# Max bit rate and guaranteed bit rate in LTE

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Is there any way to calculate the max bit rate and guaranteed bit rate in LTE, any pointer.

posted Jul 29, 2013

Assume 20 MHz channel bandwidth, normal CP, 4x4 MIMO.

First, calculate the number of resource elements (RE) in a subframe with 20 MHz channel bandwidth: 12 subcarriers x 7 OFDMA symbols x 100 resource blocks x 2 slots= 16800 REs per subframe. Each RE can carry a modulation symbol.
Second, assume 64 QAM modulation and no coding, one modulation symbol will carry 6 bits. The total bits in a subframe (1ms) over 20 MHz channel is 16800 modulation symbols x 6 bits / modulation symbol = 100800 bits. So the data rate is 100800 bits / 1 ms = 100.8 Mbps.
Third, with 4x4 MIMO, the peak data rate goes up to 100.8 Mbps x 4 = 403 Mbps.
Fourth, estimate about 25% overhead such as PDCCH, reference signal, sync signals, PBCH, and some coding. We get 403 Mbps x 0.75 = 302 Mbps.

Guaranteed bit rate is not calculated it is the operator obligation.

In order to get a raw peak data rate at the PHY layer e.g. in the DL you do the following:
1) assume we have 20MHz band, so the number of PRBs in the frequency domain is: PRBno = 100
2) assume we have 1 OFDM symbol for control region (for PHICH, PCFICH and PDCCH) in each subframe, so number of OFDM symbols per subframe for user plane data (PDSCH) is: NoOFDMSymbols = 13 (for normal CP)
3) assume we have SISO case (one antenna), so the number of Cell RS for the PDSCH per 2PRBs is: NoRS = 6
4) the number of subcarriers per PRB is: NoSubcarriers = 12
5) the number of RE (resource elements) available for carrying PDSCH per 2PRBs is: NoREs = NoOFDMSymbols * NoSubcarriers – NoRS = 13 * 12 – 6 = 150
6) the number of REs for subframe is: NoREPDSCH = NoREs * PRBno = 150 * 100 = 15000
7) for peak datarate we use 64QAM, which gives the number of bits per RE: bitsRE = 6
8) the number of bits for the whole subframe is: NoBitsPDSCH = NoREPDSCH * bitsRE = 15000 * 6 = 90 000
9) the number of subframes in one sec is: NoSFs = 1000 [SFs/Sec]
10) the max throughput then (raw, ie. without FEC) is: RawThrpt = NoBitsPDSCH [bits/SF] * NoSFs [SFs/Sec] = 90 000 * 1000 = 90 000 000 bits/sec = 90 Mbits/s
11) if you add then the typical FEC rate for good channel conditions of: FECrate = 5/6
12) you end up at: PHYThrpt = RawThrpt * FECrate = 90 Mbits/s * 5/6 = 75Mbit/s

So that’s your thrpt for DL at PHY. Then additionally you need to get out of it some % for RRC signalling and system info, and overhead of MAC/RLC/PDCP and TCP/IP stack.

For UL, the situation is similar, but you need to take into account the typical UL frame PHY overhead (DRS, SRS, PUCCH, PRACH)

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