**5.01-1**
D(y,y)-infinity b(w,w)-8,u a(v,v)-2,u c(x,x)-3,u
**5.01-2**
c-6
b-2,u
d-3u
e-inf
a-
**5.01-3**
a-uvx
c-5x
d-3u
b-2u
e-9x
**5.01-4**
a-uvxw
b-5x
c-8w
d-6,w
**5.01-5**
d-6w
c-7z
b-6x
a-uvxwz
**5.02-1**
b-1y
d-8y
a-inf
c-6y
**5.02-2**
a-yw
e-5w
c-6w
d-3w
b-9w
**5.02-3**
b-6x
c-5x
e-4x
a-ywx
d-3w
**5.02-4**
c-5x
d-4x
b-6x
a-ywxz
**5.03-1**
a-inf
b-10,E
c-4E
d-2E
**5.03-2**
c-10e
e-2e
d-4e
b-inf
a-EF
**5.03-3**
a-EFD
e-4e
c-6f
d-7d
b-8d
**5.04-1**
b-d-f-De
**5.04-2**
a-c-g-I
**5.04-3**
Node h doesn't send
**5.04-4**
Infinite amount
**5.05-1**
3,
b-4
f-5
d-1
e-0
**5.05-2**
No. Node e's distance vector does not change when the link between e and f goes down (since e's shortest path to f did not use this direct link between e and f), so e will not send out a new DV.
**5.06-1**
G-1
H-inf
**5.06-2**
H-2
I-inf
G-1
**5.06-3**
h-2
g-1
e-5
I-3
**5.06-4**
L distance
**5.07-1**
3c-eBGP
3a-iBGP
**5.07-2**
1c-eBGP
1d-iBGP
**5.08**
1d will forward along z1 since OSPF has computed the path to 1c is via z1.
**5.09**
1d will forward along z2 since 1d has a shorter intra domain path than border router 1b to border router 1c, and hot potato routing is used
**5.10**
x will not advertise to provider networks B or C that it has a path to the other provider network
**5.11**
B will advertise a route Bx to A, C and D, since A, C and D need to know how to route to B's customer network x.
**5.12**
Bx.,
Cx.,
x to w.,
x to z.
**5.13-1**
ABCz
**5.13-2**
ZW
**5.14**
The correct answers are: When executing, Dijkstra's algorithm will use the link-state database that is maintained within the SDN controller.,
When executing, Dijkstra's algorithm will run as a network control application "on top" on the SDN controller.,
If a router's forwarding table should be changed as a result of running Dijkstra's algorithm, the new flow table for that router will be updated by the SDN controller via the southbound API using the Openflow protocol.
**5.1-1**
Forwarding refers to moving packets from a router's input to appropriate router output, and is implemented in the data plane.
Routing refers to determining the route taken by packets from source to destination, and is implemented in the control plane
**5.1-2**
Per router-An individual routing algorithm components in each and every router interact in the control plane
(SDN). → A (typically) remote controller computes and installs forwarding tables in routers
**5.2-1**
Routing algorithms typically work with abstract link weights that could represent any of, or combinations of, all of the other answers
**5.2-2**
In the initialization step, the initial cost from a to each of these destinations is initialized to either the cost of a link directly connecting a to a direct neighbor, or infinity otherwise.,
Suppose nodes b, c, and d are in the set N'. These nodes will remain in N' for the rest of the algorithm, since the least-cost paths from a to b, c, and d are known.,
The values computed in the vector D(v), the currently known least cost of a path from a to any node v, will never increase following an iteration
**5.2-3**
Centralized, global routing. → All routers have complete topology, and link cost information., Static routing. → Routes change slowly over time.,
Dynamic routing. → Routing changes quickly over time.,
Decentralized routing. → An iterative process of computation, exchange of informatoin with neighbors. Routers may initially only know link costs to directly-attached neighbors.
**5.2-4(e)**
**5.2-5(4)**
**5.3-1**
Interdomain-different ASes(networks)
Intradomain-among routers within same AS(network)
**5.3-2**
OSPF implements hierarchical routing, OSPF is an intra-domain routing protocol.,
OSFP uses a Dijkstra-like algorithm to implement least cost path routing
**5.3-3**
Is an intra-domain routing protocol.
Finds a least cost path from source to destination.
Floods link state control information.
**5.4-1**
OSPF
intra-AS routing
intra-domain routing,
Driven more by performance than by routing policy
**5.4-2(B-A-X)**
**5.4-3(W) baştaki**
**5.4-4**
None of
**5.4-5**
How does router 2d learn of the path AS3, X to destination network X? → From 2c via iBGP., How does router 2c learn of the path AS3, X to destination network X? → From 3a via eBGP
**5.5-1**
SDN-controlled-C(router)
Network-A
SDN controller-B
**5.5-2**
Intent
Network graph
**5.5-3**
Statistics,
Host information,
Flow tables,
Link-state information,
Switch information
**5.5-4**
Openflow
**5.6-1**
ICMP is used by hosts and routers to communicate network-level information.,
ICMP messages are carried directly in IP datagrams rather than as payload in UDP or TCP segments.,
The TTL-expired message type in ICMP is used by the traceroute program.
**6.1-1**
Reliable data
Flow control
Coordinated
Bit-level
Multiplexing
**6.2-1**
Detect any case(2side)
Detect and correct single flip
**6.3-1**
There can be times
Channel utilization
**6.3-2**
Utilization
There can be simultaneously
**6.3.3**
Channel utilization
Utilization
**6.3-4**
TDMAandFDMA-POLLİNG
**6.3-5**
TDMA and FDMA POLLİNG
**6.3-6**
TDMA and FDMA POLLİNG
**6.3-7**
TDMA and FDMA POLLİNG
**6.4-1**
48bit
Link-layer
Same as host
**6.4-2**
32bit
Network-layer
Allocated by DHCP
**6.4-3**
Must be unique
**6.4-4**
Cylic-used to detect
Sequence-ethernet frame
Type field-used to demultiplex
Source-48bit
Data-the contents
**6.4-5**
Flood the frame
**6.4-6**
Associates MAC
Age-out
Switch frees
**6.4-7**
1
**6.4-8**
Entry for hostwill
**6.4-9**
48bit
Stays unchanged
Doesn't change
**QUESTION LIST**
1. At t=2, what two nodes communicated? Write your answer in alphabetical order as x,y (If there is only enough information for 1 node, write that, and if there's no information, write 'n/a')
2. At t=5, what two nodes communicated? Write your answer in alphabetical order as x,y (If there is only enough information for 1 node, write that, and if there's no information, write 'n/a')
3. At t=1, what two nodes communicated? Write your answer in alphabetical order as x,y (If there is only enough information for 1 node, write that, and if there's no information, write 'n/a')
4. At t=7, what two nodes communicated? Write your answer in alphabetical order as x,y (If there is only enough information for 1 node, write that, and if there's no information, write 'n/a')
**SOLUTION**
1. The transmission at t=2 was I-->L, so the answer is n/a.
2. The transmission at t=5 was K-->L, so the answer is K,L.
3. The transmission at t=1 was L-->K, so the answer is L.
4. The transmission at t=7 was G-->D, so the answer is D,G.
**QUESTION LIST**
1. At t=1, what is the source entry for switch 1? Format your answer as letter,number or 'n/a'
2. At t=1, what is the destination entry for switch 1? Format your answer as letter,number or 'n/a'
3. At t=1, what is the source entry for switch 2? Format your answer as letter,number or 'n/a'
4. At t=1, what is the destination entry for switch 2? Format your answer as letter,number or 'n/a'
5. At t=2, what is the source entry for switch 1? Format your answer as letter,number or 'n/a'
6. At t=2, what is the destination entry for switch 1? Format your answer as letter,number or 'n/a'
7. At t=2, what is the source entry for switch 2? Format your answer as letter,number or 'n/a'
8. At t=2, what is the destination entry for switch 2? Format your answer as letter,number or 'n/a'
9. At t=3, what is the source entry for switch 1? Format your answer as letter,number or 'n/a'
10. At t=3, what is the destination entry for switch 1? Format your answer as letter,number or 'n/a'
11. At t=3, what is the source entry for switch 2? Format your answer as letter,number or 'n/a'
12. At t=3, what is the destination entry for switch 2? Format your answer as letter,number or 'n/a'
13. At t=4, what is the source entry for switch 1? Format your answer as letter,number or 'n/a'
14. At t=4, what is the destination entry for switch 1? Format your answer as letter,number or 'n/a'
15. At t=4, what is the source entry for switch 2? Format your answer as letter,number or 'n/a'
16. At t=4, what is the destination entry for switch 2? Format your answer as letter,number or 'n/a'
**SOLUTION**
1. At time t=1, (C,3) is added as an entry to switch table 1.
2. At time t=1, (B,2) is added as an entry to switch table 1.
3. At time t=1, (C,8) is added as an entry to switch table 2.
4. At time t=1, switch table 2 doesn't observe this frame
5. At time t=2, (H,7) is added as an entry to switch table 1.
6. At time t=2, (D,4) is added as an entry to switch table 1.
7. At time t=2, (H,11) is added as an entry to switch table 2.
8. At time t=2, (D,8) is added as an entry to switch table 2.
9. At time t=3, (A,1) is added as an entry to switch table 1.
10. At time t=3, since the entry for computer B in switch table 1 already exists, no new table entry is made
11. At time t=3, switch table 2 doesn't observe this frame
12. At time t=3, (n/a) is added as an entry to switch table 2.
13. At time t=3, switch table 2 doesn't observe this frame
14. At time t=4, switch table 1 doesn't observe this frame
15. At time t=4, switch table 1 doesn't observe this frame
16. At time t=4, (K,14) is added as an entry to switch table 2.
17. At time t=4, since the entry for computer H in switch table 2 already exists, no new table entry is made
**QUESTION LIST**
1. What is the source mac address at point 6?
90-54-8A-61-C5-F5
1. What is the destination mac address at point 6?
0E-F7-40-E8-B7-25
1. What is the source IP address at point 6?
128.119.40.206
1. What is the destination IP address at point 6?
128.119.40.64
1. Do the source and destinaton mac addresses change at point 5? Answer with yes or no.
No, datagrams can be sent across the subnet via the link layer in one go
**QUESTION LIST 1**
1. Suppose all nodes are implementing the Aloha protocol. For each message, indicate the time at which each transmission begins. Separate each value with a comma and no spaces.
2. Which messages transmit successfully? Write your answer as a comma seperated list with no spaces using the messages' numbers
**SOLUTION (Aloha)**
1. The list of times for frame transmissions is: 0.2,0.9,1.4,1.5,1.8,2.2,2.5,2.9,3.2,4.9
2. The list of successful frames is: 10
**SOLUTION (Slotted-Aloha)**
1. The list of times for frame transmissions is: 1,1,2,2,2,3,3,3,4,5
2. The list of successful frames is: 9,10
**QUESTION LIST 2**
1. Suppose all nodes are implementing Carrier Sense Multiple Access (CSMA), but without collision detection. Suppose that the time from when a message transmission begins until it is beginning to be received at other nodes is 0.4 time units. (Thus if a node begins transmitting a message at t=2.0 and transmits that message until t=3.0, then any node performing carrier sensing in the interval [2.4, 3.4] will sense the channel busy.) For each message, indicate the time at which each message transmission begins, or indicate that message transmission does not begin due to a channel that is sensed busy when that message arrives. Separate each value with a comma and no spaces, and if the channel is sensed busy, substitute it with 's'
2. Which messages transmitted successfully? Write your answer as a comma seperated list with no spaces using the messages' numbers
**SOLUTION (CSMA)**
1. The list of times for frame transmissions is: 0.2,s,s,s,1.8,s,s,s,3.2,4.9
2. The list of successful frames is: 1,5,9,10
**QUESTION LIST 3**
1. Suppose all nodes are implementing Carrier Sense Multiple Access (CSMA), with collision detection (CSMA/CD). Suppose that the time from when a message transmission begins until it is beginning to be received at other nodes is 0.4 time units, and assume that a node can stop transmission instantaneously when a message collision is detected. (Thus if a node begins transmitting a message at t=2.0 and transmits that message until t=3.0, then any node performing carrier sensing in the interval [2.4, 3.4] will sense the channel busy.) For each message, indicate the time at which each message transmission begins, or indicate that message transmission does not begin due to a channel that is sensed busy when that message arrives. Separate each value with a comma and no spaces, and if the channel is sensed busy, substitute it with 's'
2. Which messages transmitted successfully? Write your answer as a comma seperated list with no spaces using the messages' numbers
3. At what time did each message stop transmitting due to a collision. Write your answer as a comma seperated list with no spaces using the messages' numbers in order, and if a message didn't stop, write 'x' for that message
**SOLUTION (CSMA-CD)**
1. The list of times for frame transmissions is: 0.2,s,s,s,1.8,s,s,s,3.2,4.9
2. The list of successful frames is: 1,5,9,10
3. The list of stopped packet times is: x,x,x,x,x,x,x,x,x,x
**QUESTION LIST**
1. Given a probability of transmission p = 0.21, what is the maximum efficiency?
0.25 or 25% efficiency.(PURE)
0.39 or 39% efficiency.(Slotted)
1. Given a probability of transmission p = 0.74, what is the maximum efficiency?
- %1 efficiency(PURE)
0.15 or 15% efficiency.(Slotted)
**QUESTION LIST**
1. For figure 1, compute the two-dimensional parity bits for the 16 columns. Combine the bits into one string
2. For figure 1, compute the two-dimensional parity bits for the 5 rows (starting from the top). Combine the bits into one string
3. For figure 1, compute the parity bit for the parity bit row from question 1. Assume that the result should be even.
4. For figure 2, indicate the row and column with the flipped bit (format as: x,y), assuming the top-left bit is 0,0
5. For figure 3, is it possible to detect and correct the bit flips? Yes or No
**SOLUTION**
The full solution for figure 1 is shown below:
10010100 00110101 1
00110100 11110010 0
00111110 10001001 0
01011011 10001100 0
00010011 00000110 1
11010110 11000100 0
1. The parity bits for the 16 columns is: 11010110 11000100
2. The parity bits for the 5 rows is: 10001
3. The parity bit for the parity row is: 0
4. The bit that was flipped in figure 2 is (3,4):
10010001 10110111 1
11100000 10001010 0
00011000 10010001 1
00110001 10010010 0
11101011 01111010 0
10100011 01000100 0
For figure 3, the bit that was flipped is (11,3):
00100001 01010011 0
11101101 10011010 0
01011101 01010111 0
11110101 00100110 0
00100001 11011101 0
01000101 01110101 0
1. Yes, with 2D parity, you can detect and correct the a single flipped bit