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
The Network Mapper (Nmap) commands that would be useful on this test network are:
- nmap xxx.xxx.x.x -p
- nmap xxx.xxx.x.x -A
<h3>What is a
penetration test?</h3>
A penetration test can be defined as a cybersecurity technique that simulates a cyber attack against an end user's computer system, so as to scan, identify, test and check for exploitable vulnerabilities in the following:
This ultimately implies that, a penetration test avails an tester the ability to exploit a weakness and potential security threats on a test network as a privileged user, during vulnerability assessments of the system.
<h3>The Network Mapper (Nmap) commands.</h3>
In this scenario, the Network Mapper (Nmap) commands that would be useful on this test network are:
- nmap xxx.xxx.x.x -p
- nmap xxx.xxx.x.x -A
Read more on penetration test here: brainly.com/question/25813524
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Hope this helped
