EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
Problem 1 Consider an ARQ stop-and-wait scheme between stations A and B. (a) If station A intends to transmit a single data frame, what is the average number of retransmissions required for the successful transport of data frames? (Hint: Let p denote the probability that the transmit frame is received with error. Express the average number of retransmissions in terms of p.) (b) If a station intends to transmit M data-frames, where M ≥ 2, what is the probability that the number of retransmissions NR (M ) is equal to 2? (Hint: Express the probability in terms of M and p.) Ans: (a) Let NR (1) denote the number of retransmission. Then, NR (1) follows the Geometrical distribution with parameter p. Thus, we have E[NR (1)] =
∞ X
nP (NR (1) = n) =
∞ X n=0
n=0
n(1 − p)pn =
p . 1−p
(b) Let NRk denote the number of retransmissions for the kth data-frame. Then, P NR (M ) = M k=1 NRk . Note that {NRk | k = 1, 2, . . . , M } is the set of independent, identically distributed random variables. Each random variable follows Geometrical distribution with parameter p. For NR (M ) is equal to 2, there are two possibilities: (i) One of these random variables is equal to 2 and the others are equal to 0 and (ii) Any two of these random variables are each equal to 1, and the others are equal to 0. Therefore, we have P (NR (M ) = 2) =
M X
P (NR1 = 0, NR2 = 0, NRi−1 = 0, NRi−1 = 1, NRi−1 = 0, . . . , NRM = 0)
k=1
+ P (NR1 = 1, NR2 = 1, . . . , NRi−1 = 0, . . . , NRM = 0) + . . . + P (NR1 = 0, NR2 = 0, . . . , NRi−1 = 1, . . . , NRM = 1) !
= + = =
M P (NR1 = 1, NR2 = 0, . . . , NRM = 0) 1 ! M P (NR1 = 1, NR2 = 1, NR3 = 0, . . . , NRM = 0). 2 ! ! i M M h 2 M −1 (1 − p)p (1 − p) + [(1 − p)p]2 (1 − p)M −2 1 2 M (M + 1) 2 p (1 − p)M . 2
1
EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
Problem 2 Consider an ARQ stop-and-wait scheme which is similar to the scheme described in Problem 1 but a distinct feature: suppose that an acknowledgement can be lost during its transmissions with probability q. If an acknowledgement is lost, station A will retransmit the data-frame after time-out. Such retransmission process continues until station A receives an acknowledgement. (a) If station A intends to transmit a single data frame, what is the average number of retransmissions required for the successful delivery of data frames? (Hint: Express the average number of retransmissions in terms of p and q.) (b) Find the probability that station A retransmits exactly twice. (Hint: express the probability in terms of p and q. You do not need to simplify your answer.) Ans: (a) Note that when station A transmits a data frame, it may not receive an acknowledgement if the frame is delivered to station B with errors or if the acknowledgement for the frame which is received without errors is lost. Let r denote the probability that station A does not receive an acknowledgement. Let A and B denote the events that a frame is received with errors and that an acknowledgement is lost, respectively. Then, we have r = P (A ∪ (Ac ∩ B)) = P (A) + P (Ac ∩ B) = p + (1 − p)q. Whenever an acknowledgement is not received, the data frame is retransmitted. Therefore, the number of retransmissions for a data-frame NR (1) has a Geometric distribution with parameter r. The average number of retransmissions is given by: E[NR (1)] =
p + (1 − p)q r = . 1−r (1 − p)(1 − q)
(b) The probability that station A retransmits exactly twice is given by P (NR (1) = 2) = r2 (1 − r) = [p + (1 − p)q]2 [(1 − p)(1 − q)] .
2
EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
Problem 3 Let T and S denote two independent exponentially distributed random variables with parameter λT and λS , respectively. (a) Show that the random variable T is memoryless (i.e., P (T ≤ t | T > s) = P (T ≤ t − s), ∀t ≥ s ≥ 0). (b) Set Y = min{T, S}. Show that the random variable Y is exponentially distributed, and find its parameter. (c) Let X denotes a Poisson random variable with parameter T , where T is an exponentially distributed random variable with parameter λT . Find the distribution for X (i.e., P (X = n), ∀n = 0, 1, . . . ). Ans: (a) P (T ≤ t | T > s) = P (T ≤ t − s) is given by P (s < T ≤ t) 1 − e−λT t − (1 − e−λT s ) = P (T > s) e−λT s e−λT s − e−λT t = = 1 − e−λT (t−s) = P (T ≤ t − s). e−λT s
P (T ≤ t | T > s) =
(b) Since P (min{T, S} > t) = P (T > t, S > t) = e−(λT +λS )t , which is also an exponential distribution with parameter λT + λS . (c) The distribution for X (i.e., P (X = n), ∀n = 0, 1, . . . ) is given by Z ∞ −t n e−T T n e t P (X = n) = E[P (X = n | T )] = E = λT e−λT t dt n! n! 0 Z ∞ (1 + λT )n+1 n −(1+λT )t λT = t e dt. (1 + λT )n+1 0 n! "
#
n+1
Since (1+λn!T ) tn e−(1+λT )t is the probability density function for a Gamma distribution with parameter (n + 1) and (1 + λT ), thus, we have (1 + λT )n+1 n −(1+λT )t t e dt = 1. n! 0 As a result, we conclude that Z ∞
λT 1 P (X = n) = = 1− n+1 (1 + λT ) 1 + λT
which is also a Geometrical distribution with parameter 3
1 1 + λT 1 1+λT
.
n
,
EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
Problem 4 Figure 1 illustrates the first two messages exchanged in the course of a telephone call from user A to user B. extend this diagram to include the following messages (with first two already illustrated: A Off Hook; Dial Tone to A; Dialed digits from A; Called Request signals; Call Accept signal; Conversation with B answering, A responding, and one more response from B; B On Hook, Disconnect signals (originating from location of first user to hang up); and finally A on hook. (This is an arbitrary choice of the order of actions during disconnect; for example, either user could equally well hang up first.) Local Loop User A
Telephone network
Network interface
Local loop
Network interface
User B
Off-hook signal
Dial tone
Dialed digits
Figure 1: The first two messages exchanged in the course of a telephone call from user A to user B
Ans:
4
EE132B-HW Set #4
UCLA 2014 Fall
User A
Network interface
Prof. Izhak Rubin
Network interface
User B
Off-hook signal
Dial tone
Dialed Digits
Call request
Ring
Off-hook Call respond A hears Off-hook B Answering
A responding
B responding On-hook signal Disconnect signal
A hear on-hook
On-hook signal
Figure 2: The messages exchanged in the course of a telephone call from user A to user B
5
EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
Problem 5 Compare virtual circuit and datagram service with respect to the following: (a) Ability to survive link or node failure. (b) Ability to deal with congestion in the network. (c) Applicability for use with the following types of traffic: (i) voice, (ii) interactive data traffic, and (iii) file transfer. Ans: (a) Since virtual circuits use fixed routes per connection, failure of any link or node along the route results in failure of the established virtual circuit. On the other hand, in a datagram network that employs a dynamic routing scheme, datagram packets that belong to a single flow do not use a fixed route. Datagram packets that incur at a packet switch/router node due to a failed link will typically be redirected along a different link (i.e., using an alternate route to reach the destination, if possible). (b) The alternate routing capability of a connectionless packet switching network that employs dynamic routing and congestion control mechanisms, provides for its capability to route datagram packets around (and away from) areas of congestion. In turn under a virtual packet networking operation, once congestion along the route allocated for a connection is discovered, if unacceptable, the virtual circuit connection must be completely re-configured so that a new route is discovered and configured. On the other hand, the occurrence of congestion across a virtual circuit can be limited by using, at connection setup time, effective flow control mechanisms for congestion control purposes. In this manner, the rate at which traffic flows are permitted entry to the configured virtual circuits can be controlled. This type of flow control is more difficult with datagram transmission since no direct route from a congested node back to the source of congestion is established. Thus, both types of service reveal advantages relative to each other in their capability for dealing with congestion. (c) (i) Neither virtual circuits nor datagram are ideal for voice communications. Virtual circuits have the advantage of ensuring sequenced delivery, but long delays that are occasionally encountered lead to quality degradation. Under a datagram networking service, packet delays and jitters are often less regulated and thus may be higher and subsequently lead to packet losses and quality degradations. Also, in a datagram network, packets may arrive out of order. A protocol (such as that know as Real Time Protocol) is therefore often used above the transport layer to provide ordering 6
EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
(sequencing of voice packets) as well as to attach time stamps to voice packets so that the reconstruction of the voice message at the receiving terminal can be performed in face of packet losses and time delay variations. Hence, virtual circuits offer fewer disadvantages. (ii) The choice between virtual circuit and datagram based operations for interactive data traffic should depend on the nature of the traffic. If interactions are infrequent and of short duration, the overhead incurred for connection and disconnection phases of virtual circuits mat not be justified, so datagram might be preferred. If there are frequent transmissions back and forth over extended periods of time, virtual circuits would be preferred; they would then also guarantee sequenced delivery. (iii) For file transfers, virtual circuit networking can offer more advantages than datagram networking. This is particularly the case, when quality of service (QoS) must be guaranteed for the file transport, i.e. in terms of delivery time and reliability. For long sessions, virtual circuit operation leads to reduction of overhead associated with a virtual circuit operation after a connection has been established. Otherwise, a datagram operation is acceptable.
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EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
Problem 6 (a) Compute the total delays to transfer a message 10000 (eight bit) characters long across the three hop communications path under the following timing assumptions. Assume that all links operate at 4800 bps, with a 20 msec delay per link for propagation delays plus reaction time at the receiver. (Timings given are typical, though they are so highly variable that results of this problem should be treated with caution.) (i) For circuit switching assume 3 sec per link connection delay (time hunting for an outgoing trunk, or dialing and associated delays) and 100 msec durations of call request or accept signal (times to put signals on the communications lines, for example, number of bits divided by line rate in bps). (ii) For message switching, assume 50 bytes per message of header (routing information, identification, sequencing information, and so forth) and (arbitrarily) 300 msec processing plus queuing time at each node. (iii) for packet switching, assume the message is divided into 20 equal length packets with 50 bytes of header information per packet, 300 msec processing plus queuing time at each node and 20 msec between packets. (b) Discussing your results and indicate reasonable conditions under which the relative rankings of the delays might change. Ans:
8
EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
(a) (i) B
A tc tR
D
C
Dialing and associated delays Requ est li nk
Time
tp to Requ est li n
tp k
to Requ est li nk
Conn
gnal ect si
tp
3tp
3tp
td Data tr ansmis sion
td
Figure 3: Circuit switching From Figure.3, we can obtain the total transmission time, which is given by as follows. • Dialing and associated delays = 3 sec • Times hunting for outgoing tricks = 2 × 3 = 6 sec • Durations for call requests and accepts = 4 × 0.1 = 0.4 sec • Propagation delays = 9 × 0.02 = 0.18 sec • Transmission time = 8 ×
10000 4800
= 16.67 sec
• Total time = 26.25 sec
9
EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
(ii) C
B
A
D Time
tQ tp
head
er Data
tQ tp
head er Data
tQ tp
head er Data
Figure 4: Message switching From Figure.4, we can obtain the total transmission time, which is given by as follows. • Queuing delays = 3 × 0.3 = 0.9 sec • Propagation delays = 3 × 0.02 = 0.06 sec • Transmission times = 3 × 8 ×
10050 4800
= 50.25 sec
• Total time = 51.21 sec (iii) C
B
A
D Time
tQ Head er 1 Data 1
tp
tI tQ Head er 2 Data 2
Head er 1 Data 1
tp
Head er 2 Data 2
tQ Head er 1 Data 1
tp
Head er 2 Data 2
tI
Figure 5: Packet switching with an example of 2 packets 10
EE132B-HW Set #4
UCLA 2014 Fall
Prof. Izhak Rubin
From Figure.5, we can obtain the total transmission time, which is given by as follows. The delay for this case can be computed by computing the time for the first packet to reach C then adding the time for the entire message to go from C to D. Since each packet, including header, contains 550 characters, the former time consists of: • Queuing delays = 2 × 0.3 = 0.6 sec • Propagation delays = 2 × 0.02 = 0.04 sec • Transmission times = 2 × 8 ×
550 4800
= 1.83 sec
• Total = 2.47 sec The latter time then consists of: • Queuing delays = 0.3 sec • Propagation delays = 0.02 sec • Transmission times = 20 × 8 ×
550 4800
= 18.33 sec
• Inter-packet times = 19 × 0.02 = 0.38 sec • Total = 19.03 sec Adding this to the previous value gives a total time of 21.50 sec. (b) For the parameter given, the delay incurred in message switching is far inferior to that for either of the other two approaches. This is primarily due to the extra transmission times incurred, which a separate transmissions of the complete message over each link after storage at previous node. Packet switching has the shortest delay, roughly 5 seconds less than that for circuit switching. the main source of additional delay for circuit switching is the switching time of approximately 9 seconds. This is particularly compensated for by extra transmission delays in packet switching plus almost 1 second of queuing delays. Faster circuit switching, on the order of a second or so instead of 9 seconds total switching time, would change the relative rankings so that circuit switching would yield minimum delay. Finding situations when message switching would yield minimum delay is more difficult since the main situation where this is true when there is only single link in the source-destination path and that is not situation here.
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