Refer to the exhibit. Each router must be configured with the last usable IP address in the subnet. Which configuration fulfills this requirement?
Refer to the exhibit. Each router must be configured with the last usable IP address in the subnet. Which configuration fulfills this requirement?
R7 requires the last usable IP address in the subnet 10.88.31.64/26. The range for this subnet is 10.88.31.64 to 10.88.31.127. The last usable address is 10.88.31.126. R8 requires the last usable IP address in the subnet 10.19.63.80/28. The range for this subnet is 10.19.63.80 to 10.19.63.95. The last usable address is 10.19.63.94. R9 requires the last usable IP address in the subnet 10.23.98.128/27. The range for this subnet is 10.23.98.128 to 10.23.98.159. The last usable address is 10.23.98.158. Therefore, the correct configuration for each router is provided in option D.
You can do this fairly easily by process of elimination. Starting with R7, a /26 is .192, so that leaves us with A or D. The first difference between A and D is the last octet, 127 or 126 (respectively). Do whatever process you prefer for subnetting and we figure that .126 is the last usable. .127 is the broadcast. Answer is D.
As mentioned in the question we need the last usable IP address so remember the rule Network ID Even First IP Odd Last IP Even Broadcast Address Odd So starting with Router 7 we know that the subnet is 192 so now we have 2 options A and D and we know last IP should be even so answer is D.
it's D
The right answer is B. As for R9 the last usable address should be .158.
It is D
I think its D. Check the answers again. The only difference between B and D is that in B, R9 subnet mask ends with .248, but it should be .224, because its a /27 network.
But why it has to be 224 and not 248 ?
Because a /27 is .224 A .248 would be /29. R9 is not /29.
because if 248 then i t will be /29.
it's D the subnet of first address in B is Wrong
Subnet Mask of R7 is /26 thus 255.255.255.192
10.88.31.64 - 10.88.31.127/26 (FUH 10.88.31.65-10.88.31.126 LUH) inc = 64 Last usable host for R7(fa1/0) 10.88.31.126 255.255.255.192 10.19.63.80 - 10.19.63.95/28 (FUH 10.19.63.81-10.19.63.94 LUH) inc = 16 Last usable host for R8(fa0/0) 10.19.63.94 255.255.255.240 10.23.98.128 - 10.23.98.159/27 (FUH 10.23.98.129 - 10.23.98.158 LUH) inc = 32 Last usable host for R9(fa1/1) 10.23.98.158 255.255.255.224
Just look at the subnets mask.for r7 “/26” = .192 because 128 plus 64. Then for r8 “/28 “ = .240 because 128+64+32+16 that only leaves option d
Regarding the last usable up address for r7 because it’s /26 the subnets are going to be every 64 numbers specifically r7 .128 is the network up for the next subnet .127 is the broadcast and .126 is the last useable ip
Its D the subnet masks in B are incorrect
I agree that D is the right answer
it's enough to look only R7 and answer is D
D is the correct answer.
Just by eliminating options.. the correct for R7 is option D
D is correct