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Subnetting Problem Contd

 
 
Myrt Webb
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Posts: n/a
 
      07-15-2003
Two days ago I posted the follwing question from the MS
Trng Kit for 70-216. I received several responses but none
that really explained why the answer to the question was
selection 'd.'

So, I lined out in binary form all of the eight IP
addresses. It turns out that the 'd' answer that changes
the subnet mask to 0/21 isolates the right most 3 bits of
the 3rd octet.

With the 0/21 mask there are two nets using 24 and 30 in
the second octet. Then the host addresses are changed by
one in the third octet.

But, the question remains, why does this make the routing
tables simpler for the router?

Interesting problem, don't you love that MS does not give
an explanation.

"You are the new administrator of a 2000 node network.
There is only one router on the entire network, which
provides all the computers with Internet access. The
company's ISP has assigned the following 8 network
addresses to them:

10.24.32.0/24
10.24.33.0/24
10.24.34.0/24
10.30.35.0/24
10.30.36.0/24
10.30.37.0/24
10.30.38.0/24
10.30.39.9/24

What subnet mask could you use to minimize the complexity
of the routing tables while maintaining the existing
Internet connectivity?

a. 255.255.252.0
b. 255.255.255.252
c. 255.255.255.248
d. 255.255.248.0 "


 
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Manuel
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Posts: n/a
 
      07-15-2003
Myrt,
I think is because this way all the internal PCs can be on
just one network with the same gateway ip address,

for the router is easy , you define 2 lines only:
10.24.32.0/21 my internal network
all the rest go out

for the local PCs or servers are:
10.24.32.0/21 network is local to me
all the other destinations send to the router.

IF not, the router has to know where to direct 8 internal
networks with at least 8 lines and probably assign 8 IP
addresess to the internal network card or assign other
devices as routers to help with it.

Regards,

>-----Original Message-----
>Two days ago I posted the follwing question from the MS
>Trng Kit for 70-216. I received several responses but

none
>that really explained why the answer to the question was
>selection 'd.'
>
>So, I lined out in binary form all of the eight IP
>addresses. It turns out that the 'd' answer that changes
>the subnet mask to 0/21 isolates the right most 3 bits of
>the 3rd octet.
>
>With the 0/21 mask there are two nets using 24 and 30 in
>the second octet. Then the host addresses are changed by
>one in the third octet.
>
>But, the question remains, why does this make the routing
>tables simpler for the router?
>
>Interesting problem, don't you love that MS does not give
>an explanation.
>
>"You are the new administrator of a 2000 node network.
>There is only one router on the entire network, which
>provides all the computers with Internet access. The
>company's ISP has assigned the following 8 network
>addresses to them:
>
>10.24.32.0/24
>10.24.33.0/24
>10.24.34.0/24
>10.30.35.0/24
>10.30.36.0/24
>10.30.37.0/24
>10.30.38.0/24
>10.30.39.9/24
>
>What subnet mask could you use to minimize the complexity
>of the routing tables while maintaining the existing
>Internet connectivity?
>
>a. 255.255.252.0
>b. 255.255.255.252
>c. 255.255.255.248
>d. 255.255.248.0 "
>
>
>.
>

 
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Manuel
Guest
Posts: n/a
 
      07-15-2003
Myrt,
I think is because this way all the internal PCs can be on
just one network with the same gateway ip address,

for the router is easy , you define 2 lines only:
10.24.32.0/21 my internal network
all the rest go out

for the local PCs or servers are:
10.24.32.0/21 network is local to me
all the other destinations send to the router.

IF not, the router has to know where to direct 8 internal
networks with at least 8 lines and probably assign 8 IP
addresess to the internal network card or assign other
devices as routers to help with it.

Regards,

>-----Original Message-----
>Two days ago I posted the follwing question from the MS
>Trng Kit for 70-216. I received several responses but

none
>that really explained why the answer to the question was
>selection 'd.'
>
>So, I lined out in binary form all of the eight IP
>addresses. It turns out that the 'd' answer that changes
>the subnet mask to 0/21 isolates the right most 3 bits of
>the 3rd octet.
>
>With the 0/21 mask there are two nets using 24 and 30 in
>the second octet. Then the host addresses are changed by
>one in the third octet.
>
>But, the question remains, why does this make the routing
>tables simpler for the router?
>
>Interesting problem, don't you love that MS does not give
>an explanation.
>
>"You are the new administrator of a 2000 node network.
>There is only one router on the entire network, which
>provides all the computers with Internet access. The
>company's ISP has assigned the following 8 network
>addresses to them:
>
>10.24.32.0/24
>10.24.33.0/24
>10.24.34.0/24
>10.30.35.0/24
>10.30.36.0/24
>10.30.37.0/24
>10.30.38.0/24
>10.30.39.9/24
>
>What subnet mask could you use to minimize the complexity
>of the routing tables while maintaining the existing
>Internet connectivity?
>
>a. 255.255.252.0
>b. 255.255.255.252
>c. 255.255.255.248
>d. 255.255.248.0 "
>
>
>.
>

 
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Myrt Webb
Guest
Posts: n/a
 
      07-16-2003
I agree.

With the original 0/24 subnet mask you have 8 separate
nets with one host each. By changing the subnet mask to
0/21 you have two nets with 3 hosts on the 10.24 net and 5
hosts on the 10.30 net.

I do not know a lot about routers but I imagine the two
net configuration is better than the 8 net configuration.

Thanks for your comments.

>-----Original Message-----
>Myrt,
>I think is because this way all the internal PCs can be

on
>just one network with the same gateway ip address,
>
>for the router is easy , you define 2 lines only:
>10.24.32.0/21 my internal network
>all the rest go out
>
>for the local PCs or servers are:
>10.24.32.0/21 network is local to me
>all the other destinations send to the router.
>
>IF not, the router has to know where to direct 8 internal
>networks with at least 8 lines and probably assign 8 IP
>addresess to the internal network card or assign other
>devices as routers to help with it.
>
>Regards,
>
>>-----Original Message-----
>>Two days ago I posted the follwing question from the MS
>>Trng Kit for 70-216. I received several responses but

>none
>>that really explained why the answer to the question was
>>selection 'd.'
>>
>>So, I lined out in binary form all of the eight IP
>>addresses. It turns out that the 'd' answer that changes
>>the subnet mask to 0/21 isolates the right most 3 bits

of
>>the 3rd octet.
>>
>>With the 0/21 mask there are two nets using 24 and 30 in
>>the second octet. Then the host addresses are changed by
>>one in the third octet.
>>
>>But, the question remains, why does this make the

routing
>>tables simpler for the router?
>>
>>Interesting problem, don't you love that MS does not

give
>>an explanation.
>>
>>"You are the new administrator of a 2000 node network.
>>There is only one router on the entire network, which
>>provides all the computers with Internet access. The
>>company's ISP has assigned the following 8 network
>>addresses to them:
>>
>>10.24.32.0/24
>>10.24.33.0/24
>>10.24.34.0/24
>>10.30.35.0/24
>>10.30.36.0/24
>>10.30.37.0/24
>>10.30.38.0/24
>>10.30.39.9/24
>>
>>What subnet mask could you use to minimize the

complexity
>>of the routing tables while maintaining the existing
>>Internet connectivity?
>>
>>a. 255.255.252.0
>>b. 255.255.255.252
>>c. 255.255.255.248
>>d. 255.255.248.0 "
>>
>>
>>.
>>

>.
>

 
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freak
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Posts: n/a
 
      07-16-2003

www.mcsefreak.com/subnetting.htm

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Marko
Guest
Posts: n/a
 
      07-16-2003
I posted this reply:

Jeffrey Woods has provided the best answer so far.

Your ISP may not be a total idiot for "assigning" this IP
space in the first place. It would only work on your
10.24or30.y.z network and possibly to reach the ISP's
servers / routers and that is likely it. However, if the
ISP provides mail and proxy servers (possibly other
services?) then this IP assignment is OK.

10.24.x.y to 10.30.a.b would require a mask of 255.248.0.0
to cover all those IP ranges mentioned.

The 10.24.x.y ranges could be covered by "route add
10.24.32.0 mask 255.255.253.0 (gateway IP here)"

.....Assuming 10.30.39.9/24 is a typo and should be
10.30.39.0/24 (cause the first is definitely wrong and
does not exist)...

Then route add 10.30.35.0 mask 255.255.251.0 (gateway IP)

So far, neither route table addition looks like our answer.


HOWEVER, if you concede that it is highly probable that
all the class C subnets (those finishing with /24) start
with either 10.24 OR 10.30, then 10.x.32.0 with a mask of
255.255.248.0 will cover all IPs in the range 10.x.32.0 to
10.x.39.255. Making "d" the correct choice....


It it says d is right, then I would suggest the above
paragraph is the solution.

Anyone else care to have a go?

.....care to kudos?

.....care to flame?

Marko Cosic

 
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Marko
Guest
Posts: n/a
 
      07-16-2003
I posted this reply:

Jeffrey Woods has provided the best answer so far.

Your ISP may not be a total idiot for "assigning" this IP
space in the first place. It would only work on your
10.24or30.y.z network and possibly to reach the ISP's
servers / routers and that is likely it. However, if the
ISP provides mail and proxy servers (possibly other
services?) then this IP assignment is OK.

10.24.x.y to 10.30.a.b would require a mask of 255.248.0.0
to cover all those IP ranges mentioned.

The 10.24.x.y ranges could be covered by "route add
10.24.32.0 mask 255.255.253.0 (gateway IP here)"

.....Assuming 10.30.39.9/24 is a typo and should be
10.30.39.0/24 (cause the first is definitely wrong and
does not exist)...

Then route add 10.30.35.0 mask 255.255.251.0 (gateway IP)

So far, neither route table addition looks like our answer.


HOWEVER, if you concede that it is highly probable that
all the class C subnets (those finishing with /24) start
with either 10.24 OR 10.30, then 10.x.32.0 with a mask of
255.255.248.0 will cover all IPs in the range 10.x.32.0 to
10.x.39.255. Making "d" the correct choice....


It it says d is right, then I would suggest the above
paragraph is the solution.

Anyone else care to have a go?

.....care to kudos?

.....care to flame?

Marko Cosic

 
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Myrt Webb
Guest
Posts: n/a
 
      07-16-2003
I think the "point" of the question is that you can change
a subnet mask to reduce the number of nets and thereby
make the routings simpler.

>-----Original Message-----
>I posted this reply:
>
>Jeffrey Woods has provided the best answer so far.
>
>Your ISP may not be a total idiot for "assigning" this IP
>space in the first place. It would only work on your
>10.24or30.y.z network and possibly to reach the ISP's
>servers / routers and that is likely it. However, if the
>ISP provides mail and proxy servers (possibly other
>services?) then this IP assignment is OK.
>
>10.24.x.y to 10.30.a.b would require a mask of

255.248.0.0
>to cover all those IP ranges mentioned.
>
>The 10.24.x.y ranges could be covered by "route add
>10.24.32.0 mask 255.255.253.0 (gateway IP here)"
>
>.....Assuming 10.30.39.9/24 is a typo and should be
>10.30.39.0/24 (cause the first is definitely wrong and
>does not exist)...
>
>Then route add 10.30.35.0 mask 255.255.251.0 (gateway IP)
>
>So far, neither route table addition looks like our

answer.
>
>
>HOWEVER, if you concede that it is highly probable that
>all the class C subnets (those finishing with /24) start
>with either 10.24 OR 10.30, then 10.x.32.0 with a mask of
>255.255.248.0 will cover all IPs in the range 10.x.32.0

to
>10.x.39.255. Making "d" the correct choice....
>
>
>It it says d is right, then I would suggest the above
>paragraph is the solution.
>
>Anyone else care to have a go?
>
>.....care to kudos?
>
>.....care to flame?
>
>Marko Cosic
>
>.
>

 
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MikeF
Guest
Posts: n/a
 
      07-16-2003
Right, you got it. This question is made difficult only by MS
inability to understand the meanings of words. And I am not kidding.

The question is simply: you don't want to use all these networks. You
want one network that will hold all your hosts. How do you do it?
Ans, Supernetting, i.e., borrowing network id bits to use for host
bits, increasing the number of available hosts on one network, instead
of borrowing host bits to increase the number of separate networks
which, of course, are called subnets.

So the subnet in the answer is simply the one that gives you the range
of addresses to hold 2000 or more hosts.

Mike


"Myrt Webb" <(E-Mail Removed)> wrote in message
news:0ab901c34b9a$5fe229d0$(E-Mail Removed)...
> I think the "point" of the question is that you can change
> a subnet mask to reduce the number of nets and thereby
> make the routings simpler.
>
> >-----Original Message-----
> >I posted this reply:
> >
> >Jeffrey Woods has provided the best answer so far.
> >
> >Your ISP may not be a total idiot for "assigning" this IP
> >space in the first place. It would only work on your
> >10.24or30.y.z network and possibly to reach the ISP's
> >servers / routers and that is likely it. However, if the
> >ISP provides mail and proxy servers (possibly other
> >services?) then this IP assignment is OK.
> >
> >10.24.x.y to 10.30.a.b would require a mask of

> 255.248.0.0
> >to cover all those IP ranges mentioned.
> >
> >The 10.24.x.y ranges could be covered by "route add
> >10.24.32.0 mask 255.255.253.0 (gateway IP here)"
> >
> >.....Assuming 10.30.39.9/24 is a typo and should be
> >10.30.39.0/24 (cause the first is definitely wrong and
> >does not exist)...
> >
> >Then route add 10.30.35.0 mask 255.255.251.0 (gateway IP)
> >
> >So far, neither route table addition looks like our

> answer.
> >
> >
> >HOWEVER, if you concede that it is highly probable that
> >all the class C subnets (those finishing with /24) start
> >with either 10.24 OR 10.30, then 10.x.32.0 with a mask of
> >255.255.248.0 will cover all IPs in the range 10.x.32.0

> to
> >10.x.39.255. Making "d" the correct choice....
> >
> >
> >It it says d is right, then I would suggest the above
> >paragraph is the solution.
> >
> >Anyone else care to have a go?
> >
> >.....care to kudos?
> >
> >.....care to flame?
> >
> >Marko Cosic
> >
> >.
> >



 
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Myrt Webb
Guest
Posts: n/a
 
      07-17-2003
Thanks for the confirming response.

"Supernetting" or the opposite of subnetting is not
something I have run into before.

Live and learn.

>-----Original Message-----
>Right, you got it. This question is made difficult only

by MS
>inability to understand the meanings of words. And I am

not kidding.
>
>The question is simply: you don't want to use all these

networks. You
>want one network that will hold all your hosts. How do

you do it?
>Ans, Supernetting, i.e., borrowing network id bits to use

for host
>bits, increasing the number of available hosts on one

network, instead
>of borrowing host bits to increase the number of separate

networks
>which, of course, are called subnets.
>
>So the subnet in the answer is simply the one that gives

you the range
>of addresses to hold 2000 or more hosts.
>
>Mike
>
>
>"Myrt Webb" <(E-Mail Removed)> wrote in message
>news:0ab901c34b9a$5fe229d0$(E-Mail Removed)...
>> I think the "point" of the question is that you can

change
>> a subnet mask to reduce the number of nets and thereby
>> make the routings simpler.
>>
>> >-----Original Message-----
>> >I posted this reply:
>> >
>> >Jeffrey Woods has provided the best answer so far.
>> >
>> >Your ISP may not be a total idiot for "assigning" this

IP
>> >space in the first place. It would only work on your
>> >10.24or30.y.z network and possibly to reach the ISP's
>> >servers / routers and that is likely it. However, if

the
>> >ISP provides mail and proxy servers (possibly other
>> >services?) then this IP assignment is OK.
>> >
>> >10.24.x.y to 10.30.a.b would require a mask of

>> 255.248.0.0
>> >to cover all those IP ranges mentioned.
>> >
>> >The 10.24.x.y ranges could be covered by "route add
>> >10.24.32.0 mask 255.255.253.0 (gateway IP here)"
>> >
>> >.....Assuming 10.30.39.9/24 is a typo and should be
>> >10.30.39.0/24 (cause the first is definitely wrong and
>> >does not exist)...
>> >
>> >Then route add 10.30.35.0 mask 255.255.251.0 (gateway

IP)
>> >
>> >So far, neither route table addition looks like our

>> answer.
>> >
>> >
>> >HOWEVER, if you concede that it is highly probable that
>> >all the class C subnets (those finishing with /24)

start
>> >with either 10.24 OR 10.30, then 10.x.32.0 with a mask

of
>> >255.255.248.0 will cover all IPs in the range 10.x.32.0

>> to
>> >10.x.39.255. Making "d" the correct choice....
>> >
>> >
>> >It it says d is right, then I would suggest the above
>> >paragraph is the solution.
>> >
>> >Anyone else care to have a go?
>> >
>> >.....care to kudos?
>> >
>> >.....care to flame?
>> >
>> >Marko Cosic
>> >
>> >.
>> >

>
>
>.
>

 
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