Cs455Hw5
Q1
a. 114.34.2.8
netid: 114.0.0.0
hostid: 0.34.2.8
b. 132.56.8.6
netid: 132.56.0.0
hostid: 0.0.8.6
c. 208.34.54.12
netid: 208.34.54.0
hostid: 0.0.0.12
Q2
a. 25.34.12.56/16
network mask: 255.255.0.0
Block size:256
first address: 25.34.0.0
last adderss: 25.34.255.255
b. 182.44.82.16/26
network mask: 255.255.255.192
Block size: 64
first address: 182.44.82.0/26
last address: 182.44.82.63/26
Q3
a. ::2222
0000:0000:0000:0000:0000:0000:0000:2222
b. 1111::
1111:0000:0000:0000:0000:0000:0000:0000
c. 0:1:2::
0000:0001:0002:0000:0000:0000:0000:0000
d. B:A:CC::1234:A
000B:000A:00CC:0000:0000:0000:1234:000A
Q4
a.
host bit: log2(128) = 7
32 – 7 = 25 → /25
150.80.0.0/25 - 150.80.99.128/25, 200
b.
host bit: log2(16) = 4
32 – 4 = 28 → /28
150.80.100.0/28 - 150.80.124.240/28, 400
c.
host bit: log2(4) = 2
32 – 2 = 30 → /30
150.80.125.0/30 - 150.80.150.60/30
how many addresses are still available after these allocations
Total address = 2^16 = 65536
65536 - 128 * 200 - 16 * 400 - 4 * 2000 = 25536
Thus there will have 25536 addresses left
Q5
a
R1
| Mask | Network address | Next-hop | Interface |
|---|---|---|---|
| /23 | 200.23.16.0 | — | m0 |
| /23 | 200.23.18.0 | — | m1 |
| /23 | 200.23.20.0 | — | m2 |
| /23 | 200.23.22.0 | — | m3 |
| /0 | 0.0.0.0 | R3 | m4 |
R2
| Mask | Network address | Next-hop router | Interface |
|---|---|---|---|
| /20 | 199.23.16.0 | — | m0 |
| /0 | 0.0.0.0 | R3 | m1 |
R3
| Mask | Network address | Next-hop | Interface |
|---|---|---|---|
| /20 | 200.23.16.0 | R1 | m0 |
| /20 | 199.23.16.0 | R2 | m1 |
| /0 | 0.0.0.0 | Internet | m2 |
b
R1
| Mask | Network address | Next-hop | Interface |
|---|---|---|---|
| /23 | 200.23.16.0 | — | m0 |
| /23 | 200.23.18.0 | — | m1 |
| /23 | 200.23.20.0 | — | m2 |
| /0 | 0.0.0.0 | R3 | m4 |
R2
| Mask | Network address | Next-hop | Interface |
|---|---|---|---|
| /23 | 200.23.22.0 | — | m2 |
| /20 | 199.23.16.0 | — | m0 |
| /0 | 0.0.0.0 | R3 | m1 |
R3
| Mask | Network address | Next-hop | Interface |
|---|---|---|---|
| /22 | 200.23.16.0 | R1 | m0 |
| /23 | 200.23.20.0 | R1 | m0 |
| /23 | 200.23.22.0 | R2 | m1 |
| /20 | 199.23.16.0 | R2 | m1 |
| /0 | 0.0.0.0 | I | m2 |
Q6
a. no options
b. not fragmented
c. 64 bytes
d. the value is 0 of checksum is used
e. the packet can do 32 more hops
f. identification number: 00 03 which is decimal 3
g. No
Q7
45 00 0F B4 38 E7 00 00
Total Length = 0FB4 = 4020
Needs to fragment into 3
| Fragment | Total-length (bytes) | Identification | Offset (8-byte units) | D (DF) | M (MF) |
|---|---|---|---|---|---|
| 1 | 1500 | 14567 (or 38 E7) | 0 | 0 | 1 |
| 2 | 1500 | 14567(or 38 E7) | 185 | 0 | 1 |
| 3 | 1060 | 14567(or 38 E7) | 370 | 0 | 0 |
Q8
Router A
| Destination | Next-hop | Cost |
|---|---|---|
| A | — | 0 |
| B | B | 2 |
| C | B | 6 |
| D | B | 7 |
| E | B | 11 |
| F | B | 11 |
| G | B | 12 |
Router G
| Destination | Next-hop | Cost |
|---|---|---|
| G | — | 0 |
| E | E | 1 |
| F | E | 2 |
| D | E | 5 |
| C | E | 6 |
| B | E | 10 |
| A | E | 12 |
Q9
| Destination | Distance (hops) | Next-Hop |
|---|---|---|
| Net 1 | 0 | direct |
| Net 2 | 0 | direct |
| Net 5 | 5 | Router J |
| Net 17 | 6 | Router M |
| Net 22 | 9 | Router J |
| Net 24 | 6 | Router J |
| Net 30 | 2 | Router Q |
| Net 42 | 4 | Router J |