Note technique: Throughput Analysis for different power allocation policies in a random access channel taking into account QoS constraints

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HAL Id: hal-01982968

https://hal.archives-ouvertes.fr/hal-01982968

Submitted on 16 Jan 2019

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Note technique: Throughput Analysis for different power allocation policies in a random access channel

taking into account QoS constraints

Asma Selmi, Jean Pierre Cances

To cite this version:

Asma Selmi, Jean Pierre Cances. Note technique: Throughput Analysis for different power allocation policies in a random access channel taking into account QoS constraints. [Research Report] Xlim UMR 7252; XLIM-SRI. 2018. �hal-01982968�

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Note technique: Throughput Analysis for different power

allocation policies in a random access channel taking into account QoS constraints Asma Selmi, JP Cances, Xlim UMR CNRS 7252

Summary/Introduction:

In this technical note we define a power allocation policy for IoT based user equipment which takes into account the QoS (quality of service) within a coverage cell area. Furthermore, we consider a limited number of power levels, considering the paper of Choi [1] as the basis of our new derivations. Compared to this paper, we propose new accurate lower bounds for the throughput efficiency with a deeper analysis of the throughput efficiency.

I. Transmit power adaptation to the link quality

- We use now a more realistic model for the power allocation policy, we don’t assume an infinite number of power levels as in our first contribution entitled “Capture effect at the SIC receiver, throughput analysis”, but only L power levels that are denoted by:

1 2  ... _L 0 (1)

We assume that an active user, say user k, can randomly choose one of the power levels, say _l , for random access, then the transmission power is decided as:

l k

k

P ^ ⁽²⁾

Where _k  h_k ² is the channel power gain from user k to the BS, so that the received signal power becomes _l. Assuming that the spectral density of the background noise is normalized, i.e. N₀ 1, if there are no other active users, the signal-to-noise ratio (SNR) or signal-to- interference-plus-noise ratio (SINR) at the BS becomes _l. Suppose that each power level in (1) is decided as follows [1]:

( 1)

l  Vl (3) Where is the target SINR and

1 L

l m

m l

V

 





^with^V^L ^⁰. The value of the target SINR, , can be decided depending on the desired transmission rate or quality of the link, it can be shown that:

( 1)^{L l}

l

     (4) Proof: we have successively

(3)

1 1

2 2

2 2 1

2

3 3 1 2

2 2 3 2

v ( 1)

v ( 1) (v 1) ( 1)

v ( 1) (v v 1) ( ( 1) 1) ( 2 1) ( 1)

v ( 1) (v v v 1) ( ( 1) ( 1) 1)

( ( 2 1) 1) ( 3 3 1) ( 1)

L L

L L L

L L L L

L L L L L

V V

 

  

   

    

         

                        

                   

                          ³

1 1

1

0

v ( 1) ( v 1) ( ( 1) 1)

( 1) 1

( ( 1) 1) ( 1) ( 1)

L L

L m

l l m

m l m l

L l L l

m L l

m

V ^

   

  





           

  

             



 



(5)

If there exists one active user at each power level, the SINR for the active user who chooses ₁ becomes: ¹

1 1

V  , which is from (3). Thus, when the transmission rate, denoted by R, is given by:

log (12 )

R   (6)

The signal from this user can be decoded and removed using SIC. The SINR for the active user who chooses ₂ is also: ²

2 1

 V

 . Consequently, all the L signals can be decoded using SIC in ascending order if the transmission rate is given by: Rlog (1₂  ). In other words, a total of L signals can be decoded although they are transmitted simultaneously.

Throughput calculation: A power collision happens at level l if at least two users choose the same power level l. If power collision happens at level l the signals at levels l1,...,Lcannot be decoded. For the performance of random access based on PDMA, we consider the conditional throughput that is the average number of signals that are successfully decoded for a given M. A bound on the conditional throughput can be found as follows.

1

( ; ) ( ; ) (1 )

L

m

M L M L M m

  L



 



 ⁽⁷⁾

If M 2, the bound is exact.

The proof comes from the fact that the probability that all M signals have different power levels is:

1

(1 )

L

m

 L



 , we can have (7). And, since it is also possible to decode some signals in the presence of power collision, (7) becomes a lower bound. In the case M = 2, the BS can decode two signals if two active users choose different power levels. If they choose the same power level, no signal can be decoded. Thus, the lower bound in (7) becomes exact.

However, the capture effect is more complex to define since, unlike conventional multichannel random access schemes, the power collision at each power level is not an independent event.

(4)

That is, if power collision happens at level l, the signals at levels l1,...,Lcannot be decoded, while the signals at level 1,..,l1can be decoded if there is no power collision.

In this part, we calculate the throughput efficiency conditioned on the fact that packets are colliding. Then, we consider the case of n < L colliding users at a given frequency level, we denote as l the smallest power level of colliding packets in terms of power, the packets with power levels kl can then be decoded. The associated number of non-colliding packets is given below. It is the product of the probability that n, (n2) packets are colliding on the same frequency sub-band

 

1 1

1

M n n

M n

C B B

   

 

  by the probability that packets with power levels k l can then be decoded (we have implicitly supposed that n < L in formula (8), and we have to multiply by the number of available sub-channels B):

 

1

1 1

1 1 1

M n n l

n

M n M n

l k

B k

P C l

B L

B

 



 

   

   

 

  ⁽⁸⁾ The associated average throughput efficiency with P_{M n}_ is equal to:

 

1

2 1 1

1 1

( , ) 1 1

L M n n l

n

M n

n l k

M B B C l k

B L

B



 

  

   





  

 

  ⁽⁹⁾

To obtain a more tractable expression for (9) we consider the case where L is high enough to consider: k 1

L  . First approach:

Using the notation:

1

l

k

A k

L





 





   , we have:

1

1 2 2 1

log( ) log 1 log 1 1 ... 1 1

1 2 1

log 1 log 1 ... log 1

log 1

l

k

l

m

k l l

A L L L L L

l

L L L

m L









 

        

               

      

          

     

 

   

 





If we consider high values of L (which is not a practical situation since it entails important allocated power values), we have the approximation:

1

( 1) log( )

2

l

m

m l l

A L L









   ⁽¹⁰⁾ Then, we have to calculate:

( 1) 2 1 n l l

L n

l

S le

 





⁽¹¹⁾

(5)

To find an approximate value of Sn, we can study the function:

( 1)

( ) 2

x x

f x xe L

 

 , we have :

( 1) ( 1) ( 1)

2 2 2

( 1) 2

2

2 1 ( 2 1)

'( ) [1 ]

2 2

[2 2 ]

2

x x x x x x

L L L

x x L

x x

f x e x e e x

L L

e x x L

L

  

 

   

   

   

The roots of f '( )x are : 1 / 1 16

i 4

X     L , the positive root is : ₁ 1 1 16 4

X    L  L, that means that the function is increasing over the interval [1,X₁[ and decreasing over [X₁,[ We can always write:

1 1 ( 1 1 ) _X1 1

X   X  X   X n Y , with

1 1

nX    X and Y₁ X₁   X₁ Concerning the interval

[1,n_X1], we have,

[ , 1[ with 1 _X1 1

x m m m n

      :

1 1 1 1

1 1

( 1) ( 1) ( 1)

2 2 2

( 1) 1

2

( 1) ( 1) ( 1)

2 2 2

( 1)( 2) ( 1) ( 1)

2 2 2

( 1) This yields:

1 2

( 1)

X X X X

m m x x m m

L L L

x x

L L

m m x x m m

L L L

n n x x n n

L L L

X X

me xe m e

xe e

me xe m e

n e xe n e

  

  

  

  

   

  

  

 

  

  

This yields:

1 1 ( 1) 1 ( 1) 1 1 ( 1)

2 2 2

1 1 1

( 1)

X nX X

n m m x x n m m

L L L

m m

me xe dx m e

       

 

  

  

For the interval

1 1

[n_X,X ], we have the following inequalities:

   

1 1 1 1 1 1 1

1

1 1 1

1 1 1 1 1

1 1 1

1

( 1) ( 1) ( 1)

2 2 2

1

( 1) ( 1) ( 1)

2 2 2

1 1 1

X X

X X X

X X

X n n

X X x x X X X

L L L

X

n n n

n n X x x X X

L L L

X X X

n

n e dx xe dx X e dx

X n n e xe dx X n X e

  

 

   

  



We eventually obtain for the complete interval [1,X₁]:

1 1 1 1 1 1 1

1

( 1) 1 ( 1) ( 1) ( 1) 1 ( 1)

2 2 2 2 2

1 1 1

( 1)

X X X X

n n n m m X x x X X n m m

L L L L L

X

m m

Y n e me xe dx Y X e m e

      

    

 









 



 ⁽¹²⁾

(6)

Concerning the interval [X n₁, ], we have a decreasing function, we consider once again the two intervals:

1 1

[X n, _X 1] and

[n_X11, ]n , we have :

   

1 1 1 1 1 1 1

1

1 1 1

1 1 1 1 1

1 1 1

1

1 ( 1) 1 ( 1) 1 ( 1)

2 2 2

1

( 1) 1 ( 1) ( 1)

2 2 2

1 1 1

( 1)

1 ( 1) 1

X X X X X

X X X

n n n n x x n X X

L L L

X

X X X

n n n x x X X

L L L

X X X

X

n e dx xe dx X e dx

n X n e xe dx n X X e

        

   

  

  

      

  



And  x [ ,m m1[ with n_X1  1 m n :

 1  1 1 1 

1 1

11 11

1 ( 2) ( 1) 1

2 2 2

( 1) ( 1) ( 1)

2 2 2

( 1) ( 1) ( 2)( 1)

2 2 2

( 1) ( 1) ( 1)

1

2 2 2

( 2) ( 1)

( 1)

( 1) ( 1)

X X X X

X X

n n x x n n

L L L

X X

m m x x m m

L L L

n n x x n n

L L L

m m n x x m m

n

L L L

m n n m

n e xe n e

m e xe me

ne xe n e

m e xe dx me

 

   

  

  

  

   

  

  

   

 

   

  

  

  

 

11 1

X n

n _





We eventually obtain for the complete interval [X n₁, ]:

 

1 1

1 1 1

1

1 1

1

1 1

( 1) 1 ( 1) ( 1)

2 2 2

1

( 1) 1 ( 1)

2 2

1 1

1

( 1) 1 ( 1) ( 1)

2 2 2

1

1 1

1 ( 1) ( 1)

1

1 ( 1) ( 1)

1

X X

X

X X X

X

n n n m m n x x

L L L

X X

m n X

X X n m m

L L

X

m n

n n n m m n x x

L L L

X

m n X

n X n e m e xe dx

n X X e me

Y n e m e xe dx

Y X e

   

  

 

  

 

 

   

  

 

     

   

    

 

 



 

1 1

1

( 1) 1 ( 1)

2 2

X 1

X X n m m

L L

m n

me

  

 

 





(13)

We can rewrite:

 

1 1

1

1 1

1

1 1

( 1)

( 1) ( 1)

1

2 2 2

1 1

( 1)

( 1) ( 1)

2 2 2

1 2

( 1) 1 ( 1)

1 ( 1)

X X

X

X X

X

n n

m m n x x

n

L L L

X

m n X

n n n

m m x x

n

L L L

X

m n X

m e xe dx Y n e

m e xe dx Y n e

  

   

 

  

  

 

    

   

 

Finally, we obtain:

(7)

 

₁ ¹ ¹

1 1

( 1)

( 1) ( 1)

2 2 2

1 2

1 ( 1)

X X

X

n n

m m n x x

n

L L L

X

m n X

m e xe dx Y n e

  

  

 

   

 

(14)

This yields:

 

1 1 1 1 1 1

1 1 1

1 1

1 1 1 1 1 1

1 1 1

1 1

( 1) ( 1) ( 1)

( 1) ( 1)

2 2 2 2 2

1

( 1) ( 1) (

( 1) ( 1)

2 2 2 2

1

( 1) 1 ( 1)

1 ( 1) ( 1)

X X X X X X

X

X X X X X X

X

n n n n n n n

m m x x

n

L L L L L

X X X

m n X

n n n n n n

m m n x x

n

L L L L

X X X

m n X

m e n e n e xe dx Y n e

m e xe dx Y n e n e n e

  

 

    



  

 

    



      

      

 

1 1 1 1

1 1

1) 2

( 1) ( 1)

2 2 2 2

1( 1)

X X X X

X

L

n n n n

n

m m x x

n

L L L L

X X

m n X

m e xe dx Y n e n e

 

 

   



   

 

Using (12), we have:

1 1 1 1

1

( 1)

( 1) ( 1)

2 2 2

1

1 1

X X

X n n

n m m X x x

L L L

X m

me xe dx Y n e

      



 

 

This yields:

1 1 1 1 1 1

1 1 1

1 1 1 1 1 1 1

1 1

( 1) ( 1) ( 1)

( 1) ( 1)

2 2 2 2 2

1 1

( 1) ( 1) ( 1) ( 1

( 1) ( 1)

2 2 2 2 2

1 1

( 1)

X X X X X X

X X X X X X X

n n n n n n

n

m m x x

n

L L L L L

X X X

m

n n n n n n n

n

m m x x

n

L L L L L

X X

m

me xe dx Y n e Y n e n e

me xe dx Y n e e Y e n e

  

 

    



   

 

     



    

 

     

 

 

²^L⁾ⁿ^X¹

(15)

Then, we have to calculate the integral:

( 1) 2 1

n x x

L

In xe dx

 





⁽¹⁶⁾

We use: ² ( 1) 2 u x x

L

  , this entails:

2 2

( 1)

( 1) 2 ( 1)

2 2 2

1 0 0

2 [1 ]

n n

n x x L n n n n

L

u u

L L

In xe dx L ue du L e L e

 

     









     (17)

And we obtain:

(8)

1 1 1 1 1 1 1 1

1 1

1 1 1 1 1

1 1

( 1) ( 1) ( 1) ( 1)

( 1) ( 1)

2 2 2 2 2 2

1 1

1

( 1) ( 1) (

( 1) ( 1)

2 2 2 2

1 1

1

[1 ]

[1 ] ( )

X X X X X X X X

X X X X X

n n n n n n n n

m m n n

n

L L L L L L

X X

m

n n n n n

m m n n

n

L L L L

X X

m

me L e Y n e e Y e n e

me L e Y n e e X n e

   

 

     



 

 

    



 

      

 

      



¹ 1 ¹ ¹

1 1 1 1 1 1 1 1 1 1

1 1

1

1) ( 1)

2 2

( 1) ( 1) ( 1) ( 1) ( 1)

( 1) ( 1)

2 2 2 2 2 2 2

1 1

1

( 1) ( 1)

2 2

1 1

[1 ]

[1 ] ( 1

X X X

X X X X X X X X X X

n n n

L L

X

n n n n n n n n n n

m m n n

n

L L L L L L L

X X

m

m m n n

n

L L

X m

n e

me L e Y n e e X e n e e

me L e n Y

 



    

 

      



 





   

         

   



¹ ¹ ¹ ¹ ¹ ¹

2

1 1 1 1 1

1

2 1 1

( 1) ( 1) ( 1)

2 2 2

1

( 1)

( 1) ( 1)

2 2 2 2 2 2

1 1

1

( 1) ( 1)

2 2 2

1 1

)

[1 ] ( 1) (18)

[1 ] 2 ( 1) s

X X X X X X

X X X X X

X

n n n n n n

L L L

n n n n n

m m n n

n

L L L L L L

X m

m m n n n

n

L L L

X m

e e X e

me L e n Y e e e X e

me L e n Y e

  

  

  

    



 

  



 

 

 

 

 

      

   



^h^ⁿ₂^X_L¹^^^{X e}¹ ^ⁿ^X¹⁽²ⁿ^L^X¹^¹⁾

 

For the upper bound we can conclude:

2

1 1 1

1 1

( 1)

2 2 2

1 1

1

2 ( 1) sh

2

X X X

n n n

n m m

L L X L

X m

me L n Y e n X e

L

 

  



 

    

 



⁽¹⁹⁾

And, for small values of n, we can use the following with better accuracy:

2

1 1 1

1 1

( 1)

( 1) ( 1)

2 2 2 2

1 1

1

[1 ] 2 ( 1) sh

2

X X X

n n n

m m n n

n

L L L X L

X m

me L e n Y e n X e

L

  

   



 

     

 



⁽²⁰⁾

For the lower bound, we have:

1 1 1

1

1 1 1 1

( 1) ( 1) ( 1)

2 2 2

1 1

2 1

( 1) ( 1) ( 1)

2 2 2

1 1

1

X

n m m X x x X X

L L L

m

n m m X X X x x

L L L

m

me xe dx Y X e

me Y X e xe dx

  



  



 

  

 

(21)

And, on the other side:

 

¹ ¹

11 1

( 1) ( 1) ( 1)

2 2 2

1 1

1

X

m m n x x X X

n

L L L

m n X

me xe dx Y X e



  



  

 

This yields:

(9)

 

1 1 1 1

1 1

( 1) ( 1) ( 1) ( 1)

2 2 2 2

1 1 1 1

1 1

( 1) ( 1) ( 1)

2 2 2

1

1 1

( 1) ( 1) ( 1)

2 2 2

1 1

1

1 [1 ]

m m X X n x x X X

n

L L L L

m

n

m m X X x x

n

L L L

m

m m X X n n

n

L L L

m

me Y X e xe dx Y X e

me X e xe dx

me X e L e

   



  



  



    

  

   

 



(22)

We conclude:

2

1 1 1 1 1

1 1

( 1)

( 1) ( 1) ( 1)

2 2 2 2 2

1 1 1

[1 ] 2 ( 1) sh 1 [1 ]

2

X X X

n n n

n n n n X X

L L X L L L

X n

L e n Y e n X e S L e X e

L

   

      

         

 

(23)

The accuracy of these bounds has been checked on Matlab. Plugging (51) into (37), we obtain:

 

1 1

2

1 1 1 1 1 1

1 1

( 1) ( 1)

2 2

1 2

( 1) ( 1)

( 1)

2 2 2 2 2

1 1

2

1 1

1 [1 ] 1 ( , )

1 1

... 1 [1 ] 2 sh

X X X X X X

M n n n X X

L

n L L

M n

n

n n n n n n

M n n n

L

n L L L L L

M n X X

n

B C L e X e M B

B B

B C L e Y n e e Y e n e

B B



    



 

     



 

       

 

   

   

   

          



(24)

To consider the limit obtained with a Poisson arrival law, we calculate the first terms in the lower and upper bounds of (24). We have:

 

1 1

( 1) ( 1)

2 2

1 2

2 1 ( 1)

2 2 1

( 1) ( 1)

2 2

1

1 1

1 [1 ] 1

( 1) 1 1

[ 1 [1 ] 1 ...

2

( 1)...( 1) 1 1

... 1 [1 ] 1

!

M n n n X X

L

n L L

M n

n

M X X

L L

M k k k X X

L L

k

B C L e X e

B B

B M M L e X e

B B

M M M k

L e X e

k B B

    



   

    

 

       

 

   

 

  

        

 

    

        



 

1 1

( 1) ( 1)

2 2

1

...

( 1)...( 1) 1 1

... 1 [1 ] 1 ]

!

M L L L X X

L L

L

M M M L

L e X e

L B B

    



 

    

        

Concerning the first term, we have:

1 1

2 1 ( 1)

2 2 1

1 ( 1)

2 2 1

1 (

2 2 1

( 1) 1 1

log 1 [1 ] 1

2

( 1) 1 1

log ( 2) log 1 log [1 ] 1

2

log ( 2) log 1 1 log [1 ] 1

2

M X X

L L

X X

L L

X X L

M M L e X e

B B

M M M L e X e

B B

M M L e X e

B B

   

  

 

         

    

  

 

 

    

           

   

         

1) 2L

  

 

 

(25)