instructor sub network book 2
TRANSCRIPT
-
7/28/2019 Instructor Sub Network Book 2
1/85
0011010100
10001111100
1011100101011100
101100011101001
1011110100011010
00001010010110010
1001010101100111
1111010101000101
1101001101010011
001010010101010
1010101000110010
010101001011000
110101100011010
11010100001011
001010100110
IP Addressing1001010010
and
SubnettingWorkbook
Version2.0
Instructors Edition
11111110
10010101
00011011 10000110
11010011
IPAddress ClassesC
lassA 1 127 (Network127isreserved for loopback andinternal testing)
Leading bitpattern 0 00000000.00000000.00000000.00000000Network . Host . Host . Host
ClassB 128 191Leading bitpattern 10 10000000.00000000.00000000.00000000N etwor k . Netw ork . Host . Host
ClassC192 223Leading bitpattern 110 11000000.00000000.00000000.00000000Network . N etwor k . Netw ork . Host
ClassD224 239 (Reservedformulticast)
Class E240 255 (Reserved forexperimental, used forresearch)
Private Address Space
ClassA 10.0.0.0 to 10.255.255.255
Class B172.16.0.0 to 172.31.255.255
Class C192.168.0.0 to 192.168.255.255
Default Subnet Masks
ClassA 255.0.0.0
ClassB 255.255.0.0
ClassC 255.255.255.0
Producedby:Robb Jones
[email protected] and/or [email protected]
Frederick CountyCareer &Technology Center
Cisco NetworkingAcademy
Frederick County Public Schools
Frederick, Maryland, USA
Special Thanksto MelvinBaker and JimDorsch
fortaking thetime to checkthisworkbookfor errors,
and toeveryonewho hassentin suggestions to improvethe series.
Workbooksincludedinthe series:
IPAddressing and Subnetting Workbooks
ACLs -Access Lists Workbooks
VLSMVariable-Length SubnetMask Workbooks
Instructors (andanyone elsefor thatmatter) pleasedo notpostthe Instructors version onpublicwebsites.
When you do this you aregiving everyone elseworldwidethe answers. Yes, students look foranswers this way.
Italso discourages others;myself included, frompostinghigh quality materials.
InsideCover
-
7/28/2019 Instructor Sub Network Book 2
2/85
Binary To Decimal Conversion
12864321684 21Answers Scratch Area
128 6410010010 14616 32
2 1601110111 119146 4
211111111 255
1
11911000101 197
11110110 246
00010011 19
10000001 129
00110001 49
01111000 120
11110000 240
00111011 59
00000111 7
00011011 27
10101010 170
01101111 111
11111000 248
00100000 32
01010101 85
00 111110 62
00000011 3
11101101 237
11000000 192
1
-
7/28/2019 Instructor Sub Network Book 2
3/85
Decimal To Binary Conversion
Useall 8 bits for eachproblem
1286432168421 = 255 Scratch Area
2381110111034_________________________________________
238-128 -32
11000100010_________________________________________
34 2-64
-24601111011_________________________________________
123 0-32
140 0 1 1 0 0 1 0_________________________________________
50-8
61 1 1 1 1 1 1 1_________________________________________255-4
21 1 0 0 1 0 0 0_________________________________________
200-2
00 0 0 0 1 0 1 0_________________________________________
10
1 0 0 0 1 0 1 0_________________________________________
138
0 0 0 0 0 0 0 1_________________________________________
1
0 0 0 0 1 1 0 1_________________________________________
13
1 1 1 1 1 0 1 0_________________________________________
250
0 1 1 0 1 0 1 1_________________________________________
107
1 1 1 0 0 0 0 0_________________________________________
224
0 1 1 1 0 0 1 0_________________________________________
114
1 1 0 0 0 0 0 0_________________________________________
192
1 0 1 0 1 1 0 0_________________________________________
172
0 1 1 0 0 1 0 0_________________________________________
100
0 1 1 1 0 1 1 1_________________________________________
119
0 0 1 1 1 0 0 1_________________________________________
57
0 1 1 0 0 0 1 0_________________________________________
98
1 0 1 1 0 0 1 1_________________________________________
179
0 0 0 0 0 0 1 0_________________________________________
2
2
-
7/28/2019 Instructor Sub Network Book 2
4/85
Address Class Identification
Address Class
A10.250.1.1_____
150.10.15.0 _____ B
C192.14.2.0_____
B148.17.9.1_____
193.42.1.1_____ C
A126.8.156.0 _____
220.200.23.1_____ C
D230.230.45.58 _____
B177.100.18.4_____
119.18.45.0 _____ A
E249.240.80.78 _____
199.155.77.56 _____ C
A117.89.56.45_____
C215.45.45.0 _____
199.200.15.0_____ C
A95.0.21.90_____
33.0.0.0_____ A
B158.98.80.0 _____
C219.21.56.0 _____3
-
7/28/2019 Instructor Sub Network Book 2
5/85
Network & Host Identification
Circle the net work portion Circle the host portion of
of these addresses: these addresses:
177.100.18.4 10.15.123.50
119.18.45.0 171.2.199.31
209.240.80.78 198.125.87.177
199.155.77.56 223.250.200.222
117.89.56.45 17.45.222.45
215.45.45.0 126.201.54.231
192.200.15.0 191.41.35.112
95.0.21.90 155.25.169.227
33.0.0.0 192.15.155.2
158.98.80.0 123.102.45.254
217.21.56.0 148.17.9.155
10.250.1.1 100.25.1.1
150.10.15.0 195.0.21.98
192.14.2.0 25.250.135.46
148.17.9.1 171.102.77.77
193.42.1.1 55.250.5.5
126.8.156.0 218.155.230.14
220.200.23.1 10.250.1.1
4
-
7/28/2019 Instructor Sub Network Book 2
6/85
Network Addresses
UsingtheIP addressandsubnet mask shown write outthe network address:
188 . 10.0. 0
188.10.18.2 _____________________________
255.255.0.0
10.10 . 48.0
10.10.48.80 _____________________________
255.255.255.0
192 . 149 . 24. 0
192.149.24.191_____________________________
255.255.255.0
150 . 203 . 0 . 0
150.203.23.19 _____________________________
255.255.0.0
1 0 . 0 . 0 . 0
10.10.10.10 _____________________________
255.0.0.0
186 . 13.23 . 0
186.13.23.110 _____________________________
255.255.255.0
223 . 69.0. 0
223.69.230.250_____________________________
255.255.0.0
200 . 120 . 135 . 0
200.120.135.15_____________________________
255.255.255.0
2 7 . 0 . 0 . 0
27.125.200.151_____________________________
255.0.0.0
199 . 20. 150.0
199.20.150.35 _____________________________
255.255.255.0
191 . 55. 165.0
191.55.165.135_____________________________
255.255.255.0
2 8 .2 12 .0 . 0
28.212.250.254_____________________________
255.255.0.0
5
-
7/28/2019 Instructor Sub Network Book 2
7/85
Host Addresses
Using theIPaddressandsubnetmaskshownwrite out thehost address:
0 .0 .1 8 . 2
188.10.18.2_____________________________
255.255.0.0
0 . 0 . 0 . 8 0
10.10.48.80_____________________________
255.255.255.0
0 . 0 . 0 . 1 1
222.49.49.11 _____________________________
255.255.255.0
0 .0 .2 3 0 .1 9
128.23.230.19_____________________________
255.255.0.0
0. 10 . 10.10
10.10.10.10_____________________________
255.0.0.0
0 . 0 . 0 . 1 1
200.113.123.11 _____________________________
255.255.255.0
0. 0. 23 . 20
223.169.23.20_____________________________
255.255.0.0
0 . 0 . 0 . 2 15
203.20.35.215_____________________________
255.255.255.0
0. 15 . 2 . 51
117.15.2.51_____________________________
255.0.0.0
0 . 0 . 0 . 1 35
199.120.15.135 _____________________________
255.255.255.0
0 . 0 . 0 . 1 35
191.55.165.135 _____________________________
255.255.255.0
0. 0. 25 . 54
48.21.25.54_____________________________
255.255.0.0
6
-
7/28/2019 Instructor Sub Network Book 2
8/85
Default Subnet Masks
Writethecorrect default subnetmaskforeachof thefollowingaddresses:
255 . 255 . 0 . 0
177.100.18.4_____________________________
255 . 0 . 0 . 0
119.18.45.0 _____________________________
2 5 5 .2 5 5 .0 .0
191.249.234.191 _____________________________
255 . 255 . 255 . 0
223.23.223.109_____________________________
255 . 0 . 0 . 0
10.10.250.1 _____________________________
255 . 0 . 0 . 0
126.123.23.1_____________________________
255 . 255 . 255 . 0
223.69.230.250_____________________________
255 . 255 . 255 . 0
192.12.35.105 _____________________________
255 . 0 . 0 . 0
77.251.200.51 _____________________________
2 5 5 .2 5 5 .0 .0
189.210.50.1_____________________________
255 . 0 . 0 . 0
88.45.65.35 _____________________________
2 5 5 .2 5 5 .0 .0
128.212.250.254 _____________________________
255 . 255 . 255 . 0
193.100.77.83 _____________________________
255 . 0 . 0 . 0
125.125.250.1 _____________________________
255 . 0 . 0 . 0
1.1.10.50 _____________________________
255 . 255 . 255 . 0
220.90.130.45 _____________________________
2 5 5 .2 5 5 .0 .0
134.125.34.9_____________________________
255 . 0 . 0 . 0
95.250.91.99_____________________________
7
-
7/28/2019 Instructor Sub Network Book 2
9/85
ANDING With
Default subnet masks
Every IPaddressmustbe accompanied byasubnet mask. Bynowyou shouldbeableto look
atan IPaddressandtell what classit is. Unfortunatelyyourcomputerdoesnt think thatway.
For yourcomputer todetermine thenetwork andsubnet portion ofanIPaddress it must
ANDthe IPaddress withthe subnetmask.
DefaultSubnet Masks:
ClassA255.0.0.0
ClassB255.255.0.0
ClassC255.255.255.0
ANDING Equations:
1AND1=1
1AND0=0
0AND1=0
0AND0=0
Sample:
What yousee...
IPAddress: 192.100.10. 33
What youcanfigure outinyourhead...
Address Class: C
Network Portion: 1 9 2 .1 0 0 .1 0 .33
Host Portion:192. 100. 10. 33
Inorder foryoucomputer togetthesame informationit must ANDthe IPaddress with
thesubnet mask inbinary.
NetworkHost
I P Ad dr es s: 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 1 (192 . 100 . 10 . 33)
D ef au lt S ub ne t M as k: 1 1 1 1 1 1 1 1 . 0 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 0 0 0 0 0 0 0 0 (255 . 255 . 255 . 0)
A ND : 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0. 0 0 0 0 1 0 1 0 . 0 0 0 0 0 0 0 0 (192 . 100 . 10 .0)
ANDING withthe default subnetmaskallows yourcomputer tofigureout the network
portionofthe address.
8
-
7/28/2019 Instructor Sub Network Book 2
10/85
ANDING With
Custom subnet masks
When youtake asingle networksuchas 192.100.10.0 anddivide it intofive smaller networks
(192.100.10.16,192.100.10.32, 192.100.10.48, 192.100.10.64,192.100.10.80) theoutside
worldstill seesthe networkas192.100.10.0, butthe internal computersandrouters seefive
smaller subnetworks. Eachindependent ofthe other. This canonlybe accomplished byusing
acustomsubnet mask. Acustomsubnetmaskborrowsbitsfromthe host portion ofthe
address tocreate asubnetwork addressbetweenthenetworkandhost portionsof anIP
address. In this example eachrangehas14 usableaddressesin it. Thecomputermust still
AND theIPaddressagainst thecustomsubnetmaskto seewhatthe networkportionisand
whichsubnetwork it belongsto.
IPAddress:192.100. 10. 0
Custom Subnet Mask: 255.255.255.240
AddressRanges: 192.10.10.0 to 192.100.10.15
192.100.10.16 to 192.100.10.31
192.100.10.32 to 192.100.10.47 (Range in thesample below)
192.100.10.48 to 192.100.10.63
192.100.10.64 to 192.100.10.79
192.100.10.80 to 192.100.10.95
192.100.10.96 to 192.100.10.111
192.100.10.112 to 192.100.10.127
192.100.10.128 to 192.100.10.143
192.100.10.144 to 192.100.10.159
192.100.10.160 to 192.100.10.175
192.100.10.176 to 192.100.10.191
192.100.10.192 to 192.100.10.207
192.100.10.208 to 192.100.10.223
192.100.10.224 to 192.100.10.239
192.100.10.240 to 192.100.10.255
Sub
Network Network Host
IPAddress: 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 1 (192.100. 10.33)
CustomSubnet Mask: 1 1 1 1 1 1 1 1 . 0 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 0 0 0 0(255.255.255.240)
AND: 1 1 0 0 0 0 0 0 . 0 1 1 0 0 1 0 0 . 0 0 0 0 1 0 1 0 . 0 0 1 0 0 0 0 0 (192.100. 10 . 32)
Fourbits borrowed fromthehost
portionofthe address forthe
customsubnetmask.
TheANDINGprocess ofthefour borrowed bitsshowswhich rangeofIP addressesthis
particular addresswillfall into.
Inthenext setofproblems youwill determine thenecessary information todetermine the
correct subnetmask fora varietyofIPaddresses.9
-
7/28/2019 Instructor Sub Network Book 2
11/85
How to determine the number of subnets and the
number of hosts per subnet
Twoformulascanprovidethis basicinformation:
s sNumber of subnets= 2 (Second subnet formula: Number of subnets= 2 -2 )
hNumber of hostspersubnet = 2 -2
Both formulas calculate thenumberof hostsor subnetsbasedonthenumber ofbinary bits
used.For exampleif youborrowthreebits fromthe hostportionof the address usethe
number ofsubnets formula todetermine thetotalnumberof subnets gainedby borrowing the3
threebits. Thiswouldbe2 or2x 2x2 = 8subnets
To determinethenumberofhostsper subnetyou would takethe number ofbinary bits usedin
thehost portion andapplythisto the number ofhosts per subnet formulaIf fivebitsare inthe5
hostportionoftheaddress thiswould be2 or2x 2x2 x2x 2= 32hosts.
Whendealingwith the numberof hostspersubnet youhaveto subtract two addressesfrom
therange. Thefirst addressin everyrangeis thesubnetnumber. The last addressin every
rangeis thebroadcast address. Thesetwoaddressescannot beassignedtoanydevice in
thenetworkwhich iswhyyouhave tosubtracttwo addressesto findthenumberofusable
addressesin eachrange.
For exampleif twobits areborrowed forthenetworkportionof theaddress youcaneasily
determinethe number ofsubnetsandhosts persubnetsusing thetwo formulas.
195. 223 . 50. 0 0 0 0 0 0 0 0
195. 223 . 50. 0 0 0 0 0 0 0 0
195. 223 . 50. 0 0 0 0 0 0 0 0
195. 223 . 50. 0 0 0 0 0 0 0 0
195. 223 . 50. 0 0 0 0 0 0 0 0
T he n um be r o f su bn et s T he n um be r of h os ts c re at ed b y6created by borrowing 2 l
eaving 6bits is2 - 2or2
bitsis 2 or2 x2 =4 2x2 x2x 2 x2 x2 =64-2= 62
subnets. usable hosts per subnet.
What about that second subnet formula:
sNumber of subnets= 2 -2
Insomeinstances thefirst andlast subnetrangeofaddresses arereserved.This issimilar to
thefirstand lasthostaddresses ineachrangeof addreses.
The first rangeofaddressesisthe zerosubnet . The subnet number forthe zerosubnet is
alsothe subnetnumberforthe classful subnetaddress.
The last rangeofaddressesis the broadcastsubnet . The broadcast addressforthe last
subnetin the broadcastsubnet is the sameas theclassfulbroadcast address.
10
-
7/28/2019 Instructor Sub Network Book 2
12/85
Class C Address unsubnetted:
195.223.50 . 0
195.223.50 . 0
195.223.50 . 0
195.223.50 . 0
195.223.50 . 0
195.223.50.0 to 195.223.50.255
Noticethat thesubnetand
broadcastaddressesmatch.Class CAddress subnetted (2bits borrowed):
195.223.50 . 0 0 0 0 0 0 0 0
195.223.50 . 0 0 0 0 0 0 0 0
195.223.50 . 0 0 0 0 0 0 0 0
195.223.50 . 0 0 0 0 0 0 0 0
195.223.50 . 0 0 0 0 0 0 0 0
195.223.50.0 to 195.223.50.63(
0)(Invalidrange)195.223.50.64 to 195.223.50. 127 (1)
1 95 .2 23 .5 0. 12 8 t o 1 95 .22 3. 50. 19 1(2)
195.223.50.192 to 195.223.50.255(3)(Invalidrange)
Theprimary reasonthe thezeroandbroadcast subnetswere notused hadtodo pirmarily with
the broadcastaddresses. Ifyousenda broadcast to195.223.255 areyousendingit to all255
addresses inthe classful C addressor just the62 usableaddresses inthe broadcastrange?
The CCNA and CCENT certification exams may havequestionswhichwill require youto
determine which formula touse, andwhehterornotyou canusethe first andlast subnets. Use
the chartbelowto helpdecide.
When to usewhich formula to determine the number of subnets
s sUsethe 2 -2 formula and dont use the Use the 2 formulaand use the zeroand
z er o an d br oa dc as t ra ng es i f. .. b ro ad ca st r an ge s if .. .
Cl assful rout ing is used Cl assl ess ro ut in gor VLS Mi sused
RIP version 1 is used RIP version 2, EIGRP, or OSPF is used
The noipsubnetzero command is The ip subnetzero command is
c on fi gu re d o n yo ur r ou te r c on fi gu re d on y ou r ro ut er ( de fa ul t s et ti ng )
No otherclues aregiven
Bottomlinefor theCCNA exams; if aquestion doesnot giveyouanycluesas towhetheror notto allowthese two subnets, assume youcanusethem.
sThis workbook hasyouuse thenumberofsubnets= 2 formula.
11
-
7/28/2019 Instructor Sub Network Book 2
13/85
Custom Subnet Masks
Problem 1
Numberof needed subnets 14
Numberof needed usable hosts 14
Network Address 192.10.10.0
C__________
Address class
_______________________________255.255.255.0Default subnet mask
255.255.255.240_______________________________
Custom subnet mask
___________________16Total number of subnets
16___________________
Total number of hostaddresses
14___________________
Number of usableaddresses
___________________4Numberof bits borrowed
Show yourwork for Problem 1in the space below.
Numberof
25612864 32 16 8 4 2 - H ost sNumberof
S ub ne ts - 2 4 8 1 6 3 2 6 4 1 28 2 5 6
1 28 6 4 3 2 1 6 8 4 2 1 - B i na ry v al ue s
192.10 . 10. 0 0 0 0 0 0 0 0
192.10 . 10. 0 0 0 0 0 0 0 0
192.10 . 10. 0 0 0 0 0 0 0 0
192.10 . 10. 0 0 0 0 0 0 0 0
192.10 . 10. 0 0 0 0 0 0 0 0
12816 Observethe total numberof
64Add the binaryvalue hosts.-2numbers to theleft of theline to
32create thecustom subnetmask. Subtract2forthe number of14 usablehosts.
+16
240
12
-
7/28/2019 Instructor Sub Network Book 2
14/85
Custom Subnet Masks
Problem 2
Numberof needed subnets 1000
Number of neededusable hosts 60
Network Address 165.100.0.0
B__________
Address class
_______________________________2 5 5 .2 5 5 .0 .0Default subnetmask
255.255.255.192_______________________________
Customsubnetmask
___________________1,024Total number of subnets
64___________________
Total number of host addresses
62___________________
Numberof usable addresses
___________________10Numberof bits borrowed
Show your work forProblem 2 in the spacebelow.
Numberof. 2 56 1 28 6 4 32 1 6 8 4 2Hosts-
Numberof
.Subnets - 2 4 8 16 32 64 128256
.Binaryvalues - 128 64 32 16 8 4 2 1 . . 128 64 32 16 8 4 2 1
165.100.0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0165.100.0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
165.100.0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
165.100.0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
165.100.0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
128 12864 +6432
192 64 Observe thetotalnumber of
16hosts.
-2Add the binaryvalue 8numbers to theleft of theline to Subtract 2 for thenumberof624create thecustom subnetmask. usable hosts.
2+1
255
13
-
7/28/2019 Instructor Sub Network Book 2
15/85
Custom Subnet Masks
Problem 3 /26 indicatesthe totalnumber ofbitsusedfor thenetwork and
Network Address 148.75.0.0/26 subnetwork portionof theaddress. Allbits remainingbelong
to thehostportionofthe address.
B__________
Address class
255 . 255 . 0 . 0_______________________________
Default subnetmask
255 . 255 . 255 . 192_______________________________
Customsubnetmask
1,024___________________
Total number of subnets
64___________________
Total number of host addresses
62___________________
Numberof usable addresses
10___________________
Numberof bits borrowed
Show yourwork for Problem 3in the space below.
Numberof. 2 56 1 28 6 4 3 2 1 6 8 4 2Hosts -
Numberof
.Subnets - 2 4 8 16 32 64 128256
Binaryvalues - 128 64 32 16 8 4 2 1 . ..
..
128 64 32 16 8 4 2 1
148.75 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
148.75 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
148.75 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
148.75 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
148.75 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
128 12864 +6432 192 64 Observe thetotalnumber of16 hosts.
-2Add the binaryvalue8numbers totheleft of theline to Subtract 2 for thenumberof
624create thecustom subnetmask. usable hosts.
21024
+1 Subtract2forthe total numberof-2 subnets to getthe usable numberof
255 subnets.1,022
14
-
7/28/2019 Instructor Sub Network Book 2
16/85
Custom Subnet Masks
Problem 4
Numberof needed subnets 6
Number of neededusable hosts 30
Network Address 195.85.8.0
C_______
Address class
255 . 255 . 255 . 0_______________________________
Default subnetmask
255 . 255 . 255 . 224_______________________________
Customsubnetmask
8___________________
Total number of subnets
32___________________
Total number of host addresses
30___________________
Numberof usable addresses
3___________________
Numberof bits borrowed
Show your work forProblem 5 in the spacebelow.
Number of
2 56 12 8 64 3 2 1 6 8 4 2 - H os tsNumberof
Subnets - 2 4 8 16 32 64 128256
128 64 32 16 8 4 2 1 - B i naryval ues
195 . 85. 8. 0 0 0 0 0 0 0 0
195 . 85. 8. 0 0 0 0 0 0 0 0
195 . 85. 8. 0 0 0 0 0 0 0 0
195 . 85. 8. 0 0 0 0 0 0 0 0
195 . 85. 8. 0 0 0 0 0 0 0 0
12832 864-2 -2+3230 6224
15
-
7/28/2019 Instructor Sub Network Book 2
17/85
Custom Subnet Masks
Problem 5
Numberof needed subnets 6
Numberof needed usable hosts 30
Network Address 210.100.56.0
C_______
Address class
255 . 255 . 255 . 0_______________________________
Default subnet mask
255 . 255 . 255 . 224_______________________________
Custom subnet mask
8___________________
Total number of subnets
32___________________
Total number of hostaddresses
30___________________
Number of usableaddresses
3___________________
Numberof bits borrowed
Show yourwork for Problem 4in the space below.
Numberof
2 56 1 28 6 4 3 2 1 6 8 4 2 - H os tsNumberof
Subnets - 2 4 8 16 32 64 128256
1 28 6 4 3 2 1 6 8 4 2 1 - B i na ry v al ue s
210.100.56 . 0 0 0 0 0 0 0 0
210.100.56 . 0 0 0 0 0 0 0 0
210.100.56 . 0 0 0 0 0 0 0 0
210.100.56 . 0 0 0 0 0 0 0 0
210.100.56 . 0 0 0 0 0 0 0 0
128
64 328
+32 -2-2
224 306
16
-
7/28/2019 Instructor Sub Network Book 2
18/85
Custom Subnet Masks
Problem 6
Numberof needed subnets 126
Number of neededusable hosts 131,070
Network Address 118.0.0.0
A_______
Address class
2 55 . 0 . 0 . 0_______________________________
Default subnetmask
255.254. 0 . 0_______________________________
Customsubnetmask
128___________________
Total number of subnets
131,072___________________
Total number of host addresses
131,070___________________
Numberof usable addresses
7___________________
Numberof bits borrowed
Show your work forProblem 6 in the spacebelow.
Number of. 256128 64 32 16 8 4 2-Hosts
Number of.Subnets - 2 4 8 16 32 64 128 256 .
Binary values -128 64 32 16 8 4 2 1 . .
.
.
. 1 2 8 6 4 3 2 1 6 8 4 2 1 .
.
.
.
. 128 64 32 16 8 4 2 1
1 18 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1 18 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1 18 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1 18 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1 18 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
12864
32168
128 131,0724-2 -2+2
254 126 131,070
17
-
7/28/2019 Instructor Sub Network Book 2
19/85
Custom Subnet Masks
Problem 7
Numberof needed subnets 2000
Number of neededusable hosts 15
Network Address 178.100.0.0
B__________
Address class
2 55 .2 5 5 .0 .0_______________________________
Default subnetmask
255.255.255.224_______________________________
Customsubnetmask
2,048___________________
Total number of subnets
32___________________
Total number of host addresses
30___________________
Number of usableaddresses
11___________________
Number of bits borrowed
Show yourwork for Problem 7in the space below.
Number of. 2 56 1 28 6 4 3 2 1 6 8 4 2Hosts -
Number of.S ub ne ts - 2 4 8 1 6 3 2 6 4 1 28 2 56
Binaryvalues - 12864 32 16 8 4 2 1 . . 128 64 32 16 8 4 2 1
178.100.0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
178.100.0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
178 . 100 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
12864
3216
82,048 32
4-2 -222,046 30+1
25518
-
7/28/2019 Instructor Sub Network Book 2
20/85
Custom Subnet Masks
Problem 8
Numberof needed subnets 3
Number of neededusable hosts 45
Network Address 200.175.14.0
C_______
Address class
255.255.255.0_______________________________
Default subnetmask
255.255.255.192_______________________________
Customsubnetmask
4___________________
Total number of subnets
64___________________
Total number of host addresses
62___________________
Numberof usable addresses
2___________________
Numberof bits borrowed
Show your work forProblem 8 in the spacebelow.
Number of
256 128 64 32 16 16
16
16
1 6 8 4 2 - H os ts
Numberof
S ub ne ts - 2 4 8 1 6 2 4 8 1 6
2 4 8 1 6 3 2 6 4 1 2 8 2 56
2 4 8 1 6
2 4 8 16
1 28 6 4 3 2 1 6 8 4 2 1 - B i na ry v a lu es
200 . 175 . 14. 0 0 0 0 0 0 0 0
200 . 175 . 14. 0 0 0 0 0 0 0 0200 . 175 . 14. 0 0 0 0 0 0 0 0
200 . 175 . 14. 0 0 0 0 0 0 0 0
200 . 175 . 14. 0 0 0 0 0 0 0 0
128 4 64
+64 -2 -2
240 2 62
19
-
7/28/2019 Instructor Sub Network Book 2
21/85
Custom Subnet Masks
Problem 9
Numberof needed subnets 60
Numberof needed usable hosts 1,000
Network Address 128.77.0.0
B_______
Address class
2 55 .2 5 5 .0 .0_______________________________
Default subnet mask
255.255.252.0_______________________________
Custom subnet mask
64___________________
Total number of subnets
1,024___________________
Total number of hostaddresses
1,022___________________
Number of usableaddresses
6___________________
Numberof bits borrowed
Show yourwork for Problem 9in the space below.
Number of. 2 56 1 28 6 4 3 2 16 8 4 2Hosts-
Number of
.S ub ne ts - 2 4 8 1 6 3 2 6 4 1 28 2 56
Binaryvalues - 12864 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
128.77 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
128.77 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
128.77 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
128.77 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
128.77 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
128
643216
64 1,0248-2 -2+4
252 62 1,022
20
-
7/28/2019 Instructor Sub Network Book 2
22/85
Custom Subnet Masks
Problem 10
Numberof needed usable hosts 60
Network Address 198.100.10.0
C_______
Address class
255.255.255.0_______________________________
Default subnetmask
255.255.255.192_______________________________
Customsubnetmask
4___________________
Total number of subnets
64___________________
Total number of host addresses
62___________________
Numberof usable addresses
2___________________
Numberof bits borrowed
Show your work forProblem 10in the space below.
Number of
256 128 64 32 16 16
16
16
1 6 8 4 2 - H os ts
Numberof
S ub ne ts - 2 4 8 1 6 2 4 8 1 6
2 4 8 1 6 3 2 6 4 1 2 8 2 56
2 4 8 1 6
2 4 8 16
1 28 6 4 3 2 1 6 8 4 2 1 - B i na ry v a lu es
198.100.10 . 0 0 0 0 0 0 0 0
198.100.10 . 0 0 0 0 0 0 0 0198.100.10 . 0 0 0 0 0 0 0 0
198.100.10 . 0 0 0 0 0 0 0 0
198.100.10 . 0 0 0 0 0 0 0 0
128 64 4
+64 -2 -2
192 62 2
21
-
7/28/2019 Instructor Sub Network Book 2
23/85
Custom Subnet Masks
Problem 11
Numberof needed subnets 250
Network Address 101.0.0.0
A_______
Address class
2 55 . 0 . 0 . 0_______________________________
Default subnet mask
2 5 5 .2 5 5 .0 .0_______________________________
Custom subnet mask
256___________________
Total number of subnets
65,536___________________
Total number of hostaddresses
65,534___________________
Number of usableaddresses
8___________________
Numberof bits borrowed
Show yourwork for Problem 11 in the spacebelow.
Number of. 2 56 1 28 6 4 3 2 1 6 8 4 2-Hosts
Number of
. .Subnets - 2 4 8 16 32 64128256
Binary values -128 64 32 16 8 4 2 1 . ..
.
. 128 64 32 16 8 4 2 1 .
.
... 1 28 6 4 3 2 16 8 4 2 1
1 01 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1 01 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 01 01 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1 01 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1 01 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0
1286432
1684
256 65,536 2-2 -2+1
255 254 65,534
22
-
7/28/2019 Instructor Sub Network Book 2
24/85
Custom Subnet Masks
Problem 12
Numberof needed subnets 5
Network Address 218.35.50.0
C_______
Address class
255.255.255.0_______________________________
Default subnetmask
255.255.255.224_______________________________
Customsubnetmask
8___________________
Total number of subnets
32___________________
Total number of host addresses
30___________________
Numberof usable addresses
3___________________
Numberof bits borrowed
Show your work forProblem 12in the space below.
Number of
256 128 64 32 16 16
16
16
1 6 8 4 2 - H os ts
Numberof
S ub ne ts - 2 4 8 1 6 2 4 8 1 6
2 4 8 1 6 3 2 6 4 1 2 8 2 56
2 4 8 1 6
2 4 8 16
1 28 6 4 3 2 1 6 8 4 2 1 - B i na ry v a lu es
218 . 35.50 . 0 0 0 0 0 0 0 0
218 . 35.50 . 0 0 0 0 0 0 0 0218 . 35.50 . 0 0 0 0 0 0 0 0
218 . 35.50 . 0 0 0 0 0 0 0 0
218 . 35.50 . 0 0 0 0 0 0 0 0
128
64 64 4
+32 -2 -2
224 62 2
23
-
7/28/2019 Instructor Sub Network Book 2
25/85
Custom Subnet Masks
Problem 13
Numberof needed usable hosts 25
Network Address 218.35.50.0
C_______
Address class
255.255.255.0_______________________________
Default subnet mask
255.255.255.224_______________________________
Custom subnet mask
8___________________
Total number of subnets
32___________________
Total number of hostaddresses
30___________________
Number of usableaddresses
3___________________
Numberof bits borrowed
Show yourwork for Problem 13 in the spacebelow.
Numberof
256 128 64 32 16 16
16
1 6 8 4 2 - H os ts
1
6Numberof
S ub ne ts - 2 4 8 1 6 2 4 8 16
2 4 8 16
2 4 8 16 32 64 128256
2 4 8 16
1 28 6 4 3 2 1 6 8 4 2 1 - B i na ry v al ue s
218.35 . 50. 0 0 0 0 0 0 0 0
218.35 . 50. 0 0 0 0 0 0 0 0218.35 . 50. 0 0 0 0 0 0 0 0
218.35 . 50. 0 0 0 0 0 0 0 0
218.35 . 50. 0 0 0 0 0 0 0 0
128
64 8 32
+32 -2 -2
224 6 30
24
-
7/28/2019 Instructor Sub Network Book 2
26/85
Custom Subnet Masks
Problem 14
Numberof needed subnets 10
Network Address 172.59.0.0
B_______
Address class
2 5 5 .2 5 5 .0 .0_______________________________
Default subnetmask
255.255.240.0_______________________________
Customsubnetmask
16___________________
Total number of subnets
4,096___________________
Total number of host addresses
4,094___________________
Numberof usable addresses
4___________________
Numberof bits borrowed
Show your work forProblem 14in the space below.
Numberof.256128 64 32 16 8 4 2Hosts-
Numberof
.Subnets - 2 4 8 16 32 64 128256
Binaryvalues - 128 64 32 16 8 4 2 1 . 128 64 32 16 8 4 2 1
172.59 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0172.59 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
172.59 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
172.59 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
172.59 . 0 0 0 0 0 0 0 0. 0 0 0 0 0 0 0 0
12864
16 4,096 32-2 -2+16
240 14 4,094
25
-
7/28/2019 Instructor Sub Network Book 2
27/85
-
7/28/2019 Instructor Sub Network Book 2
28/85
Custom Subnet Masks
Problem 16
Number of neededusable hosts 29
Network Address 23.0.0.0
A_______
Address class
2 55 . 0 . 0 . 0_______________________________
Default subnetmask
255.255.255.224_______________________________
Customsubnetmask
524,288___________________
Total number of subnets
32___________________
Total number of host addresses
30___________________
Numberof usable addresses
19___________________
Numberof bits borrowed
Show your work forProblem 16in the space below.
Number of. 256128 64 32 16 8 4 2-Hosts
Number of
. .Subnets - 2 4 8 16 32 64 128 256
Binaryvalues -128 64 32 16 8 4 2 1 . .
.
.
. 1 2 8 6 4 32 1 6 8 4 2 1 .
.
.
.
. 12 8 6 4 3 2 16 8 4 2 1
2 3 . 00 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 00 0 0 0 0 02 3 . 00 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 00 0 0 0 0 02 3 . 00 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 00 0 0 0 0 02 3 . 00 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 00 0 0 0 0 02 3 . 00 0 0 0 0 0 0 . 0 0 0 0 0 0 0 0 . 0 00 0 0 0 0 0
128
32 524,28864
-2 -2+3230 524,286 224
27
-
7/28/2019 Instructor Sub Network Book 2
29/85
Subnetting
Problem 1
Numberof needed subnets 14
Number of neededusable hosts 14
Network Address 192.10.10.0
__________CAddress class
255.255.255.0_______________________________
Default subnetmask
_______________________________255.255.255.240Customsubnetmask
16___________________
Total number of subnets
16___________________
Total number of host addresses
___________________14Numberof usable addresses
4___________________
Numberof bits borrowed
What is the 4th
192.10.10.48 to 192.10.10.63subnetrange? _______________________________________________
What is the subnet number
192.10 . 10.112for the 8th subnet? ________________________
What is the subnet
broadcast address for
192.10 . 10.207the 13thsubnet? ________________________
What are the assignable
addresses for the9th
192.10.10.129 to 192.10.10.142subnet? ______________________________________
28
-
7/28/2019 Instructor Sub Network Book 2
30/85
Show your work forProblem 1 in the spacebelow.
Numberof
2 56 1 28 6 4 32 1 6 8 4 2 - H os tsNumberof
S ub ne ts - 2 4 8 16 2 4 8 16
2 4 8 1 6
2 4 8 16 32 64 128256
2
4 8 1 6
1 28 64 3 2 1 6 8 4 2 1 - B i n ar y v al ue s
192.10 . 10.0 0 0 0 0 0 0 0
192.10 . 10.0 0 0 0 0 0 0 0
192.10 . 10.0 0 0 0 0 0 0 0
192.10 . 10.0 0 0 0 0 0 0 0
192.10 . 10.0 0 0 0 0 0 0 0
(
1) 0 0 0 0 192.10.10.0 to 192.10.10.15
(2) 0 0 0 1 192.10.10.16 to 192. 10.10.31
(3) 0 0 1 0 192.10.10.32 to 192. 10.10.47
(4) 0 0 1 1 192.10.10.48 to 192. 10.10.63
(5) 0 1 0 0 192.10.10.64 to 192. 10.10.79
(6) 0 1 0 1 192.10.10.80 to 192. 10.10.95
(7) 0 1 1 0 192.10.10.96 to 192. 10.10.111
(8) 0 1 1 1 192.10.10.112 to 1 92.10. 10.127
(9) 1 0 0 0 192.10.10.128 to 1 92.10. 10.143
( 10) 1 0 0 1 1 92 .10 .10 .1 44 t o 1 92 .1 0. 10 .15 9
( 11) 1 0 1 0 1 92 .10 .10 .1 60 t o 1 92 .1 0. 10 .17 5
( 12) 1 0 1 1 1 92 .10 .10 .1 76 t o 1 92 .1 0. 10 .19 1
( 13) 1 1 0 0 1 92 .10 .10 .1 92 t o 1 92 .1 0. 10 .20 7
( 14) 1 1 0 1 1 92 .10 .10 .2 08 t o 1 92 .1 0. 10 .22 3
( 15) 1 1 1 0 1 92 .10 .10 .2 24 t o 1 92 .1 0. 10 .23 9
( 16) 1 1 1 1 1 92 .10 .10 .2 40 t o 1 92 .1 0. 10 .25 5
128
64
32 16 16
+16 -2 -2Custom subnet Usable subnets Usable hosts
240 14 14mask
Thebinaryvalue of thelast bit borrowed is therange.Inthis
problemthe rangeis16.
Thefirst address in eachsubnet rangeisthe subnetnumber.
Thelast address in eachsubnet rangeisthe subnet broadcast
address.
29
-
7/28/2019 Instructor Sub Network Book 2
31/85
Subnetting
Problem 2
Numberof needed subnets 1000
Number of neededusable hosts 60
Network Address 165.100.0.0
B__________
Address class
_______________________________2 5 5 .2 5 5 .0 .0Default subnetmask
255.255.255.192_______________________________
Customsubnetmask
___________________1,024Total number of subnets
64___________________
Total number of host addresses
62___________________
Numberof usable addresses
___________________10Number of bits borrowed
Whatis the15th
165.100.3.128 to 165.100.3.191subnetrange? _______________________________________________
What is the subnet number
1 65 .1 0 0 .1 .6 4for the 6th subnet? ________________________
What is the subnet
broadcast address for
165.100.1. 127the6thsubnet? ________________________
What are the assignable
addresses for the9th
165.100.2.1 to 165.100.0.62subnet? ______________________________________
30
-
7/28/2019 Instructor Sub Network Book 2
32/85
Show your work forProblem 2 in the spacebelow.
31
-
7/28/2019 Instructor Sub Network Book 2
33/85
Subnetting
Problem 3Hint:Itispossibleto borrow
Numberof needed subnets 2 one bittocreatetwosubnets.
Network Address 195.223.50.0
C__________
Address class
255.255.255.0_______________________________
Default subnet mask
255.255.255.128_______________________________
Custom subnet mask
2___________________
Total number of subnets
128___________________
Total number of hostaddresses
126___________________
Number of usableaddresses
1___________________
Numberof bits borrowed
Whatis the2nd
195.223.50.128 - 195.223.50.255subnetrange? _______________________________________________
What is the subnet number195.223.50.128for the 2nd subnet? ________________________
What is the subnet
broadcast address for195.223.50.127the1st subnet? ________________________
What are the assignable
addresses for the1st195.223.50.1 - 195.223.50.126
subnet? ______________________________________
32
-
7/28/2019 Instructor Sub Network Book 2
34/85
Show your work forProblem 3 in the spacebelow.Numberof
2 56 1 28 6 4 3 2 1 6 8 4 2 - H os tsNumberof
S ub ne ts - 2 4 8 1 6 2 4 8 1 6
2 4 8 16
2 4 8 16 32 64128256
2 4 8 16
1 28 6 4 32 1 6 8 4 2 1 - B i na ry v a lu es
195. 223 . 50.0 0 0 0 0 0 0 0
195. 223 . 50.0 0 0 0 0 0 0 0
195. 223 . 50.0 0 0 0 0 0 0 0
195.223.50 . 0 0 0 0 0 0 0 0
195.223.50 . 0 0 0 0 0 0 0 0
(1) 195.223.50.0 to 195.223.50.127 0
(2) 1 195.223.50.128 to 195.223.50.255
33
-
7/28/2019 Instructor Sub Network Book 2
35/85
Subnetting
Problem 4
Numberof needed subnets 750
Network Address 190.35.0.0
B__________
Address class
2 55 .2 5 5 .0 .0_______________________________
Default subnet mask
255.255.255.192_______________________________
Custom subnet mask
1,024___________________
Total number of subnets
64___________________
Total number of hostaddresses
62___________________
Number of usableaddresses
10___________________
Numberof bits borrowed
Whatis the15th
190.35.3.128 to 190.35.3.191subnetrange? _______________________________________________
What is the subnet number
for the13th subnet?190.35.3.0________________________
What is the subnet
broadcast address for
the 10thsubnet?190.35.2.127________________________
What are the assignable
addresses for the6th
subnet? 190.35.1.65 to 190.35.1.126______________________________________
34
-
7/28/2019 Instructor Sub Network Book 2
36/85
Show your work forProblem 4 in the spacebelow.
35
-
7/28/2019 Instructor Sub Network Book 2
37/85
Subnetting
Problem 5
Numberof needed usable hosts 6
Network Address 126.0.0.0
A__________
Address class
2 5 5 . 0 . 0 . 0_______________________________
Default subnet mask
255.255.255.248_______________________________
Custom subnet mask
2,097,152___________________
Total number of subnets
8________________
Total number of hostaddresses
6___________________
Number of usableaddresses
21___________________
Numberof bits borrowed
Whatis the2nd
126.0.0.8 to 126.0.0.15subnetrange? _______________________________________________
What is the subnet number
126.0.0.32for the 5th subnet? ________________________
What is the subnet
broadcast address for
126.0.0.55the7thsubnet? ________________________
What are the assignable
addresses for the 10th
126.0.0.73 to 126.0.0.78subnet? ______________________________________
36
-
7/28/2019 Instructor Sub Network Book 2
38/85
Show your work forProblem 5 in the spacebelow.
37
-
7/28/2019 Instructor Sub Network Book 2
39/85
Subnetting
Problem 6
Numberof needed subnets 10
Network Address 192.70.10.0
C__________
Address class
255.255.255.0_______________________________
Default subnetmask
255.255.255.240_______________________________
Customsubnetmask
16___________________
Total number of subnets
16___________________
Total number of host addresses
14___________________
Numberof usable addresses
4___________________
Numberof bits borrowed
What is the 9th
192.70.10.128 to 192.70.10.143subnetrange? _______________________________________________
What is the subnet number192.70.10.48for the 4th subnet? ________________________
What is the subnet
broadcast address for192.70.10.191the 12thsubnet? ________________________
What are the assignable
addresses for the 10th192.70.10.145 to 192.70.10.158
subnet? ______________________________________
38
-
7/28/2019 Instructor Sub Network Book 2
40/85
Show your work forProblem 6 in the spacebelow.Numberof
256 128 64 32 16 1 6
1616
1 6 8 4 2 - Ho st s
Numberof
S ub ne ts - 2 4 8 1 6 2 4 8 1 6
2 4 8 16 32 64128256
2 4 8 16
2 4 8 1 6
1 28 6 4 3 2 1 6 8 4 2 1 - B i n ar y va lu es
192 . 70.10 .
192 . 70.10 .
192 . 70.10 .
192 . 70.10 .192 . 70.10 .
. 0 0 0 0 0 0 0 0
. 0 0 0 0 0 0 0 0
. 0 0 0 0 0 0 0 0
. 0 0 0 0 0 0 0 0
. 0 0 0 0 0 0 0 0
(1) 0 192.70.10.0 to 192.70.10.15
(2) 1 192.70.10.16 to 192.70.10.31
(3)
1 0 192.70.10.32 to 192.70.10.47
(4) 1 1 192.70.10.48 to 192.70.10.63
(5) 1 0 0 192.70.10.64 to 192.70.10.79
(6) 1 0 1 192.70.10.80 to 192.70.10.95
(7) 1 1 0 192.70.10.96 to 192.70.10.111
(8) 1 1 1 192.70.10.112 to 192.70.10.127
(9) 1 0 0 0 192.70.10.128 to 192.70.10.143
(10) 1 0 0 1 192.70.10.144 to 192.70.10.159
(11) 1 0 1 0 192.70.10.160 to 192.70.10.175
(12) 1 0 1 1 192.70.10.176 to 192.70.10.191
(13) 1 1 0 0 192.70.10.192 to 192.70.10.0207
(14) 1 1 0 1 192.70.10.208 to 192.70.10.223
(15) 1 1 1 0 192.70.10.224 to 192.70.10.239
(16) 1 1 1 1 192.70.10.240 to 192.70.10.255
128 16
+64 -2
240 14
39
-
7/28/2019 Instructor Sub Network Book 2
41/85
Subnetting
Problem 7
Network Address 10.0.0.0/16
__________AAddress class
_______________________________2 5 5 . 0 . 0 . 0Default subnet mask
_______________________________2 55 .2 5 5 .0 .0Custom subnet mask
___________________256Total number of subnets
___________________65,536Total number of hostaddresses
___________________65,534Number of usableaddresses
___________________8Numberof bits borrowed
What is the 11th
10.10.0.0 to 10.10.255.255subnetrange? _______________________________________________
What is the subnet number10.5.0.0for the 6th subnet? ________________________
What is the subnet
broadcast address for10.1.255.255the2ndsubnet? ________________________
What are the assignable
addresses for the9th10.8.0.1 to 10.8.255.254subnet? ______________________________________
40
-
7/28/2019 Instructor Sub Network Book 2
42/85
Show your work forProblem 7 in the spacebelow.
41
-
7/28/2019 Instructor Sub Network Book 2
43/85
Subnetting
Problem 8
Numberof needed subnets 5
Network Address 172.50.0.0
B__________
Address class
2 55 .2 5 5 .0 .0_______________________________
Default subnet mask
255.255.224.0_______________________________
Custom subnet mask
8___________________
Total number of subnets
8,192___________________
Total number of hostaddresses
8,190___________________
Number of usableaddresses
3___________________
Numberof bits borrowed
What is the 4th
subnetrange? _______________________________________________
172.50.96.0 to 172.50.127.255
What is the subnet number
for the 5th subnet? ________________________172.50.128.0
What is the subnet
broadcast address for
172.50.191.255the6thsubnet? ________________________
What are the assignable
addresses for the 3rd
subnet? ______________________________________172.50.64.1 to 172.50.95.254
42
-
7/28/2019 Instructor Sub Network Book 2
44/85
Show your work forProblem 8 in the spacebelow.
43
-
7/28/2019 Instructor Sub Network Book 2
45/85
Subnetting
Problem 9
Numberof needed usable hosts 28
Network Address 172.50.0.0
B__________
Address class
2 55 .2 5 5 .0 .0_______________________________
Default subnet mask
255.255.255.224_______________________________
Custom subnet mask
2,048___________________
Total number of subnets
32___________________
Total number of hostaddresses
30___________________
Number of usableaddresses
11___________________
Numberof bits borrowed
What is the 2nd
172.50.0.32 to 172.50.0.63subnetrange? _______________________________________________
Whatis thesubnetnumber
172.50.1.32for the 10thsubnet? ________________________
What is the subnet broadcast
address for
172.50.0.127the 4th subnet? ________________________
What are the assignable
addresses for the 6th
172.50.0.161 to 172.50.0.190subnet? ______________________________________
44
-
7/28/2019 Instructor Sub Network Book 2
46/85
Show your work forProblem 9 in the spacebelow.
45
-
7/28/2019 Instructor Sub Network Book 2
47/85
Subnetting
Problem 10
Numberof needed subnets 45
Network Address 220.100.100.0
C__________
Address class
255.255.255.0_______________________________
Default subnet mask
255.255.255.252_______________________________
Custom subnet mask
64___________________
Total number of subnets
4___________________
Total number of hostaddresses
2___________________
Number of usableaddresses
6___________________
Numberof bits borrowed
What is the 5th
subnetrange? _______________________________________________220.100.100.16 to 220.100.100.19
Whatis thesubnetnumber
220.100.100.12for the4thsubnet? ________________________
Whatis thesubnet
broadcast address for
the13th subnet? ________________________220.100.100.51
What are the assignable
addresses for the12th
subnet? ______________________________________220.100.100.45 to 220.100.100.46
46
-
7/28/2019 Instructor Sub Network Book 2
48/85
Show your work forProblem 10in the space below.
47
-
7/28/2019 Instructor Sub Network Book 2
49/85
Subnetting
Problem 11
Numberof needed usable hosts 8,000
Network Address 135.70.0.0
B__________
Address class
2 55 .2 5 5 .0 .0_______________________________
Default subnet mask
255.255.224.0_______________________________
Custom subnet mask
8___________________
Total number of subnets
8,192___________________
Total number of hostaddresses
8,190___________________
Number of usableaddresses
3___________________
Numberof bits borrowed
What is the 6th
135.70.160.0 to 135.70.191.255subnetrange? _______________________________________________
Whatis thesubnetnumber135.70.192.0for the7thsubnet? ________________________
Whatis thesubnet
broadcast address for135.70.95.255the 3rd subnet? ________________________
What are the assignable
addresses for the 5th
135.70.128.1 to 135.70.159.254subnet? ______________________________________
48
-
7/28/2019 Instructor Sub Network Book 2
50/85
Show your work forProblem 11in the space below.
49
-
7/28/2019 Instructor Sub Network Book 2
51/85
Subnetting
Problem 12
Numberof needed usable hosts 45
Network Address 198.125.50.0
C__________
Address class
255.255.255.0_______________________________
Default subnet mask
255.255.255.192_______________________________
Custom subnet mask
4___________________
Total number of subnets
64___________________
Total number of hostaddresses
62___________________
Number of usableaddresses
2___________________
Numberof bits borrowed
What is the 2nd
198.125.50.64 to 98.125.50.127subnetrange? _______________________________________________
Whatis thesubnetnumber
198.125.50.64for the 2nd subnet? ________________________
Whatis thesubnet
broadcast address for
198.125.50.255the 4th subnet? ________________________
What are the assignable
addresses for the3rd
198.125.50.129 to 198.125.50.190subnet? ______________________________________
50
-
7/28/2019 Instructor Sub Network Book 2
52/85
Show your work forProblem 12in the space below.
Number of
256 128 64 32 16 16
16
1 6 8 4 2 - H os ts
16Numberof
S ub ne ts - 2 4 8 1 6 2 4 8 1 6
2 4 8 16
2 4 8 1 6 3 2 6 4 1 2 8 2 56
2 4 8 1 6
1 28 6 4 3 2 1 6 8 4 2 1 - B i na ry v a lu es
198 . 125 . 50. 0 0 0 0 0 0 0 0
198 . 125 . 50. 0 0 0 0 0 0 0 0
198 . 125 . 50. 0 0 0 0 0 0 0 0
198 . 125 . 50. 0 0 0 0 0 0 0 0
198 . 125 . 50. 0 0 0 0 0 0 0 0
(1) 0 198.125.50.0 to 198.125.50.63
(2) 1 198.125.50.64 to 198.125.50.127
(3) 1 0 198.125.50.128 to 198.125.50.191
(4) 1 1 198.125.50.192 to 198.125.50.255
128 64
+64 -2
192 62
51
-
7/28/2019 Instructor Sub Network Book 2
53/85
Subnetting
Problem 13
Network Address 165.200.0.0 /26
B__________
Address class
2 55 .2 5 5 .0 .0_______________________________
Default subnet mask
255.255.255.192_______________________________
Custom subnet mask
1,024___________________
Total number of subnets
64___________________
Total number of hostaddresses
62___________________
Number of usableaddresses
10___________________
Numberof bits borrowed
Whatis the10th
165.200.2.64 to 165.200.2.127subnetrange? _______________________________________________
Whatis thesubnetnumber
165.200.2.128for the11th subnet? ________________________
Whatis thesubnet
broadcast address for
165.200.255.191the 1023rdsubnet? ________________________
What are the assignable
addresses for the1022nd
165.200.255.65 to 165.200.255.126subnet? ______________________________________
52
-
7/28/2019 Instructor Sub Network Book 2
54/85
-
7/28/2019 Instructor Sub Network Book 2
55/85
Subnetting
Problem 14
Numberof needed usable hosts 16
Network Address 200.10.10.0
C__________
Address class
255.255.255.0_______________________________
Default subnet mask
255.255.255.224_______________________________
Custom subnet mask
8___________________
Total number of subnets
32___________________
Total number of hostaddresses
30___________________
Number of usableaddresses
3___________________
Numberof bits borrowed
What is the 7th
200.10.10.192 to 200.10.10.223subnetrange? _______________________________________________
What is the subnet number200.10.10.128for the 5th subnet? ________________________
What is the subnet
broadcast address for200.10.10.127the4thsubnet? ________________________
What are the assignable
addresses for the6th200.10.10.161 to 200.10.10.190
subnet? ______________________________________
54
-
7/28/2019 Instructor Sub Network Book 2
56/85
Show your work forProblem 14in the space below.Numberof
16
16256 1286432 16
16
16 8 4 2 - HostsNumberof
S ub net s - 2 4 8 1 6 2 4 8 1 6
2 4 8 16
2 4 8 16
2 4 8 1 6 3 2 6 4 1 28 2 56
128 64 32 16 8 4 2 1 - B inaryvalues
200.10 . 10. 0 0 0 0 0 0 0 0
200.10 . 10. 0 0 0 0 0 0 0 0
200.10 . 10. 0 0 0 0 0 0 0 0
200.10 . 10. 0 0 0 0 0 0 0 0
200.10 . 10. 0 0 0 0 0 0 0 0
(1) 0 200.10.10.0 to 200.10.10.31
(2) 1 200.10.10.32 to 200.10.10.63
(3) 1 0 200.10.10.64 to 200.10.10.95
1 1(4) 200.10.10.96 to 200.10.10.127
1 0 0(5) 200.10.10.128 to 200.10.10.159
(6) 1 0 1 200.10.10.160 to 200.10.10.191
(7) 1 1 0 200.10.10.192 to 200.10.10.223
(8) 1 1 1 200.10.10.224 to 200.10.10.255
128
64 32
+32 -2
224 30
55
-
7/28/2019 Instructor Sub Network Book 2
57/85
Subnetting
Problem 15
Network Address 93.0.0.0\19
A__________
Address class
2 5 5 . 0 . 0 . 0_______________________________
Default subnet mask
255.255.224.0_______________________________
Custom subnet mask
2,048___________________
Total number of subnets
8,192___________________
Total number of hostaddresses
8,190___________________
Number of usableaddresses
11___________________
Numberof bits borrowed
Whatis the15th
93.1.192.0 to 93.1.223.255subnetrange? _______________________________________________
What is the subnet number93.1.0.0for the 9th subnet? ________________________
What is the subnet
broadcast address for93.0.223.255the7thsubnet? ________________________
What are the assignable
addresses for the 12th93.1.96.1 to 93.1.127.254
subnet? ______________________________________
56
-
7/28/2019 Instructor Sub Network Book 2
58/85
Show your work forProblem 15in the space below.
57
-
7/28/2019 Instructor Sub Network Book 2
59/85
Practical Subnetting 1
Based ontheinformation inthe graphicshown, designanetwork addressing scheme thatwill
supplythe minimum number ofsubnets ,and allowenoughextra subnetsandhostsfor
100%growth inbothareas. Circle eachsubnetonthe graphic andanswerthe questions
below.
IPAddress 172.16.0.0
F0/0
S0/0/0 S0/0/1 F0/1Router AR oute rB
F0/0
Marketing Management24Hosts 15HostsReasearch
60Hosts
BAddress class _____________________________
255.255.224.0Customsubnetmask _____________________________
4
M in im um n u mb ero f s ub ne ts n e ed ed _ __ __ __ __
+ 4Extra subnets required for 100% growth _________(
Roundup to thenextwholenumber)
= 8Tot al n u mb er o f s ub ne ts n ee de d _ __ __ __ __
Numberof hostaddresses60i n th e la rg es t s ub ne t g ro up _ __ __ __ __
Numberof addresses neededfor+ 60100% growth in thelarges t subnet _________
(
Roundup to thenextwholenumber)
Total number of address= 120n ee de d for t he l arge st s ub ne t _ __ __ __ __
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
IP address range for Research _____________________________172.16.0.0 to172.31.255
IP address range for Marketing _____________________________172.16.32.0to 172.63.255
IP address range for Management _____________________________172.16.64.0to 172.95.255
IP address range for Router AtoRouter B serialconnection _____________________________172.16.96.0 to172.127.255
58
-
7/28/2019 Instructor Sub Network Book 2
60/85
-
7/28/2019 Instructor Sub Network Book 2
61/85
Practical Subnetting 2
Based ontheinformation inthe graphicshown, designanetwork addressing scheme thatwill
supplythe minimum number ofhosts per subne t, andallowenoughextrasubnetsand
hostsfor 30%growthin allareas. Circleeachsubnetonthe graphicandanswer thequestions
below.
IPAddress 135.126.0.0
F0/0 S0/0/0S0/0/1
Router A
F0/1S0/0/1 F0/0 R oute rB
S0/0/0 Tech Ed Lab
20Hosts
F0/1Ro ute rC
Science Lab
10Hosts
EnglishDepartment
15Hosts
BAddress class _____________________________
255.255.255.224Customsubnetmask _____________________________
5
Minimum numberof subnets needed _________
+ 2Extra subnets required for 30% growth _________(
Roundup to thenextwholenumber)
= 7Total number of subnets needed _________
Numberof hostaddresses20i n t he l a rg es t s ub ne t g ro up _ __ __ __ __
Numberof addresses neededfor+ 630% growth in thelarges t subnet _________
(
Roundup to thenextwholenumber)
Total number of address= 26n ee de d for t h e l arge st s ub ne t _ __ __ __ __
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
IP address rangefor TechEd _____________________________135.126.0.0to 135.126.0.31
IP address rangefor English _____________________________135.126.0.32 to135.126.0.63
IP address rangefor Science _____________________________135.126.0.64to 135.126.0.95
IP address range for Router AtoRouter B serialconnection 135.126.0.96 to135.126.0.127_____________________________
IP address range for Router A
toRouter B serialconnection 135.126.0.128 to135.126.0.159_____________________________
60
-
7/28/2019 Instructor Sub Network Book 2
62/85
Show your work forProblem 2 in the spacebelow.
61
-
7/28/2019 Instructor Sub Network Book 2
63/85
Practical Subnetting 3
Based onthe information in the graphicshown, designa classfull networkaddressing scheme
thatwill supply the minimum number ofhosts per subne t, andallowenoughextrasubnets
andhostsfor25% growthinall areas. Circleeachsubneton thegraphic andanswerthe
questionsbelow.
IPAddress 172.16.0.0
F0/0S0/0/1
SalesF0/0Administrative RouterA 185Hosts
F0/1 S0/0/030Hosts
R oute rB
Marketing
50Hosts
BAddress class _____________________________
255.255.255.0
Customsubnetmask _____________________________
4Minimum numberof subnets needed _________
+
1Extra subnets required for 25% growth _________(
Roundup to thenextwholenumber)
= 5Total number of subnets needed _________
Numberof hostaddresses185i n t he l a rg es t s ub ne t g ro up _ __ __ __ __
Numberof addresses neededfor+ 4725% growth in thelarges t subnet _________
(
Roundup to thenextwholenumber)
Total number of address= 232n ee de d for t h e l arge st s ub ne t _ __ __ __ __
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
172.16.0.0to 172.16.0.255IP address rangefor Sales _____________________________
172.16.1.0 to 172.16.1.255
IP address range for Marketing _____________________________
172.16.2.0 to 172.16.2.255IP address range for Administrative _____________________________
IP address range for Router A172.16.3.0 to 172.16.3.255toRouter B serialconnection _____________________________
62
-
7/28/2019 Instructor Sub Network Book 2
64/85
Show your work forProblem 3 in the spacebelow.
63
-
7/28/2019 Instructor Sub Network Book 2
65/85
Practical Subnetting 4
Based ontheinformation inthe graphicshown, designanetwork addressing scheme thatwill
supplythe minimum number ofsubnets ,and allowenoughextra subnetsandhostsfor70%
growth inallareas. Circle eachsubnet onthe graphicand answer thequestionsbelow.
IPAddress 135.126.0.0
F0/0S0/0/0
S0/0/1Router A
S0/0/1 F0/0 Router B
S0/0/0
F0/0Rou ter C
F0/1Dallas
NewYork150Hosts325HostsWashingtonD.C.
220Hosts
BAddress class _____________________________
255.255.240.0Customsubnetmask _____________________________
5
Minimum numberof subnets needed _________
+ 4Extra subnets required for 70% growth _________(
Roundup to thenextwholenumber)
= 9Total number of subnets needed _________
Numberof hostaddresses325i n t he l a rg es t s ub ne t g ro up _ __ __ __ __
Numberof addresses neededfor+ 22870% growth in thelarges t subnet _________
(
Roundup to thenextwholenumber)
Total number of address= 553n ee de d for t h e l arge st s ub ne t _ __ __ __ __
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
135.126.0.0 to135.126.15.255IP address rangefor NewYork _____________________________
135.126.16.0 to135.126.31.255IP address rangefor Washington D.C. _____________________________
135.126.32.0to 135.126.47.255IP address range for Dallas _____________________________
IP address range for Router A135.126.48.0to 135.126.63.255toRouter B serialconnection _____________________________
IP address range for Router A135.126.64.0to 135.126.79.255
toRouter C serialconnection _____________________________64
-
7/28/2019 Instructor Sub Network Book 2
66/85
Show your work forProblem 4 in the spacebelow.
65
-
7/28/2019 Instructor Sub Network Book 2
67/85
Practical Subnetting 5
Based ontheinformation inthe graphicshown, designanetwork addressing scheme thatwill
supplythe minimum number ofhosts per subnet , andallowenoughextrasubnets and
hostsfor 100%growthin allareas. Circleeachsubnetonthe graphicandanswer the
questionsbelow.
IPAddress 210.15.10.0
F0/1
F0/0
Tech Ed LabScience Room18Hosts10Hosts
EnglishclassroomArt Classroom15Hosts
12Hosts
CAddress class _____________________________
255.255.255.192Customsubnetmask _____________________________
2Minimum numberof subnets needed _________
+ 2Extra subnets required for 100% growth _________(
Roundup to thenextwholenumber)
= 4Total number of subnets needed _________
Numberof hostaddresses30i n t he l a rg es t s ub ne t g ro up _ __ __ __ __
Numberof addresses neededfor+ 30100% growth in thelarges t subnet _________
(
Roundup to thenextwholenumber)
Total number of address= 60n ee de d for t h e l arge st s ub ne t _ __ __ __ __
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
210.15.10.0 to210.15.10.63IP address rangefor Router F0/0 Port _____________________________
210.15.10.64 to210.15.10.127I
P address rangefor Router F0/1 Port _____________________________
66
-
7/28/2019 Instructor Sub Network Book 2
68/85
Show your work forProblem 5 in the spacebelow.
Numberof
2 56 1 28 64 3 2 16 8 4 2 - H os tsNumberof
S ub ne ts - 2 4 8 16 2 4 8 16
2 4 8 16
2 4 8 16
2 4 8 16 32 64 128256
1
2 8 6 4 3 2 1 6 8 4 2 1 - B i na ry v al ue s
210. 15. 10 . 0 0 0 0 0 0 0 0
210. 15. 10 . 0 0 0 0 0 0 0 0
210. 15. 10 . 0 0 0 0 0 0 0 0
210. 15. 10 . 0 0 0 0 0 0 0 0
210. 15. 10 . 0 0 0 0 0 0 0 0
(
1) 210.15.10.0 to 210.15.10.630
(2) 210.15.10.64 to 210.15.10.127 1
(3) 1 0 210.15.10.128 to 210.15.10.191
(4) 1 1 210.15.10.192 to 210.15.10.255
67
-
7/28/2019 Instructor Sub Network Book 2
69/85
Practical Subnetting 6
Based ontheinformation inthe graphicshown, designanetwork addressing scheme thatwill
supplythe minimum number ofsubnets ,and allowenoughextra subnetsandhostsfor20%
growth inallareas. Circle eachsubnet onthe graphicand answer thequestionsbelow.
IPAddress 10.0.0.0S0/0/0
TechnologyS0/0/1RouterA
BuildingS0/0/1 S0/0/0F0/0 Ro ute rB320HostsF0/1
S0/0/1S0/0/0
Art & Drama Administration
75 Hosts 35 HostsRo ute rC
F0/0 F0/1
Science Building
225Hosts
AAddress class _____________________________
255.240.0.0
Customsubnetmask _____________________________
7Minimum numberof subnets needed _________
+ 2Extra subnets required for 20% growth _________(
Roundup to thenextwholenumber)
= 9Total number of subnets needed _________
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
10.0.0.0 to10.15.255.255IP address rangefor Technology _____________________________
10.16.0.0to 10.31.255.255
IP address rangefor Science _____________________________
10.32.0.0to 10.47.255.255IP address rangefor Arts &Drama _____________________________
10.48.0.0to 10.63.255.255IP Address range Administration _____________________________
IP address range for Router A10.64.0.0to 10.79.255.255toRouter B serialconnection _____________________________
IP address range for Router A
10.80.0.0to 10.95.255.255
toRouter C serialconnection _____________________________
IP address rangefor Router B10.96.0.0to 10.111.255.255toRouter C serialconnection _____________________________
68
-
7/28/2019 Instructor Sub Network Book 2
70/85
Show your work forProblem 6 in the spacebelow.
69
-
7/28/2019 Instructor Sub Network Book 2
71/85
-
7/28/2019 Instructor Sub Network Book 2
72/85
Show your work forProblem 7 in the spacebelow.
71
-
7/28/2019 Instructor Sub Network Book 2
73/85
Practical Subnetting 8
Based ontheinformation inthe graphicshown, designanetwork addressing scheme thatwill
supplythe minimum number subnets , andallowenoughextrasubnets andhosts for85%
growth inallareas. Circle eachsubnet onthe graphicand answer thequestionsbelow.
IPAddress 192.168.1.0
F0/0S0/0/0F0/1
S0/0/1Router A
F0/0 R oute rB
NewYork
8Hosts
Boston
5HostsResearch & Development
8Hosts
CAddress class _____________________________
255.255.255.224
Customsubnetmask _____________________________
3Minimum numberof subnets needed _________
+ 3
Extra subnets required for 85% growth _________(Roundup to thenextwholenumber)
= 6Total number of subnets needed _________
Numberof hostaddresses13i n t he l a rg es t s ub ne t g ro up _ __ __ __ __
Numberof addresses neededfor+ 1285% growth in thelarges t subnet _________
(
Roundup to thenextwholenumber)
Total number of address= 25n ee de d for t h e l arge st s ub ne t _ __ __ __ __
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
192.168.1.0to 192.168.1.31IP address range for Router A F0/0 _____________________________
192.168.1.32 to192.168.1.63
IP address rangefor NewYork _____________________________
IP address range for Router A192.168.1.64 to192.168.1.95toRouter B serialconnection _____________________________
72
-
7/28/2019 Instructor Sub Network Book 2
74/85
Show your work forProblem 8 in the spacebelow.Numberof
2 56 1 28 64 3 2 1 6 8 4 2 - H os tsNumberof
S ub ne ts - 2 4 8 16 2 4 8 16 2 4 8 16
2 4 8 16
2 4 8 16 32 64 128256
1
2 8 6 4 3 2 1 6 8 4 2 1 - B i na ry v al ue s
192.168.1 .0 0 0 0 0 0 0 0
192.168.1 .0 0 0 0 0 0 0 0
192.168.1 .0 0 0 0 0 0 0 0192.168.1 .0 0 0 0 0 0 0 0
192.168.1 .0 0 0 0 0 0 0 0
(1) 192.168.1.0 to 192.168.1.310
(2) 1 192.168.1.32 to 192.168.1.63
(3) 1 0 192.168.1.64 to 192.168.1.95
(4) 1 1 192.168.1.96 to 192.168.1.127
(5) 192.168.1.128 to 192.168.1.1591 0 0
(6) 192.168.1.160 to 192.168.1.11911 0 1
(7) 1 1 0 192.168.1.192 to 192.168.1.223
(8) 1 1 1 192.168.1.224 to 192.168.1.255
73
-
7/28/2019 Instructor Sub Network Book 2
75/85
Practical Subnetting 9
Based ontheinformation inthe graphicshown, designanetwork addressing scheme thatwill
supplythe minimum number ofhosts per subne t, andallowenoughextrasubnetsand
hostsfor 15%growthin allareas. Circleeachsubnetonthe graphicandanswer thequestions
below.
IPAddress 148.55.0.0
S0/0/0F0/1
S0/0/1Router A
S0/0/1 F0/0 R oute rB
S0/0/0 Dallas
1500HostsRo ute rC
F0/0
S0/0/0 S0/0/1R oute rD
Ft. Worth BAddress class _____________________________
2300Hosts
255.255.240.0Customsubnetmask _____________________________
5
Minimum numberof subnets needed _________
+ 1Extra subnets required for 15% growth _________(
Roundup to thenextwholenumber)
= 6Total number of subnets needed _________
Numberof hostaddresses2300i n t he l a rg es t s ub ne t g ro up _ __ __ __ __
Numberof addresses neededfor+ 34515% growth in thelarges t subnet _________
(
Roundup to thenextwholenumber)
Total number of address=2645n ee de d for t h e l arge st s ub ne t _ __ __ __ __
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
148.55.0.0. to148.55.15.255IP address rangefor Ft. Worth _____________________________
148.55.16.0. to148.55.31.255
IP address range for Dallas _____________________________
148.55.32.0.to 148.55.47.255IP address range for Router A _____________________________
toRouter B serialconnection
148.55.48.0.to 148.55.63.255IP address range for Router A _____________________________
toRouter C serialconnection
148.55.64.0. to148.55.79.255IP address rangefor Router C _____________________________
toRouter D serial connection74
-
7/28/2019 Instructor Sub Network Book 2
76/85
Show your work forProblem 9 in the spacebelow.
75
-
7/28/2019 Instructor Sub Network Book 2
77/85
Practical Subnetting 10
Based ontheinformation inthe graphicshown, designanetwork addressing scheme thatwill
supplythe minimum number ofsubnets ,and allowenoughextra subnetsandhostsfor
110%growth inallareas. Circle eachsubnet onthegraphicand answer thequestionsbelow.
IPAddress 172.16.0.0
MarketingSales
56Hosts115Hosts
F0/0F0/0S0/0/0
S0/0/1Ro ute rA
Ro ute rB
F0/1
ResearchManagement
35Hosts25Hosts
BAddress class _____________________________
255.255.255.240Customsubnetmask _____________________________
4
Minimum numberof subnets needed _________
+ 5Extra subnets required for 110% growth _________(
Roundup to thenextwholenumber)
= 9Total number of subnets needed _________
Numberof hostaddresses140i n t he l a rg es t s ub ne t g ro up _ __ __ __ __
Numberof addresses neededfor+ 154110% growth in thelarges t subnet _________
(
Roundup to thenextwholenumber)
Total number of address= 294n ee de d for t h e l arge st s ub ne t _ __ __ __ __
Startwith thefirst subnetand arrangeyoursub-networksfromthelargestgroupto thesmallest.
172.16.0.0 to172.16.15.255IP address range for Sales/Managemnt _____________________________
172.16.16.0to 172.16.31.255
IP address range for Marketing _____________________________
172.16.32.0to 172.16.47.255IP address range for Research _____________________________
IP address range for Router A172.16.48.0to 172.16.63.255
toRouter B serialconnection _____________________________
76
-
7/28/2019 Instructor Sub Network Book 2
78/85
Show your work forProblem 10in the space below.
77
-
7/28/2019 Instructor Sub Network Book 2
79/85
Valid and Non-Valid IP Addresses
Using thematerialin this workbook identify whichof the addresses belowarecorrectand
usable. If theyare not usable addressesexplainwhy.
IP Address:0.230.190.192________________________________ThenetworkID cannotbe 0.
Subnet Mask: 255.0.0.0 ________________________________ReferencePage Inside FrontCover
IP Address:192.10.10.1________________________________OK
Subnet Mask: 255.255.255.0 ________________________________ReferencePages28-29
IP Address:245.150.190.10 ________________________________245is reserved for
Subnet Mask: 255.255.255.0 ________________________________experimental use.ReferencePage Inside FrontCover
IP Address:135.70.191.255 ________________________________This is the broadcast address
Subnet Mask: 255.255.254.0 ________________________________for this range.ReferencePages48-49
IP Address:127.100.100.10 ________________________________127is reserved for loopback
Subnet Mask: 255.0.0.0 ________________________________testing.ReferencePagesInside FrontCover
IP Address:93.0.128.1 ________________________________OK
Subnet Mask: 255.255.224.0 ________________________________ReferencePages56-57
IP Address:200.10.10.128________________________________This isthe subnetaddressforthe
Subnet Mask: 255.255.255.224 ________________________________3rdusablerangeof 200.10.10.0ReferencePages54-55
IP Address:165.100.255.189________________________________OKSubnet Mask: 255.255.255.192 ________________________________
ReferencePages30-31
IP Address:190.35.0.10________________________________This addressis takenfromthe firstSubnet Mask: 255.255.255.192 ________________________________rangeforthis subnetwhich isinvalid.
ReferencePages34-35
IP Address:218.35.50.195________________________________This has a class B subnet
Subnet Mask: 255.255.0.0 ________________________________mask.ReferencePage Inside FrontCover
IP Address:200.10.10.175/22 ________________________________A class C address mustuse a________________________________minimum of24 bits.ReferencePages54-55and/or InsideFront Cover
IP Address:135.70.255.255 ________________________________This is abroadcastaddress.
Subnet Mask: 255.255.224.0 ________________________________ReferencePages48-49
78
-
7/28/2019 Instructor Sub Network Book 2
80/85
IP Address Breakdown
/24 /25 /26 /27 /28 /29 /30
8+8+8 8+8+8+1 8+8+8+2 8+8+8+3 8+8+8+4 8+8+8+5 8+8+8+6
2 55 .2 5 5. 25 5. 0 2 55 .2 55 .2 5 5. 12 8 2 55 .2 5 5. 25 5. 19 2 2 55 .2 55 .2 55 .2 2 4 2 5 5. 25 5. 25 5. 24 0 2 55 .2 55 .2 55 .2 48 2 55 .2 5 5. 25 5. 25 2
256 Hosts 128 Hosts 64 Hosts 32 Hosts 16 Hosts 8 Hosts 4 Hosts
0-30-7
4-70-15
8-118-15
12-15
16-1916-23
20-2316-31
24-2724-31
28-310-63
32-3532-39
36-3932-47
40-4340-47
44-47
48-5148-55
52-5548-63
56-5956-63
60-630-127
64-6764-71
68-7164-79
72-7572-79
76-79
80-8380-87
84-8780-95
88-9188-95
92-9564-127
96-9996-103
100-10396-111
104-107104-111
108-111
112-115112-119
116-119112-127
120-123120-127
124-1270-255
128-131128-135
132-135128-143
136-139136-143
140-143
144-147144-151
148-151144-159
152-155152-159
156-159128-191
160-16316-167
164-167160-175
168-171168-175
172-175
176-179176-183
180-183176-191
184-187184-191
188-191128-255
192-195192-199
196-199192-207
200-203200-207
204-207
208-211208-215
212-215208-223
216-219216-223
220-223192-255
224-227224-231
228-231224-239
232-235232-239
236-239
240-243240-247
244-247240-255
248-251248-255
252-255
79
-
7/28/2019 Instructor Sub Network Book 2
81/85
Visualizing Subnets Using
The Box Method
Thebox method is the simplest wayto visualize the breakdown of
subnets andaddresses into smaller sizes.
Start with a square. The whole square
is a single subnet comprisedof 256
addresses.
/24
255.255.255.0
256 Hosts
1 Subnet
Split theboxin half andyou get two
subnets with128 addresses,
/25
255.255.255.128
128 Hosts
2 Subnets
Dividethe box into quarters and you
get four subnets with 64 addresses,
/26255.255.255.192
64 Hosts
4 Subnets
80
-
7/28/2019 Instructor Sub Network Book 2
82/85
Split each individualsquare and you
geteight subnets with 32 addresses,
/27
255.255.255.224
32 Hosts
8 Subnets
Split the boxes in half again and you
getsixteen subnets with sixteen
addresses,
/28
255.255.255.240
16 Hosts
16 Subnets
Thenext split gives youthirty two
subnets with eight addresses,
/29
255.255.255.248
8 Hosts
32 Subnets
Thelast splitgives sixtyfour subnets
with four addresses each,
/30255.255.255.252
4 Hosts
64 Subnets
81
-
7/28/2019 Instructor Sub Network Book 2
83/85
Class A Addressing Guide# of Bits Subnet Total # of Total # of Usable # of
CIDR Borrowed Mask Subnets Hosts Hosts______________________________________________________________________________________________/
8 0 255.0.0.0 1 16,777,216 16,777,214_____________________________________________________________________________________________
/9 1 255.128.0.0 2 8,388,608 8,
388,606_____________________________________________________________________________________________
/10 2 255.192.0.0 4 4,194,304 4,194,302__________________________________________________________________________________________________
/11 3 255.224.0.0 8 2,097,152 2,
097,150______________________________________________________________________________________________
/12 4 255.240.0.0 16 1,048,576 1,048,574_____________________________________________________________________________________________
/13 5 255.248.0.0 32 524,288 52
4,286________________________________________________________________________________________________
/14 6 255.252.0.0 64 262,144 262,142______________________________________________________________________________________________
/
15 7 255.254.0.0 128 131,072 131,070__________________________________________________________________________________________________/16 8 255.255.0.0 256 65,536 65,5
34___________________________________________________________________________________________________
/17 9 255.255.128.0 512 32,768 3
2,766_______________________________________________________________________________________________
/18 10 255.255.192.0 1,024 16,384 16,382_____________________________________________________________________________________________________
/19 11 255.255.224.0 2,048 8,192 8,
190______________________________________________________________________________________________
/20 12 255.255.240.0 4,096 4,096 4,094__________________________________________________________________________________________________
/2
1 13 255.255.248.0 8,192 2,048 2,046_________________________________________________________________________________________________
/22
14 255.255.252.0 16,384 1,024 1,022________________________________________________________________________________________________
/23 15 255.255.254.0 32,768 512 510____________________________________________________________________________________________________
/24 16 255.255.255.0 65,536 256 254_____________________________________________________________________________________________________
/25 17 255.255.255.128 131,072 128 126____________________________________________________________________________________________________
/26 18 255.255.255.192 262,144 64 62___________________________________________________________________________________________________
/27 19 255.255.255.224 524,288 32 30____________________________________________________________________________________________________
/28 20 255.255.255.240 1,048,576 16 14____________________________________________________________________________________________________
/2
9 21 255.255.255.248 2,097,152 8 6________________________________________________________________________________________________
/
30 22 255.255.255.252 4,194,304 4 2
Class B Addressing Guide
# of Bits Subnet Total # of Total # of Usable # of
CIDR Borrowed Mask Subnets Hosts Hosts______________________________________________________________________________________________/16 0 255.255.0.0 1 65,536
65,534_____________________________________________________________________________________________
/17 1 255.255.128.0 2 32,768 32,766_____________________________________________________________________________________________
/18 2 255.255.192.0 4 16,384 16,3
82__________________________________________________________________________________________________
/19 3 255.255.224.0 8 8,192 8,190______________________________________________________________________________________________
/20 4 255.255.240.0 16 4,096 4,
094_____________________________________________________________________________________________
/21 5 255.255.248.0 32 2,048 2,046________________________________________________________________________________________________
/22 6 2
55.255.252.0 64 1,024 1,022______________________________________________________________________________________________/23 7 255.255.254.0 128 512 51
0__________________________________________________________________________________________________
/24 8 255.255.255.0 256 256 254___________________________________________________________________________________________________
/25 9 255.255.255.128 512 128
126_______________________________________________________________________________________________
/26 10 255.255.255.192 1,024 64 62_____________________________________________________________________________________________________
/27 11 255.255.255.224 2,048 32
30______________________________________________________________________________________________
/28 12 255.255.255.240 4,096 16 14______________________________________________________________________________________________
/2
9 13 255.255.255.248 8,192 8 6________________________________________________________________________________________________
/
30 255.255.255.252 414 16,384 2
Class C Addressing Guide
# of Bits Subnet Total # of Total # of Usable # of
CIDR Borrowed Mask Subnets Hosts Hosts______________________________________________________________________________________________/2
4 0 255.255.255.0 1 256 254_____________________________________________________________________________________________/25 1 255.255.255.128 2 128
126_____________________________________________________________________________________________/26 2 255.255.255.192 4 64 62__________________________________________________________________________________________________/27 3 255.255.255.224 8 32
30______________________________________________________________________________________________/28 4 255.255.255.240 16 16 14_____________________________________________________________________________________________/2
9 5 255.255.255.248 32 8 6____________________________________________________________________________________________