Exam 200-301 topic 1 question 904 discussion

C is correct. 252(/30) ip subnet have only 2 usable ip address. 248(/29) ip subnet have 6 usable ip address.

A is incorrect, it does not leave any usable addresses. C is correct

C is correct

it is C

it´s C

6 usable ip address

We have to configure the link which will need 2 IP addresses, 1 for each port on each Router. We also need 2 spare IPs for future growth, so overall we need 4 usable IP addresses. If we consider using the /30 (255.255.255.252) mask, it will give us 2^2 (=4) i.e., total 4 IPs and 2 usable IPs, which doesn’t fulfil the given requirements. So, we can consider using the next /29 (255.255.255.248) mask, which gives us 2^3 (=8) i.e., total 8 IP address and 6 usable IP addresses, which perfectly fulfil the given requirements.

C. R1(config-if)#ip address 10.10.10.1 255.255.255.248 R2(config-if)#ip address 10.10.10.4 255.255.255.248 This will give you 6 usable ip addresses which is enough for the 4 ip adresses required and will only waste 2 ip addresses.

all is correct except A

Why not B? wouldn't that leave 14 usable addresses?

As per the exhibit, there's a requirement to configure the link and also keep 2 spare IP addresses for future use. So, calculate like this: 2 IPs for Network ID and broadcast IP for whatever subnet we use. 2 IPs for connected interfaces on each router i.e. 1 for R1 gi0/0/0 and 1 for R2 gi0/0/0. 2 IPs reserved for future use. In this way, we need atleast 6 IPs here and the subnet that can provide a closest value in terms of total required IPs is /29 (255.255.255.248). The /29 subnet provides a total of 8 IPs addresses per subnet and out of that 6 are host usable IPs. So, for given subnet 10.10.10.X/29 we can have IP range from 10.10.10.0 - 10.10.10.7. Where 10.10.10.0 will be network ID and 10.10.10.7 will be broadcast IP. Rest 10.10.10.1 - 10.10.10.6 can be used to assign to any hosts of links.

With a /30 subnet mask, you can have a total of 4 IP addresses, 2 of which can be used for hosts. The reason for this is that the /30 subnet mask has 30 bits set to 1, leaving 2 bits for host addresses. With a /29 subnet mask, you can have a total of 8 IP addresses, 6 of which can be used for hosts. The reason for this is that the /29 subnet mask has 29 bits set to 1, leaving 3 bits for host addresses.

A is correct. A 30-bit subnet mask allows for four IPv4 addresses: two host addresses, one all-zeros network, and one all-ones broadcast address. A point-to-point link can only have two host addresses. There is no real need to have the broadcast and all-zeros addresses with point-to-point links. Even A 31-bit subnet mask allows for exactly two host addresses, and eliminates the broadcast and all-zeros addresses, thus it conserves the use of IP addresses to the minimum for point-to-point links.

C is correct

2 additional!!

Definitely "C" is the correct answer

Is it not .248? .252 would not leave any free ip addresses