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(a) Use the uniformization method to compute E( ), ( ) and P { > t} for t = 2, 5 and 10 when = 1, = 10 and the number of standby units is varied as R = 1, 2 and 3 (b) Extend the analysis in (a) for the case that the repair time has a Coxian-2 distribution and investigate how sensitive the results in (a) are to the second moment of the repair-time distribution 418 Messages arrive at a node in a communication network according to a Poisson process with rate Each arriving message is temporarily stored in an in nite-capacity buffer until it can be transmitted The messages have to be routed over one of two communication lines each with a different transmission time The transmission time over the communication line is i exponentially distributed with mean 1/ i (i = 1, 2), where 1/ 1 < 1/ 2 and 1 + 2 > The faster communication line is always available for service, but the slower line will be used only when the number of messages in the buffer exceeds some critical level Each line is only able to handle one message at a time and provides non-pre-emptive service With the goal of minimizing the average sojourn time (including transmission time) of a message in the system, the following control rule with switching level L is used The slower line is turned on for transmitting a message when the number of messages in the system exceeds the level L and is turned off again when it completes a transmission and the number of messages left behind is at or below L Show how to calculate the average sojourn time of a message in the system This problem is taken from Lin and Kumar (1984) 419 Two communication lines in a packet switching network share a nite storage space for incoming messages Messages of the types 1 and 2 arrive at the storage area according to two independent Poisson processes with respective rates 1 and 2 A message of type j is destined for communication line j and its transmission time is exponentially distributed with mean 1/ j , j = 1, 2 A communication line is only able to transmit one message at a time The storage space consists of M buffer places Each message requires exactly one buffer place and occupies the buffer place until its transmission time has been completed A number Nj of buffer places are reserved for messages of type j and a number N0 of buffer places are to be used by messages of both types, where N0 + N1 + N2 = M That is, an arriving message of type j is accepted only when the buffer is not full and less than N0 + N1 other messages of the same type j are present; otherwise, the message is rejected Discuss how to calculate the optimal values of N0 , N1 and N2 when the goal is to minimize the total rejection rate of both types of message Write a computer program and solve for the numerical data M = 15, 1 = 2 = 1 and 1 = 2 = 1 This problem is based on Kamoun and Kleinrock (1980) 420 A traf c source is alternately on and off, where the on- and off-times are exponentially distributed with respective means 1/ and 1/ During on-periods the traf c source generates messages for a transmission channel according to a Poisson process with rate The transmission channel can handle only one message at a time and the transmission time of a message has an exponential distribution with mean 1/ The on-times, off-times and transmission times are independent of each other Further, it is assumed that /[ ( + )] < 1 Let the states (i, 0) and (i, 1) correspond to the situation that there are i messages at the transmission channel and the traf c source is off r on respectively (a) Verify for the numerical values = 1, = 1, = 2, = 05 that the system of linear equations (446) is given by 1 3z 2z 05z z2 25z + 1 G0 (z) G1 (z) = (1 z)p00 (1 z)p01 .



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In this section we discuss a generally applicable up- and downcrossing technique that, in onjunction with the PASTA property, can be used to establish relations between customer-average and time-average probabilities in queueing systems To illustrate this, we consider the so-called GI /M/1 queue In this single-server system, customers arrive according to a renewal process and the service times of the customers have a common exponential distribution The single server can handle only one customer at a time and there is ample waiting room for customers who nd the server busy upon arrival The service times of the customers are independent of each other and are also independent of the arrival process Denoting by the average arrival rate (1/ = the mean interarrival time) and by the service rate (1/ = the mean service time), it is assumed that < The continuous-time stochastic process {X(t), t 0} and the discrete-time stochastic process {Xn , n = 1, 2, } are de ned by X(t) = the number of customers present at time t, and Xn = the number of customers present just prior to the nth arrival epoch The stochastic processes {X(t)} and {Xn } are both regenerative The regeneration epochs are the epochs at which an arriving customer nds the system empty It is stated without proof that the assumption of / < 1 implies that the processes have a nite mean cycle length Thus we can de ne the time-average and the customer-average probabilities pj and j by pj = the long-run fraction of time that j customers are present and j = the long-run fraction of customers who nd j other customers present upon arrival for j = 0, 1, Time averages are averages over time, and customer averages t are averages over customers To be precise, pj = limt (1/t) 0 Ij (u) du and j = limn (1/n) n Ik (j ), where Ij (t) = 1 if j customers are present at k=1 time t and Ij (t) = 0 otherwise, and In (j ) = 1 if j other customers are present just before the nth arrival epoch and In (j ) = 0 otherwise The probabilities pj and j are related to each other by j 1 = pj , j = 1, 2, (271).





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ISAKMP payload N PDF417 generation on net generate, create 128 image on microsoft word using barcode implement for The protocols are identified by protocol identifiers and may include IPsec ESP, IPsec AH, and TLS depending on the type of protection eeded for the data communications (authentication, encryption, or both) In order to facilitate the negotiation, each party may include multiple proposal payloads along with each exchange message The proposal payload includes a proposal number that is used in a clever manner: when a party needs to use a bundle (multiple) of security protocols to protect the data, for instance both ESP and AH, then that party uses the same proposal number within the proposal payloads for each of those protocols In other words, using the same proposal number in multiple payloads has a logical-AND meaning.

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other customers present The process {X(t)} makes a downcrossing from state j to state j 1 if the service of a customer is completed and j 1 other customers are left behind Observation 1 Since customers arrive singly and are served singly, the long-run average number of upcrossings from j 1 to j per time unit equals the long-run average number of downcrossings from j to j 1 per time unit This follows by noting that in any nite time interval the number of upcrossings from j 1 to j and the number of downcrossings from j to j 1 can differ at most by 1 Observation 2 The long-run fraction of customers seeing j 1 other customers upon arrival is equal to the long-run average number of upcrossings from j 1 to j per time unit the long-run average number of arrivals per time unit for j = 1, 2, In other words, the long-run average number of upcrossings from j 1 to j per time unit equals j 1 The latter relation for xed j is in fact a special case of the Little relation (241) by assuming that each customer nding j 1 other customers present upon arrival pays $1 (using this reward structure observation 2 can also be obtained directly from the renewal-reward theorem) Observations 1 and 2 do not use the assumption of exponential services and apply in fact to any regenerative queueing process in which customers arrive singly and are served singly Observation 3 For exponential services, the long-run average number of downcrossings from j to j 1 per time unit equals pj with probability 1 for each j 1 The proof of this result relies heavily on the PASTA property To make this clear x j and note that service completions occur according to a Poisson process with rate as long as the server is busy Equivalently, we can assume that an exogenous Poisson process generates events at a rate of , where a Poisson event results in a service completion only when there are j customers present Thus, by part (a) of Theorem 241, E[Ij (t)] = E[Dj (t)] for t > 0 (272).

The message to the other peer is I need to use ESP AND AH to protect the data On the other hand, using different proposal numbers for each proposal has a logical-OR function and means I need to use either ESP or TLS The proposal payload also includes a number of transform field and a sending entity SPI , whose size and type depends on the protocol that is being offered Transform payload: Each proposal payload is a proposal to use a specific security protocol such as IPsec ESP As we know, each security protocol can use a number of transforms (security algorithms) The transform payload is used within the security association negotiation exchange to propose a transform such as DES or SHA1 The transform payload includes a transform number field.

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Figure 45 Available Hatch Brush Styles Shown with Black Foreground and White Background in Visual Studio .NET Make QR Code ISO EC18004 in Visual Studio .NET Figure 45 Available Hatch Brush Styles Shown with Black Foreground and White Background. QR-Code In VS .NET Using Barcode reader for .A LinearGradientBrush is used to draw a smooth blending between two end points and between two colors The gradations are drawn at a specified angle, as defined either by passing a float or by passing one of four LinearGradientMode values: enum LinearGradientMode { Horizontal, // 0 degrees Vertical, // 90 degrees ForwardDiagonal, // 45 degrees BackwardDiagonal, // 135 degrees } The angle is used to set up a blend , which governs the transition between colors over the area of the brush along the angle of the line You can set this blend either directly or indirectly In the direct technique, you use Blend property, which determines positions and factors of fallout between the two colors To set the blend indirectly, you use a focus point for the end color and a fall-out rate toward the start color, as shown in Figure 46.Related: 

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The responder chooses a transform within those proposed in conjunction with each proposal number to indicate its choice of security algorithm Key exchange (KE) payload: As mentioned in 3, to avoid overuse of the limited bandwidth resources, IKE facilitates the negotiation of Diffie Hellman (or Oakley) parameters between the two parties, without the need for transfer of the complete parameter data IKE accomplishes this by using the ISAKMP KE payload defined in [ISAKMP2408] This way only the group numbers, rather than the parameters that define the group, need to be sent It should be noted that the actual group parameters are however defined by IKE specification [IKE2409] rather than by ISAKMP specification Identification payload: This payload is used to provide the identity of the communicating peer We will refer this payload again when describing the use of certificates for IKE in 9Related: Codabar Generating NET , .

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