09-06-2023 09:20 PM - last edited on 09-08-2023 03:13 AM by Translator
Taking the exam soon.
The first chapter first two questions:
Question 1 answer is b ,d
Question 2 answer is a, c
I'm trying to figure out why.
Here's what I think.
Question 1
10.1.1.27/28
is a 2^4 is a 16 subnet
so it goes in blocks of 10.1.1.15, 10.1.1.31, 10.1.1.47, 10.1.1.63, 10.1.1.79 and so forth.
Since 10.1.1.18 and 10.1.1.27 are in the same sumbnets the device is broadcasting on the devices on the local subnet for a mac address, and the receiving device (PC2) should be on this subnet.
In question 2 since the subnets are in blocks of 6, 10.1.1.18 and 10.1.1.27 so the PC is forwarding to the
default gateway
Am i correct in assuming this?
Thanks
Solved! Go to Solution.
09-06-2023 09:47 PM - last edited on 09-08-2023 03:27 AM by Translator
Hello @hfakoor222,
You correctly identified that
10.1.1.27/28
represents a subnet with 16 addresses (from 10.1.1.16 to 10.1.1.31).
Both 10.1.1.18 and 10.1.1.27 are within this subnet, so they are on the same local subnet. When a device in this subnet wants to communicate with another device within the same subnet, it uses
ARP
to find the MAC address of the destination device. So, PC1 broadcasts for the MAC address of 10.1.1.27, which PC2 responds to because it's in the same subnet.
As concerned Q2, you're also correct.
10.1.1.18 and 10.1.1.27 are in different subnets. In this scenario, if PC1 wants to communicate with PC2 (10.1.1.27), it will send the packet to its
default gateway
(the router) using
ARP
to get gateway MAC.
The router will then forward the packet to the appropriate subnet (10.1.1.24/29) where PC2 resides.
09-06-2023 09:47 PM - last edited on 09-08-2023 03:27 AM by Translator
Hello @hfakoor222,
You correctly identified that
10.1.1.27/28
represents a subnet with 16 addresses (from 10.1.1.16 to 10.1.1.31).
Both 10.1.1.18 and 10.1.1.27 are within this subnet, so they are on the same local subnet. When a device in this subnet wants to communicate with another device within the same subnet, it uses
ARP
to find the MAC address of the destination device. So, PC1 broadcasts for the MAC address of 10.1.1.27, which PC2 responds to because it's in the same subnet.
As concerned Q2, you're also correct.
10.1.1.18 and 10.1.1.27 are in different subnets. In this scenario, if PC1 wants to communicate with PC2 (10.1.1.27), it will send the packet to its
default gateway
(the router) using
ARP
to get gateway MAC.
The router will then forward the packet to the appropriate subnet (10.1.1.24/29) where PC2 resides.
09-06-2023 11:11 PM
Thanks
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