Andrew D Boston.
TCP/IP Class, Fall 2008.
Due Tuesday, September 23, 2008 – 6:00pm.
Assignment #04, Chapter #05 / Exercises #1 – 12.
5.6 Practice Set
#1). In a block of addresses, we know the IP address of one
host is 25.34.1256/16. What is the first address (network address) and the last
address (limited broadcast address) in
the block?
In a block of
addresses, we know the IP address of the host is 25.34.12.56/16
One host, first address: 25.34.0.1
Network address: 25.34.0.0
Last address : 25.34.255.255
Limited address : 25.34.255.255
In the block.
#2). In a block of addresses, we know the IP address of one
host is 182.44.82.16/26. What is the first address (network address) and the
last address (limited broadcast address)
in the block?
In a block of
addresses, we know the IP address of the host is 25.34.12.56/16
One host, first address: 182.44.82.1
Network address: 182.44.82.0
Last address : 182.44.82.254
Limited address : 182.44.82.255
In the block.
#3). In fixed-length subnetting, find the number of 1s that
must be added to the mask if the number of desired subnets is
a.
2 subnetting, subnet bits = 1.
b.
62 subnetting, cant, in powers of 2, nearest
value 64.
c.
122 subnetting, cant, in powers of 2, nearest
number 122.
d.
256 subnetting, subnet bits = 8.
#4). What is the maximum number of subnets if the prefix
length of block is?
a.
18 = /18, 16382 addresses, max subnets is 4.
b.
10 = /10, 4194304 addresses, max subnets is 2.
c.
27 = /27, 30 addresses max subnets is 2.
d.
31 = /31, cant mask /31 no addresses,
impossible.
#5). An organization is granted the block 16.0.0.0/8. The
administrator wants to create 500 fixed-lenth subnets.
a.
Find the subnet mask.
255.0.0.0
b.
Find the number of addresses in each subnet.
1 subnet with 16777214 host/addresses.
c.
Find the first and last address in the first
subnet.
Network Address 10.0.0.0
First Address : 16.0.0.1
Last Address : 16.255.255.254
d.
Find the first and the last address in the last
subnet (subnet 500).
Impossible for 500, in only 1
subnet.
#6). An organization is granted the block 130.56.0.0/16. The
administrator wants to create 1024 subnets.
a.
Find the subnet mask.
255.255.0.0
b.
Find the number of addresses in each subnet.
1 subnet with 65534 host/addresses.
c.
Find the first and the last address in the first
subnet.
Network Number is 130.56.0.0.
First Address is 130.56.0.1.
Second Address is 130.56.255.254.
d.
Find the first and last address in the last
subnet (subnet 1024).
Incredibly would be the exact same
results with a subnet 1024, as follows again.
Network Number is 130.56.0.0.
First Address is 130.56.0.1.
Second Address is 130.56.255.254.
#7). An organization
is granted the block 211.17.180.0/24. The administrator wants to create 32
subnets.
a.
Find the subnet mask.
255.255.255.0
b.
Find the number of addresses, in each subnet.
254 addresses.
c.
Find the first and the last address in the first
subnet:
First Addresses: 211.17.180.1.
Last Addresses: 211.17.180.254.
d.
Find the first and last address in the last
subnet (subnet 32).
Incredibly would be the exact same results with a subnet 32, as follows
again.
First Addresses: 211.17.180.1.
Last Addresses: 211.17.180.254.
8). Write the following mask in slash notation (/n):
a.
255.255.255.0 = /24 Mask Slash.
b.
255.0.0.0 = /8 Mask Slash.
c.
255.255.224.0 = /19 Mask Slash.
d.
255.255.240.0 = /20 Mask Slash.
#9). Find the range of addresses in following blocks:
a. 123.56.77.32/29.
Range of addresses = 123.56.77.33 to 123.56.77.38
b. 200.17.21.128/27.
Range of addresses = 206.17.21.129 to 206.17.21.158
c. 17.34.16.0/23.
Range of addresses = 17.34.16.1 to 17.34.17.254
d. 180.34.64.64/30.
Range of addresses = 180.34.64.65 to 180n.34.64.66
#10). An ISP is granted a block of addresses starting with
150.80.0.0/16. The ISP wants to distribute these blocks to 1000 Customers as
follows.
a.
The first group has 200 medium-size business
each needs 128 addresses.
GROUP 1: 200 medium businesses, with 128
addresses.
Need 128 addresses.
7(2^7 = 128)
32 – 7 = /25
Usable addresses: 126.
Total number of addresses: 128.
Mask: 255.255.255.128.
Network Id: 150.80.0.0.
First 150.80.0.0 / 25.
150.80.0.255 /25.
Last 150.80.127.0 / 25.
150.80.127.255 / 25.
Total 200 x 128: = 25,600 available .addresses.
b.
The second group has 400 small businesses each
needs 16 addresses.
GROUP 2 : 400 Small Business, with 16
addresses.
Need 16 addresses.
4(2^4=16)
32-4=/28
Usable addresses 14.
Total number of addresses: 16.
Mask: 255.255.255.240.
Network ID: 150.80.0.0
First 150.80.0.0 / 25.
150.80.0.255 /25.
Last 150.80.15.0 / 25.
150.80.127.255 / 25.
Total 400 x 16 : =6,400 available addresses.
c.
The third group has 2000 households: each needs
4 addresses.
GROUP 3 : 2000 households, with 4
addresses.
Need 4 addresses.
2(2^2=16).
32-2=/30.
Usable addresses 2.
Total number of addresses: 4.
Mask: 255.255.255.252.
Network ID: 150.80.0.0
First 150.80.0.0 / 16.
to
150.80.0.255 /16.
Last 150.80.15.0 / 16.
to 150.80.15.255 / 16.
Total 2000 x 4 : =8,000 available addresses.
Design the subblocks and give the slash notation for each
subblock. -did
Find out how many addresses are still available after these
allocation. -did
#11). An ISP is granted a block of addresses starting with
120.60.4.0/20. The wants to distribute these blocks to 1000 organizations with
each organization receiving 8 addresses only. Design the sub blocks and give
the slash notation for each sub block. Find out how many addresses are still
available after these allocations.
a.
The first group has 100 organizations of
business, each needs 8addresses.
GROUP 1: 100 organizations, with 8
addresses each.
Need 8 addresses.
3(2^3 = 8)
32 – 3 = /29
Usable addresses: 6.
Total number of addresses: 8.
Mask: 255.255.255.248.
Network Id: 120.60.4.0.
First 120.60.4.0 / 29.
150.80.4.255 / 29.
Last 120.60.12.0 / 29.
120.60.12.255 / 29.
Total 100 x 8 = 800 available addresses.
#12). An ISP has a block of 1024 addresses. It needs to
divide the addresses to 1024 customers, does it need subnetting? Explain your answer?
For 1024 Customers:
Subnetting the 1024 addresses, I believe “no” is the answer.
Reason – Subnetting remember,
is for classless addresses, used only for a organization of some kind, to break
up into distinguished divisions, for very important reasons, here I feel we do
not need to break up the 1024 address amount, there’s 1024 customers, 1024
addresses available. Remember in subnetting, a network is broken up into
smaller workable networks (or subnets).
Right now the amount is the exact
amount, in the near future possibly, due to aggressive growth, to make more
addresses for customer accounts, and divide the areas by geographic portions,
from the total whole address, that will be necessary, till then “no”.
When there are too many customer
accounts, getting too big, subnetting will be definitely be utilized. Breaking
the total amount of addresses into smaller divisions or groupings, 1024 value is
not that big right now, pretty small number compared to like 62,000 addresses,
is like a big number, break that kind of number down into smaller groups to be
worked.
Andrew Boston.
