Computer Network Fundamentals  Homework - 1

CS 542 – Computer Network Fundamentals 
Homework - 1 (70 points)
Instructions:
• Please submit soft copies on the blackboard.
• Team submissions are accepted.
• A team of 1 – 3 is accepted. Please mention all the team members' names and
CWID (A number) on the first page.
• For team submissions, one submission is sufficient. Anyone from the team
can submit—no need for everyone in the group to submit.
• All should submit typewritten documents, not handwritten ones.
• Show the complete step-by-step solution for full credits. If there is no
proper explanation and justification for your final answer, only partial
credits are given, even though your answer is correct.
• Please contact Viswatej Kasapu (vkasapu@hawk.iit.edu) if something is
not clear.
• The due date is Friday, Apr 02nd, 2021, at 11:59 PM (Central Time).
• Submissions after the due date are not accepted.
1. What is the range of addresses that can assign to users in the 2021 block of
class C? (3 points)
2. Convert the number C0514019 in the hexadecimal base to the dotted-decimal
notation. What is the class of this address? (consider classful addressing). (5
points)
3. Define the 1202 block of class B? (Give first and last address in the block) (3
points)
4. Convert the decimal number 5141.01568603515625 to the base 256 number
system. (5 points)
5. What is the value of 60.63.12.12.0
20.21.04.04
? Give results in 256 base system. (Given
numbers are in 256 base system) (4 points)
6. An organization is granted the block 142.200.208.0/21. The administrator wants
to create 16 subnets.
a. Find the subnet mask (1 point)
b. Find the number of addresses in each subnet (1 point)
c. Find the subnet address and the direct broadcast address for the first
subnet. (2 points)
d. Find the 4th and 99th addresses in the last subnet. (4 points)
7. Give the mask in the dotted-decimal notation:
a. For a block of Class A which results in 128 subnets (1 point)
b. Which combines 128 blocks of Class C into a supernet (1 point)
8. Convert an IP address 254.128.64.32 to the binary notation (2 points)
9. The 14th address of a block assigned to a specific organization is 120.65.89.141.
The organization needs 120 addresses to give to its 120 users. Find the mask
and define this block of addresses. Is there any wastage of the IP addresses? If
yes, how many? (Note: The number of router interfaces is 2)(4 points)
10. A block of addresses 120.200.240.0/20 granted to an ISP. These addresses are
allocated between two groups of customers. The first group has 20 customers,
each of which needs 64 addresses, the second group has 20 customers, each of
which needs 128 addresses. Show the subblocks and range of addresses for the
10th customer of the first group and the 10th customer of the second group. How
many addresses are still available after this allocation? (5 points)
11. Find first address, last address, and number of addresses in the block, if one of
the addresses in a block is 140.240.90.25/20 (3 points)
12. Consider the following routing table (the next-hop address is omitted):
Mask Network address Interface
/27 144.56.55.0 M0
/26 123.80.97.0 M1
/25 123.80.97.128 M2
/24 118.114.132.0 M3
Default Default M4
Give the interface number for a packet whose destination IP address is:
i) 144.56.55.31 (1 point)
ii) 144.56.56.31 (1 point)
iii)123.80.97.60 (1 point)
iv) 123.80.97.200 (1 point)
v) 123.80.97.88 (1 point)
vi) 118.114.133.1 (1 point)
13. The routing table of routers R1, R2, and R3 are given. Draw the possible network
configuration with all 3 routers, not separate configurations corresponding to
each routing table. Indicate the next-hop addresses in the figure. (10 points)
R1:
Mask Network Address Next-Hop Address Interface Number
/24 80.70.56.0 100.160.32.67 M2
/24 130.135.7.0 150.137.45.78 M1
/16 180.170.0.0 ----------- M0
/16 100.160.0.0 ----------- M2
/16 150.137.0.0 ----------- M1
Default Default 180.170.4.6 M0
R2:
Mask Network Address Next-Hop Address Interface Number
/24 80.70.56.0 ----------- M0
/16 100.160.0.0 ----------- M1
Default Default 100.160.56.7 M1
R3:
Mask Network Address Next-Hop Address Interface Number
/24 130.135.7.0 ----------- M0
/16 150.137.0.0 ----------- M1
Default Default 150.137.72.48 M1
14. Consider the network configuration below. A packet arrived at the router R3
with the destination address 150.14.8.56. Show how it is forwarded. (Assume
classless addressing and mask of each network is /24) Create a routing table for
R1 and R3. (10 points)
150.14.0.0 133.79.0.0
129.101.0.0
190.180.0.0
Rest of the
Internet
150.14.5.165
133.79.7.11
129.101.19.20
129.101.31.18
129.101.17.32
190.180.7.9
M0
M0
M0
M2
M1
M1
M1
R2
R1
R3