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    <title>topic Hi,let write 2nd octet in in Technical Documentation Ideas</title>
    <link>https://community.cisco.com/t5/technical-documentation-ideas/summary-static-route/m-p/2857519#M1085</link>
    <description>&lt;P&gt;Hi,&lt;BR /&gt;let write 2nd octet in binary form.&lt;/P&gt;
&lt;P&gt;&lt;BR /&gt;172.20.0.0/16 - 172.0001 0100.0.0/16&lt;BR /&gt;172.21.0.0/16 - 172.0001 0101.0.0/16&lt;BR /&gt;172.22.0.0/16 - 172.0001 0110.0.0/16&lt;BR /&gt;172.23.0.0/16 - 172.0001 0111.0.0/16&lt;/P&gt;
&lt;P&gt;You see that first 6 bits in 2nd octet are same. So you can write summary route (same bits from left):&lt;BR /&gt;172.0001 0100.0.0 (rest are zero) and length of subnet mask is as long as bits are same, so in this example it is 14 bits (8 bits from 1st octet and 6 bits from 2nd octet).&lt;BR /&gt;Summary route in decimal form is: 172.20.0.0/14&lt;/P&gt;
&lt;P&gt;I hope that this help you.&lt;/P&gt;</description>
    <pubDate>Mon, 18 Apr 2016 05:43:50 GMT</pubDate>
    <dc:creator>Milos Megis</dc:creator>
    <dc:date>2016-04-18T05:43:50Z</dc:date>
    <item>
      <title>Summary Static Route</title>
      <link>https://community.cisco.com/t5/technical-documentation-ideas/summary-static-route/m-p/2857518#M1084</link>
      <description>&lt;P&gt;Hi,&lt;/P&gt;
&lt;P&gt;How to write / calculate the summary route? See below example, here I can understand how the summary route was determined.&lt;/P&gt;
&lt;P&gt;&lt;IMG src="https://community.cisco.com/legacyfs/online/media/summaryroute.png" class="migrated-markup-image" /&gt;&lt;/P&gt;</description>
      <pubDate>Thu, 21 Mar 2019 02:45:48 GMT</pubDate>
      <guid>https://community.cisco.com/t5/technical-documentation-ideas/summary-static-route/m-p/2857518#M1084</guid>
      <dc:creator>amh4y0001</dc:creator>
      <dc:date>2019-03-21T02:45:48Z</dc:date>
    </item>
    <item>
      <title>Hi,let write 2nd octet in</title>
      <link>https://community.cisco.com/t5/technical-documentation-ideas/summary-static-route/m-p/2857519#M1085</link>
      <description>&lt;P&gt;Hi,&lt;BR /&gt;let write 2nd octet in binary form.&lt;/P&gt;
&lt;P&gt;&lt;BR /&gt;172.20.0.0/16 - 172.0001 0100.0.0/16&lt;BR /&gt;172.21.0.0/16 - 172.0001 0101.0.0/16&lt;BR /&gt;172.22.0.0/16 - 172.0001 0110.0.0/16&lt;BR /&gt;172.23.0.0/16 - 172.0001 0111.0.0/16&lt;/P&gt;
&lt;P&gt;You see that first 6 bits in 2nd octet are same. So you can write summary route (same bits from left):&lt;BR /&gt;172.0001 0100.0.0 (rest are zero) and length of subnet mask is as long as bits are same, so in this example it is 14 bits (8 bits from 1st octet and 6 bits from 2nd octet).&lt;BR /&gt;Summary route in decimal form is: 172.20.0.0/14&lt;/P&gt;
&lt;P&gt;I hope that this help you.&lt;/P&gt;</description>
      <pubDate>Mon, 18 Apr 2016 05:43:50 GMT</pubDate>
      <guid>https://community.cisco.com/t5/technical-documentation-ideas/summary-static-route/m-p/2857519#M1085</guid>
      <dc:creator>Milos Megis</dc:creator>
      <dc:date>2016-04-18T05:43:50Z</dc:date>
    </item>
    <item>
      <title>Hi,</title>
      <link>https://community.cisco.com/t5/technical-documentation-ideas/summary-static-route/m-p/2857520#M1086</link>
      <description>&lt;P&gt;Hi,&lt;/P&gt;
&lt;P&gt;Thanks for reply and it helped.&lt;/P&gt;</description>
      <pubDate>Tue, 19 Apr 2016 08:32:48 GMT</pubDate>
      <guid>https://community.cisco.com/t5/technical-documentation-ideas/summary-static-route/m-p/2857520#M1086</guid>
      <dc:creator>amh4y0001</dc:creator>
      <dc:date>2016-04-19T08:32:48Z</dc:date>
    </item>
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