amirbiomedica
دوشنبه 08 فروردین 1390, 19:22 عصر
Assignment 2 Solution
NO WORK SHOWN, NO MARKS
Note: Bits coloured in red are related to the subnets
Problem 1 Solution
Number of needed usable subnets: 126
Number of needed usable hosts per subnet: 131,070
Network Address: 121.0.0.0
Answer
Show Your Work Here for Marks
Address class
A
*** No work needed here ***
Default subnet mask
255.0.0.0
1111 1111.0000 0000.0000 0000.0000 0000
Custom subnet mask
255.254.0.0
1111 1111.1111 1110.0000 0000.0000 0000
Total number of subnets
128
2 7 = 128
Number of usable subnets
126
2 7 - 2 = 126
Total number of host addresses/subnet
128 k
2 17 = ( 27 ) ( 210 ) = 128 k
Number of usable host addresses/subnet
128 k - 2
128 k – 2
Number of bits borrowed
7
27 yields a total of 128 subnets
Problem 2 Solution
Number of needed usable subnets: 14
Network Address: 179.50.0.0
Answer
Show Your Work Here for Marks
Address class
B
*** No work needed here ***
Default subnet mask
255.255.0.0
1111 1111.1111 1111.0000 0000.0000 0000
Custom subnet mask
255.255.240.0
1111 1111.1111 1111.11110000.0000 0000
Total number of subnets
16
24 = 16
Number of usable subnets
14
2 4 – 2 = 14
Total number of host addresses/subnet
4k
212 = (22)(210) = 4 k
(Positions of host bits are coloured in green above)
Number of usable host addresses/subnet
4k - 2
212 – 2 = 4k - 2
Number of bits borrowed
4
24 yields a total of 16 subnets
What is the 4rd usable subnet range, including the invalid host addresses
179.50.64.0 to
179.50.79.255
Usable subnet 3rd Octet 4th Octet
1st 00010000 0000 0000
2nd 0010 0000 0000 0000
3rd 0011 0000 0000 0000
4th 0100 0000 0000 0000
Thus, the 1st invalid host in the 4th usable subnet is:
Usable subnet 3rd Octet 4th Octet
4th 0100 0000 0000 0000
(In decimal) 64 0
and the last invalid host in the 4th usable subnet is:
Usable subnet 3rd Octet 4th Octet
4th 0100 1111 1111 1111
(In decimal) 79 255
What is the subnet number for the 5th usable subnet?
179.50.80.0
Usable subnet 3rd Octet 4th Octet
5th0101 0000 0000 0000
(Note: “Subnet number” or “network number” requires that the host be all zeros)
What is the subnet broadcast address for the 6th usable subnet?
179.50.111.255
Usable subnet 3rd Octet 4th Octet
6th 0110 0000 0000 0000
Since broadcast address is the address with the host being all ones. Thus, broadcast address is:
Usable subnet 3rd Octet 4th Octet
6th 0110 1111 1111 1111
(In decimal) 111 255
What are the assignable addresses for the 5th usable subnet?
179.50.80.1
to
179.50.95.254
Usable subnet 3rd Octet 4th Octet
5th 0101 0000 0000 0000
First assignable address is:
Usable subnet 3rd Octet 4th Octet
5th 0101 0000 0000 0001
(In decimal) 80 1
Usable subnet 3rd Octet 4th Octet
5th 0101 1111 1111 1110
(In decimal) 95 254
Problem 3 Solution
Number of needed usable subnets: 30
Number of needed usable hosts per subnet: 6
Network Address: 199.168.100.0
Answer
NO WORK SHOWN, NO MARKS
Note: Bits coloured in red are related to the subnets
Problem 1 Solution
Number of needed usable subnets: 126
Number of needed usable hosts per subnet: 131,070
Network Address: 121.0.0.0
Answer
Show Your Work Here for Marks
Address class
A
*** No work needed here ***
Default subnet mask
255.0.0.0
1111 1111.0000 0000.0000 0000.0000 0000
Custom subnet mask
255.254.0.0
1111 1111.1111 1110.0000 0000.0000 0000
Total number of subnets
128
2 7 = 128
Number of usable subnets
126
2 7 - 2 = 126
Total number of host addresses/subnet
128 k
2 17 = ( 27 ) ( 210 ) = 128 k
Number of usable host addresses/subnet
128 k - 2
128 k – 2
Number of bits borrowed
7
27 yields a total of 128 subnets
Problem 2 Solution
Number of needed usable subnets: 14
Network Address: 179.50.0.0
Answer
Show Your Work Here for Marks
Address class
B
*** No work needed here ***
Default subnet mask
255.255.0.0
1111 1111.1111 1111.0000 0000.0000 0000
Custom subnet mask
255.255.240.0
1111 1111.1111 1111.11110000.0000 0000
Total number of subnets
16
24 = 16
Number of usable subnets
14
2 4 – 2 = 14
Total number of host addresses/subnet
4k
212 = (22)(210) = 4 k
(Positions of host bits are coloured in green above)
Number of usable host addresses/subnet
4k - 2
212 – 2 = 4k - 2
Number of bits borrowed
4
24 yields a total of 16 subnets
What is the 4rd usable subnet range, including the invalid host addresses
179.50.64.0 to
179.50.79.255
Usable subnet 3rd Octet 4th Octet
1st 00010000 0000 0000
2nd 0010 0000 0000 0000
3rd 0011 0000 0000 0000
4th 0100 0000 0000 0000
Thus, the 1st invalid host in the 4th usable subnet is:
Usable subnet 3rd Octet 4th Octet
4th 0100 0000 0000 0000
(In decimal) 64 0
and the last invalid host in the 4th usable subnet is:
Usable subnet 3rd Octet 4th Octet
4th 0100 1111 1111 1111
(In decimal) 79 255
What is the subnet number for the 5th usable subnet?
179.50.80.0
Usable subnet 3rd Octet 4th Octet
5th0101 0000 0000 0000
(Note: “Subnet number” or “network number” requires that the host be all zeros)
What is the subnet broadcast address for the 6th usable subnet?
179.50.111.255
Usable subnet 3rd Octet 4th Octet
6th 0110 0000 0000 0000
Since broadcast address is the address with the host being all ones. Thus, broadcast address is:
Usable subnet 3rd Octet 4th Octet
6th 0110 1111 1111 1111
(In decimal) 111 255
What are the assignable addresses for the 5th usable subnet?
179.50.80.1
to
179.50.95.254
Usable subnet 3rd Octet 4th Octet
5th 0101 0000 0000 0000
First assignable address is:
Usable subnet 3rd Octet 4th Octet
5th 0101 0000 0000 0001
(In decimal) 80 1
Usable subnet 3rd Octet 4th Octet
5th 0101 1111 1111 1110
(In decimal) 95 254
Problem 3 Solution
Number of needed usable subnets: 30
Number of needed usable hosts per subnet: 6
Network Address: 199.168.100.0
Answer