Saturday, April 21, 2007

Skeletal Flow Diagram (Thesis/Dissertation)

Posted by Ian Mars Acut at 12:23 AM 0 comments
Labels:

Wednesday, April 4, 2007

IPv6 compatible IPv4

Structure of IPv4:

32 bits dotted decimal

e.g

100.25.200.100

***********************************************************************
***********************************************************************

Structure of IPv6:

128 bits colon hexadecimal

e.g

0000:0000:0000:0000:0000:0000:0000:0000
or
0000::0000 (loopback address)


Conversion:

IPv4
172.163.1.200

Binary
1010 1100.1010 0011.0000 0001.1100 1000

IPv6
0000::ACA3:01C8


For a more detailed info click on the list on the right side or click me
Posted by Ian Mars Acut at 7:00 AM 0 comments
Labels:

Monday, April 2, 2007

IPv4 Classfull Subnet Addressing - (Class A, Class B, Class C)

1st octet . 2nd octet . 3rd octet . 4th octet




Class A = 1 - 126.0.0.0/8

0

Internet address space

N.H.H.H

Default Subnet Mask = 255.0.0.0/8

Number of Networks = 2^7 = 128-2 = 126

Hosts/Network(usable addresses) = 2^24 = 16,777,216-2 = 16,777,214

Class B = 128 - 191.0.0.0/16

10

Company address space

N.N.H.H

Default Subnet Mask = 255.255.0.0/16

Number of Networks = 2^14 = 16,384-2 = 16,382

Hosts/Network(usable addresses) = 2^16 = 65,536-2 = 65,534

Class C = 192 - 223.0.0.0/24

110

Arbitrary address space

N.N.N.H

Default Subnet Mask = 255.255.255.0/24

Number of Networks = 2^21 = 2,097,152-2 = 2,097,150

Hosts/Network(usable addresses) = 2^8 = 256-2 = 254

Class D = 224 - 239

1110

Multicasting

Class E = 240 - 254

11110

Military



Private IP Address: (RFC 1918)

Class A

10.x.x.x/8

Class B

172.30.x.x/16

Class C

192.168.x.x/24



************************************************************************************
************************************************************************************

Class A IP Address

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 10.0.200.100
Problem: I need a total of 126 subnets

Solution:


::2^7=128-2=126
Total Network=126

::2^7 = 128-2=126
Total Subnets=126

::2^17=131,072-2=131,070
Total Host/Subnet=131,070

You need to borrow 7 bits from Host Bits in Network Address


Find the Network Address:
10.0.200.100 --> Given IP Address
AND
255.0.0.0/8 --> Default Subnet Mask (Class A)
-------------------
10.0.0.0/8 --> Network Address(Subnet 0 = unusable -->wire address)

Then borrow 7 bits from Host bits in Network Address:
10.11111110.0.0/15
10.254.0.0/15 --> Last Subnet=subnet 127(unusable)

Last usable subnet:
-----------------------------
Last usable subnet = Last unusable subnet - Lowest Significant Bit
10.254.0.0/15 - 2 = 10.252.0.0/15



255.254.0.0/15 --> Custom Subnet Mask


Subnet Address <---> Broadcast Address <--->Host Range
0) 10.0.0.0 <----->10.1.255.255 <----->10.0.0.1~10.1.255.254
1) 10.2.0.0 <----->10.3.255.255 <----->10.2.0.1~10.3.255.254
2) 10.4.0.0 <----->10.5.255.255 <----->10.4.0.1~10.5.255.254
3) 10.6.0.0 <----->10.7.255.255 <----->10.6.0.1~10.7.255.254
4) 10.8.0.0 <----->10.9.255.255 <----->10.8.0.1~10.9.255.254
5) 10.10.0.0 <---> 10.11.255.255 <---> 10.10.0.1~10.11.255.254
6) 10.12.0.0 <---> 10.13.255.255 <---> 10.12.0.1~10.13.255.254
7) 10.14.0.0 <---> 10.15.255.255 <---> 10.14.0.1~10.15.255.254
8) 10.16.0.0 <---> 10.17.255.255 <---> 10.16.0.1~10.17.255.254
9) 10.18.0.0 <---> 10.19.255.255 <---> 10.18.0.1~10.19.255.254
10) 10.20.0.0 <-> 10.21.255.255 <---> 10.20.0.1~10.21.255.254
11) 10.22.0.0 <--> 10.23.255.255 <--> 10.22.0.1~10.23.255.254
12) 10.24.0.0 <--> 10.25.255.255 <--> 10.24.0.1~10.25.255.254
13) 10.26.0.0 <--> 10.27.255.255 <--> 10.26.0.1~10.27.255.254
14) 10.28.0.0 <--> 10.29.255.255 <--> 10.28.0.1~10.29.255.254
15) 10.30.0.0 <--> 10.31.255.255 <--> 10.30.0.1~10.31.255.254
16) 10.32.0.0 <--> 10.33.255.255 <--> 10.32.0.1~10.33.255.254
17) 10.34.0.0 <--> 10.35.255.255 <--> 10.34.0.1~10.35.255.254
18) 10.36.0.0 <--> 10.37.255.255 <--> 10.36.0.1~10.37.255.254
19) 10.38.0.0 <--> 10.39.255.255 <--> 10.38.0.1~10.39.255.254
20) 10.40.0.0 <--> 10.41.255.255 <--> 10.40.0.1~10.41.255.254
*
*
*
*
126)10.252.0.0 <-> 10.253.255.255 <-> 10.252.0.1~10.253.255.254
127)10.254.0.0 <-> 10.254.255.255 <----------> unusable

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 10.60.12.250
Problem: 9 Subnets

Solution:

::2^7=128-2=126
Total Networks=126

::2^4=16-2=14
Total Subnets=14

::2^20=1,048,576-2=1048574
Total Hosts/Subnet=1048574

Borrow 4 bits from Host bits in N.A.


10.60.12.250
AND
255.0.0.0/8 -->DSM
------------------
10.0.0.0/8 -->NA

10.11110000.0.0/12
10.240.0.0/12 --->Subnet 15=last subnet(unusable)

Last usable subnet:
-----------------------------
Last usable subnet = Last unusable-Lowest Significant Bit
10.240.0.0/12 - 16 = 10.224.0.0/12



255.240.0.0/12 -->CSM



Subnet Address <---> Broadcast Address <--->Host Range
0)10.0.0.0 <------->10.15.255.255<-----> 10.0.0.1~10.15.255.254
1)10.16.0.0<------>10.31.255.255 <----> 10.16.0.1~10.31.255.254
2)10.32.0.0<------>10.47.255.255 <----> 10.32.0.1~10.47.255.254
3)10.48.0.0<------>10.63.255.255 <----> 10.48.0.1~10.63.255.254
4)10.64.0.0<------>10.79.255.255 <----> 10.64.0.1~10.79.255.254
5)10.80.0.0<------>10.95.255.255 <----> 10.80.0.1~10.95.255.254
6)10.96.0.0<------>10.111.255.255 <--> 10.96.0.1~10.111.255.254
7)10.112.0.0<----> 10.127.255.255 <--> 10.112.0.1~10.127.255.254
8)10.128.0.0<---->10.143.255.255 <---> 10.128.0.1~10.143.255.254
9)10.144.0.0<---->10.159.255.255 <---> 10.144.0.1~10.159.255.254
10)10.160.0.0<--> 10.175.255.255<---> 10.160.0.1~10.175.255.254
11)10.176.0.0<--> 10.191.255.255<---> 10.176.0.1~10.191.255.254
12)10.192.0.0<--> 10.207.255.255<---> 10.192.0.1~10.207.255.254
13)10.208.0.0<--> 10.223.255.255<---> 10.208.0.1~10.223.255.254
14)10.224.0.0<--> 10.239.255.255<---> 10.224.0.1~10.239.255.254
15)10.240.0.0 <-> 10.240.255.255 <--> unusable

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 125.100.10.50
Problem: 65,534 Subnets


Solution:

::2^7=128-2=126
Total Networks

::2^16=65,536-2=65,534
Total Subnets

::2^8=256-2=254
Total Hosts/Subnets

Borrow 16 bits from Host Bits in Network Address

125.100.10.50
AND
255.0.0.0/8 --> DSM
----------------
125.0.0.0/8 --> NA

125.11111111.11111111.0/24
125.255.255.0/24 --> Last unusable subnet=subnet 65,535(unusable)

Last usable subnet:
--------------------------
65,535-1 = subnet 65,534
Last usable subnet=125.255.255.0/24 - 1 = 125.255.254.0/24

255.255.255.0/24 --> CSM

Subnet Address <---> Broadcast Address <--->Host Range
0)125.0.0.0<---->125.0.0.255<-------> 125.0.0.1~125.0.0.254
1)125.0.1.0<---->125.0.1.255 <------> 125.0.1.1~125.0.1.254
2)125.0.2.0<---->125.0.2.255 <------>125.0.2.1~125.0.2.254
3)125.0.3.0<---->125.0.3.255 <------>125.0.3.1~125.0.3.254
4)125.0.4.0<---->125.0.4.255 <------>125.0.4.1~125.0.4.254
5)125.0.5.0<---->125.0.5.255 <------>125.0.5.1~125.0.5.254
6)125.0.6.0<---->125.0.6.255 <------>125.0.6.1~125.0.6.254
7)125.0.7.0<---->125.0.7.255 <------>125.0.7.1~125.0.7.254
8)125.0.8.0<---->125.0.8.255 <------>125.0.8.1~125.0.8.254
9)125.0.9.0<---->125.0.9.255 <------>125.0.9.1~125.0.9.254
10)125.0.10.0<->125.0.10.255 <---> 125.0.10.1~125.0.10.254
11)125.0.11.0<-->125.0.11.255 <-> 125.0.11.1~125.0.11.254
12)125.0.12.0<-->125.0.12.255 <-> 125.0.12.1~125.0.12.254
13)125.0.13.0<-->125.0.13.255 <-> 125.013.1~125.0.13.254
14)125.0.14.0<-->125.0.14.255 <--> 125.1.14.1~125.1.14.254
15)125.0.15.0<-->125.0.15.255 <-> 125.0.15.1~125.0.15.254
* *
* *
* *
* *
65,534) 125.255.254.0 <-> 125.255.254.255 <-> 125.255.254.1~125.255.254.254
65,535) 125.255.255.0 <--> 125.255.255.255 <---> unusable

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 120.200.168.10
Problem: I need 262,142 subnets

Solution:

::2^7=128-2=126
Total Network

::2^18=262,144-2=262,142
Total Subnets

::2^6=64-2=62
Total Hosts

Borrow 18 bits from Host bits in NA

120.200.168.10
AND
255.0.0.0
---------------------
120.0.0.0/8

120.11111111.11111111.11000000/26
120.255.255.192/26 --> Last unusable subnet=262,143 subnets(unusable)

Last usable subnet:

120.255.255.192 - 64 = 120.255.255.128/26

255.255.255.192/26 --> CSM

Subnet Address <---> Broadcast Address <--->Host Range
0)120.0.0.0 <--->120.0.0.63<-----> 120.0.0.1~120.0.0.62
1)120.0.0.64<--> 120.0.0.127 <--> 120.0.0.65~120.0.0.126
2)120.0.0.128<-> 120.0.0.191 <--> 120.0.1.129~120.0.0.190
3)120.0.0.192 <-> 120.0.0.255 <--> 120.0.0.193~120.0.0.254
4)120.0.1.0 <-----> 120.0.1.63 <---> 120.0.1.1~120.0.1.62
5)120.0.1.64 <---> 120.0.1.127<--> 120.0.1.65~120.0.1.126
6)120.0.1.128 <->120.0.1.191<---> 120.0.1.129~120.0.1.190
7)120.0.1.192 <->120.0.1.255<--->120.0.1.193~120.0.1.254
8)120.0.2.0 <---->120.0.2.63 <---->120.0.2.1~120.0.2.62
9)120.0.2.64 <-->120.0.2.127<--->120.0.2.65~120.0.2.126
10)120.0.2.128 <-->120.0.2.191<-->120.0.2.129~120.0.2.190
11)120.0.2.192 <-->120.0.2.255<-->120.0.2.193~120.0.2.254
12)120.0.3.0 <----->120.0.3.63<---->120.0.3.1~120.0.3.62
13)120.0.3.64 <--->120.0.3.127<-->120.0.3.65~120.0.3.126
14)120.0.3.128 <->120.0.3.191<-->120.0.3.129~120.0.3.190
15)120.0.3.192 <->120.0.3.255<-->120.0.3.193~120.0.3.254
16)120.0.4.0 <---->120.0.4.63<---->120.0.4.1~120.0.4.62
17)120.0.4.64 <-->120.0.4.127<-->120.0.4.65~120.0.4.126
18)120.0.4.128 <-->120.0.4.191<-->120.0.4.129~120.0.4.190
19)120.0.4.192 <-->120.0.4.255<-->120.0.4.193~120.0.4.254
20)120.0.5.0 <-----> 120.0.5.63 <--->120.0.5.1~120.0.5.62
*
*
*
262,142)120.255.255.128<>120.255.255.191<> 120.255.255.129~120.255.255.190
262,143)120.255.255.192<>120.255.255.255<> unusable


///////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////

Given: 110.200.100.100
Problem: 131,070 subnets

Solution:

::2^7=128-2=126
Total Networks

::2^17=131,072-2=131,072
Total Subnets

::2^7=128-2=126
Total Hosts/Subnet

Borrow 17 bits from Host bits in the NA

110.200.100.100
AND
255.0.0.0/8 --> DSM
-----------------------
110.0.0.0/8 --> NA

110.11111111.11111111.10000000/17
110.255.255.128/17 --> Last unusable subnet=subnet 131,073


Last usable subnet:
---------------------------
110.255.255.128/17 - 128 = 110.255.255.0/17
110.255.255.0/17 --> Last usable subnet = subnet 131,072


255.255.255.128/17 --> CSM


Subnet Address <---> Broadcast Address <--->Host Range
0)110.0.0.0<------> 110.0.0.127<---->110.0.0.1~110.0.0.126
1)110.0.0.128 <-->110.0.0.255 <---> 110.0.0.129~110.0.0.254
2)110.0.1.0 <-----> 110.0.1.127 <---> 110.0.1.1~110.0.1.126
3)110.0.1.128 <-> 110.0.1.255 <---> 110.0.1.129~110.0.1.254
4)110.0.2.0 <----> 110.0.2.127 <--->110.0.2.1~110.0.2.126
5)110.0.2.128 <-> 110.0.2.255 <--> 110.0.2.129~110.0.2.254
6)110.0.3.0 <----> 110.0.3.127 <--> 110.0.3.1~110.0.3.126
7)110.0.3.128 <-> 110.0.3.255 <-> 110.0.3.129~110.0.3.254
8)110.0.4.0 <----> 110.0.4.127 <--> 110.0.4.1~110.0.4.126
9)110.0.4.128 <--> 110.0.4.255 <-> 110.0.4.129~110.0.4.254
10)110.0.5.0 <---> 110.0.5.127 <--> 110.0.5.1~110.0.5.126
11)110.0.5.128 <--> 110.0.5.255 <--> 110.0.5.129~110.0.5.254
12)110.0.6.0 <-----> 110.0.6.127 <--> 110.0.6.1~110.0.6.126
13)110.0.6.128 <--> 110.0.6.255 <--> 110.0.6.129~110.0.6.254
14)110.0.7.0 <-----> 110.0.7.127 <---> 110.0.7.1~110.0.7.126
15)110.0.7.128 <-> 110.0.7.255 <---> 110.0.7.129~110.0.7.254
16)110.0.8.0 <----> 110.0.8.127 <---> 110.0.8.1~110.0.8.126
17)110.0.8.128 <--> 110.0.8.255 <--> 110.0.8.129~110.0.8.254
18)110.0.9.0 <-----> 110.0.9.127 <--> 110.0.9.1~110.0.9.126
19)110.0.9.128 <-> 110.0.9.255 <--> 110.0.9.129~110.0.9.254
20)110.0.10.0 <--> 110.0.10.127 <-> 110.0.10.1~110.0.10.126
*
*
*
131,072)110.255.255.0<>110.255.255.127<>110.255.255.1~110.255.255.126
131,073)110.255.255.128<>110.255.255.255 <----> unusable

////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 10.16.5.3
Problem: I need 4,194,302 subnets

Solution:

::2^7=128-2=126
Total Networks

::2^22=4,194,304-2=4,194,302
Total Subnets

::2^2=4-2=2
Total Hosts/Subnet

Borrow 22 bits from Host bits in the NA

10.16.5.3
AND
255.0.0.0/8
-------------
10.0.0.0/8

10.11111111.11111111.11111100/22
10.255.255.252/22 --> Last unusable subnet = subnet 4,194,303

Last usable subnet:
--------------------------
10.255.255.248/22 --> Last usable subnet = subnet 4,194,302


255.255.255.248/22 --> CSM


Subnet Address <---> Broadcast Address <--->Host Range
0)10.0.0.0 <--> 10.0.0.3 <--------> 10.0.0.1~10.0.0.2
1)10.0.0.4 <--> 10.0.0.7 <--------> 10.0.0.5~10.0.0.6
2)10.0.0.8 <--> 10.0.0.11 <------> 10.0.0.9~10.0.0.10
3)10.0.0.12 <---> 10.0.0.15 <---> 10.0.0.13~10.0.0.14
4)10.0.0.16 <--> 10.0.0.19 <----> 10.0.0.17~10.0.0.18
5)10.0.0.20 <--> 10.0.0.23 <----> 10.0.0.21~10.0.0.22
6)10.0.0.24 <--> 10.0.0.27 <----> 10.0.0.25~10.0.0.26
7)10.0.0.28 <--> 10.0.0.31 <----> 10.0.0.29~10.0.0.30
8)10.0.0.32 <--> 10.0.0.35 <----> 10.0.0.33~10.0.0.34
9)10.0.0.36 <--> 10.0.0.39 <----> 10.0.0.37~10.0.0.38
10)10.0.0.40 <--> 10.0.0.43 <--> 10.0.0.41~10.0.0.42
11)10.0.0.44 <--> 10.0.0.47 <--> 10.0.0.45~10.0.0.46
12)10.0.0.48 <-> 10.0.0.51 <--> 10.0.0.49~10.0.0.50
13)10.0.0.52 <-> 10.0.0.55 <--> 10.0.0.53~10.0.0.54
14)10.0.0.56 <-> 10.0.0.59 <--> 10.0.0.57~10.0.0.58
15)10.0.0.60 <-> 10.0.0.63 <--> 10.0.0.61~10.0.0.62
16)10.0.0.64 <-> 10.0.0.67 <--> 10.0.0.65~10.0.0.66
17)10.0.0.68 <-> 10.0.0.71 <--> 10.0.0.69~10.0.0.70
18)10.0.0.72 <-> 10.0.0.75 <--> 10.0.0.73~10.0.0.74
19)10.0.0.76 <-> 10.0.0.79 <--> 10.0.0.77~10.0.0.78
20)10.0.0.80 <-> 10.0.0.83 <--> 10.0.0.81~10.0.0.82
*
*
*
4,194,302)10.255.255.248<>10.255.255.251<>10.255.255.249~10.255.255.250
4,194,303)10.255.255.252<>10.255.255.255<> unusable


##########################################################################
##########################################################################
##########################################################################
##########################################################################
##########################################################################



Class B IP Address
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 130.17.12.160
Problem: I need 4094 subnets

Solution:

::2^14 = 16,384-2 = 16,382
Total Networks

::2^12 = 4,096-2 = 4,094
Total Subnets

::2^4 = 16-2 = 14
Total Hosts/Subnet


Borrow 12 bits from Host bits in the Network Address

130.17.12.160
AND
255.255.0.0 /16 --> DSM
-------------------
130.17.0.0 /16 --> NA

130.17.11111111.11110000/28
130.17.255.240/28 --> Last unusable subnet = subnet 4,095

Last usable subnet:
-------------------------
130.17.255.240/28 - 16 = 130.17.255.224/28

255.255.255.240/28 --> CSM


Subnet Address <---> Broadcast Address <--->Host Range
0)130.17.0.0 <----->130.17.0.15 <----> 130.17.0.1~130.17.0.14
1)130.17.0.16 <---> 130.17.0.31 <---> 130.17.0.17~130.17.0.30
2)130.17.0.32 <---> 130.17.0.47 <---> 130.17.0.33~130.17.0.46
3)130.17.0.48 <---> 130.17.0.63 <---> 130.17.0.49~130.17.0.62
4)130.17.0.64 <---> 130.17.0.79 <---> 130.17.0.65~130.17.0.78
5)130.17.0.80 <---> 130.17.0.95 <---> 130.17.0.81~130.17.0.94
6)130.17.0.96 <---> 130.17.0.111 <--> 130.17.0.97~130.17.0.110
7)130.17.0.112 <-> 130.17.0.127 <--> 130.17.0.113~130.17.0.126
8)130.17.0.128 <-> 130.17.0.143 <--> 130.17.0.129~130.17.0.142
9)130.17.0.144 <-> 130.17.0.159 <--> 130.17.0.145~130.17.0.158
10)130.17.0.160 <-> 130.17.0.175 <--> 130.17.0.161~130.17.0.174
11)130.17.0.176 <-> 130.17.0.191 <--> 130.17.0.177~130.17.0.190
12)130.17.0.192 <-> 130.17.0.207 <-> 130.17.0.193~130.17.0.206
13)130.17.0.208 <-> 130.17.0.223 <-> 130.17.0.209~130.17.0.222
14)130.17.0.224 <--> 130.17.0.239 <--> 130.17.0.225~130.17.0.238
15)130.17.0.240 <--> 130.17.0.255 <--> 130.17.0.141~130.17.0.254
16)130.17.1.0 <-----> 130.17.1.15 <----> 130.17.1.1~130.17.1.14
17)130.17.1.16 <---> 130.17.1.31 <----> 130.17.1.17~130.17.1.30
18)130.17.1.32 <---> 130.17.1.47 <---> 130.17.1.33~130.17.1.46
19)130.17.1.48 <---> 130.17.1.63 <---> 130.17.1.49~130.17.1.62
20)130.17.1.64 <--> 130.17.1.79 <----> 130.17.1.65~130.17.1.78
*
*
*
4,094)130.17.255.224<>130.17.255.239<>130.17.255.225~130.17.255.238
4,095)130.17.255.240

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 144.1.0.0
Problem: I need 14 subnets

Solution:

::2^14 = 16,384-2 = 16,382
Total Networks

::2^4 = 16-2 = 14
Total Subnets

::2^12 = 4,096-2 = 4,094
Total Hosts/Subnet

Borrow 4 bits from Host bits in Network Address

144.1.0.0
AND
255.255.0.0/16 --> DSM
----------------
144.1.0.0/16 --> NA

144.1.11110000.00000000/20
144.1.240.0/20 --> Last Unusable Subnet = subnet 15

Last usable subnet:
-------------------------
144.1.224.0/20 --> Last usable subnet = subnet 14

Subnet Address <----> Broadcast Address <--->Host Range
0)144.1.0.0 <----> 144.1.15.255 <----> 144.1.0.1~144.1.15.254
1)144.1.16.0 <---> 144.1.31.255 <--->144.1.16.1~144.1.31.254
2)144.1.32.0 <---> 144.1.47.255 <---> 144.1.32.1~144.1.47.254
3)144.1.48.0 <---> 144.1.63.255 <---> 144.1.48.1~144.1.63.254
4)144.1.64.0 <---> 144.1.79.255 <---> 144.1.64.1~144.1.79.254
5)144.1.80.0 <---> 144.1.95.255 <--->144.1.80.1~144.1.95.254
6)144.1.96.0 <---> 144.1.111.255 <--> 144.1.96.1~144.1.111.254
7)144.1.112.0 <--> 144.1.127.255 <---> 144.1.112.1~144.1.127.254
8)144.1.128.0 <--> 144.1.143.255 <---> 144.1.128.1~144.1.143.254
9)144.1.144.0 <--> 144.1.159.255 <---> 144.1.144.1~144.1.159.254
10)144.1.160.0 <-> 144.1.175.255 <---> 144.1.160.1~144.1.175.254
11)144.1.176.0 <-> 144.1.191.255 <---> 144.1.176.1~144.1.191.254
12)144.1.192.0 <-> 144.1.207.255 <---> 144.1.192.1~144.1.207.254
13)144.1.208.0 <-> 144.1.223.255 <---> 144.1.208.1~144.1.223.254
14)144.1.224.0 <-> 144.1.239.255 <---> 144.1.224.1~144.1.239.254
15)144.1.240.0 <---> 144.1.255.255 <----> unused

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 190.150.12.2
Problem: I need 16, 382 subnets

Solution:

::2^14 = 16,384-2 = 16,382
Total Networks

::2^14 = 16,384-2 = 16,382
Total Subnets

::2^2 = 4-2=2
Total Hosts/Subnet

Borrow 14 bits from Host bits in the NA

190.150.12.2
AND
255.255.0.0/16 --> Default Subnet Mask
------------------
190.150.0.0/16 --> Network Address

190.150.11111111.11111100/30
190.150.255.252/30 --> Last unusable subnet = subnet 16,383


Last usable subnet:
--------------------------
190.150.255.252/30 - 4 = 190.150.255.248/30


255.255.255.252/30 --> Custom Subnet Mask


Subnet Address<--->Broadcast Address<--->Host Range
0)190.150.0.0 <------->190.150.0.3 <----> 190.150.0.1~190.150.0.2
1)190.150.0.4 <-------> 190.150.0.7 <---> 190.150.0.5~190.150.0.6
2)190.150.0.8 <-------> 190.150.0.11 <--> 190.150.0.9~190.150.0.10
3)190.150.0.12 <------> 190.150.0.15 <--> 190.150.0.13~190.150.0.14
4)190.150.0.16 <------> 190.150.0.19 <--> 190.150.0.17~190.150.0.18
5)190.150.0.20 <------> 190.150.0.23 <--> 190.150.0.21~190.150.0.22
6)190.150.0.24 <------> 190.150.0.27 <--> 190.150.0.25~190.150.0.26
7)190.150.0.28 <------> 190.150.0.31 <--> 190.150.0.29~190.150.0.30
8)190.150.0.32 <------> 190.150.0.35 <--> 190.150.0.33~190.150.0.34
9)190.150.0.36 <------> 190.150.0.39 <---> 190.150.0.37~190.150.0.38
10)190.150.0.40 <----> 190.150.0.43 <---> 190.150.0.41~190.150.0.42
11)190.150.0.44 <----> 190.150.0.47 <---> 190.150.0.45~190.150.0.46
12)190.150.0.48 <----> 190.150.0.51 <---> 190.150.0.49~190.150.0.50
13)190.150.0.52 <----> 190.150.0.55 <---> 190.150.0.53~190.150.0.54
14)190.150.0.56 <----> 190.150.0.59 <---> 190.150.0.57~190.150.0.58
15)190.150.0.60 <----> 190.150.0.63 <---> 190.150.0.61~190.150.0.62
*
*
*
16,382)190.150.255.248 <---> 190.150.255.251 <---> 190.150.255.249~190.150.255.250
16,383)190.150.255.252 <----> 190.150.255.255 <----> unusable


////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 172.168.10.10
Problem: I need 254 subnets

Solution:

::2^14 = 16,384-2 = 16,382
Total Networks

::2^8 = 256-2 = 254
Total Subnets

::2^8 = 256-2 = 254
Total Hosts/Subnet

Borrow 8 bits from the Host bits in NA

172.168.10.10
AND
255.255.0.0/16 --> Default Subnet Mask
-------------------
172.168.0.0/16--> Network Address

172.168.11111111.0/24
172.168.255.0/24 --> Last unusable subnet = subnet 255


Last usable subnet:
-------------------------
172.168.254.0/24


255.255.255.0/24 --> Custom Subnet Mask

Subnet Address <---> BroadCast Address <---> Host Address
0)172.168.0.0 <--> 172.168.0.255 <---> 172.168.0.1~172.168.0.254
1)172.168.1.0 <--> 172.168.1.255 <---> 172.168.1.1~172.168.1.254
2)172.168.2.0 <--> 172.168.2.255 <---> 172.168.2.1~172.168.2.254
3)172.168.3.0 <--> 172.168.3.255 <---> 172.168.3.1~172.168.3.254
4)172.168.4.0 <--> 172.168.4.255 <---> 172.168.4.1~172.168.4.254
5)172.168.5.0 <--> 172.168.5.255 <---> 172.168.5.1~172.168.5.254
6)172.168.6.0 <--> 172.168.6.255 <---> 172.168.6.1~172.168.6.254
7)172.168.7.0 <--> 172.168.7.255 <---> 172.168.7.1~172.168.7.254
8)172.168.8.0 <--> 172.168.8.255 <---> 172.168.8.1~172.168.8.254
9)172.168.9.0 <--> 172.168.9.255 <----> 172.168.9.1~172.168.9.254

*
*
*
254)172.168.254.0 <---> 172.168.254.255 <---> 172.168.254.1~172.168.254.254
255)172.168.255.0 <----> 172.168.255.255 <---> unusable


/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 130.230.110.100
Problem: I need 1,022 subnets

Solution:

::2^14 = 16,384-2 = 16,382
Total Network

::2^10 = 1,024-2 = 1,022
Total Subnets

::2^6 = 64-2 = 62
Total Hosts/Subnet

Borrow 10 bits from Host bits in NA


130.230.110.100
AND
255.255.0.0/16--> DSM
------------------------
130.230.0.0/16 --> NA

130.230.11111111.11000000/26
130.230.255.192/26 --> Last unusable subnet = subnet 1,023

Last usable subnet:
-------------------------
130.230.255.192/26 - 64 = 130.230.255.128/26


255.255.255.192/26 --> CSM


Subnet Address <---> Broadcast Address <---> Host Range
0)130.230.0.0 <--> 130.230.0.63 <----> 130.230.0.1~130.230.0.62
1)130.230.0.64 <-> 130.230.0.127 <--> 130.230.0.65~130.230.0.126
2)130.230.0.128 <-> 130.230.0.191 <--> 130.230.0.129~130.230.0.190
3)130.230.0.192 <--> 130.230.0.255 <--> 130.230.0.193~130.230.0.254
4)130.230.1.0 <----->130.230.1.63 <----> 130.230.1.1~130.230.1.62
5)130.230.1.64 <---> 130.230.1.127 <--> 130.230.1.65~130.230.1.126
6)130.230.1.128 <--> 130.230.1.191 <-> 130.230.1.129~130.230.1.190
7)130.230.1.192 <--> 130.230.1.255 <-->130.230.1.193~130.230.1.254
8)130.230.2.0 <---> 130.230.2.63 <--> 130.230.2.1~130.230.2.62
9)130.230.2.64 <--> 130.230.2.127 <--> 130.230.2.65~130.230.2.126
10)130.230.2.128 <-> 130.230.2.191 <-> 130.230.2.129~130.230.2.190
11)130.230.2.192 <-> 130.230.2.255 <-> 130.230.2.193~130.230.2.254
12)130.230.3.0 <-> 130.230.3.63 <-> 130.230.3.1~130.230.3.62
13)130.230.3.64 <-> 130.230.3.127 <-> 130.230.3.65~130.230.3.126
14)130.230.3.128 <-> 130.230.3.191 <-> 130.230.3.129~130.230.3.190
15)130.230.3.192 <--> 130.230.3.255 <-> 130.230.3.193~130.230.3.254
*
*
*
1,022)130.230.255.128<---> 130.230.255.191 <---> 130.230.255.129~130.230.255.190
1,023)130.230.255.192 <--> 130.230.255.255 <---> unusable


//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 172.16.0.0
Problem: 4 subnets
Sample Laboratory

Solution:

::2^14 = 16,384-2 = 16,382
Total Network

::2^4 = 16-2 = 14
Total Subnets

::2^12 = 4,096-2 = 4,094
Total Hosts/Subnet

********************************************************************************************
********************************************************************************************
* Note on borrowing host bits:
*
* In creating a subnet, borrow bits from the host portion of the IP address
*
* Network.Host
* after borrowing -->Network.Subnet.Host
*
* Bits borrowed must be >= subnets required (b^L =N)
* Never < subnet required (b^L =N)
*

Class A

Maximum bits borrowed from hosts = 22 bits
Minimum bits borrowed from hosts = 2 bits(same with Class B and Class C)
Class B

Maximum bits borrowed from hosts = 14 bits
Class C

Maximum bits borrowed from hosts = 6 bits

********************************************************************************************
********************************************************************************************

Borrow 4 bits from the host bits in the NA

172.16.0.0
AND
255.255.0.0/16
----------------
172.16.0.0/16

172.16.11110000.00000000/20
172.16.240.0/20 --> Last unusable subnet = subnet 15

Last usable subnet:
-------------------------
172.16.240.0/20 - 16 = 172.16.224.0/20


255.255.240.0/20 --> CSM

Subnet Address <---> Broadcast Address <---> Host Range
0)172.16.0.0 <---> 172.16.15.255 <-->172.16.0.1~172.16.15.254
1)172.16.16.0 <--> 172.16.31.255 <--> 172.16.16.1~172.16.31.254
2)172.16.32.0 <--> 172.16.47.255 <--> 172.16.32.1~172.16.47.254
3)172.16.48.0 <--> 172.16.63.255 <--> 172.16.48.1~172.16.63.254
4)172.16.64.0 <--> 172.16.79.255 <--> 172.16.64.1~172.16.79.254
5)172.16.80.0 <--> 172.16.95.255 <--> 172.16.80.1~172.16.95.254
6)172.16.96.0 <--> 172.16.111.255 <-> 172.16.96.1~172.16.111.254
7)172.16.112.0 <-> 172.16.127.255 <-> 172.16.112.1~172.16.127.254
8)172.16.128.0 <-> 172.16.143.255 <-> 172.16.128.1~172.16.143.254
9)172.16.144.0 <-> 172.16.159.255 <-> 172.16.144.1~172.16.159.254
10)172.16.160.0 <-> 172.16.175.255 <-> 172.16.160.1~172.16.175.254
11)172.16.176.0 <-> 172.16.191.255 <-> 172.16.176.1~172.16.191.254
12)172.16.192.0 <-> 172.16.207.255 <-> 172.16.192.1~172.16.207.254
13)172.16.208.0 <-> 172.16.223.255 <-> 172.16.208.1~172.16.223.254
14)172.16.224.0<-->172.16.239.255<-->172.16.224.1~172.16.239.254
15)172.16.240.0 <---> 172.16.255.255 <---> unusable


////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 172.16.10.53
Problem: 510 subnets

Solution:

::2^14 = 16,384-2 = 16,382
Total Networks

::2^9 = 512-2 = 510
Total Subnets

::2^7=128-2 = 126
Total Host/Subnet

Borrow 9 bits from the host bits in NA

172.16.10.53
AND
255.255.0.0/16 --> DSM
--------------------
172.16.0.0/16 --> NA

172.16.1111111.10000000/25
172.16.255.128/25 --> Last unusable subnet = subnet 511

Last subnet usable:
--------------------------
172.16.255.128/25 - 128 = 172.16.255.0/25

255.255.255.128/25 --> CSM


Subnet Address <--> Broadcast Address <--> Host Range
0)172.16.0.0 <-----> 172.16.0.127 <---> 172.16.0.1~172.16.0.126
1)172.16.0.128 <--> 172.16.0.255 <--> 172.16.0.129~172.16.0.254
2)172.16.1.0 <--> 172.16.1.127 <--> 172.16.1.1~172.16.1.126
3)172.16.1.128 <-> 172.16.1.255 <-> 172.16.1.129~172.16.1.254
4)172.16.2.0 <--> 172.16.2.127 <--> 172.16.2.1~172.16.2.126
5)172.16.2.128 <--> 172.16.2.255 <--> 172.16.2.129~172.16.2.254
6)172.16.3.0 <--> 172.16.3.127 <--> 172.16.3.1~172.16.3.126
7)172.16.3.128 <--> 172.16.3.255 <--> 172.16.3.129~172.16.3.254
8)172.16.4.0 <--> 172.16.4.127 <--> 172.16.4.1~172.16.4.126
9)172.16.4.128 <--> 172.16.4.255 <--> 172.16.4.129~172.16.4.254
10)172.16.5.0 <--> 172.16.5.127 <--> 172.16.5.1~172.16.5.126
11)172.16.5.128 <--> 172.16.5.255 <--> 172.16.5.129~172.16.5.254
12)172.16.6.0 <--> 172.16.6.127 <--> 172.16.6.1~172.16.6.126
13)172.16.6.128 <--> 172.16.6.255 <--> 172.16.6.129~172.16.6.254
14)172.16.7.0 <--> 172.16.7.127 <--> 172.16.7.1~172.16.7.126
15)172.16.7.128 <--> 172.16.7.255 <--> 172.16.7.129~172.16.7.254
16)172.16.8.0 <--> 172.16.8.127 <--> 172.16.8.1~172.16.8.126
17)172.16.8.128 <-> 172.16.8.255 <-> 172.16.8.129~172.16.8.254
18)172.16.9.0 <--> 172.16.9.127 <-> 172.16.9.1~172.16.9.126
19)172.16.9.128 <-> 172.16.9.255 <--> 172.16.9.129~172.16.9.254
20)172.16.10.0 <-> 172.16.10.127 <--> 172.16.10.1~172.16.10.126
*
*
*
510)172.16.255.0 <--> 172.16.255.127 <--> 172.16.255.1~172.16.255.126
511)172.16.255.128


/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 172.16.12.200
Problem: 7 subnets

Solution:

::2^14 = 16,384-2 = 16,382
Total Networks

::2^8 = 256-2 = 254
Total Subnets

::2^8 = 256-2 = 254
Total Hosts/Subnet

Borrow 8 bits from the Host bits in NA

172.16.12.200
AND
255.255.0.0/16 --> DSM
--------------------
172.16.0.0/16 --> NA

172.16.11111111.0/24
172.16.255.0/24 --> Last Subnet unusable = subnet 255


Last usable subnet:
--------------------------
172.16.255.0/24 - 1 = 172.16.254.0/24


255.255.255.0/24 --> CSM

Subnet Address <--> Broadcast Address <--> Host Range
0)172.16.0.0 <---> 172.16.0.255 <--> 172.16.0.1~172.16.0.254
1)172.16.1.0 <---> 172.16.1.255 <--> 172.16.1.1~172.16.1.254
2)172.16.2.0 <---> 172.16.2.255 <--> 172.16.2.1~172.16.2.254
3)172.16.3.0 <---> 172.16.3.255 <--> 172.16.3.1~172.16.3.254
4)172.16.4.0 <---> 172.16.4.255 <--> 172.16.4.1~172.16.4.254
5)172.16.5.0 <---> 172.16.5.255 <--> 172.16.5.1~172.16.5.254
6)172.16.6.0 <---> 172.16.6.255 <--> 172.16.6.1~172.16.6.254
7)172.16.7.0 <---> 172.16.7.255 <--> 172.16.7.1~172.16.7.254
8)172.16.8.0 <---> 172.16.8.255 <--> 172.16.8.1~172.16.8.254
9)172.16.9.0 <--> 172.16.9.255 <--> 172.16.9.1~172.16.9.254
10)172.16.10.0 <--> 172.16.10.255 <--> 172.16.10.1~172.16.10.254
11)172.16.11.0 <--> 172.16.11.255 <--> 172.16.11.1~172.16.11.254
12)172.16.12.0 <--> 172.16.12.255 <--> 172.16.12.1~172.16.12.254
13)172.16.13.0 <--> 172.16.13.255 <--> 172.16.13.1~172.16.13.254
14)172.16.14.0 <--> 172.16.14.255 <--> 172.16.14.1~172.16.14.254
15)172.16.15.0 <--> 172.16.15.255 <--> 172.16.15.1~172.16.15.254
*
*
*
254)172.16.254.0
255)172.16.255.0


/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 130.191.3.3
Problem: Find the 200th subnet

Solution:

::2^14 = 16,384-2 = 16,382
Total Networks

::2^8 = 256-2 = 254
Total Subnets

::2^8 = 256-2 = 254
Total Hosts/Subnet

130.191.3.3
AND
255.255.0.0/16 --> DSM
----------------
130.191.0.0/16 --> NA

130.191.11111111.0/24
130.191.255.0/24 --> Last unusable subnet = subnet 255

Last usable subnet = 130.191.254.0/24

255.255.255.0/24 --> CSM

Subnet Address <--> Broadcast Address <--> Host Range
0)130.191.0.0 <---> 130.191.0.255 <--> 130.191.0.1~130.191.0.254
1)130.191.1.0 <---> 130.191.1.255 <--> 130.191.1.1~130.191.1.254
2)130.191.2.0 <---> 130.191.2.255 <--> 130.191.2.1~130.191.2.254
3)130.191.3.0 <---> 130.191.3.255 <--> 130.191.3.1~130.191.3.254
4)130.191.4.0 <---> 130.191.4.255 <--> 130.191.4.1~130.191.4.254
5)130.191.5.0 <---> 130.191.5.255 <---> 130.191.5.1~130.191.5.254
6)130.191.6.0 <---> 130.191.6.255 <---> 130.191.6.1~130.191.6.254
7)130.191.7.0 <---> 130.191.7.255 <---> 130.191.7.1~130.191.7.254
*
*
*
254)130.191.254.0 <--> 130.191.254.255
255)130.191.255.0 <--> 130.191.255.255



//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 150.150.0.0
Problem: 6 subnets

Solution:

::2^14 = 16,384-2 = 16,382
Total Network

::2^3 = 8-2 = 6
Total Subnets

::2^13 = 8192-2 = 8190
Total Hosts/Subnet

150.150.0.0
AND
255.255.0.0/16 --> DSM
------------------
150.150.0.0/16 --> NA

150.150.11100000.0/19
150.150.224.0/19 --> Last unusable subnet = subnet 7

150.150.192.0/19 --> Last usable subnet

255.255.224.0/19 --> Custom Subnet Mask


Subnet Address <---> Broadcast Address <---> Host Range
0)150.150.0.0 <-->150.150.31.255 <--> 150.150.0.1~150.150.21.254
1)150.150.32.0<->150.150.63.255<->150.150.32.1~150.150.63.254
2)150.150.64.0<->150.150.95.255<->150.150.64.1~150.150.95.254
3)150.150.96.0<->150.150.127.255<->150.150.96.1~150.150.127.254
4)150.150.128.0<->150.150.159.255<->150.150.128.1~150.150.159.254
5)150.150.160.0<->150.150.191.255<->150.150.160.1~150.150.191.254
6)150.150.192.0<->150.150.223.255<->150.150.192.1~150.150.223.254
7)150.150.224.0<-->150.150.255.255<-->UNUSABLE



########################################################################
########################################################################
########################################################################

Class C IP Address


////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 200.133.175.0
Problem: 14 subnets

Solution:

::2^21 = 2,097,152-2 = 2,097,150
Total Network

::2^4 = 16-2 = 14
Total Subnets

::2^4 = 16-2 = 14
Total Hosts/Subnet


200.133.175.0
AND
255.255.255.0/24 --> DSM
--------------------
200.133.175.0/24 --> NA


200.133.175.11110000/28
200.133.175.240/28 --> Last unusable subnet = subnet 15


Last usable subnet:
-------------------------
200.133.175.224/28 --> Last usable subnet


255.255.255.240/28 --> CSM


Subnet Address <--> Broadcast Address <--> Host Range
0)200.133.175.0 <-> 200.133.175.15 <-> 200.133.175.1~200.133.175.14
1)200.133.175.16<->200.133.175.31<->200.133.175.17~200.133.175.30
2)200.133.175.32<->200.133.175.47<->200.133.175.33~200.133.175.46
3)200.133.175.48<->200.133.175.63<->200.133.175.49~200.133.175.62
4)200.133.175.64<->200.133.175.79<->200.133.175.65~200.133.175.78
5)200.133.175.80<->200.133.175.95<->200.133.175.81~200.133.175.94
6)200.133.175.96<->200.133.175.111<->200.133.175.97~200.133.175.110
7)200.133.175.112<->200.133.175.127<->200.133.175.113~200.133.175.126
8)200.133.175.128<->200.133.175.143<->200.133.175.129~200.133.175.142
9)200.133.175.144<->200.133.175.159<->200.133.175.145~200.133.175.158
10)200.133.175.160<->200.133.175.175<->200.133.175.161~200.133.175.174
11)200.133.175.176<->200.133.175.191<->200.133.175.177~200.133.175.190
12)200.133.175.192<->200.133.175.207<->200.133.175.193~200.133.175.206
13)200.133.175.208<->200.133.175.223<->200.133.175.209~200.133.175.222
14)200.133.175.224<->200.133.175.239<->200.133.175.225~200.133.175.238
15)200.133.175.240 <--> 200.133.175.255 <--> unusable


////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Given: 207.5.3.56
Problem: 14 subnets

Solution:

::2^21 = 2,097,152-2 = 2,097,150
Total Network

::2^4 = 16-2 = 14
Total Subnet

::2^4 = 16-2 = 14
Total Hosts/Subnet

207.5.3.56
AND
255.255.255.0/24 --> DSM
------------------------
207.5.3.0/24 --> NA


207.5.3.11110000/28
207.5.3.240/28 --> Last unusable subnet = subnet 15


207.5.3.224/28 --> Last usable subnet


255.255.255.240/28 --> CSM


Subnet Address <--> Broadcast Address <--> Hosts Range
0)207.5.3.0 <-> 207.5.3.15 <->207.5.3.1~207.5.3.14
1)207.5.3.16<->207.5.3.31<->207.5.3.17~207.5.3.30
2)207.5.3.32<->207.5.3.47<->207.5.3.33~207.5.3.46
3)207.5.3.48<->207.5.3.63<->207.5.3.49~207.5.3.62
4)207.5.3.64<->207.5.3.79<->207.5.3.65~207.5.3.78
5)207.5.3.80<->207.5.3.95<->207.5.3.81~207.5.3.94
6)207.5.3.96<->207.5.3.111<->207.5.3.97~207.5.3.110
7)207.5.3.112<->207.5.3.127<->207.5.3.113~207.5.3.126
8)207.5.3.128<->207.5.3.143<->207.5.3.129~207.5.3.142
9)207.5.3.144<->207.5.3.159<->207.5.3.145~207.5.3.158
10)207.5.3.160<->207.5.3.175<->207.5.3.161~207.5.3.174
11)207.5.3.176<->207.5.3.191<->207.5.3.177~207.5.3.190
12)207.5.3.192<->207.5.3.207<->207.5.3.193~207.5.3.206
13)207.5.3.208<->207.5.3.223<->207.5.3.209~207.5.3.222
14)207.5.3.224<->207.5.3.239<->207.5.3.225~207.5.3.238
15)207.5.3.240<->207.5.3.255<---> unused
Posted by Ian Mars Acut at 7:28 PM 0 comments
Labels:

Visual Basic Programming/.NET

C&E Corporation Phils. were I was once working as an Application Developer late 2005 now Chiyoda Phils. sponsored training from Mapua Institute of Technology - IT Center.



AMA Computer Learning Center, Inc. Training Certificate in my local area


VBA Applicaitons, Object based VB programming and object oriented VB programming.
Posted by Ian Mars Acut at 12:34 AM 0 comments
Labels:

Sunday, April 1, 2007

Certificates

My array of certificates while still in the University(University of San Carlos-Technological Center)






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My training certificate while I was still teaching in Royal Oaks International School(www.royaloaks.edu.ph)

I also spearheaded the launching of the school site.
Posted by Ian Mars Acut at 3:48 AM 0 comments
Labels:

Birth Certificates

Office of the City Civil Registrar: Cebu City 6000 Philippines


Birth Certificate.me


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Birth Certificate Bro



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Birth Certificate Mae


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Birth Certificate Ermat



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Birth Certificate Erpat
Posted by Ian Mars Acut at 3:35 AM 0 comments
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