Let assume are lettered A to E in that order. Thus, there
will be 10 potential lines: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. Each of
these potential lines has 4 possibilities. Therefore, the total number of
topologies is 4¹⁰=1,048,576. 1,048,576. At 100ms <span>it will take 104,857.6 seconds which is slightly above 29 hours to inspect
each and one of them.</span>
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:
See the components in explaination
Explanation:
In order to make it as IPv6, few key components should be supported, those components are given below:
The infrastructure must support the enhanced protocol StateLess Address Auto-Configuration (SLAAC).
Static addressing with DHCPv6, dynamic addressing with DHCPv6 and SLAAC are the methods used to configure the IPv6. The network administrator should able to understand and implement the IPv6 through the DHCPv6.
Other than the implementation, working of IPv4 and IPv6 are same. Therefore, the administrator need not to learn new information for its working.
As the IPv6 address length is 128-bit and purpose is for everything on line to have an IP address. It must allow the internet to expand faster devices to get internet access quickly.
The DHCPv6 is not supported by all windows. Therefore, network administrator should check the corresponding Operating system (OS) would support the DHCPv6 for IPv6.
The network administrator must have good knowledge and skills on the IPv6.
The above mentioned key components should be verified by the network administrator in order to support for IPv6 project with DHCPv6.
Answer:g
public static int addOddMinusEven(int start, int end){
int odd =0;
int even = 0;
for(int i =start; i<end; i++){
if(i%2==0){
even = even+i;
}
else{
odd = odd+i;
}
}
return odd-even;
}
}
Explanation:
Using Java programming language:
- The method addOddMinusEven() is created to accept two parameters of ints start and end
- Using a for loop statement we iterate from start to end but not including end
- Using a modulos operator we check for even and odds
- The method then returns odd-even
- See below a complete method with a call to the method addOddMinusEven()
public class num13 {
public static void main(String[] args) {
int start = 2;
int stop = 10;
System.out.println(addOddMinusEven(start,stop));
}
public static int addOddMinusEven(int start, int end){
int odd =0;
int even = 0;
for(int i =start; i<end; i++){
if(i%2==0){
even = even+i;
}
else{
odd = odd+i;
}
}
return odd-even;
}
}
