CCE Usage formula

Hi Expert…any idea how those fixed numbers in the formula:

CCE Usage = 100% x (Used CCE UL + Used CCE DL) / ( ( 60 x 60 x 200 ) x (88 + 55 + 84 + 88) )

Maybe total no of CCE available in 1 hour…

I want to understand the calculation.

60 x 60 is: 60 minutes x 60 seconds
200 x (… + … + … +…) is: for e.g. every 5 ms you have ( … + …) CCE, it depend on cell configuration.
Then in 1s = 200 x 5s

@Manoj_deka, below is detailed explanation:

Here denom indicates number of CCEs available in TDD 20 Mz carrier in 1 hour, with 2CRS configuration.
(88+55+84+88) → Indicated number of CCEs available in 5 ms (Half frame of TDD: DSUDD)

88-> Number of CCEs in subframe without containing PHICH (2 subframes in 5ms)
55-> Number of CCEs in spacial subframe (1 subframe in 5ms)
85-> Number of CCEs in subframe with containing PHICH (1 subframes in 5ms)

Above numbers are coming out of calculation as follow:
“Available CCE for PDCCH = Total REs - CRS REs - PCFICH REs - PHICH REs”
For spacial subframe only 2 symbols are available.

(60 x 60 x 200) Indicates number of CCEs in 1 hour.
(60 x 60 x200 x 5 = 1 Hr)

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Wonderful explaination I must say @sadanandk2 !

Really excellent explanation
I was not getting the technique behind this calculation. So:
For first DL, available no of CCE for CFI 3 = (3600 - 400 - 16) / 36 = 88 CCEs
For S, available no of CCE for CFI 2 = (2400 - 400 - 16) / 36 = 55 CCEs
For 3rd DL ( SF=3, which carry ACK/NACK), available no of CCE for CFI = 3 = ( 3600 - 400 - 12 - 156) / 36 = 84 CCE.
Similarly, for 4th DL sf, available no of CCE for CFI 3 = (3600 - 400 - 16) / 36 = 88 CCEs.

My question is what will happen if we use SSF configuration 2-5 (3/9/2).
In this case, will the special sf carry any CCE since we don’t scheduled any data here…

In case of 2-5 CCEs cannot be used for scheduling, so available CCEs will be reduced.

I think in 2-5, special sf is only used for carrying PSS.
So no CCE?

No CCEs no DL data.