Utility-Based Resource Allocation in
Wireless Networks
Wen-Hsing Kuo and Wanjiun Liao
Abstract— In this paper, we study utility-based maximization for resource allocation in the downlink direction of centralized wireless networks. We consider two types of traffic, i.e., best effort and hard QoS, and develop some essential theorems for optimal wireless resource allocation. We then propose three allocation schemes. The performance of the proposed schemes is evaluated via simulations. The results show that optimal wireless resource allocation is dependent on traffic types, total available resource, and channel quality, rather than solely dependent on the channel quality or traffic types as assumed in most existing work.
Index Terms— Utility-based maximization, resource allocation, wireless networks.
I. INTRODUCTION
R
ESOURCE allocation is an important research topic in wireless networks [1-10, 12-16]. In such networks, radio resource is limited, and the channel quality of each user may vary with time. Given channel conditions and total amount of available resource, the system may allocate resource to users according to some performance metrics such as throughput and fairness [1], [2] or according to the types of traffic [3]. “Throughput” and “fairness,” however, are conflicting perfor-mance metrics. To maximize system throughput, the system will allocate more resource to the users in better channel con-ditions. This may cause radio resource monopolized by a small number of users, leading to unfairness. On the other hand, to provide fairness to all users, the system tends to allocate more resource to the users in worse channel conditions so as to compensate for their shares. As a result, the system throughput may be degraded dramatically. The work in [4-5] show that the system can behave either “throughput-oriented” or “fairness-oriented” by adjusting certain parameters. However, they do not describe how to determine and justify the value of these parameters, leaving this trade-off unsolved.In this paper, we focus on “user satisfaction” for resource allocation to avoid such a “throughput-fairness” dilemma. Since it is unlikely to fully satisfy the different demands of all users, we turn to maximize the total degree of user satisfaction. The degree of user satisfaction with a given amount of resource can be described by the utility function U (r), a non-decreasing function with respect to the given amount of resource r. The more the resource is allocated, the
Manuscript received December 1, 2005; revised September 15, 2006; accepted December 22, 2006. The associate editor coordinating the review of this paper and approving it for publication was J. Hou. This work was supported by National Science Council (NSC), Taiwan, under a Center Excellence Grant NSC95-2752-E-002-006-PAE, and under Grant Number NSC95-2221-E-002-066.
W.-H. Kuo and W. Liao (corresponding author) are with the Department of Electrical Engineering and the Graduate Institute of Communication Engineer-ing, National Taiwan University, Taipei, Taiwan (email: wjliao@ntu.edu.tw).
Digital Object Identifier 10.1109/TWC.2007.05942.
more the user is satisfied. The marginal utility function defined by u(r) = dU(r)dr is the derivative of the utility function U (r) with respect to the given amount of resource r. The exact expression of a utility function may depend on traffic types, and can be obtained by studying the behavior and feeling of users. We leave the work of finding utility functions to psychologists and economists, and focus on maximizing the total utility for a given set of utility functions.
We are not the pioneer to study utility-based resource management in wireless networks. Proposals [6-10] are the examples which are based on the utility functions of different objectives for wireless networks. In [6], a utility-based power control scheme with respect to channel quality is proposed. In that scheme, users with higher SIR values have higher utilities, and thus are more likely to transmit packets. Therefore, the wireless medium can be better utilized and the transmission power can be conserved. The work in [7] propose a utility-based bandwidth allocation scheme, which can adapt to chan-nel conditions and guarantee the minimum utility requested by each user. In [8-9], the authors design a utility-based fair allocation scheme to ensure the same utility value for each user. However, letting users with different traffic demands to achieve an identical level of satisfaction may not be an efficient way of using wireless resource. Worse, traffic which is difficult to be satisfied tends to consume most of the system resource, leading to another kind of unfairness. In [10], a utility-based scheduler together with a Forward Error Correction (FEC) and an ARQ scheme is proposed. That work gives lagging users more resource and thus results in a similar performance level (i.e., fixed utility value) for each user. The work in [19-20] targets at multi-hop wireless networks.
Utility functions have also been widely used in Internet pricing and congestion control [11]. Many bandwidth pricing schemes have been proposed for wireless networks [12-16]. The typical approach is to set a price to radio resource and to allocate tokens to users. The objective is then to maximize the “social welfare” through a bidding process. These kinds of bidding schemes, while useful for Internet pricing and congestion control, may not be practical for wireless networks. In wireless environments, the types of traffic, the number of users, and channel conditions are all time-varying. It would be very expensive to implement a wireless bidding process because the users would have to keep exchanging control messages for real-time bidding, and the control protocols of the wireless system would also have to be modified to accom-modate this process. Finally, the complexity and efficiency of wireless bidding have not been analyzed. It is hard to estimate the time elapsed to achieve the Nash equilibrium.
In this paper, we study the wireless resource allocation
problem in the downlink of a wireless network with a central control system, such as a cellular base station, and attempt to maximize the total utility of all users without a bidding process. We consider two common types of traffic: hard QoS and best effort traffic, and propose three allocation algorithms1
for these two types of traffic, namely, 1) the HQ allocation for hard-QoS traffic, 2) the elastic allocation for best effort traffic, and 3) the mixed allocation for the co-existence of both types of traffic. These three allocation schemes are all polynomial time solutions and proved to be optimal under certain conditions, and in any case, the difference between the total utilities obtained by our solutions and the optimal utility are bounded. We also develop some theorems as the gen-eral design guidelines for utility-based resource allocation in wireless networks. The performance of the proposed schemes is validated via simulations. The results show that optimal wireless resource allocation depends on the traffic demand, total available resource, and wireless channel quality, rather than solely dependent on channel quality or traffic type as assumed in most existing work.
The rest of the paper is organized as follows. In Sec. II, three wireless allocation schemes are proposed and proved to be optimal under certain conditions. Some theorems are devel-oped as design guidelines for utility-based resource allocation schemes. In Sec. III, the performance of the proposed schemes is validated via simulations. Finally, the paper is concluded in Sec. IV.
II. UTILITY-BASEDALLOCATIONSCHEMES FOR
TWOTYPES OFTRAFFIC
A. Problem Statement and Definitions
Suppose that there are n users served by a base station. Let rtotal denote the total amount of radio resource available at the base station, and ri, the amount of resource to be allocated to user i. Users with the same kind of traffic may not feel the same way by given the same amount of resource because the wireless channel quality for each user may not be identical. Let qi denote the channel quality2 of user i, 0 ≤ qi ≤ 1, and i = 1, 2, · · · n. The smaller the value of qi, the worse the channel quality. Given an amount of resource ri and channel quality qi, the amount of resource actually beneficial to user i is given by θi = ri· qi. Let T (i) denote the type of traffic of user i. The utility function of user i is expressed by Ui(ri) = UT (i)(riqi), where UT (i)(.) is the utility function of traffic T (i) and Ui(.) is the utility function for the type of traffic described by UT (i)(.) but taking into account the channel quality of user i. The marginal utility
1These algorithms can help determine and validate the parameter settings
of existing schedulers. Most of the existing wireless schedulers focus on the design of scheduling systems which provide such performance guarantees as delay bounds of packets or fairness among users, but leave the weights of the queues (or users) undecided. This missing component can be provided by optimal resource allocation schemes as proposed in this paper.
2The channel quality parameter q
i is provided by the lower layers to indicate the proportion of effective transmission with a given amount of radio resource. Sinceqi may be time-variant, the proposed algorithm should be operated periodically to adapt to the current channel condition. For simplicity, but without loss of generality, we will treatqias a constant for each useri in the rest of the paper.
Bandwidth M U M r U
(a) Hard QoS
U
Bandwidth
(b) Best effort Fig. 1. The utility functions of two types of traffic.
u
Bandwidth
M
r
(a) Hard QoS
u
Bandwidth (b) Best effort Fig. 2. The marginal utility functions of two types of traffic.
function of Ui(.) is dUT (i)dr(riiqi)= qi· uT (i)(riqi), and that of
UT (i)(.) is uT (i)(.).
Our objective is to maximize n i=1Ui (ri), subject to n i=1ri ≤ rtotal and ∀ri ≥ 0. An optimal allocation for n users with total available resource rtotal is defined as follows. Note that the optimal allocation may not be unique in the system.
Definition 2.1: A resource allocation ∗ =
{r1, r2, · · · , rn} for n users is an optimal allocation if for all feasible allocations a = {r1, r2, · · · , rn},
U (∗) ≥ U(a), where U(∗) =
n i=1 Ui(ri) and U (a) = n j=1Uj (r j).
Definition 2.2: A unit-step utility function Ustep(r) refers to a utility function whose ustep(rM) = ∞ if r = rM, and
ustep(r) = 0, otherwise, where ustep(r) = dUstepdr(r).
Definition 2.3: A concave utility function Uconcave(r)
refers to a utility function whose uconcave(r) > 0 and
uconcave(r) < 0 for all r, where uconcave(r) = dUconcavedr (r) and uconcave(r) = duconcavedr (r).
By definition, a unit-step function is a discrete function, and a concave utility function is a non-decreasing and continuous function with respect to resource r. Fig. 2 plots the marginal utility functions for the two types of traffic shown in Fig. 1 More terminology used in this paper is defined as follows.
Definition 2.4: = {r1, r2, · · · rn} is a full allocation if
∀ri∈
ri= rtotal.
Definition 2.5: For = {r1, r2, · · · rn}, all users i with
ri> 0 (i.e., those who are allocated resources) are referred to as allocated users and all users j with rj= 0 (i.e., those who are not allocated resources) are unallocated users.
Definition 2.6: = {r1, r2, · · · rn} is marginally fair if it satisfies the following two conditions:
1) Each allocated user i (i.e., with ri> 0) in must have the same marginal utility value.
2) For each unallocated user j (i.e., with rj= 0) in , its marginal utility value uj(0) cannot exceed ui(ri), which is the marginal utility of each allocated user i.
Definition 2.7: For a marginally fair allocation = {r1, r2, · · · rn}, the marginal utility value of each allocated user in is equal, and is referred to as the allocated marginal utility um() . Thus, for each allocated user i in , ui(ri) =
um(), and for each unallocated user j, uj(0) < um().
B. HQ Allocation for Hard QoS Traffic
Suppose that there are n users in the queue, all with hard QoS traffic. Let rresidue denote the residual resource in the system. The resource allocation algorithm designed for users whose utility functions are all unit-step functions is referred to as the HQ allocation and the output is denoted byHQ= {r1, r2, · · · rn}. Given the total available resource in the system rtotal, the channel quality qi and utility function
UT (i)(.) for all user i, HQ can be obtained as follows. 1) Initialize ri← 0, i = 1, 2, · · · , n; rresidue← rtotal. 2) Sort all users i in the queue in descending order of
UMiqi
rMi .
3) Repeat Steps (4) and (5) until the queue becomes empty. 4) Pop out user i who is now at the head of the queue. 5) If rresidue> rMiqi , then
ri= rMiqi ; rresidue= rresidue− ri.
The utility function for user i with hard QoS traffic is described by UT (i)(r) = UMi × fu(qir − rMi), where fu(.) is a unit-step function, qi is the channel quality of this user,
Mi is the kind of QoS traffic, rMi is the preferred amount of
resource to be allocated.
The allocation rule of the HQ allocation is to assign resources to users in descending order of UrMiqi
Mi , subject to
k i=1
ri≤rtotal, where k is the largest value satisfying this constraint, as illustrated in Fig. 3. The allocation problem for users with arbitrary unit-step utility functions can be proved to be NP-complete. The performance of the HQ allocation can be proved to be close to the optimum. When the utility functions of all users are identical, the HQ allocation can be proved to be optimal.
Lemma 2.1: The allocation problem is NP-complete if the utility functions for all users are arbitrary unit-setp functions. Proof: This problem can be reduced from the 0/1 knap-sack problem, an NP-complete problem. Consider a knapknap-sack with capacity c > 0 and n items. Each item has a value of vi > 0 and a weight of wi > 0. The problem is then to find a selection of items that maximize n
i=1δivi subject to n
i=1δiwi
≤ c, where δi = 1 if the item is selected, and
0, otherwise. Therefore, any instance of the knapsack can be reduced to an instance of our problem by substituting rtotal = c, Ui = vi × fstep(qir − wi) and qi = 1, for
i = 1, 2, · · · , n. Since an optimal solution to our problem is also a solution to the given knapsack problem, the knapsack problem is a special case of our problem, it follows that our problem is NP-hard.
Next, it can be observed that our problem is an NP problem because any given solution can be verified as a feasible solution and bounded at a given utility value u in polynomial
U
1U
2U
k-1U
kAssignment
1 1 1 M M r q U 2 2 2 M M r q U 1 1 1 k k M k M r q U k k M k M r q UFig. 3. Allocation ordering ofk users in the HQ allocation
time. Since our problem is both in NP and NP-hard, it is an NP-complete problem.
Theorem 2.2: U (HQ) ≥ U(op) − Umax, where HQ
and op are the proposed solution and the optimal solution, respectively, of this HQ allocation problem, U (x) is the total utility of all users for solution x, and Umax is the maximum utility value taken over all users, i.e., Umax= max1≤k≤n{UMk} .
Proof: Let p denote the original HQ allocation prob-lem, which is NP-complete as proved in Lemma 2.1, and let p denote the problem by relaxing the integrity constraint of problem p (i.e., each user in p can be served fractionally). To better indicate the optimal solution to each problem, we let pop and pop represent the optimal solutions to problems p and p, respectively. Since p is a maximization problem, we have U (p
op) ≥ U(pop) ≥ U(pHQ), where p
HQ means
HQ.
The optimal solution to p, p
op, is obtained by sorting QoS users in the queue in decreasing order of UMiqi/rMi
as in pHQ. The difference between p
op and pHQ is only in those users fractionally served in p
op. It follows that
U (pHQ) + Ux> U (p
op), where Uxis the utility value of the unallocated user whose UrMiqi
Mi in
p
HQ is the largest. Thus,
Ux ≤ Umax. Since U (HQ) + Umax ≥ U(pHQ) + Ux >
U (pop) ≥ U(pop), we obtain U(pHQ) > U(pop) − Umax.
Theorem 2.3: The HQ allocation solution HQ is an
optimal allocation if the unit-step utility functions Ustep(r) of all n users in the queue are identical.
Proof: Let UM denote the unit-step utility function for all users. The inequality in Theorem 2.2 can be rewritten as nUM + UM > U (p op) ≥ U(pop) ≥ nUM, i.e., n + 1 > U(pop) UM ≥ U(pop) UM ≥ n. Since both U(pop)
UM and n are integral,
U (p
op) = nUM. Therefore,HQ must be optimal.
R1 R2 Rtotal Ubp ) ( 1 u ui ( ) 1 u ui iui u u u1( ) 1( ) iui u u u1( ) 1( )
Fig. 4. An example of theelastic allocation. First, the marginal utility functions of usersi and j are inverted (i.e., u−1i (r) and u−1j (r)). Then, the two inverted marginal utility functions are summed, i.e., u−1Σ (u) =
i u
−1
i (u). Finally, the relationship among uBE, rtotal, r1, and r2 is decomposed.
Proof: The time complexity of HQ can be expressed
by a function of the number of users in the network. Since the complexity of Step (2) (sorting) is O(n log n) and this iteration dominates the operation, the overall complexity of HQ is O(n log n).
C. Elastic Allocation for Best Effort Traffic
We next consider the best effort traffic. The resource al-location algorithm for users with concave utility functions Ui(r) is referred to as the elastic allocation and the output is denoted by elastic = {r1, r2, · · · rn}. Given the total available resource rtotal, the channel quality qi and marginal utility function ui(r) for each user i, elastic can be obtained as follows.
1) For each user i, derive u−1i (u), the inverted function of ui(r).
2) Derive u−1Σ (u) by summing up u−1i (u) over all users i, i.e., u−1Σ (u) =
i
u−1i (u).
3) Find uΣ(r), the inverted function of u−1Σ (u).
4) Find uBE, which is equal to uΣ(rtotal). 5) For all ri, i = 1, 2, · · · , n,
if uBE < ui(0), then ri= u−1i (uBE) ; else ri= 0.
The allocation rule of this scheme is to 1) derive the aggregated utility function from the inverse functions of all users, 2) calculate the allocated marginal utility from the aggregated utility function, and 3) determine ri for each user. As an example, Fig. 4 illustrates the elastic allocation algorithm with two best effort traffic .
Theorem 2.5: The elastic allocationelastic is optimal if the utility functions for all users are concave utility functions.
Proof: In the elastic allocation, 1) uBE is the marginal utility value at which point the total resource has been fully allocated, and 2) for all allocated users i in elastic,
uBE < ui(0) and ui(ri) = uBE; for all unallocated users
j, uBE≥ uj(0). With concave utility functions, i.e., function with u(r) > 0 for all r, it can be easily proved by contradiction that an optimal allocation for elastic users, i.e.,elastic, must be full. Similarly, by using contradiction again, we can prove that elastic must also be marginally fair. Since all users’ utility functions are increasing, only one allocation can be both full and marginally fair. Therefore, if an allocation is full and marginally fair, it must beelastic.
Theorem 2.6: The time complexity of the algorithm elastic allocation is O(n).
Proof: Since the operation at each step at most takes time O(n), the time complexity of the elastic allocation algorithm is O(n).
D. A Mixture of Hard QoS and Best Effort Traffic
Finally, we consider the co-existence of QoS and best effort traffic in the system, which is referred to as mixed allocation and the output of which is denoted bymix= {r1, r2, · · · rn}. Let rBE denote the amount of residual resource to be given to best effort traffic, and ΔUi, the utility gain by allocating resource ri to QoS user i. Other notations remain the same as in the HQ and the elastic allocations. Given the total available resource rtotal, the channel quality qi and marginal utility function ui(r) for each user i, mix can be obtained as follows.
1) Initialize ri← 0, i = 1, 2, · · · , n ; and rBE ← rtotal. 2) Sort all QoS users i in descending order of UrMiMiqi, and
store them in the queue.
3) For each best effort user j, derive u−1j (u) from uj(r); Find u−1Σ (u) by summing up u−1j (u) over all users j; Find uΣ(r), the inverted function of u−1Σ (u).
4) If the queue is not empty, then
pop out the QoS user i at the head of the queue; else go to Step (8).
5) For the popped user i:
if rBE ≥rqMii , then ri =rqMii ; else ri= 0; go to Step (4). 6) ΔUi= UMi− rBE rBE−ri uΣ(r)dr. 7) If (ΔUi> 0), then rBE = rBE− ri; go to Step (4); else ri = 0; go to Step (8); 8) If uΣ(rBE) < uj(0), then rj = u−1j (uΣ(rBE)); else rj= 0.
The allocation rule of this mixed allocation is to: 1) allocate resource to the first k QoS users at the sorted queue, and 2) then allocate the residual bandwidth (i.e., rtotal− rQoS) to all best effort users based on the elastic allocation. The value of k is determined based on the requirement that there is sufficient resource for this QoS user and the utility gain ΔUkis positive (i.e., rBE− rk> 0) .
We have proved in Lemma 2.1 that the allocation problem for hard QoS traffic is NP-complete. It follows that the
Bandwidth QoS users 2 U k U U1 total BE r r ,0 1 2 k 1 , BE r 2 , BE r k BE r , rBE,k 1 u Best Effort users
Fig. 5. An example of mixed allocation.
allocation problem for the co-existence of best effort and hard QoS traffic is also NP-complete. Again, the proposed mixed allocation algorithm can achieve a performance lower bounded by U (op) − Umax. When all QoS users have an identical utility function, the mixed allocation can be proved to be optimal.
Theorem 2.7: For the mixed allocation problem,
U (mix) ≥ U(op) − Umax, where mix is the proposed solution,op is the optimal solution, U (x) is the total utility of all users for solution x, and Umax is the maximum utility value taken over all users, i.e., Umax= max
1≤k≤n{UMk}.
Proof: This theorem can be proved by relaxing the con-straint in the original problem as in Theorem 2.2. Therefore, the different in the total utility betweenmixand the relaxed problem is bounded by Umax, i.e., U (mix) ≥ U(op) −
Umax.
Theorem 2.8: The mixed allocation mix is an optimal
allocation for traffic mixed with identical unit-step functions and arbitrary concave functions.
Proof: Consider the first k QoS users in the sorted
queue, where k is the maximum possible value satisfying k
i=1ri
≤ rtotal. The set of all possible optimal allocations is then:0, 1, · · · , i, · · · , k, wherei is the allocation in which the first i QoS users are allocated resource in an amount of rj = rM/qj, j = 1, 2, · · · , i, and the residual bandwidth is all allocated to best effort users. The utility gain ΔUi,
i = 1, 2, · · · , k, is expressed by ΔUi= U(i) − U(i−1) =
UMi−
rBE
rBE−ri
uΣ(r)dr.
Since QoS users are sorted in decreasing order of their qi, this leads to that ri is allocated in increasing order of qi, resulting in ΔUi ≥ ΔUi+1 (as shown in Fig. 5). Thus, the allocation i in [0, 1, 2, · · · , i, · · · , k] is the optimal allocation, where i is the largest value satisfying ΔUi> 0.
E. Implementation Issue
In practice, we can avoid such operations as inverting nonlinear functions, inverting the summations of nonlinear functions, and inverting non-linear functions by a series of summations and simple interpolations.
Suppose umax is the maximum marginal utility value allowed by the system. We can partition the range of the
0.02 0.09 0.10 0.17 0.20 0.30 0.34 0.43 0.55 0.56 0.62 0.76 0.86 0.90 0.99 1.00 0 20 40 60 80 100 120
r
iq
ir
total (a)ri 0.02 0.09 0.10 0.17 0.20 0.30 0.34 0.43 0.55 0.56 0.62 0.76 0.86 0.90 0.99 1.00 0 10 20 30 40 ir
total qi (b)θiFig. 6. Resource allocation inHQ= {r1, r2, · · · rn}.
values in the y-axis for any (continuous) marginal utility function, say u(.), into values spaced equally apart, i.e., uk = umax− kumaxM , where M is a tunable input parameter
which determines the number of steps in total, and k = 0, 1, · · · , M. We then let rk = u−1(uk), k = 0, 1, · · · , M, in a one dimensional fixed length array Γ = [r0, r1, r2, ....rM]. Since u(.) is a decreasing function, there is a one-to-one correspondence between rk and uk. This method eliminates the inversion process and can obtain any required values (i.e., u(r) and u−1(u)) by interpolation. For a given value r, the system can find the value k satisfying rk ≤ r < rk+1. By interpolation, u(r) ∼= uk − umaxM · rk+1r−r−rkk. Similarly, for any given value u, u−1(u) ∼= rk + (rk+1− rk) · uk+1u−u−ukk,
where k is the value satisfying uk+1 < u ≤ uk. This approach can also reduce the computational overhead of the summation function uΣ(u), which can be calculated by RΣ=
[j i=1u −1 i (u0), j i=1u −1 i (u1), · · · , j i=1u −1
i (uM)] given that there are j utility functions (u1, u2, · · · , uj) to be aggregated. Like-wise, the overhead for the integral of uΣ(u) can also be
reduced.
In this way, the computational complexity of finding u(r) and u−1(u) is only O(M) because it takes at most M steps to find the value k for rk ≤ k < rk+1. The complexity of the summation is O(jM ) because for k = 0, 1, · · · , M , it takes j steps to calculate j i=1u −1 i (uk).
0.02 0.11 0.12 0.19 0.30 0.37 0.41 0.51 0.51 0.70 0.71 0.80 0.86 0.87 0.91 0.92 0 20 40 60 80 100 120 140
r
iq
ir
total (a)ri 0.02 0.11 0.12 0.19 0.30 0.37 0.41 0.51 0.51 0.70 0.71 0.80 0.86 0.87 0.91 0.92 0 10 20 30 40 50r
total iq
i (b)θiFig. 7. Resource allocation inelastic= {r1, r2, · · · rn}.
III. PERFORMANCE EVALUATION
In this section, we conduct simulations to evaluate the performance of our allocation algorithms. We consider QoS traffic, best effort traffic, and the con-existence of both. The simulation parameters are described as follows. For QoS traffic, the utility function is a unit-step function with ra = 10 and UM = 1, i.e., UQoS(r) = fu(r−10); for best effort traffic,
UBE(r) = 1 − er/10. The value of qi is randomly generated by a uniform distribution over [0, 1]. We then measure the distributions of ri and θi under different values of rtotal.
Fig. 6 shows the values of ri and θi for QoS traffic in the
HQ allocation. Fig. 6(a) depicts that when rtotal is small, the system tends to allocate more resource to the users in better channel conditions; but as rtotal increases, the amount of resource allocated to users is still fixed either at ra or 0, because the utility function is a unit-step function. Fig. 6(b) shows that the actual amount of resource obtained by each user, if allocated, is identical, i.e., at their θi.
Fig. 7 shows the values of ri and θi for best effort traffic in the elastic allocation. Fig. 7(a) depicts that when rtotal is small, the system tends to allocate more resource to the users in better channel conditions. Since the utility function of best-effort traffic satisfies u(r) > 0 for all r, all users will
0 250 500 750 1000 1250 1500 1750 50 250 450 650 850 1050 1250 1450 1650 1850 rtotal A llo ca te d B a n d w id th r_QoS r_BE
Fig. 8. An example of mixed allocation.
demand resource all the time. Thus, the value of ri for each user i increases with the value of rtotal. Fig. 7(b) shows that users with larger qi always result in larger θi.
Fig. 8 shows the values of rQoS and rBE, as a function of rtotal for mixed traffic in the mixed allocation. Since the QoS traffic is allocated resources in discrete amounts, when one more QoS user is served, the value of rBE will drop. This effect becomes more pronounced when rtotal is large, implying that the remaining unallocated QoS users have lower qi and need more resources to compensate for their bad channel qualities. Due to the space limitations, we do not include the distributions of ri and θi for the mixed
allocation in this paper. The results can be found in [18], the characteristics of which are similar to those in Figs. 6 and 7. In Fig. 9, different resource allocation schemes are com-pared with the proposed allocation schemes. The comparison is based on the scheme propsoed in [4], which allocates radio resource proportionally based on factor qα
i. Depending on the setting of the value α, the system can be tuned to work with different performance metrics. The curve denoted ”throughput” is for α = 1, which gives more resources to the users in better channel conditions, thereby leading to a larger system throughput. The curve denoted ”fairness” is for α = −1, giving all users an identical value of θi = ri · qi. The curve denoted ”fixed” is for α = 0, which provides the same amount of resource to all users. Note that the schemes proposed in [8-9] are the examples of the ”fairness” scheme (i.e., α = −1), and the GR+ scheme in [1] is an example of the ”throughput” scheme.
Fig. 9(a) compares the proposed HQ allocation with differ-ent allocation schemes, and Fig. 9(b) compares the proposed elastic allocation with different allocation schemes. Note that the axis of rtotal in Fig. 9(b) is in the logarithmic scale. The results show that the ”throughput-first” scheme has a higher total utility when rtotalis small, but the ”fixed” allocation one is closer to the proposed scheme as rtotal increases. Finally, when rtotal becomes very large, the ”fairness-first” scheme can achieve the highest utility.
IV. CONCLUSION
In this paper, we study utility-based maximization for resource allocation in infrastructure-based wireless networks. We develop some essential theorems for utility-based resource
0 2 4 6 8 10 12 14 16 0 200 400 600 800 1000 1200 1400 1600 1800 2000 rtotal Utility HQ Throughput Fixed Fairness
(a) QoS traffic
0 2 4 6 8 10 12 14 16 100 1000 10000 rtotal Utility Elastic Throughput Fixed Fairness
(b) Best effort traffic
Fig. 9. Utility comparison with different resource allocation schemes.
management. Then, three polynomial time resource allocation algorithms are proposed for two types of utility functions. We prove that, in any case, the difference between the total utilities obtained by our proposed solutions and the optimal utility is bounded, and under certain conditions, all these three schemes can achieve the maximum total utility (i.e., optimal). From the simulation results, we find that different types of traffic require different kinds of schemes to achieve optimal allocation. In addition, when rtotal is small, the system tends to allocate more resources to the users in better channel conditions, i.e., ”throughput-oriented;” however, when rtotal is abundant, the system becomes ”fairness-oriented,” meaning that even with the same traffic, the preference tendency between throughput and fairness can still differ. This leads us to conclude that existing channel-dependent-only resource schemes and sched-ulers cannot provide optimal allocation in wireless networks. To the best of our knowledge, this is the first work to address all the three issues together.
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Wen-Hsing Kuo was born on February 6, 1980,
in Taichung, Taiwan. He received the B.S. degree in electrical engineering from National Taiwan Univer-sity, Taipei, Taiwan, in 2002, where he is currently working toward the Ph.D. degree in the Graduate Institute of Electronics Engineering.
He joined Internet Research Laboratory, NTU, in 2002. His research interests include wireless re-source management, network economics and 802.16 WiMAX networks.
Wanjiun Liao received the BS and MS degrees
from National Chiao Tung University, Taiwan, in 1990 and 1992, respectively, and the Ph.D. degree in Electrical Engineering from the University of Southern California, Los Angeles, California, USA, in 1997. She joined the Department of Electrical Engineering, National Taiwan University (NTU), Taipei, Taiwan, as an Assistant Professor in 1997. Since August 2005, she has been a full professor. Her research interests include wireless networks, multimedia networks, and broadband access net-works.
Dr. Liao is currently an Associate Editor of IEEE Transactions on Wireless
Communications and IEEE Transactions on Multimedia. She served as the
Technical Program Committee (TPC) chairs/co-chairs of many international conferences, including the Tutorial Co-Chair of IEEE Infocom 2004, the Technical Program Vice Chair of IEEE Globecom 2005 Symposium on Au-tonomous Networks, and the Technical Program Co-Chair of IEEE Globecom 2007 General Symposium. Dr. Liao has received many research awards. Papers she co-authored with her students received the Best Student Paper Award at the First IEEE International Conferences on Multimedia and Expo (ICME) in 2000, and the Best Paper Award at the First IEEE International Conferences on Communications, Circuits and Systems (ICCCAS) in 2002. Dr. Liao was the recipient of K. T. Li Young Researcher Award honored by ACM in 2003, and the recipient of Distinguished Research Award from National Science Council in Taiwan in 2006. She is a Senior member of IEEE.
