Question

In a class B subnet, we know the IP address of one of the hosts and the subnet mask as given below: IP Address: 188.48.82.176 Subnet mask: 255.255.255.248 1) What is the first valid host address in the subnet? (in dotted decimal) 2) What is the last valid address in the subnet? (in dotted decimal) 3) What is the subnet network address? (in dotted decimal) 4) What is the subnet broadcast address? (in dotted decimal) 5) What is the number of valid hosts addresses in this subnet? 6) How many bits from host Id were used for subnetting? 7) How many possible subnets can be created with the given subnet mask: 255.255.255.248? 8) What is the network address for the given network that contains the given IP Address 188.48.82.176? (in dotted decimal) 9) What is the broadcast address for the given network that contains the given IP Address 188.48.82.176? (in dotted decimal) 10) Represent the IP Address: 188.48.82.176 in given subnet using classless address format?

Answer 1

these are formulas i think that might help you to understand the solutions.

Given IP Address = 188.68.82.176 In Binary depresentation, we 10111100.0011 0000.01010010.10110000 Given subnet mask=255.255.255.248 In binary representations- 11111111.11111111.11111111.11111000 Now, need to first find the Subnet network address by Just bitwise "and" oresation. In IP address and scesnet mask. network.io Host 1011 1100.00110000.01010010.1011 Ooo 11111111. 11111111.1111 1111.1111 1000 101 1100.00 H 0000.0 10100 10.10 1 0 0 0 0 C188 so, The subnet network address 188.48.82.176. (30d answer) bus 682 5176 Now, the first host ID is next adaress of scebnet network. TOIT110000110000010100101011 0001 188. 48.82.177 C 1st answer).

The subnet subnet broadcast address al is in the most part in the scegn et network. 10111100, 00110000. 01010010.10110111 datos so, 188.48.82.183 is the subnet broced cast address 4 Now, the last valid address in the sebnet previous address of the subnet broadcast address. so, just change last one with Zreso. 10111100.00110000.01010010.10110110 last valid host address in the scebnet 188.48.82.182

13bit Host bit In 5) The subnet mash is 255.255.255.248. binary. 11111111.11111111.11111111.11111000 T Subsretwone bit so, we have 3 host bit. No. of valid nast address in this Sebnet 2-2-8-2²6 sou sorbits were wood It is a class B networm and for class B в 16 bit were used for network Id and 16 bit for nost ID. Here. No. of Ia last bit was B. so 13 bit from Host ID ID were used for subnetting for In Host ID port 13 bit used sebnetting, so Possible scebnet can be generated : 213=8192

LIFE INSURANCE CON Question : is some , NO. solution of ③ of ③ and . 10) to reduce + the wastage of IP address in a block, we use scebnexting what we do is that we are host ID bits as nelwork ID bits of classful IP address. we give the IP address ond Desine no. of bits for mask along with it followed by a cysymbol. here given sebnet mash as 255.255.255.248. we CUTITIT.IIIIIIIl. II III111. 11 11 1 000 29 bit for mash so given go IP address 186.48.82.176 represent format 188. 48. 82.176/29 in classless
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