Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
Answer:
Setting of short lease time for IP addresses in order to enhance quicker access from clients
Answer:
True
Explanation:
If s=abcd is a string defined over {a,b,c,d}, it corresponds to a regular expression which can be represented using a finite automata. Then the reverse of the string essentially corresponds to another finite automata where the starting state becomes the accepting state and vice versa.Moreover all the directions of state transitions will be reversed for each of the transitions in the original automata.
With these modifications, the new finite automata will accept a string which is reverse of the original string ,namely, dcba and this string will ne part of the reverse language.
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