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So here's some practice for subletting.
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We're going to start practicing right away with a sample question.
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And this question is.
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A host has the IP address one nine eight eight two two four five six eight one seven three two six.
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So what do you suppose are the network addresses, the broadcast addresses and the valid host addresses
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for the IP subnet, of which the host is a member of.
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Oh, and there's a second part to this question.
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What is this subnet mask and the dotted decimal notation?
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OK, don't worry, I'm not leaving you on your own.
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We're going to have a look at how we're going to figure out the answer.
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So we're going to do the second part for us because it's easier and it's actually more logical the answer
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to that question that way.
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So the second part of the question was what's going to be the sub mask slash 26 in dotted decimal?
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So the way that you actually do this, when you first learning's of netting is to write out the bit
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pattern at the top of a piece of paper.
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Then later on, you will be able to do it in your head, but when you're first learning, it's a whole
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lot easier to just write it all out.
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OK, so take a piece of paper.
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Start at the top and from right to left right out.
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One two four eight 16, 32 64 128.
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Make sure you put a dot between each number.
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OK, so it's going to look like this.
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Court now proceed to the next octet until you've written down the bit pattern for the four octet in
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a subnet mask like this.
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Good, so the subnet mask and our example question is Slash 26.
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So under your bed pattern at the top of the paper, right out 26 ones underneath.
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Starting from the left side.
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Then you can fill in the remainder with zeros for the host portion of the address.
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You got it.
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Now, draw a line in after the 26 ones to indicate the demarcation point between the network and the
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host portions of the address.
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Remember, the network portion is to the left of the line and the host portion is to the right of it.
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Good, so we can just add up the ones to find the subnet mask in dotted decimal notation.
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So the first octet is all ones 128 plus 64 plus 32 plus 16 plus eight plus four plus two plus one equals
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255.
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So 255 is the answer.
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2ND and third octet are also old ones, adding up to 255, so now we've got two five five two five five
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two five five.
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And for the fourth octet, the ones are under 128 and 64, so 128 plus 64 equals 192.
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Therefore, the subnet mask was slashed 26 in dotted decimal notation is two five five two five five
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two five five not one nine two.
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Now, what was the other part of the question again?
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Yeah.
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What is the network addressed the broadcast address and a valid host address for the IP address.
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One nine two two eight two two eight four five one seven three eight two six.
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Well, I'll tell you some again, piece of paper is going to come in handy.
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So we're going to write out the IP address and dotted decimal notation now as well.
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So the IP address in our example question is one nine eight Dot two to Dot four or five dot one seven
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three.
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So when we start with the first octet one nine eight.
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Starting from the one to eight bit on the left, right down to one under the bit, if it adds up to
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less than the number in the IP address, he yeah, let me just explain it quick.
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The first octet is one nine eight, so one two eight is less than one 980, so we write down or one
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under the one two eight.
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OK, so the next bit after one to eight is 64, 128 plus 64 is 192.
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OK, so one nine two is less than one 980, so we write a one under the 64.
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The next bit after 64 is 32.
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One two, eight six four plus three two equals two to four.
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Now 224 is more than 198, so that means we'll write a zero, not a one under the 32.
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And the next bit after 32 is 16.
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All right, so far we have 128 plus 64 equals, 192 192 plus 16 equals 208.
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And of course, 208 is more than 198.
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So we write in zero under the 16.
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The next bit after the 16 is eight one, 92 plus eight equals 200 and 200 is more than 198.
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So we write a zero under the eight.
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Next bit after the eight is four one nine two plus four is one nine six one nine six is less than one
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nine eight, so we write a one under the for next bit after four is two, so one nine six plus two equals
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198 198 is exactly 198.
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So we write our one under the two.
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So we got to the exact number, right?
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That means that we've finished calculating this octet so we can write a zero for all the remaining bits.
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You get it.
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Which is actually here in this example, just the final one bit at the end.
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So now that we got a written down, we can quickly add up the ones to check, our answer adds up to
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the one nine eight first octet in our one nine eight two two four five eight one seven three IP address.
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OK, we'll do it together.
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128 plus 64 plus four plus two equals one hundred ninety eight.
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Then we just repeat the process for the other three tests in order to write out the entire IP address.
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One nine eight two two four or five one seven three and dotted decimal notation.
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Now we're working with a slash 24 subnet mask, so the network portion of this address is a first 26
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bit.
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The first 24 bits, the first three octet are one nine eight, not two, two, not four or five.
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We can easily see that right just by looking at the IP address.
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Then we just count whatever it is in the fourth octet, first to bid to get all 26 bits.
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So I've got a one for one 28 and a zero for 64.
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So that's going to add up to 128.
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All right.
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So the network address is one nine eight two two eight four five eight one two eight.
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Cool.
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So that's the first part of the question.
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That's right.
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Did you forget about the second part?
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Well, next up is to figure out the broadcast addresses.
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So the broadcast address is the final IP address in this IP subnet, which is one less than the network
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address of the next IP subnet.
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So the network address of this IP subnet is one nine eight two, two four or five Dot one 28.
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Right, because we just calculated.
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Now, looking at the line, which demarcates the network and host partitions of the address, we can
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see that after it is 64.
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So that means the IP subnets are going up in increments of 64.
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And as you can also see from the Red Line, we are subletting on the 4th Octet.
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So this IP subnet is one nine, eight, two, two, four or five.
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Got one to eight.
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Now, to calculate the next I.P. step that we had 64 to the octet we are submitting on, which is the
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fourth octet in this case, one two eight plus 64 equals 192.
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You still with me.
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So see, the network address of the next IP subnet is one less than one nine two two two two four five
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eight one nine two.
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And that would be one nine eight nine eight two two four five eight one nine one.
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So that makes a broadcast address of this one, nine, eight, two, 2.4 or 5.1 to eight subnet.
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One nine eight two two four five eight one nine one.
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Cool, so now we've got the network address and the broadcast address.
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So finally, we just need to determine the possible host addresses.
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Now they happen to be all the addresses between the network address of one nine eight two two 2.4 or
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5.1 one two eight.
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And the broadcast address of one nine eight two two four or five eight one nine one.
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So that gives us one, nine, eight, two, two, four or five, not one to nine all the way through
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one nine eight talked to two, not four or five, not one nine zero.
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Now a final and important point to tell you here is notice that all our calculations were done on the
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octet where the subnet line is.
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So this example is a slash 26 that is in the fourth octet.
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So we don't actually need to worry about the first three octet.
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Now you get it because they're always going to be unchanged.
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One nine, eight, two, two, four or five, in our example.
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So when you're working on a sub that in question, you only need to write out the octet that this sub
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netting is actually happening on because all the other architects are always going to just be there
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unchanged, you don't have to write out the whole IP address in full every time.
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I know that's a load off.
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So that's it, right?
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That's how you practice up netting.
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So take it away.
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It's all yours, you can practice that all you want.
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But.
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We're going to move on to IPv6 in the next lesson.
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I'll see you then.
13303
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