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Open Advanced Search. Each node has one or several client interfaces. Labels at SDU levels identify client flows, whereas labels at PDU level are used to identify the flows between source and destination nodes. Control PDUs carry both general information e.
Incoming PDUs marked with an unknown label are passed transparently. For example, a source node shall insert PDUs on the ring; the desti- nation node of a unicast flow shall receive the PDU and erase it in order to facilitate spatial reuse; on the other hand, a destination node of a multicast flow shall only receive the PDU while letting it pass for downstream destinations; a multicast extract node shall erase the PDU and possibly not receive it, if it is e.
The cost of QoS in Transport Networks A transport network relies on physical resources that have been provisioned, based on conservative traffic predictions, as making new resources available involves lengthy processes such as deploying new optical fibres.
In the past, trans- port networks relied on sets of permanent, fixed rate circuits established between each source, destination pair. A MAN build with POADM rings still relies on optical fibres, but replaces fixed rate circuits by fixed rate virtual circuits, which potentially allows optimizing bandwidth usage. However, the need to provision resources based on predicted peak traffic rates should still be taken into account. As long as client layers conform with T conformance can be enforced at each node by a simple access control procedure , the performance delivered by POADM rings can be shown to comply to QoS objectives set for Carrier Ethernet in a MAN [7].
Indeed, latency is mostly due to geographical distance, as POADM nodes are optically trans- parent for transit data PDUs; one can safely assume that the distance between nodes in a MAN is less than km, which yields a latency less than 10ms. Jitter, on the other hand, is due to the insertion process. Even if a flow has to cross 3 POADM rings, and therefore be inserted 3 times, the end-to-end jitter is still less than 2ms. Bidirectional rings are necessary; a protected flow is allocated a single label on both directions of the ring.
Failure detection is very rapid, as it is detected when a control packet is not received; failure notification is directly sent by the two nodes adjacent to the failure, and all nodes are made aware of the failure in a few ms depending on the ring size. Each node stores in a local Protection Information Table PIT the set of flows that are affected by each potential failure.
When a failure notification is received, each node can then apply the appropriate protection mode for each impacted flow. With these procedures, the unavailability duration for a given flow due to a fibre failure is less than the round trip time between source and destination nodes, i.
When dimensioning a POADM ring with protected labelled flow, one should derive from T and the protection mode for each labelled flow each matrix Tl corresponding to the traffic matrix in case of failure of link l. Jitter in nodes X and Y [6]. J L'". Odrediti polu- 7 '" Dat je red J!. Dat je red xn aER. Specijalno: a ako je J x parna, tada je a x Dokazati da konvergiraju integrali:. J cos X2 dx. Da II funkCIja fex.. Fourierovi redovi I x S l X Proizvod O je apsolutno ili uslovno ne apsolutno konvergentan zavisno od toga da li red 2 konvergira apsolutno ili IT l-an.
Posl ' da e pokazah Dokazati primerom. Ii' dokazati da je I x, y neprekidna. Parcijalni izvodi i diferencijali. Prvi parcijalni izvod po promenIjivoj ,;, e limes oj. OX2 ox oy Oy2. Ako su x i y dovoJno mal'l, d O k azatl. Izraziti d 3 u preko parcijalnih izvoda funkcije I x, y i dx i dy.
P dx, dy, dz. F, ;,. Funkcionalne determinante. Diferenciranje implicitnih funkcija. Ako su funkcije Ui, iz 3 , dlferencljabllne tada se diferenCIJal! Dokazati da sistem Da Ii sistem Izraziti ovaj izraz II polarnim koor- Ox oy x or2 az az Taylorova formula. Ekstremumi funkcija. X' oy;. Neka je f x neprekidna na segmentu [a, bl.
O Primenom formule za diferenciranje pod znakom integrala odrediti Fou- gde je rierove koeficijente al i b! JJ l dx l y2-x2. Neka su Fek i E k potpuni integrali v. Nepravi integral kao funkcija parametra. Uniformna konvergencija integrala 10 Nepravi integral Ovaj kriterijum je poznat kao Cauchyev kriterijum.
Pokazati da se uniformno konvergentni integral Odrediti oblast konvergencije integrala: 1 1 2 I dx x" if sin2 x I oo sin :;X2 dx. Dokazati da je integral O 1 Zameoa promenljivih u nepravim integralima.
J l-e-XY dx. J arc tg xy dx. J oo sin aXxsin px dx. Jo x 2n e- x' cos 2 bx dx, nEN. Jrt' cos mt dt, mER. Jcos X2 cos 2 ax dx. Dat je integral ' Dokazati formule: l e Ispitati konvergenciju integrala I m. Neka je funkcija J x apsolutno integrabilna na intervalu - oo, oo J. Dokazati da integral Eulerovi integrali J Vl-x 4 O J VI-x 3 Jsin 6 x COS'; X dx. J oo sin2 X2 X dx. Jx 2n e- x1 dx, nEN. J Vx X2 dx. J xalnx dx. J xme-xndx.
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