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:
The configuration of the R1 is as follows
Explanation:
Router>enable
Router#show running-config
Router#show startup-config
Router#configure terminal
Router(config)#hostname R1
R1(config)#line console 0
R1(config-line)#password letmein
R1(config-line)#login
R1(config-line)#exit
R1(config)#enable password cisco
R1(config)#enable secret itsasecret
R1(config)#service password-encryption
R1(config)#banner motd #Unauthorized access is strictly prohibited#
R1(config)#end
R1#exit
R1>enable
R1#copy running-config startup-config
R1#show flash
R1#copy startup-config flash
Kobe was the best basketball player R.I.P
Answer:
<u>d. Statement A is true and Statement B is false</u>
Explanation:
Indeed, when using duplex transmission either node can transmit while the other node can receive data from the network. Also, in half-duplex transmission, both the nodes can transmit as well as receive data.
However, in half-duplex transmission, the nodes <em>cannot </em>transmit and receive data at the same time. Hence, this makes Statement B false, while Statement A is true.
