Power Allocation and Coverage for a Relay-Assisted ...

Power Allocation and Coverage for a Relay-Assisted Downlink with Voice Users Junjik Bae, Randall Berry, and Michael L. Honig

Department of Electrical Engineering and Computer Science Northwestern University, Evanston, IL 60208 USA [email protected],{rberry,mh}@ece.northwestern.edu

Abstract— We study the downlink coverage of a base station terminal (BST), which has access to a relay node. Continuing a previous study in which the BST is assumed to provide a variable-rate data service, here we assume that each active user requires a target data rate, corresponding to a voice type of service. The relay is assumed to serve a separate set of (noncellular) users, corresponding to a WiFi Access Point (AP). A one-dimensional model is considered in which cellular and noncellular users are uniformly distributed along a line. The BST and AP jointly allocate available power across users and the BST-AP link to maximize the total number of users served. We characterize the optimized set of active cellular users served by the BST directly and the AP relay, and the non-cellular users served by the AP. We also give a closed-form upper bound on the increase in the total number of users provided by the relay as a function of user densities and path loss exponents. Our results show that depending on the distance between the BST and the AP, the addition of a relay gives a modest increase in the total number of active users.

I. I NTRODUCTION Adding fixed relays to a cellular network can potentially increase the network capacity and extend network coverage [1]–[5], [7]. This has motivated the introduction of relays in emerging standards such as 802.16j [8] and 4G mobile systems [9]. Rather than deploying new cellular relays, in some cases, it may be possible for a cellular service provider to use an existing wireless LAN (WLAN) access point (AP) as a relay for cellular traffic [6]. In that case, although the AP potentially brings additional resources to the cellular network, those resources must be allocated across both WLAN and cellular user requests. Previous studies on relay-assisted cellular systems have generally assumed that the relays are dedicated to serving only cellular users (e.g., see [3]–[5], and [6], where relays are proposed to relieve congestion in a particular cell by siphoning off traffic to neighboring cells). This paper is continuation of previous work [7] in which we evaluate the increase in downlink capacity (sum data rate over all users) associated with the addition of a relay to a single cell in a cellular network. Whereas [7] evaluates a sum data rate, which corresponds to a data type of service, here we assume a voice type of service in which each user requires a target data rate. The system objective is then to maximize the total number of users served by both the cellular system and the AP. 1 This work was supported by the Motorola-Northwestern Center for Telecommunications.

As in [7], we consider a single one-dimensional cell with two relays (AP nodes) symmetrically placed on either side of the Base STation (BST), and a static user population. Both the cellular users and non-cellular users, served by the AP, are assumed to be continuously and uniformly distributed along the line with given densities. This corresponds to the large system model presented in [10]. All users (both cellular and AP) are assumed to be orthogonal in time, frequency, and/or signature space, and do not interfere. Furthermore, the AP is assumed to use a different band from the cellular band, so that cellular users do not interfere with non-cellular users. Also, here we consider only the power allocation problem subject to a total power constraint. We assume that each user receives a single unit of bandwidth (i.e., channel, time slot, or signature), and that the total number of users does not exceed a bandwidth constraint. Finally, in the model considered here path loss is determined only by distance. Random propagation effects, such as shadowing, are not explicitly modeled, but can be incorporated in the model by changing the distance metric. 1 Our problem is to allocate available BST and AP power across both cellular and AP users, as well as to the link from the BST to the AP, to maximize the total number of active users. We solve this problem for the following two scenarios: (i) the information flows to the cellular users served by the relay are jointly encoded and transmitted from the BST to the AP; and (ii) the preceding information flows are transmitted in parallel (separately encoded) from the BST to the AP. Joint encoding requires less power to transmit the data destined for relayed users from the BST to the AP, but requires that the received data at the AP be demultiplexed. We also give an upper bound on the increase in total number of users provided by the AP, based on the scenario in which the BST and AP have a wireline connection, so that the BST does not expend any power to communicate with the AP. The optimization of power across users determines intervals of cellular users served by the BST directly, relayed cellular users, and non-cellular users. The relative position of these intervals depends on the BST and AP power constraints, the path loss exponents associated with the cellular, non-cellular, and BST-AP links, and the user densities. We give numerical examples, which show that the total number of active users can 1 This change in distance metric effectively amounts to changing the distribution of users along the line.

max γP BST

for Total flow conservation

PB (x) for Individual Flow Conservation

PB(r)

P’A (x)

PA (x) r

x

Distance d Fig. 1.

One-dimensional model of a cell with relay nodes (APs).

be increased by more than 15% if the position of the relay is optimized. If the AP has no users to serve, then the increase in the number of the cellular users can be more than 35%. Jointly encoding the information from the BST to the AP is shown to result in only a small performance gain.

II. S YSTEM M ODEL The one-dimensional cellular model is illustrated in Fig. 1. We assume a symmetric cell in which two APs are placed at the same distance d from the BST. As in [7], [10], we assume a static set of users, which are uniformly and continuously distributed along the line. The density of the cellular users, served by the BST, is ρB , and the density of the AP users is ρA . This corresponds to a large system limit in which the number of users in the system tends to infinity in proportion with the available bandwidth (i.e., fixed users per Hz). Given a finite total power constraint, the BST and AP each serve a finite number of users. In what follows, we assume that each active user receives a fixed unit of bandwidth, and that the corresponding number of available channels (and/or time slots and/or signatures) exceeds the total number of active users. All users are therefore assumed to be orthogonal, i.e., they do not interfere. We first consider the situation in which the BST does not use the AP as a relay. From symmetry, we need only consider the power allocation and set of active users on one side of the BST. Because each active user receives a target rate, our objective is to maximize the number of active users for a given power constraint. Assuming a path loss of 1/r a , where a is the path-loss exponent, and optimal coding, the power needed to transmit at rate RB to a cellular user at distance r is PB (r) = RB
N0 WB e WB −1 ra ≡ Bra , where N0 is the noise density and WB is the cellular bandwidth per user. The maximum number max of cellular users that the BST can support with power PBS is then ρB C0 , where C0 is the length of the interval containing

RC max the active users and satisfies 0 0 ρB PB (r)dr = PBS , i.e.,2   1 a + 1 max a+1 C0 = . (1) P ρB B BS Similarly, assuming non-cellular and cellular users have the same path-loss exponents, the minimum power needed by the AP to transmit at rate RA to a user at distance x is PA (x) = RA
N0 WA e WA − 1 xa ≡ Axa , where WA is the AP bandwidth per user. The maximum number of non-cellular users that the max AP can support with power PAP is therefore 2ρA D0 , where max 1 P AP a+1 D0 = ( ρa+1 is the one-sided AP coverage (i.e., for 2 ) AA users at distance d + x, with x ≥ 0).3 Therefore, without relaying, the total number of active cellular and non-cellular users (on one side of the BST) is ρB C0 + 2ρA D0 . Now we consider the case where the BST uses the AP to relay voice traffic to cellular users. We assume that a cellular user receives data either from the BST directly or through AP, but not simultaneously.4 The presence of the relay adds a flow conservation constraint, namely, the total rate that the AP provides to cellular (relayed) users must be the same as the total rate it receives from the BST. With target rates RB and RA for BST and AP users, respectively, the power allocation problem is then to maximize the total number of BST and AP users, namely, max ρB |C| + 2ρA |D|, (2) {C,D}

subject to the power and flow conservation constraints. Here, C and D are the regions of active cellular and one-sided non-cellular users, respectively, and | · | denotes the size (i.e. Lebesgue measure) of the corresponding region. It can be easily shown that D is always a single connected interval. However, since each cellular user can receive data from either the BST or AP, the cellular coverage set C can consist of two or more disjoint intervals Cd and Cr , where Cd = ∪i Cd,i is the union of non-overlapping intervals corresponding to the direct (non-relayed) cellular users and similarly, Cr corresponds to the relayed users. We assume that Cr is one (connected) segment. In what follows, we consider two different techniques for coding the streams transmitted from the BST to the AP for the relayed users. Joint coding or total flow rate conservation assumes that the BST jointly encodes all of the data destined for relayed users, and transmits the resulting stream to the AP. The AP then decodes and demultiplexes the individual data flows. In that case, the total data rate from the BST to the AP is equal to the sum rate of the flows to relayed users. In the second technique, individual flow rate conservation, data flows, which are to be relayed by the AP, are transmitted in parallel from the BST to the AP. The AP then decodes the 2 The

max . total BST power for users on both sides of the BST is then 2PBS we are assuming that the AP coverage region is symmetric about the max ≤ AP location. This will only be the case if D0 ≤ d, which implies PAP A A a+1 d . Otherwise the coverage would be bounded on the left at zero 2 ρa+1 (since the other users would be covered by the AP to the left of the BST). 4 This is in contrast to the model for data service considered in [7] where the BST and AP can simultaneously transmit to a user. 3 Note

packets for each data flow separately. Hence the code rate used by the BST to transfer the data to the AP is the same as that used by the AP to transmit to the relayed user. Total flow rate conservation generally requires less power for the BSTAP link than individual flow rate conservation, but is more complex, since the relayed users must be multiplexed at the BST and demultiplexed at the AP.

Given α and γ, to maximize the total coverage this bound will be met with equality. max The BST allocates (1 − γ)PBS for the direct-path cellular users. Consider the two possible inequalities Z d+x max ρB PB (r) dr ≷ (1 − γ)PBS . (10)

III. T OTAL F LOW R ATE C ONSERVATION

The left side of the inequality is the BST power required to activate users in the interval (0, d + x). This can be rewritten as 1  a+1  a+1 max (1 − γ)PBS x≷x ˜= − d, (11) ρB B

The optimization problem in this case is to maximize (2) subject to Z max max ρB PB (r) dr + γPBS ≤ PBS , (3) Cd

2

Z

ρA PA (x) dx +

D

Z

Cr

max ρB PA0 (x) dx ≤ PAP ,

max γPBS WBA log 1 + N0 WBA db





= RB ρB |Cr |,

C = Cr ∪ Cd , Cr ∩ Cd = ∅, γ ∈ [0, 1]

(4) (5) (6)

where WBA is the bandwidth between the BST and the AP, and γ is a variable to be optimized over indicating the fraction of BST power allocated to the relay channel. The AP power allocation for a relayed cellular user at distance x from the RB
AP is PA0 (x) = N0 WA e WA − 1 |x|a = A0 |x|a . The pathloss exponent between the BST and the AP is b, which can differ from the direct path-loss exponent a. That is, through proper placement of the AP, we may have b < a. Constraints (3) and (4) are the total power constraints for the BST and AP, respectively, and (5) is the total flow rate conservation. To solve this optimization problem, we first assume that max the AP allocates power αPAP for relayed cellular users, where α ∈ [0, 1], and subsequently optimize over α. With this assumption the preceding problem decomposes into two independent optimization problems for the BST users and the AP users. Given α, the AP optimization problem is to maximize the number of active non-cellular users. The solution max 1 (1−α)PAP is 2ρA |D(α)|, where |D(α)| = ( ρa+1 ) a+1 . The 2 AA BST optimization problem is to maximize the coverage of the cellular users |C(α)|. To solve the BST optimization problem, we first write constraints on the interval containing relayed users, which max follow from the AP power allocation αPAP and the BST max power allocation γPBS . Denote the interval of relayed users ¯], where x ≥ −d. Given x, from (4) x ¯≤x ¯A (x), by [d+x, d+ x where x ¯A (x) is the value of x¯ that satisfies Z x¯ max ρB PA0 (x) dx = αPAP . (7) x

Likewise, from (5), we have x ¯ ≤ x¯B (x), where   max γPBS WBA log 1 + x¯B (x) = x + . RB ρB N0 WBA db

(8)

Combining these, we must have that x ¯ ≤ x¯(x) ≡ min{¯ xA (x), x ¯B (x)}.

(9)

0

where [0, d + x ˜] is the largest interval of users which can be served directly by the BST given the available power. For a given interval of relayed users [d + x, d + x ¯], if x > x˜, then there is a gap between the cellular users served directly by the BST and the cellular users served by the AP. The total cellular user coverage is then C(γ, α, x) = d + x ˜+x ¯ − x.

(12)

If x ≤ x ˜, then there is no gap in the coverage of cellular users. That is, the BST power needed to activate users in [0, d + x] is max less than (1 − γ)PBS . Hence, there will be a second interval of cellular users served directly by the BST given by [d + x ¯, C(γ, α, x)], where   1 C(γ, α, x) = (d + x˜)a+1 + (d + x ¯)a+1 − (d + x)a+1 a+1 (13) is the total cellular coverage. Given α, the BST optimization problem reduces to optimizing x, which determines the interval of relayed users, and γ, i.e., C(α) = max max C(α, γ, x). (14) 0≤γ≤1 x≥−d

Finally, the maximum total number of users served by both the BST and AP is given by max ρB |C(α)| + 2ρA |D(α)|.

0≤α≤1

(15)

Although we are unable to obtain a closed-form solution to the preceding optimization problem, we can follow the approach in [7] to derive a closed-form upper bound on the total number of users served by the BST and AP. Namely, we assume that the connection between the BST and AP is “free”, i.e., does not require any expenditure of power. This could correspond to the situation in which there is a wired connection between the BST and the AP. In that case, the intervals of users served by the BST and AP have lengths |C

W L,∞

  1  1 a + 1 max a+1 a + 1 max a+1 P PAP |= +2 ρB B BS 2 1  ρ ρA − a+1 ρB B + , × A01/a A1/a A01/a (16) 

1  a+1 1  ρ ρA − a+1 1 B + 1/a 01/a A A A1/a (17) provided that the distance d between the BST and the AP is large enough so that there is no coverage overlap between the users directly served by the BST and the relayed users. The relative increase in the total number of users due to the addition of the relay is then

|DW L,∞ | =



a + 1 max PAP 2

ρB |C W L,∞ | + 2ρA |DW L,∞ | ρB C0 + 2ρA D0 1  a  a+1  1/a  max P ρA B 1/a a+1 1 B + 1 + 2 AP max 2 PBS ρB A1/a A01/a . = 1 a   a+1  a+1  max B PAP 1 A 1 + 2 ρρB A 2 P max

(18)

BS

If the AP does not have a separate set of non-cellular users to serve (ρA = 0), then this ratio becomes 1   a+1 max B PAP ρB |C W L,∞ | + 2ρA |DW L,∞ | 1 . = 1+2 max ρB C0 + 2ρA D0 A0 2 PBS (19) Even with a wired connection between the BST and AP, the preceding bound is not tight unless the distance d between the BST and AP is large enough so that the intervals covered by BST and AP do not overlap. With small d this bound can be improved by optimizing the total number of users served over x and α as before, assuming that the BST-AP link requires no power consumption (since the BST-AP link requires no power, we no longer have to optimize over γ). IV. I NDIVIDUAL F LOW R ATE C ONSERVATION Next we consider individual flow rate conservation. The BST individually encodes all data destined to the relayed users and transmits these individual streams in parallel to the AP. max Assuming the AP allocates power αPAP to the relayed users, we can again decompose the optimization problem (2) into two separate problems. The AP problem is still the same as the one for total flow rate conservation. The BST problem for given user density ρB now becomes maximizing the cellular user coverage |C| = |Cd | + |Cr | subject to Z Z max ρB PB (r) dr + , (20) ρB PB (d) dx ≤ PBS Cd

Cr

Z

Cr

max ρB PA0 (x) dx ≤ αPAP .

(21)

Note in this case, (20) combines the BST power constraint and the total flow constraint; namely, PB (d) = Bda is the power required from the BST to deliver each relayed flow to the AP. As before, Cd and Cr are also constrained to be disjoint sets. To simplify our discussion, first we consider the case where b = a, i.e., the path-loss between the BST and the AP is the same as the path-loss between the BST and each cellular user. Then for a cellular user at distance r < d, the BST will not receive any power savings by using the relay. Therefore, if max B B a+1 d , the BST will is less than ρa+1 the BST power PBS not use the relay even when α > 0. In this case, the total

max a+1 PBS ) coverage is C(α) = C0 = ( ρa+1 < d. However, if BB ρ B max B PBS > a+1 da+1 , using the relay reduces the power needed by the BST. Again, denote the interval of relayed users by ¯]. Given x and α, x ¯ is again constrained by x ¯≤ [d + x, d + x x ¯A (x), where xA (x) is the value of x ¯ that satisfies (7). The following proposition shows that when the relay is used, the optimal relay user interval starts at x = 0. max Proposition 1: Given AP power αPAP and b = a, if ρB B a+1 max PBS > a+1 d , then the optimal relay user interval is given by [d, d + x], where x ≤ xA (0). We omit the proof to save space. Note the constraint x ≤ xA (0) may not be tight if the BST does not have enough power to relay sufficient traffic to utilize all the allocated AP power. In this case, x can be calculated from (20). Next, we consider the case where the path-loss between the BST and the AP is 1/r b , where b < a. In this case, the power required by the BST to transmit directly to a user at distance r at rate RB is PB (r) = Bra and the power required by the BST to transmit a relayed packet to the AP for this user is PB (d) = Bdb . Hence, the BST can save the power by using the relay if r > d1 = db/a . When the BST power is small enough so that it can only serve users at distance less than d1 , it will not usethe relay. As its power increases, it will begin to serve the relayedusers. Initially, the set of relayed users will lie in a symmetric interval around the AP. Given AP power max αPAP , let [d − ∆, d + ∆] be the largest symmetric interval of relyed usersmaxthat can be supported around the AP, i.e., 1 αPAP ∆ = ( ρa+1 ) a+1 . As the BST power further increases, 0 2 BA the resulting coverage will follow two possible evolutions, depending on whether or not d−∆ ≥ d1 . Two examples of this evolution are shown in Figure 2. On the left is the case where d − ∆ ≥ d1 . In this case the relay always serves a symmetric set of cellular users, and as the BST power increases, it will eventually serve some users to the right of the relayed set. On the right is the case where d−∆ < d1 . In this case, when there is sufficient relayed traffic, the relayed set is not symmetric, but extends further in the positive direction. This is because the users to the left are served directly by the BST. For the former case (d − ∆ ≥ d1 ), the coverage as a function of the BST power is given in table I; we have derived a similar table for the latter case, but omit it due to space considerations. max In Table I, PBS (critical) represents the minimum power max max such that when PBS > PBS (critical), the BST serves users to the right of the relayed interval. In this case, the cellular coverage as a function of x becomes  a + 1 max P + (d + xA (x))a+1 C(α, x) = ρB B BS 1  a+1 b a+1 − (a + 1)d (xA (x) − x) − (d + x) , 1

(22)

where x is constrained to be no smaller than min{−∆, d1 −d}. Therefore, the optimal cellular coverage for a given α is C(α) = C(α, min{x∗ , x}), ˙

(23)

TABLE I C ELLULAR COVERAGE AS A

FUNCTION OF

max PBS

max BST Power PBS

A B C D

0
PBS

=

b < a AND d − ∆ ≥ d1 .

BST coverage C(α) ” 1 “ a+1 P max C(α) = C0 = ρa+1 B BS

ρB B a+1 d a+1 1

max PBS

WHEN

B

· 2∆ · 2∆

C(α) = d1 + C(α) =

max . On the left is the Fig. 2. Cellular user coverage as a function of PBS case where d − ∆ ≥ d1 and on the right the case where d − ∆ < d1 .

∗ where x∗ and x˙ are obtained by solving ∂C(α,x) ∂x |x=x = 0 and ˙ respectively. C(α, x) ˙ = d + xA (x), As in Section III, given C(α), the optimal number of cellular and AP users can then be found by searching for the optimal α∗ ∈ [0, 1].

V. N UMERICAL R ESULTS We next give some numerical results to illustrate the gains from relaying under the two flow rate conservation constraints. First, we consider total flow rate conservation. Figure 3 shows the ratio of the total number of users (BST+AP) with relaying to those without relaying as a function of distance d between the BST and the AP. Two different path loss scenarios are considered: a = 4, b = 3 and a = b = 4, with the remaining parameters indicated in the caption. For each case the ratio of the total number of cellular users is also shown (BST only). Note there is clearly an optimal location for maximizing each of these ratios. When the path-loss exponent b between the BST and the AP is the same as the path-loss exponent a between the BST and each user, the increase in the total number of active users with the relay is very small (less than ∼ 7% when a = b = 4). As the path-loss between the BST and the AP improves, however, the ratio of the total number of users increases more significantly. The relay achieve more than a 17.8% increase in total number of users served when a = 4

n

a+1 ρB B

`

max ρB B a+1 PBS − a+1 d1

ρB Bdb

max − ρ Bdb · 2∆ PBS B

1 ´o a+1

+ 2∆

C(α) − See eq. (22) and (23)

and b = 3. In this case the cellular users served increase by more than 34.9%. (This is offset by a decrease in the number of AP users.) It follows that the increase in the total number of cellular user is larger than 34.9% when the AP only serves as a relay (i.e., ρA = 0). The upper bound we obtained by assuming the wired connection between the BST and the AP for d = 4.5 is 1.20 (∼ 1.30 for d = ∞), and the loss due to the power consumption on the back-haul link between the BST and the AP is small in this example. This is partially because we only consider one AP. In a two dimensional model that has more APs, the back-haul wireless communication loss should be larger. Fig.max 4 shows the ratio of total number of users as a P function of BS N0 when a = 4 and b = 3, assuming the AP is located d = 4.5 from the BST. It can be seen that there is an optimal power which maximizes this gain; in this case this is approximately the power used in Figure 3. Figures 5 and 6 show the analogous results under the individual flow conservation constraint. These figures are very similar to those with total flow conservation. Interestingly, the maximum gains occur at similar power levels and AP locations in the two cases. In this case, the gains are slightly lower due to the increased power needed for the backhaul link. For example, in the case of a = 4, b = 3, the maximum percentage increase in the total number of users served is about 16.6% (v.s. 17.8% in the total flow case). In these examples the gain from joint encoding is very small, and may not warrant the additional implementation complexity required. VI. C ONCLUSION We have studied the total number of downlink voice users in a one dimensional model of a single cell site when two APs located symmetrically around a BST are available as relays. In addition to relay traffic, we assumed that the AP has its own customers to serve. We studied two different flow conservation schemes which correspond to whether or not the relay traffic is multiplexed when sent to the AP. Our results show that when the sets of active users are optimized, the total number of cellular users increases significantly under both flow rate conservation schemes, but at the cost of reducing the AP users. This corresponds to a coverage extension of the cellular network. Although the relay schemes considered in this paper increase the coverage of the BST, it is only possible when the AP is cooperative. Relay cooperation can be achieved either

1.4

1.4

BST+AP (a=4,b=3) BST only (a=4,b=3)

BST+AP (a=4,b=3) BST only (a=4,b=3) BST+AP (a=4,b=4) BST only (a=4,b=3)

1.35

1.35

1.3 1.3

Ratio of users

Ratio of users

1.25

1.2

1.25

upper bound when Pmax =500 BS

1.2

1.15 1.15 1.1 1.1

1.05

1

2

2.5

3

3.5

4 Distance d

4.5

5

5.5

6

Fig. 3. Ratio of the total number of users served under total flow rate max PAP P max = 500, = conservation as a function of distance d. Here BS N0 N0 5, WB = WA = 1, WBA = 3, RB = RA = 1, ρB = ρA = 1.

Ratio of users

1.3

1.25 upper bound when Pmax =500 BS

400

450

500 BST power

550

600

650

700

Fig. 6. Ratio of the total number of users served under individual flow max PAP P max . Here, = 5, WB = WA = conservation as a function of BS N0 N0 1, RB = RA = 1, ρB = ρA = 1, d = 4.4.

R EFERENCES

BST+AP (a=4,b=3) BST only (a=4,b=3)

1.2

1.15

1.1 300

350

400

450

500 BST power

550

600

650

700

Fig. 4. Ratio of the total number of users served under total flow rate P max P max . Here AP = 5, WB = WA = conservation as a function of BS N0 N0 1, WBA = 3, RB = RA = 1, ρB = ρA = 1, d = 4.5.

1.4 BST+AP (a=4,b=3) BST only (a=4,b=3) data3 (a=4,b=4) data4 (a=4,b=4)

1.35

1.3

1.25 Ratio of users

350

when the BST owns the AP or when the BST and AP have an agreement on usage of AP resources, for example, through a bargaining process. Bargaining between two non-cooperative agents, such as the BST and the AP, is an interesting topic for future work.

1.35

1.2

1.15

1.1

1.05

1

1.05 300

2

2.5

3

3.5

4 Distance d

4.5

5

5.5

6

Fig. 5. Ratio of the total number of users served under individual flow max P max PAP conservation as a function of distance d. Here BS = 500, = N0 N0 5, WB = WA = 1, RB = RA = 1, ρB = ρA = 1.

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