Algorithms for Minimizing Response Time in Broadcast Scheduling

DOI: 10.1007/s00453-003-1058-x

Algorithmica (2004) 38: 597–608 Algorithmica© 2003 Springer-Verlag New York Inc.

Algorithms for Minimizing Response Timein Broadcast Scheduling1

Rajiv Gandhi,2 Samir Khuller,3 Yoo-Ah Kim,4 and Yung-Chun (Justin) Wan4

Abstract. In this paper we study the following problem. There are n pages which clients can request at anytime. The arrival times of requests for pages are known in advance. Several requests for the same page mayarrive at different times. There is a server that needs to compute a good broadcast schedule. Outputting a pagesatisfies all outstanding requests for the page. The goal is to minimize the average waiting time of a client.This problem has recently been shown to be NP-hard. For any fixed α, 0 < α ≤ 1

2 , we give a 1/α-speed,polynomial time algorithm with an approximation ratio of 1/(1− α). For example, setting α = 1

2 gives a2-speed, 2-approximation algorithm. In addition, we give a 4-speed, 1-approximation algorithm improving theprevious bound of 6-speed, 1-approximation algorithm.

Key Words. Approximation algorithms, Scheduling, Broadcasting.

1. Introduction. There has been a lot of interest lately in data dissemination services,where clients request information from a source. Advances in networking and the needto provide data to mobile and wired devices have led to the development of large-scaledata dissemination applications (election results, stock market information, etc.). Whilethe WWW provides a platform for developing these applications, it is hard to providea completely scalable solution. Hence researchers have been focusing their attention onData Broadcasting methods.

Broadcasting is an appropriate mechanism to disseminate data since multiple clientscan have their requests satisfied simultaneously. A large amount of work in the databaseand algorithms literature has focused on scheduling problems based on a broadcastingmodel (including several Ph.D. theses from Maryland and Brown) [7], [9], [4], [1],[5], [2], [8], [21], [6]. Broadcasting is used in commercial systems, including the IntelIntercast System [15] and the Hughes DirecPC [13]. There are two primary kinds ofmodels that have been studied—the first kind is a push-based scheme, where someassumptions are made on the access probability for a certain data item and a broadcastschedule is generated [3], [9], [7], [16], [6]. We focus our attention on the second kind,namely pull-based schemes, where clients request the data that they need (for example,via phone lines) and the data is delivered on a fast broadcast medium (often usingsatellites) [5]. This model is motivated by wireless web applications. This work deals

1 This research was supported by NSF Awards CCR-9820965 and NSF CCR-0113192. A preliminary versionof this work appeared at the 9th Integer Programming and Combinatorial Optimization Conference (2002).2 Department of Computer Science, Rutgers University, Camden, NJ 08102, USA. [email protected] Department of Computer Science and Institute for Advanced Computer Studies, University of Maryland,College Park, MD 20742, USA. [email protected] Department of Computer Science, University of Maryland, College Park, MD 20742, USA. {ykim,ycwan}@cs.umd.edu.

Received February 25, 2002; revised July 23, 2003. Communicated by H. N. Gabow.Online publication December 19, 2003.

598 R. Gandhi, S. Khuller, Y.-A. Kim, and Y.-C. Wan

entirely with the pull-based model, where requests for data arrive over time and a goodbroadcast schedule needs to be created.

A key consideration is the design of a good broadcast schedule. The challenge is indesigning an algorithm that generates a schedule that provides good average responsetime. Several different scheduling policies have been proposed for online data broadcast.Aksoy and Franklin [5] proposed one such online algorithm. Their algorithm, calledRxW , takes the product of the number of outstanding requests for page with the longestwaiting time to compute the “demand” for a page. The page with the maximum demandis then broadcast. Another reasonable heuristic is to take the sum of waiting times of alloutstanding requests for a page to compute its demand. Tarjan et al. [20] have recentlyshown how to implement this algorithm efficiently. Not surprisingly, the worst casecompetitive ratio of these algorithms is unbounded, as shown in [17].

While the practical problem is clearly online, it is interesting to study the complexity ofthe offline problem as well. In trying to evaluate the performance of online algorithms, itis useful to compare them with an optimal offline solution. In addition, when the demandsare known for a small window of time into the future (also called the look-ahead modelin online algorithms) being able to compute an optimal offline solution quickly canbe extremely useful. Many kinds of demands for data (e.g., web traffic) exhibit goodpredictability over the short term, and thus knowledge of requests in the immediate futureleads to a situation where one is trying to compute a good offline solution.

One could also view the requests in the offline problem as release times of jobs, andone is interested in minimizing average (weighted) flow time. While this problem hasbeen very well studied (see [11] and references therein), the crucial difference betweenour problem and the problem that has been studied is the fact that scheduling a jobsatisfies many requests simultaneously. (The term “overlapping jobs” has also been usedto describe such scheduling problems in the past.)

The informal description of the problem is as follows. There are n data items, 1, . . . , n,called pages. Time is broken into “slots.” A time slot is defined as the unit of time totransmit one page on the wireless channel. A request for a page j arrives at time t andthen waits. When page j has been transmitted, this request has been satisfied. Arrivaltimes of requests for pages are known, and we wish to find a broadcast schedule thatminimizes the average waiting time. There has been a lot of work on this problem whenwe assume some knowledge of a probability distribution for the demand of each page[6], [7], [16].

In the same model, the paper by Bartal and Muthukrishnan [8] studies the problemof minimizing the maximum response time. The offline version has a simple 2 approx-imation, and there is a 2-competitive algorithm for the online version [8]. (They alsocredit Charikar, Khanna, Motwani, and Naor for some of these results that were obtainedindependently.)

The recent paper by Kalyanasundaram et al. [17] studies this problem as well. Theyshowed that for any fixed ε, 0 < ε ≤ 1

3 , it is possible to obtain a 1/ε-speed 1/(1− 2ε)-approximation algorithm for minimizing the average response time, where a k-speedalgorithm is one where the server is allowed to broadcast k pages in each time slot [18]. Forexample, by putting ε = 1

3 they obtain a 3-speed, 3-approximation. The approximationfactor bounds the cost of the k-speed solution compared with the cost of an optimal 1-speed solution. (This kind of approximation guarantee is also referred to as a “bicriteria”

Algorithms for Minimizing Response Time in Broadcast Scheduling 599

bound in many papers.) Note that they cannot set ε = 12 to get a 2-speed, constant

approximation. Their algorithm is based on rounding a fractional solution for a “network-flow” like problem that is obtained from an integer programming formulation. Theproblem of minimizing the average response time has recently been shown to be NP-hard by Erlebach and Hall [14].

Our Results. We consider a different Integer Program (IP) for this problem,4 and showthat by relaxing this IP, for any fixed α ∈ (0, 1

2 ], we can obtain a 1/α-speed solutionthat is a 1/(1− α)-approximation. For example, by setting α = 1

2 we obtain a 2-speed,2-approximation. By setting α = 1

3 we obtain a 3-speed, 1.5-approximation. Note thatour algorithm improves both the speed and approximation guarantee of the algorithm in[17]. This can be viewed as a step towards ultimately obtaining a 1-speed algorithm. Therounding method is of independent interest and quite different from the rounding methodproposed by Kalyanasundaram et al. [17]. Moreover, our formulation draws on roundingmethods developed for the k-medians problem [10]. This connection with scheduling isperhaps a surprising aspect of our work. Erlebach and Hall [14] showed that one can geta 1-approximation via a 6-speed algorithm. Here we show how to use the method in thispaper to get a 1-approximation via a 4-speed algorithm.

2. Problem. The problem is formally stated in [17], but for the sake of completenesswe describe it here. There are n possible pages, P = {1, 2, . . . , n}. We assume that timeis discrete and at time t , any subset of pages can be requested. Let (p, t) represent arequest for page p at time t . Let r p

t denote number of requests (p, t). Let T be the time oflast request for a page. Without loss of generality, we can assume that T is polynomiallybounded as a function of n and the number of distinct times at which requests are made.This is because at each time when a request is made, we only need to broadcast as manypages after that as the number of distinct pages requested. A time slot t is the window oftime between time t − 1 and time t . We also have a k-speed server that can broadcast upto k pages at any time t . We say that a request (p, t) is satisfied at time S p

t , if S pt is the

first time instance after t when page p is broadcast. In this paper, we work in the offlinesetting in which the server is aware of all the future requests. Our goal is to schedule thebroadcast of pages in a way so as to minimize the total response time of all requests. Thetotal response time is given by

∑p

∑t

r pt (S

pt − t).

Consider the example shown in Figure 1. The table on the left shows requests for the threepages A, B, and C at different times. An optimal schedule for this instance broadcastspages B,C, A, B,C at time slots 1, 2, 3, 4, 5, respectively. The table on the right of thefigure shows the response time for each request in the optimal schedule. Adding up theresponse time of each request gives us the total response time of 25.

4 Recently, we have been able to show that the LP relaxations considered in [17] and this work are equivalent.


Input:r pt Response time:r p

t (Spt − t)

t=0 t=1 t=2 t=3 t=4 t=0 t=1 t=2 t=3 t=4page A 3 2 2 0 0 page A 9 4 2 0 0page B 2 0 2 0 0 page B 2 0 4 0 0page C 0 2 0 0 2 page C 0 2 0 0 2

Fig. 1. The table on the left is an example input and the table on the right shows the response time for eachrequest in an optimal schedule of broadcasting pages B,C, A, B,C at times 1, 2, 3, 4, 5, respectively.

3. Integer Programming Formulation. The Broadcast Scheduling Problem can beformulated as an integer program as follows. The binary variable y p

t ′ = 1 iff page p isbroadcast at time t ′. The binary variable x p

tt ′ = 1 iff a request (p, t) is satisfied at timet ′ > t , i.e., y p

t ′ = 1 and y pt ′′ = 0, t < t ′′ < t ′. The constraints (2) ensure that whenever a

request (p, t) is satisfied at time t ′, page p is broadcast at t ′. Constraints (3) ensure thatevery request (p, t) is satisfied at some time t ′ > t . Constraints (4) ensure that at mostone page is broadcast at any given time.

min∑

p

∑t

T+n∑t ′=t+1

(t ′ − t) · r pt · x p

tt ′(1)

subject to

y pt ′ − x p

tt ′ ≥ 0, ∀p, t, t ′ > t,(2)T+n∑

t ′=t+1

x ptt ′ = 1, ∀p, t,(3)

∑p

y pt ′ ≤ 1, ∀t ′,(4)

x ptt ′ ∈ {0, 1}, ∀p, t, t ′,(5)

y pt ′ ∈ {0, 1}, ∀p, t ′.(6)

The corresponding linear programming (LP) relaxation can be obtained by letting thedomain of x p

tt ′ and y pt ′ be 0 ≤ x p

tt ′ , y pt ′ ≤ 1. For the example in Figure 1, running an LP

solver produces a fractional schedule that broadcasts pages {A, B}, {A,C}, {A, B}, {A,B}, {C} at times 1, 2, 3, 4, 5, respectively, where broadcasting {P1, P2} at any time tmeans that exactly half of page P1 and half of page P2 are broadcast at time t . The costof this fractional solution is 24.5.

4. Outline of the Algorithm. Let I be the given instance of the problem. The algorithmsolves the LP for I to obtain an optimal (fractional) solution. It uses the LP solution tocreate a simplified instance I. A 1/α-speed (1/α is an integer) fractional solution isconstructed for instance I, which is converted into a 1/α-speed integral solution for Iusing a min-cost flow computation. The integral solution for I is then converted into aschedule for I .

For any page p, let Np = {t1, t2, . . . , t fp } denote the times at which requests for pagep are made in instance I .


REQUEST CONSOLIDATION(page p, α)1 Np ← {t fp }2 l ← fp

3 r ′ptl ← r ptl

4 for k ← l − 1 down to 1 do5 f t (α, p, tk)← mint ′ {

∑t ′t=tk+1 x p

tk t ≥ α}6 if ( f t (α, p, tk) ≤ tl) then7 Np ← Np ∪ {tk}8 l ← k9 r ′ptl ← r p

tl10 else11 g(p, tk)← tl12 r ′ptl ← r ′ptl + r p

tk13 r ′ptk ← 014 return Np

Fig. 2. Algorithm for consolidating requests.

5. Algorithm

Step I. Let I be the given instance of the problem. We first solve the LP for I toobtain an optimal fractional solution (x, y). Let f t (α, p, t) be the first time instancewhen an α-fraction of request (p, t) gets satisfied in the LP solution, i.e., f t (α, p, t) =min{t ′′|∑t ′′

t ′=t+1 x ptt ′ ≥ α}. We consolidate the requests in I , transforming the instance I

into a simplified instance I which has the following property. IfNp represents the timesof positive requests for page p in I, then for any times {t ′u, t ′v} ⊆ Np, such that t ′u < t ′v ,we have f t (α, p, t ′u) ≤ t ′v , where α is any fixed fraction in (0, 1

2 ]. For every request (p, t)that is grouped with a request (p, g(p, t)), g(p, t) ≥ t , we have f t (α, p, t) > g(p, t).Let r ′pg(p,t) denote the number of requests (p, g(p, t)) in I. This transformation is doneseparately for each page. The pseudo-code for this transformation is given in Figure 2.This step is illustrated in Figure 3 for α = 1

2 : Figure 3(a) shows the LP solution for pageA and Figure 3(b) shows the requests for page A in the simplified instance.

Step II. Now we find a 1/α-speed fractional solution to instanceI. LetNp = {t ′1, t ′2, . . .}be the times of positive requests for page p in I. Note that at least an α-fraction of eachrequest (p, t ′i ) ∈ I gets satisfied before t ′i+1 in the LP solution (x, y), i.e., f t (α, p, t ′i ) ≤t ′i+1. Consider the solution (xα, y) for I, where

x pαt t ′ =

x ptt ′ if t ′ < f t (α, p, t),

α −t ′−1∑

t ′′=t+1

x ptt ′′ if t ′ = f t (α, p, t),

0 otherwise.


t=0 1 2 3 4 5 6A 4A 10A

7 8 9 10 115AA2A8A

A.2A.2A.3A.4A .5A

t=0 1 2 3 4 5 6

.7A

5A 12A7 8 9 10 11

8A8A

A.2A.2A.3A.4A .5A

2A

.3A

.7A.3A .2A

.2A

t=0 1 2 3 4 5 6

5A 12A7 8 9 10 11

8A8A

A.2A.2A.3A.4A .5A

A.2A.2A.3A.4A .5A

.7A.3A

.7A.3A

.2A

.2A

(.6A) (.4A) (.8A) (.2A) (.6A) (.4A) (1A)

(a)

(b)

(c)

1

2

3

4

5

6

7

8

9

10

11

(A,1)

(A,4)

(A,6)

(A,10)

[0,1,10]

(d)

Fig. 3. (a) The LP solution for a particular page A. (b) The output of REQUEST-CONSOLIDATION(A, 12 ). (c) The

2-speed fractional solution. (d) The flow network induced on requests for page A. The source and sink are notshown for clarity in exposition.

By scaling all the y values and the xα values by 1/α we obtain a feasible 1/α-speed fractional solution, ((1/α)xα, (1/α)y). This step is illustrated in Figure 3(c). Thenumbers in the parentheses denote the (1/α)xα values for α = 1

2 .

Step III. To obtain a 1/α-speed integer solution to instance I, we construct a min-imum cost flow network N . N is the flow network that consists of a bipartite graphG = (X, Y, E). Each node in X corresponds to a request (p, t ′) ∈ I. Each node inY corresponds to a time instance at which a page can be broadcast. There is an edgebetween (p, t) ∈ X and t ′ ∈ Y if xαt t ′ > 0. This edge has a lower capacity of 0 and anupper capacity of 1 and a cost of r ′pt (t

′ − t). Note that for any two nodes in X , say (p1, t ′1)and (p2, t ′2), that share a neighbor in Y , p1 �= p2. In addition, we have a source and asink in N . There is an edge of 0 cost from the source to a node (p, t) ∈ X with a lowerand upper capacity of 1. Each edge from a node t ∈ Y to the sink is of 0 cost and has a


source sink

[1,1,0][0,1/α,0]

(p1,t’j)

(p2,t’j)

X Y

[0,1,r’p1

t’j(tx-t’j)] tx

(p3,t’m) ty

(p3,t’k)

[0,1,r’p3

t’m(ty-t’m)]

(p4,t’k)

Fig. 4. The minimum cost flow network N . Each edge e ∈ N has a label of the form [l, u, c], where l and udenote the lower and the upper bounds, respectively, on the amount of flow through e and c denotes the costper unit flow.

lower capacity of 0 and an upper capacity of 1/α. Figure 3(d) shows the flow networkinduced on requests for page A. For clarity in exposition, the source and sink are notshown in Figure 3(d). Figure 4 illustrates the min-cost flow network N . A 1/α-speedintegral schedule can be obtained by setting y p

t ′ = f ((p, t), t ′), where f (a, b) denotesthe flow along edge (a, b) ∈ N .

6. Analysis

LEMMA 1. In Step I, recall that each request (p, t) ∈ I is grouped with a request(p, g(p, t)) to form instance I, where g(p, t) ≥ t . If S and S′ are integral solutions toinstance I and I, respectively, then

cost(S) ≤ cost(S′)+∑(p,t)∈I

r pt (g(p, t)− t).

PROOF. The inequality follows easily because the additional cost of converting asolution for request (p, g(p, t)) ∈ I to a solution for request (p, t) ∈ I isexactly r p

t (g(p, t) − t). Summing this over all requests will give us the requiredinequality.

Recall that (x, y) is an optimal LP solution for I . Define C(p,t) to be the cost of sat-isfying one request (p, t) ∈ I in the LP solution (x, y), i.e., C(p,t) =

∑T+nt ′=t+1(t

′ −t)x p

tt ′ . For any request (p, g(p, t)) ∈ I, define Cθ(p,g(p,t)) as the cost of satisfying

exactly a θ -fraction of request (p, g(p, t)) ∈ I, i.e., Cθ(p,g(p,t)) =∑T+n

t ′=g(p,t)+1(t′ −

g(p, t))x pθg(p,t)t ′ . Note that the cost of the 1/α-speed fractional solution for instance I

is (1/α)∑

(p,g(p,t))∈I r ′pg(p,t)Cα(p,g(p,t)) = (1/α)∑

(p,t)∈I r pt Cα(p,g(p,t)) and the cost of the


optimal fractional solution for instance I is∑

(p,t)∈I r pt C(p,t). Lemma 2 relates the two

costs.

LEMMA 2. The cost of the 1/α-speed fractional solution in I and the cost of an optimalfractional solution for instance I are related as follows:

1

α

∑(p,t)∈I

r pt Cα(p,g(p,t)) ≤

1

1− α∑(p,t)∈I

r pt C(p,t) −

∑(p,t)∈I

r pt (g(p, t)− t).

PROOF. For the case when t = g(p, t), the inequality follows because (g(p, t)− t) = 0and C(p,t) is the cost of satisfying a complete request whereas Cα(p,g(p,t)) is the cost ofsatisfying an α-fraction of the request. For the case when t < g(p, t), we know thatf t (α, p, t) > g(p, t). This means that at least a (1−α)-fraction of the request (p, t) ∈ Igets satisfied after g(p, t). In other words, a broadcast of page p between g(p, t) andf t (1−α, p, g(p, t)) partially satisfies request (p, t) and (p, g(p, t)) in I . Thus we have

C1−α(p,g(p,t)) + (1− α)(g(p, t)− t) ≤ C(p,t).(7)

Multiplying (7) by 1/(1− α) and rearranging the terms gives us

1

1− α C1−α(p,g(p,t)) ≤

1

1− αC(p,t) − (g(p, t)− t).(8)

The left side of (8) represents the average cost of satisfying a (1−α)-fraction of request(p, g(p, t)) ∈ I. The cost of satisfying exactly an α-fraction of the request, Cα(p,g(p,t)),is upper bounded5 by (α/(1− α))C1−α

(p,g(p,t)), which is obtained by multiplying (8) by α,thus giving us

Cα(p,g(p,t)) ≤α

1− α C1−α(p,g(p,t)) ≤

α

1− αC(p,t) − α(g(p, t)− t).(9)

Multiplying (9) by (1/α)r pt and summing over all requests gives us the cost of the

1/α-speed fractional solution as required.

LEMMA 3. For any feasible flow in the minimum cost flow network, N , there is a 1/α-speed feasible fractional solution for instance I of the same cost.

PROOF. Let f (a, b) denote the flow along edge (a, b) ∈ N . Set x pαt t ′ = α f ((p, t), t ′)

and y pt ′ = x p

αt t ′ . Since we know that for any node (p, t) ∈ X ,∑

t ′ f ((p, t), t ′) = 1, wehave (1/α)

∑t ′ x

pαt t ′ = 1. We can show that we have a 1/α-speed solution as follows.

We know that∑

(p,t) f ((p, t), t ′) ≤ 1/α, which implies that (1/α)∑

(p,t) x pαt t ′ ≤ 1/α

and hence (1/α)∑

p y pt ′ ≤ 1/α. Thus feasibility is ensured. The cost of the 1/α-speed

solution equals

1

α

∑(p,t)∈I

∑t ′>t

r ′pt (t′ − t)x p

αt t ′ =∑

(p,t)∈X

∑t ′∈Y

r ′pt (t′ − t) f ((p, t), t ′).

This proves that the 1/α-speed solution has the same cost as the flow in N .

5 This holds as long as α ≤ 1− α, hence our assumption that α ≤ 12 .


LEMMA 4. There is a flow in N of the same cost as the 1/α-speed feasible fractionalsolution for instance I.

PROOF. Let the flow along an edge ((p, t), t ′) equal (1/α)x pαt t ′ . By definition of xα , the

capacity constraint of each edge is satisfied. Since we have a 1/α-speed feasible solution,(1/α)

∑t ′ x

pαt t ′ = 1 and hence the flow is conserved at every node in X which has an

incoming flow of 1 unit. Since our solution is 1/α-speed, (1/α)∑

p y pt ′ ≤ 1/α. Set the

flow along any edge from node t ′ to the sink equal to (1/α)∑

p y pt ′ . Note that the flow

is also conserved at a node t ′ ∈ Y as we know that y pt ′ ≥ x p

αt t ′ . The cost of the solutionequals ∑

(p,t)∈X

∑t ′∈Y

r ′pt (t′ − t) f ((p, t), t ′) = 1

α

∑(p,t)∈I

∑t ′

r ′pt (t′ − t)x p

αt t ′ .

Thus the cost of the flow in N is the same as the cost of the 1/α-speed solution.

LEMMA 5. There exists a 1/α-speed integral solution for I of the same cost as the1/α-speed fractional solution, ((1/α)xα, (1/α)y).

PROOF. From Lemmas 3 and 4 we know that the minimum cost flow in N equals thecost of the 1/α-speed fractional solution ((1/α)xα, (1/α)y). By the integrality theorem[12], we can determine in polynomial time a minimum cost integral flow f ∗ in N thatsatisfies the capacity constraints. Using this integral flow f ∗ we can derive a 1/α-speedintegral solution for instance I using Lemma 3.

THEOREM 6. There is a 1/α-speed, 1/(1− α)-approximation solution for the Broad-cast Scheduling Problem.

PROOF. We will prove that the algorithm in Section 5 gives a 1/α-speed, 1/(1− α)-approximate solution. From Lemmas 2 and 5, we know that the cost of the 1/α-speedintegral solution equals

1

α

∑(p,t)∈I

r pt Cα(p,g(p,t)) ≤

1

1− α∑(p,t)∈I

r pt C(p,t) −

∑(p,t)∈I

r pt (g(p, t)− t).

Substituting this expression for cost(S′) in Lemma 1, we get

cost(S) ≤ 1

1− α∑(p,t)∈I

r pt C(p,t) ≤ 1

1− αOPT.

COROLLARY 7. There is a 2-speed, 2-approximation solution and a 3-speed, 1.5-approximation solution for the Broadcast Scheduling Problem.

PROOF. The proof follows easily from Theorem 6 by setting α = 12 and α = 1

3 ,respectively, for a 2-speed and a 3-speed solution.


7. 4-Speed 1-Approximation Algorithm. Erlebach and Hall [14] showed that onecan use a 6-speed algorithm to get a 1-approximation. This is done by taking a fractionalsolution obtained by solving a linear program, and then applying randomized roundingto generate schedules for two channels. On the remaining four channels they use the4-speed algorithm of [17]. They are able to prove that the expected cost of each requestis upper bounded by its fractional cost.

In this section we show that one can obtain a random fractional schedule using theapproach outlined in our paper. We can establish an upper bound on the cost of thisfractional schedule. Finally, we convert this fractional schedule to an integral scheduleusing the network flow approach.

We obtain a 4-speed fractional solution as follows. On two channels, we select twopages (independently) with probability y p

t at each time t . For the remaining two channels,we take our 2-speed fractional solution constructed in Step II of Section 5. Note that inthis 2-speed fractional solution, a request (p, t) is satisfied by the same broadcast of pas (p, g(p, t)) because we construct a 2-speed fractional solution after merging (p, t) to(p, g(p, t)). Therefore the cost of satisfying one request (p, t) in the 2-speed fractionalsolution is 2C1/2

(p,g(p,t)) + (g(p, t)− t). Let Cost(p, t) denote 2C1/2(p,g(p,t)) + (g(p, t)− t).

LEMMA 8. The expected cost of satisfying a request (p, t) by the 4-speed fractionalschedule is at most the optimal LP cost for a request (p, t).

PROOF. In our 4-speed fractional schedule, each request (p, t) could be satisfied byeither of the two random channels and if it is not satisfied by time g(p, t), the 2-speedfractional schedule will satisfy it fractionally. In other words, the probability that arequest (p, t) is satisfied by either of the random channels at time t ′ (t < t ′ ≤ g(p, t))is (∏t ′−1

t ′′=t+1 (1− y pt ′′)

2)(1 − (1− y p

t ′ )2) and the probability that it is satisfied after time

g(p, t) is∏g(p,t)

t ′=t+1 (1− y pt ′ )

2. Thus the expected cost of satisfying a request (p, t) by the4-speed fractional schedule is at most

A =g(p,t)∑t ′=t+1

((t ′−1∏

t ′′=t+1

(1− y pt ′′)

2)(1− (1− y p

t ′ )2

)(t ′ − t)

)

+(

g(p,t)∏t ′=t+1

(1− y pt ′ )

2

)Cost(p, t).

The optimal LP cost for a request (p, t) is at least

B =g(p,t)∑t ′=t+1

y pt ′ · (t ′ − t)+

(1−

g(p,t)∑t ′=t+1

y pt ′

)· Cost(p, t).

Because∑g(p,t)

t ′=t+1 y pt ′ <

12 we can get A ≤ B using the same proof as [14].

LEMMA 9. We can find a 4-speed fractional schedule of the cost at most OPT in poly-nomial time.


PROOF. By Lemma 8 and the linearity of expectation, the expected cost of this 4-speedfractional schedule is at most OPT . Using standard derandomization techniques [19],we can obtain a deterministic polynomial-time algorithm that yields a 4-speed fractionalschedule of cost at most OPT .

Now we convert the 4-speed fractional schedule to a 4-speed integral schedule. Be-cause the schedule on two random channels is already integral, we only need to convertthe 2-speed fractional schedule to an integral schedule without losing any cost.

LEMMA 10. We can convert the 4-speed fractional schedule to a 4-speed integral sched-ule of the same cost.

PROOF. By constructing the minimum cost flow network only for the requests satisfiedby the fractional schedule, we can convert the 2-speed fractional solution to 2-speedintegral solution of the same cost. Combining with two random channels, we have a4-speed integral solution.

THEOREM 11. There is a 4-speed, 1-approximation solution for the Broadcast Schedul-ing Problem.

References

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[13] DirecPC website, http://www.direcpc.com


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Algorithms for Minimizing Response Time in Broadcast Scheduling

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