Routing Table Question

Kenrick

I don't know how to do the above network question , please guide me,thanks in advance.

routing table question.jpg

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Comments

OfWolfAndMan

There are a couple components to learn when understanding this question:

1. You must know your subnetting
2. You must know what the default route does
3. You must understand what localhost means

Let's examine the first option, 142.93.45.15. 142.93.45.0 with a subnet mask of 255.255.255.128 ranges from 142.93.45.1-142.93.45.126 (Without broadcast address), therefore the destination would be this interface (0).

Now that you understand that, the next two are the same. Choose the matching (or most specific prefix) and switch it out the appropriate interface. Localhost means it resides locally off of one of the routers interface.

As for the last one, 142.93.43.65, obviously there is no specific prefix in the routing table, therefore the default route (0.0.0.0 0.0.0.0) is used and the packet is forwarded to the next hop of R3

Kenrick

Can I say that 142.93.44.12 will be routed to interface (2) because 142.93.44.1 - 142.93.44.126 and 142.93.45.208 will be routed to interface(1) because 142.93.45.129-142.93.45.254?

OfWolfAndMan

Kenrick wrote: »

Can I say that 142.93.44.12 will be routed to interface (2) because 142.93.44.1 - 142.93.44.126 and 142.93.45.208 will be routed to interface(1) because 142.93.45.129-142.93.45.254?

Right on the money

Kenrick

255.255.255.192
However, I have one more question, what does the 192 in the table mean?

OfWolfAndMan

Remember these numbers:128, 192, 224, 240, 248, 252, 254, 255. When the subnet mask is 255.255.255.0, this indicates available hosts is 254 available. For example, 10.0.1.0 255.255.255.0 has a range of 10.0.1.1-10.0.1.254. As you add bits to the subnet mask, you reduce your available hosts. 255.255.255.0 uses 24, 255.255.255.128 uses 25, 255.255.255.192 uses 26, etc. Now take the difference of 26 bits from 32 (which is 6), and put it in the exponent 2^n, where n is the remaining bits available. So 2^6 is 64. Subtracting 2 from that, one for the network ID and one from the broadcast address, you get 62 available hosts. So 10.0.1.0/26 has a range from 10.0.1.1 to 10.0.1.62. Get it?