1.
A Class 5 switch is a big expensive telephony switch, located in CO (central office)
2.
ATM allows multiple logical connections to be multiplexed over a single physical interface.
3.
Which layers does the Frame Relay operate? => Layer 1 and 2 (physical and datalink)
4.
List the three traditional switching methods. What are the most common today? =>Circuit switching, Packet
switching, Message switching. (2 cái đầu dùng cho ngày nay)
5.
What design has got these characteristics? “A logical connection is set up between 2 stations. Packets are
labeled with a circuit number and a sequence number. Packets arrive in sequence?” => virtual circuit (VC)
6.
How many bytes are there in an ATM cell? => 53 bytes (48 bytes payload and a 5-byte header)
7.
In circuit switching, the resources need to be reserved during the setup phase; the resources remain
dedicated for the entire duration of data transfer phase until the teardown phase.
8.
Choose the wrong statement about circuit switching
a.
Channel capacity is dedicated for the duration of a connection, even if no data are being transferred. (True)
b.
Circuit switching can be rather inefficient. (True)
c.
Circuit switching was developed just for handling voice traffic.
d.
Once the circuit is established, the network is effectively transparent to the users.(True)
9.
Based on the Clos criteria, if N=200, then the minimum number of crosspoints is greater than or equal to:
4 N  2 N 

12
 1  15200

10.
Which protocol in Frame Relay that provides a multicast mechanism to allow a local end system to send
frames to more than one remote end system. => LMI
11.
A cell is defined as a small, fixed-size block of information.
12.
In ATM, end devices such as routes use all three layers, while switches inside the ATM network use only
the bottom two layers.
13.
What are the two approaches to packet-switching?
=> Virtual-circuit approach and Datagram approach.
14.
A crossbar switch connects n inputs to m outputs in a grid, using electronic microswitches (transistors) at
each crosspoint requires mxn crosspoints.
15.
Which routing strategies do not require the use of network information? Flooding routing and Random
routing
16.
A switch in a datagram network uses a routing table that is based on the destination address.
17.
Briefly describe the ATM layers and their functions.
ATM defines three layers: ATM adaptation layer (AAL); ATM layer, roughly corresponding to the OSI data link
layer; and physical layer, equivalent to the OSI physical layer. The ATM layer provides routing, traffic
management, switching, and multiplexing services. It processes outgoing traffic by accepting 48-byte segments
from the AAL sublayers and transforming them into 53-byte cells by the addition of a 5-byte header.
18.
Softswitch refers to an architecture for a device that supports the integration of IP telephony and the PSTN
(T or F) => True
19.
Circuit switching takes place at the physical layer.
20.
How many methods of switching are there that are important, traditionally?
=> The three traditional switching methods are circuit switching, packet switching, and message switching
21.
How many digital voice calls can be transmitted over five T3 lines?
=> 672x5=3360 (digital voice calls)
22.
The local loop has twisted-pair cable that connects the subscriber telephone to the nearest end office.
23.
Traditionally, three methods of switching have been important.
24.
We can divide today's networks into three broad categories.
25.
Packet-switched networks can also be divided into two subcategories: virtual-circuit networks and datagram
networks
26.
A circuit-switched network is made of a set of switches connected by physical links, in which each link is
divided into n channels.
27.
In datagram switching, there is no resource allocation for a packet.
28.
In datagram switching, resources are allocated on demand.
29.
In datagram switching, each packet is treated independently of all others.
30.
In datagram switching, there are no setup or teardown phases.
31.
A packet-switched network is a cross between a circuit-switched network and a datagram network. It has
some characteristics of both.
32.
We can say that a packet switch has four types of components.
33.
A switched WAN is normally implemented as a virtual-circuit network.
34.
In a virtual-circuit network, two types of addressing are involved: global and local.
35.
The network layer in the Internet is designed as a datagram network.
36.
The destination address in the header of a packet in a datagram network normally remains the same during
the entire journey of the packet.
37.
What is the hexadecimal equivalent of the Ethernet address 01011010 00010001 01010101 00011000
10101010 00001111? => 5A:11:55:18:AA:0F
If an Ethernet destination address is 07:01:02:03:04:05, then this is a multicast address.
38.
If an Ethernet destination address is 08:07:06:05:44:33, then this is a broadcast address.
39.
Which of the following could not be an Ethernet unicast destination? => 43:7B:6C:DE:10:00
40.
Which of the following could not be an Ethernet multicast destination? => 7C:56:21:1A:DE:F4
41.
Ethernet is the most widely used local area network protocol.
42.
The IEEE 802.3 Standard defines 1-persistent CSMA/CD as the access method for first-generation 10-Mbps
Ethernet.
The data link layer of Ethernet consists of the LLC sublayer and the MAC sublayer.
43.
The MAC sublayer is responsible for the operation of the CSMA/CD access method and framing.
44.
Each station on an Ethernet network has a unique 48-bit address imprinted on its network interface card
(NIC).
45.
The maximum frame length for 10-Mbps Ethernet is 1518 bytes.
46.
10Base5 uses thick coaxial cable.
47.
10Base2 uses thin coaxial cable.
48.
10Base-F uses fiber-optic cable.
49.
Fast Ethernet has a data rate of 100 Mbps.
50.
In Fast Ethernet, autonegotiation allows two devices to negotiate the mode or data rate of operation.
51.
100Base-TX uses two pairs of twisted-pair cable.
52.
100Base-FX uses two fiber-optic cables.
53.
100Base-T4 uses four pairs of voice-grade, or higher, twisted-pair cable.
54.
Gigabit Ethernet has a data rate of 1000 Mbps.
CHAPTER 1
55.
What is the value of the single digital voice call?
=> 64kbps  f  2 f  8KHz (with 8bits / sample  64Kpbs) 
s
max
56.
How many voice calls can be transmitted over DS-3?
=> DS-0: 1 voice channel, T1: 24 DS-0, DS-3: 28 T1s (28*24=672 voice channels)
57.
In PSTN network architechture, which switches are connected to customers?
=> A class 5 switch is an end office switch that is located at CO (central office)
58.
What is the best characteristics of the cicuit switching technology?
=> A dedicated-circuit is a circuit that is switched on once and stays on.
59.
What is the best characteristics of the packet switching technology?
=> The advantage of the connectionless packet model is that packets are forwarded independent of other packets.
Packets are forwarded on-the-fly by routers, based on the most current best path to a destination. If a link or
router fails, packets are quickly diverted along another path. Since routers don’t maintain information about
virtual circuits, their job is greatly simplified.
60.
The following characteristics belong to the circuit switching technology?
a.
The services are shared in the same link.
b.
Data is transmitted in time slots.
c.
The signaling protocol is SS7
d.
The resources need to be reserved during the setup phase; the resources remain dedicated for the entire
duration of data transfer phase until the teardown phase.
e.
Two technologies: the space-division switch or the time-division switch.
f.
A dedicated communication cahnnel between two end systems.
g.
Circuit switching takes place at the physical layer.
61.
The following characteristics belong to the packet switching technology?
a.
There is no resource allocation for a packet
b.
Two subcategories: virtual-circuit networks and datagram networks
c.
The resources need to be reserved during the setup phase
d.
Two technologies: the space-division switch or t he time-division switch.
e.
A vitual-circuit network is a cross between a circuit-switched network and a data-gram network. It has
some characteristics of both.
62.
Compare transmission of data in a circuit switched network and a packet-switched network.
Circuit switching
Datagram Packet Switching
Virtual
Circuit
Packet
Switching
Dedicated tranmission path
No dedicated path
No dedicated path
Continuous tranmission of data
Transmission of packets
Transmission of packets
Fast enough for interactive
Fast enough for interactive
Fast enough for interactive
Messages are not stored
Packets maybe stored until Packets stored until deliveried
deliveried
The path is established for entire Route established for each packet The path is established for entire
conversation
conversation
Call setup delay; negligible Packet transmission delay
Call setup delay; negligible
transmission delay
transmission delay
Busy signal if called party busy
Sender maybe notified if packet Sender notified of connection
not deliveried
denial
Overload may block call setup; no Overload increases packet delay
Overload may block call setup;
delay for established calls
increases packet delay
Electromechanical
or Small switching nodes
Small switching nodes
computerized switching nodes
User responsible for message loss Network maybe responsible for Network maybe responsible for
protection
individual packets
packet sequences
63.
Which layer is the virtual-circuit network normally implemented in? => Data link layer
64.
Which layer is the circuit-switched network implemented in? => physical layer
65.
Which layer is the datagram switching normally done at? => Network layer
66.
In the packet-switched networks. What problems will can happen if the buffer of router is mall?=> Drop
pakets.
67.
Distinguish the media gateway from the media gateway controller?
The media gateway is just the switch portion of the Class 5 switch.
This component runs the call control and application-level calling features in a server that supports easy
software upgrades and wxpansion of features. The media gateway translates between circuit-switched voice
traffic and packet-based traffic. It is also called the “call agent” and it controls the media gateway.
CHAPTER 2
68.
Which technologies are usually used in the node-node links of the circuit-switched network? => FDM,
TDM
69.
Which dynamic routing algorithms are usually used for the circuit switching of the circuit-switched
network? Let’s describe.
Alternate Routing: Định tuyến luân phiên theo thứ tự xác định trước.
o
Dùng môi trường full mesh
o
Truyền ưu tiên đường trực tiếp từ các mesh với nhau
o
Nếu đường này bị crop thì chạy đường khác luân phiên.
Adaptive Routing:
o
Tính toán tuyến đường
o
Phân bổ lưu lượng truy cập cho 1 nhóm lớn các đường dẫn thay thế, không có thứ tự xác định trước.
70.
Distinguish between blocking network and a nonblocking network.
Blocking
Non-blocking
- Tất cả các đường giữa 2 trạm đều bị chiếm dụng - Cho phép tất cả các trạm đều được kết nối bất kể
(không thể sử dụng ngay) => Mạng không thể kết nối thời điểm nào.
các trạm
- Được dùng cho các kết nối dữ liệu.
- Mạng có thể bị nghẽn
- Được dùng trong hệ hống thoại
71.
What are the advantages of packet switching?
Flexibility: Nếu nghẽn có thê định tuyến đi qua đường khác, (Còn chuyển mạch thì blocking)
Resource sharing: Trên 1 link có thể có nhiều gói tin chuyển cùng 1 lúc (còn chuyển mạch mạch thì
dedicate, khi nào có tín hiệu giải phóng thì mới được phân bổ khác)
Robustness: Kèm theo độ tin cậy: Do có nhiều node chuyển mạch (multi-point) nên ít khi bị nghẽn, và
được định tuyến nên chuyển hướng rất nhanh.
72.
So sánh Diagram và Virtual-circuit trong packet switching.
Diagram
Virtual-circuit
Switch trong diagram sử dụng bảng routing table dựa Tất cả các gói tin thuộc về cùng 1 nguồn và đích sẽ
trên địa chỉ đích
được truyền trên cùng 1 đường.
Nhưng các gói tin có thể chuyển đến đích vs các độ trễ
khác nhau, nếu phân bổ nguồn được yêu cầu.
73.
Which are the routing technologies which are usually in a packet-switching network? Let’s describe
Định tuyến Fixed: Tuyến đường đi sẽ được cố định , xác định sẵn tuyến đường từ nguồn -> đích.
Định tuyến Flooding: tại các node sẽ broadcast gói tin đến tất cả các tuyến đường đi mà nó có, không yêu cầu
thông tin mạng.
Định tuyến ngẫu nhiên: Tương tự như Flooding, không cần thông tin mạng, bởi vì tuyến đường được chọn
ngẫu nhiên, không xét cost thấp nhất cũng như số hop mà nó đi qua.
Định tuyến thích nghi: Tính toán tuyến đường đi ngắn nhất => Nhanh hơn nhưng phức tạp hơn nếu nodes
mạng nhiều.
Frame Relay
74.
Why is Frame Relay more suitable for the realiable media?
1. Frame Relay operates at a higher speed (1.544 Mbps and recently 44.376 Mbps). This means that it can easily
be used instead of a mesh of T-1 or T-3 lines.
2. Frame Relay operates in just the physical and data link layers. This means it can easily be used as a backbone
network to provide services to protocols that already have a network layer protocol, such as the Internet.
3. Frame Relay allows bursty data.
4. Frame Relay allows a frame size of 9000 bytes, which can accommodate all localarea network frame sizes.
5. Frame Relay is less expensive than other traditional WANs.
6. Frame Relay has error detection at the data link layer only. There is no flow control or error control. There
is not even a retransmission policy if a frame is damaged; it is silently dropped. Frame Relay was designed in
this way to provide fast transmission capability for more reliable media and for those protocols that have flow
and error control at the higher layers.
75.
In Frame Relay, which protocols are used to handle frames which arrive from other protocols? => Frame
relay assembler/disassembler (FRAD) and Voice over Frame Relay (VOFR)
76.
In a Frame relay network, can we use the same DLCIs to connect two devices? Why? => DLCI là duy nhất
cho 1 interface cụ thể. Một con switch gắn 1 DLCI cho mỗi kết nối ảo trong mỗi interface. Do đó 2 kết nối khác
nhau thuộc về 2 interface khác nhau có thể cùng DLCI.
CHAPTER 6 MPLS
77.
MPLS: Layer 2 (swith) + 3 (Tra bảng đinh tuyến)
78.
Vừa dùng Hop-by-hop vừa dùng Explicit routing.
79.
Labels can be created based on the: => Traffic, Topology, request.
80.
What is label merging? => Merging of traffic for the same destination on different interfaces.
81.
The creation of MPLS tunnels is for controlling the entire path a packet takes to destination: => Needs to
explicity specify the intermediate routers and can span multiple networks segments.
CHAPTER 8 SWITCHING
Q8-1. Describe the need for switching and define a switch. => Switching provides a practical solution to the
problem of connecting multiple devices in a network. It is more practical than using a bus topology; it is more
efficient than using a star topology and a central hub. Switches are devices capable of creating temporary
connections between two or more devices linked to the switch.
Q8-2. List the three traditional switching methods. Which are the most common today?=> The three
traditional switching methods are circuit switching, packet switching, and message switching. The most common
today are circuit switching and packet switching.
Q8-3. What are the two approaches to packet switching? => There are two approaches to packet switching:
datagram approach and virtualcircuit approach.
Q8-5. What is the role of the address field in a packet traveling through a datagram network? => The address
field defines the end-to-end (source to destination) addressing.
Q8-6. What is the role of the address field in a packet traveling through a virtualcircuit network? => The
address field defines the virtual circuit number (local) addressing.
Q8-7. Compare space-division and time-division switches. => In a space-division switch, the path from one
device to another is spatially separate from other paths. The inputs and the outputs are connected using a grid of
electronic microswitches. In a time-division switch, the inputs are divided in time using TDM. A control unit sends
the input to the correct output device.
Q8-8. What is TSI and what is its role in time-division switching? => TSI (time-slot interchange) is the most
popular technology in a time-division switch. It used random access memory (RAM) with several memory
locations. The RAM fills up with incoming data from time slots in the order received. Slots are then sent out in an
order based on the decisions of a control unit.
Q8-9. Compare and contrast the two major categories of circuit switches. => TSI (time-slot interchange) is the
most popular technology in a time-division switch. It used random access memory (RAM) with several memory
locations. The RAM fills up with incoming data from time slots in the order received. Slot are then sent out in an
order based on the decisions of a control unit.
Q8-10. List four major components of a packet switch and their functions. => A packet switch has four
components: input ports, output ports, the routing processor, and the switching fabric. An input port performs the
physical and data link functions of the packet switch. The output port performs the same functions as the input port,
but in the reverse order. The routing processor performs the function of table lookup in the network layer. The
switching fabric is responsible for moving the packet from the input queue to the output queue
P8-1. A path in a digital circuit-switched network has a data rate of 1 Mbps. The exchange of 1000 bits is required
for the setup and teardown phases. The distance between two parties is 5000 km. Answer the following questions
if the propagation speed is 2 × 108 m:
a. What is the total delay if 1000 bits of data are exchanged during the datatransfer phase?
b. What is the total delay if 100,000 bits of data are exchanged during the data-transfer phase?
c. What is the total delay if 1,000,000 bits of data are exchanged during the data-transfer phase?
d. Find the delay per 1000 bits of data for each of the above cases and compare them. What can you infer?
Giả sử setup phase giao tiếp 2 đường và teardown phase giao tiếp 1 đường. Delay cho setup và teardown phase là:
3
5000000
1000
 3
 0.075  0.003  0.078s  78ms
8
2  10
1000000
Total delay = Delay for setup and teardown + propagation delay + transmission delay
1000
= 104 ms
1000
100000
b. 78 + 25 +
= 203 ms
1000
1000000
c. 78 + 25 +
= 1103 ms
1000
a. 78 + 25 +
d. Delay for per 1000 bits data: In case a, we have 104 ms. In case b we have 203/100 = 2.03 ms. In case c, we
have 1103/1000 = 1.101 ms. The ratio for case c is the smallest because we use one setup and teardown phase to
send more data.
Packet 1: 2
Packet 2: 3
Packet 3: 3
Packet 4: 2
Packet 1: 2, 70
Packet 2: 1, 45
Packet 3: 3, 11
Packet 4: 4, 41
P8-12. We need a three-stage space-division switch with N = 100. We use 10 crossbars at the first and third stages
and 4 crossbars at the middle stage.
a. Draw the configuration diagram.
b. Calculate the total number of crosspoints.
c. Find the possible number of simultaneous connections.
d. Find the possible number of simultaneous connections if we use a single crossbar (100 × 100).
e. Find the blocking factor, the ratio of the number of connections in part c and in part d.
b. Number of crosspoints = 10 (10 × 4) + 4 (10 × 10) + 10 (4 × 10) = 1200
c. Only four simultaneous connections are possible for each crossbar at the first stage. This means that the total
number of simultaneous connections is 40.
d. If we use one crossbar (100 × 100), all input lines can have a connection at the same time, which means 100
simultaneous connections.
e. The blocking factor is 40/100 or 40 percent.
P8-13. Repeat Problem 8-12 if we use 6 crossbars at the middle stage
b. Number of crosspoints = 10 (10 × 6) + 6 (10 × 10) + 10 (6 × 10) = 1800
c. Only six simultaneous connections are possible for each crossbar at the first stage. This means that the total
number of simultaneous connections is 60.
d. If we use one crossbar (100 × 100), all input lines can have a connection at the same time, which means 100
simultaneous connections.
e. The blocking factor is 60/100 or 60 percent.
P8-14. Redesign the configuration of Problem 8-12 using the Clos criteria.
According to Clos, n = (N/2)1/2 = 7.07. We can choose n = 8. The number of crossbars in the first stage can be 13
(to have similar crossbars). Some of the input lines can be left unused. We then have k = 2n - 1 = 15. Figure 8.3
shows the configuration.
We can calculate the total number of crosspoints as
13 (8 × 15) + 15 (13 × 13) + 13 (15 × 8) = 5655
The number of crosspoints is still much less than the case with one crossbar (10,000). We can see that there is no
blocking involved because each 8 input line has 15 intermediate crossbars. The total number of crosspoints here is
a little greater than the minimum number of crosspoints according to Clos using the formula 4N[(2N)1/2 - 1],
which is 5257.
P8-15. We need to have a space-division switch with 1000 inputs and outputs. What is the total number of
crosspoints in each of the following cases?
a. Using a single crossbar.
b. Using a multi-stage switch based on the Clos criteria.
a. Total crosspoints = N2 = 10002 = 1,000,000
b. Total crosspoints ≥ 4Ν[(2Ν)1/2 -1] ≥ 174,886. With less than 200,000 crosspoints we can design a three-stage
switch. We can use n = (N/2)1/2 =23 and choose k = 45. The total number of crosspoints is 178,200.
P8-16 We need a three-stage time-space-time switch with N = 100. We use 10 TSIs
at the first and third stages and 4 crossbars at the middle stage.
a. Draw the configuration diagram.
b. Calculate the total number of crosspoints.
c. Calculate the total number of memory locations we need for the TSIs.
BT. A circuit-switching network with space-division switches that has 5000 inputs and 5000 outputs. Find:
a. The possible number of simulataneous connections if we use 1 single crossbar?
b. Draw the configuration diagram for using a multi-stage switch based on the Clos criteria. Find the total number
of crosspoints for this case.
=> Trong vở
CHAPTER 18
1. There are no sequence numbers in Frame Relay. Why? => Frame Relay does not use flow or error control,
which means it does not use the sliding window protocol. Therefore, there is no need for sequence numbers.
2. Can two devices connected to the same Frame Relay network use the same DLCIs? => DLCIs are unique only
for a particular interface. A switch assigns a DCLI to each virtual connection in an interface. This way two
different connections belonging to two different interfaces may have the same DLCI.
3. Why is Frame Relay a better solution for connecting LANs than T-1 lines? => T-lines provide point-to-point
connections, not many-to-many. In order to connect several LANs together using T-lines, we need a mesh with
many lines. Using Frame Relay we need only one line for each LAN to get connected to the Frame Relay network
4. Compare an SVC with a PVC. => In a PVC, two end systems are connected permanently through a virtual
connection. In a SVC, a virtual circuit needs to be established each time an end system wants to be connected
with another end system.
5. Discuss the Frame Relay physical layer. => Frame Relay does not define a specific protocol for the physical
layer. Any protocol recognized by ANSI is acceptable.
6. Why is multiplexing more efficient if all the data units are the same size? => If data packets are different sizes
there might be variable delays in delivery
7. How does an NNI differ from a UNI? => A UNI (user network interface) connects a user access device to a
switch inside the ATM network, while an NNI (network to network interface) connects two switches or two ATM
networks.
8. What is the relationship between TPs, VPs, and VCs? =>A TP (transmission path) is the physical connection
between a user and a switch or between two switches. It is divided into several VPs (virtual paths), which provide
a connection or a set of connections between two switches. VPs in turn consist of several VCs (virtual circuits)
that logically connect two points together.
9. How is an ATM virtual connection identified? =>An ATM virtual connection is defined by two numbers: a
virtual path identifier (VPI) and a virtual circuit identifier (VCI).
10. Name the ATM layers and their functions. => The Application Adaptation Layer (AAL) allows existing
networks to connect to ATM facilities by mapping packet data into fixed-sized ATM cells. The ATM layer
provides routing, traffic management, switching, and multiplexing services.
11. How many virtual connections can be defined in a UNI? How many virtual connections can be defined in an
NNI? => In an UNI, the total length of VPI+VCI is 24 bits. This means that we can define 224 virtual circuits in
an UNI. In an NNI, the total length of VPI+VCI is 28 bits. This means that we can define 228 virtual circuits in
an NNI
12. Briefly describe the issues involved in using ATM technology in LANs.
We can briefly summarize the most important issues:
a. Traditional LANs are connectionless protocols; ATM is a connection-oriented protocol.
b. Traditional LANs define the route of a packet through source and destination addresses; ATM defines the
route of a cell through virtual connection identifiers.
c. Traditional LANs can do unicast, multicast, and broadcast transmission; ATM is designed only for unicast
transmission.
13. The address
(in decimal)?
field
of
a
Frame
Relay
frame
is
1011000000010111.
What
is
the
DLCI
14. The address field of a Frame Relay frame is 101100000101001. Is this valid? => The address field in Frame
Relay
is
16
bits.
The
address
given
is
only
15
bits.
It
is
not
valid
15. Find the DLCI value if the first 3 bytes received is 7C 74 E1 in hexadecimal.
16. Find the value of the 2-byte address field in hexadecimal if the DLCI is 178. Assume no congestion.
17. In Figure 18.30 a virtual connection is established between A and B. Show the DLCI for each link.
18. In Figure 18.31 a virtual connection is established between A and B. Show the corresponding entries in the
tables of each switch.
19. An AAL1 layer receives data at 2 Mbps. How many cells are created per second by the ATM layer? => In
AAL1, each cell carries only 47 bytes of user data. This means the number of cells sent per second can be
calculated as [(2,000,000/8)/47] ≈ 5319.15
20. What is the total efficiency of ATM using AAL1 (the ratio of received bits to sent bits)? => In AAL1, each
53-byte cell carries only 47 bytes of user data. There are 6 bytes of overhead. The efficiency can be calculated as
47/ 53 ≈ 89%.
21. If an application uses AAL3/4 and there are 47,787 bytes of data coming into the CS, how many padding
bytes are necessary? How many data units get passed from the SAR to the ATM layer? How many cells are
produced?
a. In AAL3/4, the CS layer needs to pass 44-byte data units to SAR layer. This means that the total length of the
packet in the CS layer should be a multiple of 44. We can find the smallest value for padding as follows:
H + Data + Padding + T = 0 mod 44
4 + 47,787 + Padding + 4 = 0 mod 44
Padding = 33 bytes
b. The number of data unit in the SAR layer is (4 + 47787 + 33 +4) / 44 = 1087
c. In AAL3/4, the number of cells in the ATM layer is the same as the number of data unit in the SAR layer. This
means we have 1087 cells.
22. Assuming no padding, does the efficiency of ATM using AAL3/4 depend on the size of the packet? Explain
your answer. => If we assume that there is no need for padding, the efficiency of the AAL3/4 depends on the size
of the packet because of the 8 bytes of overhead in the CS layer. A larger packet is more efficient than a smaller
packet. A packet of size 8 bytes has an efficiency of 8/16 = 50% while a packet of size 1000 bytes has an
efficiency of 1000/1008 ≈ 99%.
23. What is the minimum number of cells resulting from an input packet in the AAL3/4 layer? What is the
maximum number of cells resulting from an input packet?
a. The minimum number of cells is 1. This happens when the data size ≤ 36
bytes. Padding is added to make it exactly 36 bytes. Then 8 bytes of header creates a data unit of 44 bytes at the
SAR layer.
b. The maximum number of cells can be determined from the maximum number of data units at the CS sublayer.
If we assume no padding, the maximum size of the packet is 65535 + 8 = 65543. This needs 65543 / 44 ≈
1489.61. The maximum number of cells is 1490. This happens when the data size is between 65,509 and 65,535
(inclusive) bytes. We need to add between 17 to 43 (inclusive) bytes of padding to make the size 65552 bytes.
The 8 byte overhead at the CS layer makes the total size 65560 which means 1490 data units of size 44.
24. What is the minimum number of cells resulting from an input packet in the AAL5 layer? What is the
maximum number of cells resulting from an input packet?=> a. The minimum number of cells is 1. This happens
when the data size ≤ 40 bytes. Padding is added to make it exactly 40 bytes. Then 8 bytes of header creates a data
unit of 48 bytes at the SAR layer. => b. The maximum number of cells is 1366. It can be determined from the
maximum number of data units at the CS sublayer. If we assume no padding, the maximum size of the packet is
65535 + 8 = 65543. This needs 65543 / 48 ≈ 1365.48 or 1366 cells. This happens when the data size is between
65,513 and 65,535 (inclusive) bytes. We need to add between 25 to 47 (inclusive) bytes of padding to make the
size 65560 bytes. The 8 byte overhead at the CS layer makes the total size 65568 which means 1366 data unit of
size 48.
25. Explain why padding is unnecessary in AAL1, but necessary in other AALs. => AAL1 takes a continuous
stream of bits from the user without any boundaries. There are always bits to fill the data unit; there is no need for
padding. The other AALs take a bounded packet from the upper layer.
26. Using AAL3/4, show the situation where we need _______ of padding.
a. 0 bytes (no padding) => When user (4 + user data + 4) mod 44 = 0
b. 40 bytes => When user (4 + user data + 40 + 4) mod 44 = 0.
c. 43 bytes =>When user (4 + user data + 43 + 4) mod 44 = 0
27. Using AAL5, show the situation where we need _______ of padding.
a. 0 bytes (no padding) => When user (user data + 8) mod 48 = 0
b. 40 bytes => When user (user data + 40 + 8) mod 48 = 0
c. 47 bytes => When user (user data + 43 + 8) mod 48 = 0.
28. In a 53-byte cell, how many bytes belong to the user in the following (assume no
padding)?
a. AAL1 → 53 - 5 - 1 = 47
b. AAL2 → 53 - 5 - 1 - (CS header) < 47
c. AAL3/4 (not the first or last cell)→ 53 - 5 - 4 - (CS header) < 44
d. AAL5 (not the first or last cell) → 53 - 5 - (CS header) < 48
Vẽ hình crossbar, chuẩn Clos (1,5đ) 5 ý
Minimunhop, least cost: 1đ
DLCI: 1đ
SAR (1,5đ – 2 ý)
Tóm tắt lại lý thuyết
Find the DLCI value if the first 3 bytes received is 7874C1 in hexadecimal.
The address field of a Frame Relay is 1011110000010111. What is the DLCI (in decimal)?
Why does the efficiency of ATM using AAL3/4 depend on the size of the packet if a assuming no padding.
If an