A GENERALIZED COUPON COLLECTOR PROBLEM

Transcription

1 J. Appl. Prob. 48, (20) Printe in Englan Applie Probability Trust 20 A GENERALIZED COUPON COLLECTOR PROBLEM WEIYU XU an A. KEVIN TANG, Cornell University Abstract This paper presents an analysis of a generalize version of the coupon collector problem, in which the collector receives coupons each run an chooses the least-collecte coupon so far. In the asymptotic case when the number of coupons n goes to infinity, we show that, on average, (n log n)/ +(n/)(m ) log log n+o(mn) runs are neee to collect m sets of coupons. An exact algorithm is also evelope for any finite case to compute the exact mean number of runs. Numerical examples are provie to verify our theoretical preictions. Keywors: Coupon collector problem; expecte run; state-space representation; wireless communication; opportunistic scheuling 200 Mathematics Subject Classification: Primary 60G70 Seconary 60C05. Introuction The classic coupon collector problem asks the following question: Given that a collector ranomly receives a coupon each run, how many runs are necessary to collect a complete set of n ifferent coupons? The answer is nh n, where H n = n k= /k is the harmonic number [2]. One can further ask how many runs are necessary to collect m complete sets of coupons. This question has been aresse by Newman an Shepp [9]. The coupon collector problem an its variants are of traitional an recurrent interest [3], [4], [5], [7], [8]. Besies their rich theoretical structures an implications, they have various applications, incluing ynamic resource allocation, hashing, an online loa balancing [], to name just a few. In particular, these problems also serve as basic moels to analyze the elay for opportunistic scheuling in broacast wireless faing channels [0]. For example, to maximize system throughput, we shoul serve to the user whose channel conition is the best at every time slot. In orer to evaluate performance, we coul etermine the expecte number of time slots neee for all users to be serve at least once. Assuming that all channels are inepenent an ientically istribute, this is equivalent to the classic coupon collector problem. In this paper we investigate a natural generalization of the coupon collector problem. Instea of receiving one coupon, the collector receives ( n) istinct coupons ranomly each run an chooses the least-collecte coupon so far. Formally, we enote the number of runs (a run is often referre to as a time slot or simply as a unit of time in this paper) necessary to collect m sets of coupons as Dm,n. We are intereste in characterizing the mean value of the ranom variable Dm,n, especially in the asymptotic region when n is large. Clearly, when Receive 25 November 200; revision receive June 20. Postal aress: School of Electrical an Computer Engineering, Cornell University, Ithaca, NY 4853, USA. aress: wx42@cornell.eu aress: atang@ece.cornell.eu 08

2 082 W. XU AND A. K. TANG =, we go back to the classic cases; when = n, there is no ranomness an Dm,n n = mn. In the scheuling transmission context iscusse above, can be viewe as a parameter that controls the trae-off between efficiency (high throughput) an fairness among all users, with = purely focusing on efficiency while = n correspons to perfect fairness. Essentially, we are intereste in investigating the power of multiple choices in speeing up the wireless scheuling; it is worth mentioning that the power of two choices has been investigate in the ranomize loa balancing setting, where two choices can exponentially reuce the maximum loa; see, for example, [6]. The remainer of this paper is organise as follows. We first briefly review existing relate results in Section 2. Although they are all special cases of the general problem, the techniques use to erive them cannot be applie irectly to the general case. Instea, we evelop a new technique to characterize E(Dm,n ), an provie upper an lower bouns for E(D m,n ) in Section 3 an Section 4. An asymptotic analysis shows that the upper boun an lower boun match in the asymptotic regime of n in Section 5. Furthermore, for any finite n, an algorithm is motivate an propose in Section 6 to calculate E(Dm,n ) exactly. We use numerical examples to valiate our theoretical preictions in Section 7. Concluing remarks an future work are given in Section Existing results Existing results for special cases are liste below. If = then the problem is solve for all m. If > then only the m = case is known. = an m = (see [2]). It is clear that the number of runs neee to obtain the (i + )th coupon after obtaining the ith coupon follows a geometric istribution with parameter (n i)/n. Therefore, E(D,n ) = nh n = n For large n, E(D,n ) = n log n + no(). We see that the ranomness cost is expresse approximately by a factor log n. = an m (see [9]). We have where For fixe m an large n, n k= k. E(Dm,n ) = n ( ( S m (t)e t ) n ) t, 0 S m (t) = m k=0 t k k!. E(Dm,n ) = n log n + n(m ) log log n + no(). It is interesting to note that although collecting the first set of coupons nees about n log n runs, all later sets nee only n log log n runs per set.

3 A generalize coupon collector problem 083 an m = (see [0]). This has applications in the scheuling of ata packet transmission over wireless channels. Here n E(D,n ) = ( i ( ) / n ), where ( i ) = 0ifi<. For fixe m an large n, E(D,n ) n log n. This shows that, for the m = case, choosing coupons ranomly each time ecreases the expecte number of runs, with the greatest reuction occurring between = an = 2. an m. In the context of scheuling, a transmitter who wants to sen m packets to each of the n users can only transmit one packet to one user chosen from the users who have the best wireless communication channels. Owing to the time varying nature of the wireless channels, it is natural to assume that, for each time inex, the users who have the best communication channels are uniformly istribute among the n users. So E(Dm,n ) gives an estimate on the total elay in elivering these m packets, an, in this paper, we will offer a characterization of E(Dm,n ). 3. Lower boun on E(D m,n ) We will first lower boun E(Dm,n ) by consiering a ifferent coupon collecting process. In this new process, each time, we uniformly select istinct coupons out of n coupons, instea of keeping only one coupon out of these selecte coupons, we woul keep all the coupons. The expecte time of collecting m sets of coupons in this way will be no larger than the process in which we keep only one coupon each. However, it is not so straightforwar to irectly obtain an estimate for this new process. This motivates us to consier another process in which each time we collect uniformly, inepenently chosen (allowing repetition) coupons we keep all of them. This process stops when m sets of coupons are fully collecte. Lemma. Let t be the expecte time to collect m sets of coupons for the process in which each time uniformly chosen istinct coupons are kept. Let t 2 be the expecte time to collect m sets of coupons for the process in which each time uniformly chosen (allowing repetition) coupons are kept. Then ( n t ) n t 2. () Proof. We simulate the process of choosing istinct coupons through an expurgate process of choosing inepenent coupons (allowing repetition). If the coupons we inepenently choose (allowing repetition) are not istinct, we will iscar this group of coupons; if the coupons are all istinct, we will keep them. The kept coupons from the expurgate process follow the same istribution as the chosen istinct coupons. However, the expecte time to collect a group of istinct coupons is clearly n / ( n). So in the worst case, t2 n t / ( n ). In summary, in orer to give a lower boun on E(Dm,n ), we first nee a lower boun on t 2 for the process of keeping uniformly, ranomly chosen coupons (allowing repetition). To o this, we follow the approach of generating functions in [9].

4 084 W. XU AND A. K. TANG Let p i be the probability of failing to obtain m sets of coupons when we have kept i coupons. Let P x,...,x n be a power series, an let {P x,...,x n } be the power series when all terms having exponents greater than or equal to m have been remove. Thus, t 2 = j=0 p j an with x,...,x n all equal to. In aition, we know that p j = {(x + +x n ) j } n j, E(Dm,n ) = q=0 j=0 p j+q = n where S m (t) = m k=0 tk /k! [9]. We also note that p i is nonincreasing as i grows, so t 2 = p j ( ) p j n j=0 j=0 0 0 ( ( S m (t)e t ) n ) t, ( ( S m (t)e t ) n ) t. So by () we know that ( n ) E(Dm,n ) E(Dm,n ) n. 4. Upper boun on E(Dm,n ) In this section we upper boun the expecte time necessary to collect m complete sets of coupons. To achieve this, we upper boun the expecte time necessary to collect m complete sets of coupons in a suboptimal process. In this new process, each time, we uniformly an inepenently choose coupons (allowing repetition). Among this group of coupons, if the ith ( i ) coupon is the first coupon to so far have fewer than m copies then we will keep this coupon an iscar the remaining i coupons. First, we observe that istinct coupons are favorable in terms of minimizing the collection time compare to coupons with possible repetition. Theorem. The minimize expecte time of collecting m sets of coupons when the collector is given uniformly chosen, istinct coupons each run, but is allowe to keep only one coupon, is no larger than the minimize expecte time of collecting m sets of coupons when the collector is given uniformly chosen coupons (allowing repetition) each run, but is allowe to keep only one coupon. Proof. We consier two couple coupon collecting processes: process P, which receives inepenent, uniformly chosen coupons (allowing repetition) each run, an process Q, which receives istinct coupons each run. For each run, process Q sequentially receives istinct coupons in this way: if the ith ( i ) receive coupon of process P is a repetition of one of the first i receive coupons of process Q, the ith receive coupon of process Q will be

5 A generalize coupon collector problem 085 uniformly chosen among the unreceive n (i ) coupons of process Q in this run; otherwise, the ith ( i ) receive coupon of process Q will be the same as the ith ( i ) receive coupon of process P. Since in each run all types of coupon receive in process P are also receive in process Q, the coupon collector using process Q can keep the same coupon as in process P an, thus, has no larger collecting time than using process P. Secon, we show that it is an optimal strategy for the coupon collector to keep the leastcollecte coupon out of the incoming coupons (whether allowing repetition or not). Theorem 2. The expecte time to collect m sets of coupons is minimize when the collector keeps the least-collecte coupon each run, if the collector is allowe to keep only one out of the offere coupons. Proof. Suppose that we have c coupons of type j ( j n) an c 2 coupons of type l ( l n), where j = l an c <c 2 <m. Now consier two coupon collectors P an Q. From an incoming set of coupons which contain both type j an type l, P keeps a coupon of type l, an we use state A, represente by the tuple {c,c 2 }=(c,c 2 + ), to recor the numbers of type-j an type-l coupons kept by P. Collector Q instea keeps a coupon of type j, an we corresponingly use state B, represente by {c,c 2 }=(c +,c 2 ). Now we nee only to argue that, to collect m sets of coupons, on average, collector Q, who starts from state B, will take no longer than collector P, who starts from state A. In each run, P an Q receive the same set of coupons, an a collector can keep a coupon only if he/she has fewer than m copies of that type. We consier the following-an-upating process escribe below. We let P make his/her optimal keeping ecision each run to minimize the expecte collecting time. Then we let collector Q keep the same type of coupon as collector P. For a run where collectors P an Q ecie to keep a coupon of type j or type l, we upate their states A an B. There are two possible cases. In the first case, c = c 2 an it is the coupon of type j that both collectors ecie to keep in that run. Then P an Q are in states {c +,c 2 } an {c +,c 2 }, respectively. We upate state A to {c +,c 2 }, but instea symmetrically upate the state of Q to B ={c 2,c + }, which has the same optimize expecte collecting time as state {c +,c 2 }.Ifc + = c 2, A an B are two equivalent states (note that c + c 2 = c + c 2 ); thus, we achieve our objective. In the secon case, after keeping a new coupon of type j or type l, we simply upate states A an B to recor the new numbers of type-j an type-l coupons. Before the states of the two collectors become equivalent, the collectors repeat the followingan-upating process escribe above, starting from new states A an B each iteration. With a little abuse of notation, we still use {c,c 2 } an {c,c 2 } to represent states A an B before each iteration, even though the values of c, c 2, c, an c 2 may have change over iterations. It is not har to see that the following conitions always hol before each iteration: c + c 2 = c + c 2, c c 2, c c 2, c 2 c >c 2 c, c,c 2,c,c 2 m. Because of these conitions, in each run, when P keeps a certain type of coupon, Q can also follow P s ecision an keep the same type of coupon. In each iteration, we always increase c + c 2 an c + c 2, an because c,c 2,c,c 2 m, we will eventually obtain a pair of equivalent states in some iteration. Since we can always arrive at two equivalent states for the two coupon collectors, starting from B to collect m sets of coupons will not take any longer than starting from A.

6 086 W. XU AND A. K. TANG Now we prepare a final lemma before presenting the upper boun on E(D m,n ). Lemma 2. The function f(i)= n i ( i/n) is nonincreasing in i [,n], an / f(i) 0 for i n. Proof. We nee to show that, for i [,n], the following erivative is nonpositive: f (i) = Let g(x) = ( x) + x( x) ( )/2,so ( i/n) n( ( i/n) ) 2 n i 2. g (x) = 2 ( x)( 3)/2 (x + x + 2( x) (+)/2 2). Also, let h(x) = x + x + 2( x) (+)/2 2, so an h(x) h(0) = 0 for x>0. Because h (x) = + ( + )( x) ( )/2 0, g (x) = 2 ( x)( 3)/2 h(x) 0 for 0 <x, we have g(x) g(0) = for 0 <x. This translates into ( x) x( x) ( )/2 for 0 <x, so ( x) n( ( x) ) 2 nx 2 for 0 <x. Plugging in x = i/n,wehave ( i/n) n( ( i/n) ) 2 n i 2 for i n. So f (i) 0 for i n. Calculating f (n),wehave n i ( i/n) for i n. At this point, we are reay to present the following upper boun for E(D m,n ). Theorem 3. Suppose that the coupon collector is given uniformly ranomly chosen istinct coupons an that he/she is allowe to keep only one out of these istinct coupons. Then the expecte time E(Dm,n ) E(D m,n )/ + mn( /). Proof. Inspire by Theorems an 2, we consier the process in which each run inepenently chosen coupons (allowing repetition) are presente an among them the coupon collector keeps only the first useful coupon which he/she has fewer than m copies of.

7 A generalize coupon collector problem 087 We focus on upper bouning the expecte finishing time, conitioning on a specific sequence of mn kept useful coupons, specifie by their type an the orer in which they are kept. Suppose that immeiately after the rth kept coupon in this keeper sequence, there are s types of coupon for which the coupon collector has m copies. Conitioning on the specifie keeper coupon sequence, unless an incoming coupon belongs to the s types of coupon for which the collector alreay has m copies, this incoming coupon must be the (r + )th keeper coupon. So, given that the keeper sequence is known, with probability ( (n s)/n), none of the uniformly chosen (allowing repetition) coupons is the known (r + )th keeper. So, conitioning on the specifie keeper sequence, the expecte time to collect the (r + )th keeper coupon is E = ( (n s)/n). However, if the coupon collector is offere only one instea of coupons each run, the expecte time to collect the (r + )th coupon is E = n/(n s). By Lemma 2, E E / / for any s n. Since a whole keeper sequence S has mn coupons, the total expecte time E S to collect them, when coupons (allowing repetition) are offere each run, an the total expecte time E, when one coupon is offere each run, satisfy E S E S / mn( /). We note that the probability that the coupon collector follows a specific sequence of keeper coupons is the same no matter whether = or not. This is because, among a batch of offere coupons, the collector still checks them one by one an keeps only the first coupon that is useful. So this implies that E(Dm,n ) E(D m,n )/ + mn( /) for any. We remark that the upper boun in Theorem 3 can also be obtaine by consiering the classic coupon collector process an iviing the incoming coupons into batches of coupons: after each coupon is accepte, the collector throws away the next coupons to ensure that the next kept coupon belongs to a fresh batch of coupons. The proof of Theorem 3 reuces to an exact result for the m = case, an can lea to tighter upper bouns for general m if a finer analysis is applie (for example, using the exact formula instea of the upper boun in Lemma 2, which is possible at least for collecting the last coupon). 5. An asymptotic analysis (n ) In this section we provie an asymptotic analysis for the upper an lower bouns of E(Dm,n ), an we see how this analysis behaves asymptotically for fixe an m as n goes to. We will begin with an asymptotic analysis through an exact expression for E(D,n ). Theorem 4. When n is large enough an >, ( ) log n E(D,n ) = n + O(). Proof. We have n E(D,n ) = ( i ( ) / n ) = n i(i ) (i + )/n(n ) (n + ) S

8 088 W. XU AND A. K. TANG n + ((i + )/(n + )) i= n = + ( i/(n + )). i= Since ( x) x for 0 x, E(D,n ) n i= n + i = n + n n ( n n i = ) E(D,n ). n Thus, n E(D,n ) = i(i ) (i + )/n(n ) (n + ) = n i= (i/n) n ( i/n) n ( n i ) + i= = E(D,n ) + n ( ). Theorem 5. When m is fixe, for any >, E(Dm,n lim ) n log(n)/ n (n(m ) log log n)/ =. Proof. From the lower boun an upper boun for E(Dm,n ) in Section 3 an Section 4, we know that E(Dm,n lim ) n E(Dm,n ) =. Then the asymptotic expression emerges immeiately by recalling the asymptotic expression for E(D m,n ). 6. An algorithmic approach (for any finite n) In this section we give an algorithm which calculates exactly E(D m,n ) for specifie m, n, an base on a state-space representation of the Markov process of collecting the coupons. For each n 0,n,n 2,...,n m 0 satisfying n 0 + n + +n m = n, efine S m = (n 0,n,...,n m ) to be the state, where n i (0 i m) is the number of coupons that the coupon collector has

9 A generalize coupon collector problem 089 collecte i times. Hence, E(Dmn ) is the expecte number of runs necessary for the coupon collector to go from state (n, 0,...,0) to state (0,...,0,n). We now provie an algorithm to calculate E(Dm,n ). Define N m (S m) to be, starting from state S m, the number of runs after which m-complete sets of coupons have been collecte, i.e. the number of runs from state S m to (0,...,0,n). Clearly, N m (n, 0,...,0) = D m,n an Nm (0,...,0,n)= 0. Suppose that we are in state S m = (n 0,n,...,n m ). After one run, the transition probability from S m to the following two states are as follows: ( )/( ) nm n (n 0,n,...,n m ) with probability, (n 0,...,n i,n i+ +,...,n m ) with probability p i, 0 i<m, where Therefore, we have (( mt=i ) n t p i = E(N m (n 0,...,n m )) = + (( nm m + ( mt=i+ n t ))/( ) n. )/( )) n E(Nm (n 0,...,n m )) p i E(N m (n 0,...,n i,n i+ +,...,n m )). So (( )/(( ) n n E(Nm (n 0,...,n m )) = m (( mt=i ) n t ( nm ))) ( mt=i+ n t ))/(( ) n ( nm )) E(N m (n 0,...,n i,n i+ +,...,n m )). (2) Define the map : {(n 0,...,n m ): n 0,n,...,n m 0, n 0 + n + +n m = n} N, where (n 0,n,n 2,...,n m ) = Obviously, is an injection an m ( + n) m i n i. (n, 0,...,0) = n( + n) m, (0,...,0,n)= n.

10 090 W. XU AND A. K. TANG Since (n 0,...,n m ) (n 0,...,n i,n i+ +,...,n m ) = (( + n) m i n i + ( + n) m i n i+ ) (( + n) m i (n i ) + ( + n) m i (n i+ + )) = ( + n) m i ( + n) m i > 0, by (2), the expecte number of runs from state S epens only on the expecte number of runs from the states S with (S )< (S). Therefore, we can orer all the states (n 0,...,n m ) accoring to the value of (n 0,...,n m ), an compute E(N m (n 0,...,n m )) one by one, from the starting state (0,...,0, n) to the last state (n, 0,...,0). The algorithm is escribe below. Algorithm. (Calculating E(Dm,n ).) for n 0 = 0 to n for n = 0 to n n 0. for n m = 0 to n m 2 n i o n m = n m n i if n m = n then E(N m (n 0,...,n m )) = 0 else use (2) to compute E(N m (n 0,...,n m )) ( Since the number of nonnegative integer solutions to the equation n 0 + +n m = n is n+m ) ( n, the number of states is n+m ) ( n, an the complexity of Algorithm is O( n+m ) n ). To conclue this section, we now use a simple example (n = 6 an m = 2) to illustrate Algorithm. When m = 2, each state has three parameters: n 0, n, an n 2. Since n 0 + n + n 2 = n, we coul raw the state transition iagram as in Figure. Algorithm computes E(N2 (n 0,n,n 2 )) for each state (n 0,n,n 2 ) by the orer shown in Figure 2. Note that the expecte number of runs from any state epens only on the number of runs from its escents in Figure, 0,0,6,0,5 0,,5 2,0,4,,4 0,2,4 3,0,3 2,,3,2,3 0,3,3 4,0,2 3,,2 2,2,2,3,2 0,4,2 5,0, 4,, 3,2, 2,3,,4, 0,5, 6,0,0 5,,0 4,2,0 3,3,0 2,4,0,5,0 0,6,0 Figure : State transition iagram for n = 6 an m = 2. Self-loops are omitte. The noes are labele with the values of n 0, n, an n 2.

11 A generalize coupon collector problem Figure 2: State transition iagram for n = 6 an m = 2. The noes are labele with the computation orer. The highest noe, which represents (0, 0, 6), is labele 0 because N2 (0, 0, 6) is known to be Figure 3: State transition iagram for n = 6 an m = 2. The noes are labele with the value of E(N 2 2 (n 0,n,n 2 )). the computation of E(N2 ( )) for any state is carrie out after the computations for its escents by Figure 2. The values of E(N 2 (n 0,n,n 2 )) for each state (n 0,n,n 2 ) is shown in Figure Numerical examples We now engage in numerical exercises to support the results of the last two sections, i.e. the correctness of Algorithm an the erive upper an lower bouns of E(D m,n ). 7.. Algorithm First, we give numerical results for the expecte collection time when n = 00 an m =, 2, 3, respectively, in Tables 3. The results show that Algorithm gives an expecte elay consistent with the simulation results.

12 092 W. XU AND A. K. TANG Table : Numerical results for the expecte collection time when m = an n = Algorithm Simulation Table 2: Numerical results for the expecte collection time when m = 2 an n = Algorithm Simulation Table 3: Numerical results for the expecte collection time when m = 3 an n = Algorithm Simulation Asymptotic results Two cases are consiere: = 3 an m =, an = 3 an m = 2. For each case, in Figures 4 an 5 we plot the lower boun from Theorem 2, the upper boun from Theorem, an the result compute from Algorithm for n from 00 to 500. The results show that the upper an lower bouns boun the expecte collecting time very well. In fact, when m an are fixe, the upper an lower bouns both scale as (/) E(Dm,n ) as n. Expecte collecting time Lower boun Upper boun Exact value n Figure 4: Asymptotics for m = an = 3.

13 A generalize coupon collector problem 093 Expecte collecting time Lower boun Upper boun Exact value n Figure 5: Asymptotics for m = an = Conclusion an future work In this paper we have consiere a generalize coupon collector problem in which the coupon collector nees to collect m sets of coupons an has the freeom to keep one coupon out of the coupons offere each time. We obtaine asymptotically matching upper an lower bouns for the expecte collection time. We also provie an algorithm to calculate the expecte collection time exactly base on a state representation for the coupon collecting process. We shoul note that, asymptotically, even if the coupon collector is allowe to keep only one coupon out of the coupons, the neee time will still be shortene by a factor of, compare to if the coupon collector is allowe to keep all the coupons offere each time. There is much scope for future work on this problem. First, one coul attempt to obtain a close-form expression for E(Dm,n ). Secon, one coul attempt to improve Algorithm. Algorithm has a run time of ( n+m) n. To take avantage of this run time requires constant time inexing. The irect approach is to inex the states in an n-imensional matrix of size (n+) m +. However, since there are a total of ( n+m) n states, a large fraction of the matrix space is not require. Hence, it woul be helpful to fin an algorithm which carries out triangular inexing in constant time. This woul reuce the memory requirements an increase the range of parameters over which the problem is computationally feasible. One coul further observe that although there are ( n+m) ( n states, only n+m ) m are actually neee at any time. So, with constant time triangular inexing, one coul reuce the memory requirements further, although the gain from the secon reuction is minimal. Acknowlegements The authors woul like to thank the anonymous reviewers for insightful comments an suggestions. The authors woul also like to thank the input of Xiaojie Gao an Wuhan Desmon Cai. References [] Azar, Z., Broer, A. Z., Karlin, A. R. an Upfal, E. (999). Balance allocations. SIAM J. Comput. 29, [2] Feller, W. (950). An Introuction to Probability Theory an Its Applications. John Wiley, New York.

14 094 W. XU AND A. K. TANG [3] Foata, D. an Zeilberger, D. (2003). The collector s brotherhoo problem using the Newman-Shepp symbolic metho. Algebra Universalis 49, [4] Holst, L. (200). Extreme value istributions for ranom coupon collector an birthay problems. Extremes 4, [5] Kan,N.D.(2005). Martingale approach to the coupon collection problem. J. Math. Sci. 27, [6] Mitzenmacher, M. D. (996). The power of two choices in ranomize loa balancing. Doctoral Thesis, University of California, Berkeley. [7] Myers, A. N. an Wilf, H. S. (2003). Some new aspects of the coupon-collector s problem. SIAM J. Discrete Math. 7, 7. [8] Neal, P. (2008). The generalise coupon collector problem. J. Appl. Prob. 45, [9] Newman, D. J. an Shepp, L. (960). The ouble ixie cup problem. Amer. Math. Monthly 67, [0] Sharif, M. an Hassibi, B. (2006). Delay consierations for opportunistic scheuling in broacast faing channels. IEEE Trans. Wireless Commun. 6,