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diff --git a/jeff-subnetting-chart.md b/jeff-subnetting-chart.md
index 939b2e3..bbcf53e 100644
--- a/jeff-subnetting-chart.md
+++ b/jeff-subnetting-chart.md
@@ -10,3 +10,70 @@ Each IPv4 address has 8 bits and each bit has a place, each place has a value.
 
 This is positional notation.
 
+If we explode the third octate, x.x._ _ _ _ _ _ _ _.x
+
+The values for each place in an octet are 128 64 32 16 8 4 2 1
+
+Adding these values together will get us: 128 192 224 240 248 252 254 255 which is the decimal mask based upon position.
+
+Pretending we are in the third octet for example, if we were to list out masks in slash notation we would have:
+
+/17 /18 /19 /20 /21 /22/ 23 /24
+
+Putting those lines together get us a lot of information such as, a mask in decimal, block size, and decimal mask in slash.
+
+128 192 224 240 248 252 254 255 (mask in decimal)
+128  64  32  16   8   4   2   1 (positional notation and block size)
+/17 /18 /19 /20 /21 /22 /23 /24 (mask in slash notation)
+
+That is an excelent chart to use, we can add the slash notation depending upon which octet we are working in.
+
+For example:
+
+Given Class B 172.25.0.0 we need 0 hosts over 6 networks
+
+Figuring they gave you the network ID an dsince it is class B the default mask is /16 or 255.255.0.0
+
+We'd start from 172.25.0.0/16, the first two octets are network bits but based on the mask the 3rd and 4th octets are completly available. We need to subnet to meet their
+requirements of 80 hosts per network and at least 6 networks. Since the last two octets are completely avilable we have 16 bits available to do that.
+
+Using the the `2^h-2` and `2^n` formulas, where h=number of hosts bits left over and n=number of hosts bits borrowed.
+
+2^6-2=62 but 2^7-2=126, so we need to ensure however many bits we borrow we leave at least 7 to meet the hosts requirements.
+
+They want over 6 new networks so we need to borrow all of the rest of the bits from the 16 we had available, which is 9. We already set aside 7 for the host requirements
+(16-7=9). We add the 9 bits (locking bits in place from left to right) to the bits we already have representing network bits (/16, the first two octets), which is 16+9 (16+9=25)
+and that gives us a new mask of /25. 
+
+We can double check by subtracting the total number of bits from our 25 to see if we have the amount of bits needed for the required hosts, 32-25=7.
+
+Another forumla we can use, 2^n, can be used to find the number of new networks. 2^9=512. We have lots of networks and should definietely be over 6.
+
+Our subnets are going to start at the address we were given, 172.25.0.0, and go up from there by increments of 128.
+
+New subnets:
+1. 172.25.0.0/25
+2. 172.25.0.128/25
+3. 172.25.1.0/25
+4. 172.25.1.128/25
+5. 172.25.2.0/25
+6. 172.25.2.128/25
+7. 172.25.3.0/25
+8. 172.25.3.128/25
+
+# Finding the Network Address
+
+Given 192.168.23.12/29
+
+We're in the fourth octet so adding slash notation as well for the fourth octet
+
+128 192 224 240 248 252 254 255
+128  64  32  16   8   4   2   1
+/25 /26 /27 /28 /29 /30 /31 /32
+
+/29 gives a block size of 8.
+
+Subnet ID: 192.168.23.8/29
+First Usable: 192.168.23.9
+Last Usable: 192.168.23.14
+Broadcast: 192.168.23.15 (The final address before the next block)
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