350-001 Cisco Certified Internetworking Expert Brain Dump

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350-001 Cisco Certified Internetworking Expert

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Demo Question 5.

The EliteCertify WAN consists of a frame relay network using ANSI LMI. What is the maximum theoretical number of DLCI's that can be advertised on a Frame-Relay interface with an MTU of 1500 bytes when using ANSI LMI?

A. 1024
B. 1023
C. 992
D. 297
E. 186
F. 796

Display Answer

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Answer: D

Explanation: The formula for finding the maximum number of DLCI's for ANSI is (1500-13)/5=max DLCIs = 297.4. See below for the specifics for how this formula is generated: Analysis In a PVC information packet, the Report Type (RT) portion is one byte long and the KeepAlive (KA. portion is two bytes long. For the ANSI and Q933a LMIs, the PVC information is 3 bytes long, whereas for the Cisco LMI it is 6 bytes long due to the additional "bw" (for BandWidth) value. The "bw" value represents the Committed Information Rate (CIR); the actual bw value will only be seen if the frame relay switch is configured to forward this information. The static overhead in each case is 13 bytes [Entire LMI packet minus IEs (10 bytes) + RT (1 byte) + KA (2 bytes)]. We can subtract this number from the Maximum Transmission Unit (MTU) to get the total available bytes for DLCI information. We then divide that number by the length of the PVC IE (5 bytes for ANSI and Q933a, 8 bytes for Cisco) to get the maximum theoretical number of DLCIs for the interface: For ANSI or Q933a, the formula is: (MTU - 13) / 5= max DLCIs. For Cisco, the formula is (MTU - 13) / 8= max DLCIs.