 256 combination and in that 250 for two address then remaining 256–250=6 are for one address and as 12 bits are for address so total one address instructions are 6*2^12=24,756