Given,
Destination IP address: 128.171.17.5
Anding with Subnet mask gives destination network or subnet.
ANDing with 255.255.0.0 (/16)
ANDing with 255.255.255.0 (/24)
ANDing with 255.0.0.0 (/8)
Therefore, It matches with the following destination networks:
Hence, Router will choose interface 3 and route the packet to destination address or subnet 128.171.17.0 because it has the longest prefix match with it compared to 128.17.0.0
Question 48 10 pts Row Mask (/Prefix) Interface Destination Network or Subnet Metric (Cost) Next- Hop...
Row Network/Subnet 0.0.0.0 1 128.171.0.0 172.30.33.0 Mask (/Prefix) Metric (Cost) Interface Next Hop Router 0.0.0.0 (/0) 255.255.0.0/16) 47 255.255.255.0 (/24) Local 255.255.255.0/24) 12 255.255.255.0 (/24) 55 255.255.255.0 (/24) 34 Local 255.255.255.0 (/24) 20 192.168.6.0 128.171.17.0 172.29.8.0 172.29.8.0 For the routing table above, if the arriving packet has a destination IP address 128.171.17.56, it will match with row 1 5 3
Test Your Understanding 10. a) Distinguish between Step 1 and Step 2 in the routing process. b) If any row other than the default row matches an IPv4 address, why will the router never choose the default row? c) Which rows in Figure 8-11 match 128.171.17.13? (Don't forgst the default row.) Show your calculations for rows that match. d) Which of these is the best-match row? Justify your answer. e) What rows match 172.40.17.6? Show your calculations for rows that...
Subnet mask 255.255.255.0 can never be used for class B network. Question 1 options: True False Question 2 (1 point) In CIDR notation, /22 indicates there are 22 ones in the subnet mask. Question 2 options: True False Question 3 (1 point) When deploy a network inside a commercial building, there must be only one LAN. Question 3 options: True False Question 4 (1 point) When a packet is going to be routed to a computer in a LAN, other...
Addressing Table IPv4 Address Subnet Mask Device Interface Default Gateway IPv6 Address/Prefix IPv6 Link-local N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A N/A G0/O 2001 DB8:ACAD:: 1/64 FE80::1 G0/1 2001:DB8:ACAD:1:1/64FE80::1 R1 G0/2 2001:DB8:ACAD:21/64FE80::1 172.16.1.2 2001:DB8:2::1/64 209.165.200.226 2001:DB8:1::1164 172.16.1.1 2001:DB8:2:2/64 255.255.255.252 FE80::1 255.255.255.252 FE80::2 255.255.255.252 FE80:2 S0/0/1 S0/0/0 Central S0/0/1 S1 S2 S3 VLAN 1 VLAN 1 VLAN 1 Staff NIC 2001:DB8:ACAD::2/64FE80::2 FE80::1 Sales 2001 DB8:ACAD:1:2/64FE80::2 FE80::1 IT 2001 DB8:ACAD:2: 2/64 FE80::2 64.100.0.3 2001 DB8 CAFE::3/64 192.168.0.196 2001:DB8...
PLEASE ATTACH A SCREENSHOT OF YOUR SUCCESSFUL PING IN PACKET TRACER FOR THE END!!!!!!!!!!! THANK YOU! Objectives Part A: Configure a simple static routing . Part B: Configure a simple RIP routing Part A: Configure a simple static routing 1. Create the following network topology on Packet Tracer Router-PT Router-PT Addressing Table Device Interface IP Address Subnet Mask Default Gatewa 10.0.0.1 20.0.0.1 30.0.0.1 20.0.0.2 0.0.0.10 30.0.0.10 N/A 255.0.0.0 255.0.0.0 255.0.0.0 255.0.0.0 255.0.0.0 255.0.0.0 Routero 2/0 NIA NIA NIA 10.0.0.1 30.0.0.1...
A/ Given the following IP address from the Class B address range using the default subnet mask: 100.110.0.0. Your network plan requires no more than 64 hosts on a subnet. When you configure the IP address in Cisco IOS software, which value should you use as the subnet mask? 255.255.0.0 255.255.128.0 255.255.255.128 255.255.255.252 B/ Identify how many valid host addresses can you have on 192.168.27.32 network with a subnet mask of 255.255.255.240. (2^4) - 2 (2^3) – 2 (2^2)...
PART A 21 MARKS SHORT ANSWER QUESTIONS Answer ALL questions from this part. Write your answers in the Examination Answer Booklet. Each question is worth 1.5 marks (14 x 1.5 = 21 marks). Question 1 An organisation has been granted a block of addresses with the mask /22. If the organisation creates 8 equal-sized subnets, how many addresses (including the special addresses) are available in each subnet? Show your calculations. Question 2 Give an example of a valid classful address...