RE: csu-dsu [7:11233]

2001-07-07 Thread Joshua Beining

Did you specify the clock source?  

service-module t1 clock source internal
service-module t1 clock source line 

where internal uses the adapters internal clock and line gets the clocking
from your service provider.

HTH,

-Joshua

-Original Message-
From: Moahzam Durrani [mailto:[EMAIL PROTECTED]]
Sent: Friday, July 06, 2001 9:08 PM
To: [EMAIL PROTECTED]
Subject: csu-dsu [7:11233]


At present I have a 7500 that has aN external DSU -CSU (RAD FCD-1) .
Providing a T1 conectivity netween two of our sites. We are going to be
using a 2600 for this same connection .The 2600 has a built in CSU/DSU. My
problem is I do not get a coneectivity between the two sites when I use the
2600.  I know the card works as I tested it with another T1 conection. No
problem. the carrier is using SF Framing , linecode  AMI . the encapsulation
is HDLC. the 7500 with the external CSU/DSU works fine. however with the
2600 I keep on seeing Serial up , line down.  I know on our existing 2600
with built in csu/dsu I have chanel-group 0 , time slot1-24 and speed 64
configured on the controllers.  one thing I noticed on the 7500 csu/dsu  it
was set up with clk_master : LBT .is this specific to the carrier ? is there
something im missing on my configs. The route statementsconfigured on teh
2600 are exactly the same as the 7500 , so  there is no routing issue.




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Will the NM-4T work on 2600 series routers [7:3056]

2001-05-03 Thread Joshua Beining

It seems that there is some confusion wrt network modules and 2600/3600
routers.  Searching on Ciscos site, they say that the NM-4T is not supported
for the 2600 series but I see cisco resellers, and others advertising the
NM-4T for 2600/3600 routers.  Am I missing the doc from Cisco that says the
NM-4T is now supported on the 2600 sereis?  Can anyone provide some imput.
Thanks.

-Joshua




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RE: difficult ccna question

2001-03-22 Thread Joshua Beining

Just remember the following fomulas:

(2^# of masked bits) - 2 = Total # of subnets
(2^# of unmasked bits - 2 = Total # of hosts

Based on this the correct answer is A.

-Original Message-
From: George [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 22, 2001 12:16 AM
To: [EMAIL PROTECTED]
Subject: difficult ccna question


if you have a class B network with a 10-bit subnet mask, how many subnet and
how many hosts do you have?

a. 1022 subnets, 62 hosts
b. 62 subnets, 8190 hosts
c. 8190 subnets, 254 hosts
d. 254 subnets , 126 hosts


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RE: difficult ccna question

2001-03-22 Thread Joshua Beining

Based on the subnet mask (255.255.192.0) subtract 256 from 192 which is 64.
Therefore 64 is your first subnet.  To get the second subnet, add 64 to the
first subnet.  To get the third add 64 to the second subnet and so on.
Continue in this fashion until you reach 192.  Remember that you cannot use
the ranges 172.16.0.1  -  172.16.0.62 and 172.16.255.193  -  172.16.255.254
(network and broadcast respectively) unless your router is configured to do
so.  

Subnet  Host Range
1   172.16.0.65  -  172.16.0.126
2   172.16.0.129  -  172.16.0.190
3   172.16.0.193  -  172.16.0.254
4   172.16.1.1  -  172.16.1.62
5   172.16.1.65  -  172.16.1.126
6   172.16.1.129  -  172.16.1.190
7   172.16.1.193  -  172.16.1.254
8   172.16.2.1  -  172.16.2.62
9   172.16.2.65  -  172.16.2.126
   10   172.16.2.129  -  172.16.2.190
.
.
.
.
.
.

 1015   172.16.253.193  -  172.16.253.254
 1016   172.16.254.1  -  172.16.254.62
 1017   172.16.254.65  -  172.16.254.126
 1018   172.16.254.129  -  172.16.254.190
 1019   172.16.254.193  -  172.16.254.254
 1020   172.16.255.1  -  172.16.255.62
 1021   172.16.255.65  -  172.16.255.126
 1022   172.16.255.129  -  172.16.255.190

-Joshua
-Original Message-
From: Lowell Sharrah [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 22, 2001 8:14 AM
To: [EMAIL PROTECTED]
Subject: RE: difficult ccna question


how do you know where the first subnet begins?

 Joshua Beining [EMAIL PROTECTED] 03/22/01 10:34AM 
Just remember the following fomulas:

(2^# of masked bits) - 2 = Total # of subnets
(2^# of unmasked bits - 2 = Total # of hosts

Based on this the correct answer is A.

-Original Message-
From: George [mailto:[EMAIL PROTECTED]] 
Sent: Thursday, March 22, 2001 12:16 AM
To: [EMAIL PROTECTED] 
Subject: difficult ccna question


if you have a class B network with a 10-bit subnet mask, how many subnet and
how many hosts do you have?

a. 1022 subnets, 62 hosts
b. 62 subnets, 8190 hosts
c. 8190 subnets, 254 hosts
d. 254 subnets , 126 hosts


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RE: difficult ccna question

2001-03-22 Thread Joshua Beining

Technically yes.  But like most things you cannot get something for free.
There is a price for subnetting which is that you loose the first subnet
(network - all zeros) and the last subnet (broadcast - all ones).  Note that
with some routers you can configure them to use these subnets.  But be
careful, because some OS's and devices do not react well to using then.
HTH.

-Joshua

-Original Message-
From: David A. Lauer [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 22, 2001 11:14 AM
To: [EMAIL PROTECTED]
Subject: RE: difficult ccna question



Wouldn't the first subnet be 172.16.0.0?

-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of
Joshua Beining
Sent: Thursday, March 22, 2001 11:43 AM
To: 'Lowell Sharrah'
Cc: '[EMAIL PROTECTED]'
Subject: RE: difficult ccna question


Based on the subnet mask (255.255.192.0) subtract 256 from 192 which is 64.
Therefore 64 is your first subnet.  To get the second subnet, add 64 to the
first subnet.  To get the third add 64 to the second subnet and so on.
Continue in this fashion until you reach 192.  Remember that you cannot use
the ranges 172.16.0.1  -  172.16.0.62 and 172.16.255.193  -  172.16.255.254
(network and broadcast respectively) unless your router is configured to do
so.

Subnet  Host Range
1   172.16.0.65  -  172.16.0.126
2   172.16.0.129  -  172.16.0.190
3   172.16.0.193  -  172.16.0.254
4   172.16.1.1  -  172.16.1.62
5   172.16.1.65  -  172.16.1.126
6   172.16.1.129  -  172.16.1.190
7   172.16.1.193  -  172.16.1.254
8   172.16.2.1  -  172.16.2.62
9   172.16.2.65  -  172.16.2.126
   10   172.16.2.129  -  172.16.2.190
.
.
.
.
.
.

 1015   172.16.253.193  -  172.16.253.254
 1016   172.16.254.1  -  172.16.254.62
 1017   172.16.254.65  -  172.16.254.126
 1018   172.16.254.129  -  172.16.254.190
 1019   172.16.254.193  -  172.16.254.254
 1020   172.16.255.1  -  172.16.255.62
 1021   172.16.255.65  -  172.16.255.126
 1022   172.16.255.129  -  172.16.255.190

-Joshua
-Original Message-
From: Lowell Sharrah [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 22, 2001 8:14 AM
To: [EMAIL PROTECTED]
Subject: RE: difficult ccna question


how do you know where the first subnet begins?

 Joshua Beining [EMAIL PROTECTED] 03/22/01 10:34AM 
Just remember the following fomulas:

(2^# of masked bits) - 2 = Total # of subnets
(2^# of unmasked bits - 2 = Total # of hosts

Based on this the correct answer is A.

-Original Message-
From: George [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 22, 2001 12:16 AM
To: [EMAIL PROTECTED]
Subject: difficult ccna question


if you have a class B network with a 10-bit subnet mask, how many subnet and
how many hosts do you have?

a. 1022 subnets, 62 hosts
b. 62 subnets, 8190 hosts
c. 8190 subnets, 254 hosts
d. 254 subnets , 126 hosts


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