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same exponential distribution as the time elapsed between epoch 0 and the epoch of the rst arrival after epoch 0 Next mimic the proof of Theorem 111 To illustrate the foregoing, we give the following example Example 111 A taxi problem Group taxis are waiting for passengers at the central railway station Passengers for those taxis arrive according to a Poisson process with an average of 20 passengers per hour A taxi departs as soon as four passengers have been collected or ten minutes have expired since the rst passenger got in the taxi (a) Suppose you get in the taxi as rst passenger What is the probability that you have to wait ten minutes until the departure of the taxi (b) Suppose you got in the taxi as rst passenger and you have already been waiting for ve minutes In the meantime two other passengers got in the taxi What is the probability that you will have to wait another ve minutes until the taxi departs To answer these questions, we take the minute as time unit so that the arrival rate = 1/3 By Theorem 111 the answer to question (a) is given by P {less than 3 passengers arrive in (0, 10)}

e 10/3

.

(10/3)k = 03528 k!

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over the PDP context runs through a gprsSSF, which is a logical entity that resides in the SGSN and serves as a relay between the PDP context process in the SGSN and the gsmSCF The gprsSSF may in many aspects be compared with the gsmSSF that resides in the MSC or GMSC Protocol rules for the gprsSSF are further explained in Section 5334 The interworking between the SGSN and the gprsSSF is re ected in 3GPP TS 23060 [81] The GPRS procedures are de ned in 3GPP TS 23060 [81] by means of sequence diagrams The sequence diagrams for the various procedures contain CAMEL interaction boxes These interaction boxes may be compared with the DPs from the BCSMs form CS call control One example of such sequence diagram is the PDP context activation procedure for UMTS; see Figure 537 The rectangular boxes 1 and 2 in Figure 537 constitute the DPs PDP context establishment and DP PDP context establishment acknowledgement, respectively It is at these points that the corresponding CAMEL procedure may be invoked These CAMEL procedures are speci ed in 3GPP TS 23078 [83] The CAMEL service invocation request (ie the IDP-GPRS information ow) contains various elements that may be used by the CAMEL service for its internal processing Unlike CS call control, not all information elements that are needed for PDP context handling are available Some of the required elements are reported only at PDP context establishment acknowledgement If the CAMEL service allows the PDP context establishment to continue, then it sends a CAP CUE-GPRS or CAP CON-GPRS operation to the gprsSSF: Continue GPRS this operation indicates to the SGSN that PDP context establishment will continue with the information that was available in the SGSN just prior to CAMEL service invocation Connect GPRS this operation indicates to the SGSN that PDP context establishment will continue, using the APN that is included in CAP CON-GPRS If an APN was already provided by the UE, then the APN that is provided by the CAMEL service will overwrite the UE-provided APN Both the NI part and the OI part of the APN will be overwritten When the CAMEL service has instructed the gprsSSF to continue the PDP context establishment, the SGSN runs the APN and GGSN selection procedure This procedure is speci ed in 3GPP.

The answer to question (b) follows from the memoryless property stated in Theorem 112 and is given by P { 5 > 5} = e 5/3 = 01889 In view of the lack of memory of the Poisson process, it will be intuitively clear that the Poisson process has the following properties: (A) Independent increments: the numbers of arrivals occurring in disjoint intervals of time are independent (B) Stationary increments: the number of arrivals occurring in a given time interval depends only on the length of the interval A formal proof of these properties will not be given here; see Exercise 18 To give the in nitesimal-transition rate representation of the Poisson process, we use 1 e h = h h3 h2 + = h + o(h) 2! 3! as h 0

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