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A large number of consecutive IP addresses are available starting at 198.16.0.0. Suppose
that four organizations, A, B, C, and D, request 4000, 2000, 4000, and 8000 addresses,
respectively, and in that order. For each of these, give the first IP address assigned,
the last IP address assigned, and the mask in the w.x.y.z/s notation.

2 Answers

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IP addresses will be allocated in blocks of power of 2. So the four organizations will be allocated IPs as A-4096, B-2048, C-4096 and D-8192. Remaining unused IPs are wasted. IPs will be allocated to the organizations contiguously

A has $2^{12}$ hosts. So lower order 12 bits will denote host ID and higher order 32-12=20 bits denotes network ID

A's first IP=198.16.0.0 (Host IP part contains all 0s)

A's last IP=11000110.00010000.00001111.11111111 (Host ID part contains all 1s

=198.16.15.255

B's Mask=198.16.15.255

_____________________________________________________________________

B has $2^{11}$ hosts. So lower order 11 bits will denote host ID and higher order 32-11=21 bits denotes network ID

B's first IP=198.16.16.0

B's last IP=11000110.00010000.00010111.11111111=198.16.23.255

B's Mask=198.16.16.0/21

_____________________________________________________________________

C has $2^{12}$ hosts. So lower order 12 bits will denote host ID and higher order 32-12=20 bits denotes network ID

C's first IP=198.16.24.0

C's last IP=11000110.00010000.00011111.11111111=198.16.31.255

C's Mask=198.16.24.0/20

_____________________________________________________________________

D has $2^{13}$ hosts. So lower order 13 bits will denote host ID and higher order 32-13=19 bits denotes network ID

D's first IP=198.16.32.0

D's last IP=11000110.00010000.00111111.11111111=198.16.63.255

D's Mask=198.16.32.0/19
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for 4 organization,we need 4 subnet by 2 bit;

198.18.11000000.00000000 = subnetid

00,10,01,11 are poosible subnet,then it is easy to find the answer

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