On the Value of Job Migration in Online Makespan Minimization

Abstract

Makespan minimization on identical parallel machines is a classical scheduling problem. We consider the online scenario where a sequence of n jobs has to be scheduled non-preemptively on m machines so as to minimize the maximum completion time of any job. The best competitive ratio that can be achieved by deterministic online algorithms is in the range [1.88, 1.9201]. Currently no randomized online algorithm with a smaller competitiveness is known, for general m. In this paper we explore the power of job migration, i.e. an online scheduler is allowed to perform a limited number of job reassignments. Migration is a common technique used in theory and practice to balance load in parallel processing environments. As our main result we settle the performance that can be achieved by deterministic online algorithms. We develop an algorithm that is \(\alpha _m\)-competitive, for any \(m\ge 2\), where \(\alpha _m\) is the solution of a certain equation. For \(m=2\), \(\alpha _2 = 4/3\) and \(\lim _{m\rightarrow \infty } \alpha _m = W_{-1}(-1/e^2)/(1+ W_{-1}(-1/e^2)) \approx 1.4659\). Here \(W_{-1}\) is the lower branch of the Lambert W function. For \(m\ge 11\), the algorithm uses at most 7m migration operations. For smaller m, 8m to 10m operations may be performed. We complement this result by a matching lower bound: No online algorithm that uses o(n) job migrations can achieve a competitive ratio smaller than \(\alpha _m\). We finally trade performance for migrations. We give a family of algorithms that is c-competitive, for any \(5/3\le c \le 2\). For \(c= 5/3\), the strategy uses at most 4m job migrations. For \(c=1.75\), at most 2.5m migrations are used.

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Correspondence to Susanne Albers.

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A preliminary version of this paper has appeared in Proc. 20th Annual European Symposium on Algorithms (ESA), 2012. Susanne Albers’ work supported by the German Research Foundation, project Al 464/7-1.

Appendix

Appendix

Proof of Lemma 1

Fix \(m\ge 2\). We first evaluate \(f_m(2)\) and \(f_m(1+1/(3m))\). For \(\alpha =2\), we have \(\lceil (1-1/\alpha )m\rceil \ge m/2\). Hence \(\lceil (1-1/\alpha )m\rceil \alpha /m \ge 1\) and \(f_m(2) \ge 1\). For \(\alpha =1+ 1/(3m)\), there holds \(\lceil (1-1/\alpha )m\rceil =1\). Thus \(f_m(1+1/(3m)) = 1/(3m) H_{m-1} +1/m + 1/(3m^2)< 1/3 + 1/2 + 1/12 < 1\). It remains to show that \(f_m(\alpha )\) is continuous and strictly increasing. To this end we show that, for any \(\alpha >1\) and small \(\epsilon >0\), \(f_m(\alpha ) - f_m(\alpha -\epsilon )\) and \(f_m(\alpha +\epsilon ) - f_m(\alpha )\) converge to 0 as \(\epsilon \rightarrow 0\). Moreover \(f_m(\alpha +\epsilon ) - f_m(\alpha )\) is strictly positive.

First consider an \(\alpha >1\) such that \((1-1/\alpha )m\notin \mathbb {N}\). In this case we choose \(\epsilon > 0\) such that \(\lceil (1-1/(\alpha -\epsilon ))m\rceil = \lceil (1-1/(\alpha +\epsilon ))m\rceil = \lceil (1-1/\alpha )m\rceil \). We have

$$\begin{aligned} f_m(\alpha )= & {} (\alpha -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil \alpha /m\\ f_m(\alpha -\epsilon )= & {} (\alpha -\epsilon -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil (\alpha -\epsilon )/m\\ f_m(\alpha +\epsilon )= & {} (\alpha +\epsilon -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil (\alpha +\epsilon )/m. \end{aligned}$$

Thus \(f_m(\alpha ) - f_m(\alpha -\epsilon ) = f_m(\alpha +\epsilon ) - f_m(\alpha ) = \epsilon (H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil \epsilon /m\). Hence \(f_m(\alpha )-f_m(\alpha -\epsilon )\) and \(f_m(\alpha +\epsilon ) - f_m(\alpha )\) tend to 0 as \(\epsilon \rightarrow 0\). Since \(\alpha >1\) there holds \(\lceil (1-1/\alpha )m\rceil \ge 1\) and thus \(f_m(\alpha +\epsilon ) - f_m(\alpha ) >0\).

Next let \(\alpha >1\) such that \((1-1/\alpha )m\in \mathbb {N}\). In this case we choose \(\epsilon > 0\) such that \(\lceil (1-1/(\alpha -\epsilon ))m\rceil = \lceil (1-1/\alpha )m\rceil \) and \(\lceil (1-1/(\alpha +\epsilon ))m\rceil = \lceil (1-1/\alpha )m\rceil +1\). There holds

$$\begin{aligned} f_m(\alpha )= & {} (\alpha -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil \alpha /m\\ f_m(\alpha -\epsilon )= & {} (\alpha -\epsilon -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil (\alpha -\epsilon )/m\\ f_m(\alpha +\epsilon )= & {} (\alpha +\epsilon -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil }) + (\lceil (1-1/\alpha )m\rceil +1) (\alpha +\epsilon )/m. \end{aligned}$$

As above \(f_m(\alpha ) - f_m(\alpha -\epsilon ) = \epsilon (H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil \epsilon /m\) and the latter expression tends to 0 as \(\epsilon \rightarrow 0\). Taking into account that \((1-1/\alpha )m\in \mathbb {N}\) we obtain

$$\begin{aligned} f_m(\alpha +\epsilon ) - f_m(\alpha )= & {} -(\alpha -1)\cdot 1/((1-1/\alpha )m) + \epsilon (H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil })\\&+ (\lceil (1-1/\alpha )m\rceil +1) \epsilon /m + \alpha /m\\= & {} \epsilon (H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil }) + (\lceil (1-1/\alpha )m\rceil +1) \epsilon /m. \end{aligned}$$

Again, \(f_m(\alpha +\epsilon ) - f_m(\alpha )\) is strictly positive and tends to 0 as \(\epsilon \rightarrow 0\). \(\square \)

Proof of Lemma 2

We first prove that \((\alpha _m)_{m\ge 2}\) is non-decreasing. A first observation is that \(\alpha _m \le m\) because \(f_m(m)\ge 1\). We will show that, for any \(m\ge 3\) and \(1<\alpha \le m\), there holds \(f_{m-1}(\alpha ) \ge f_m(\alpha )\). This implies \(1= f_{m-1}(\alpha _{m-1}) \ge f_m(\alpha _{m-1})\). By Lemma 1, \(f_m\) is strictly increasing and thus \(\alpha _m\ge \alpha _{m-1}\). Consider a fixed \(\alpha \) with \(1<\alpha \le m\). We study two cases depending on whether or not \(\lceil (1-1/\alpha )(m-1)\rceil = \lceil (1-1/\alpha )m\rceil \).

If \(\lceil (1-1/\alpha )(m-1)\rceil = \lceil (1-1/\alpha )m\rceil \), then

$$\begin{aligned} f_m(\alpha )= & {} (\alpha -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil \alpha /m\\ f_{m-1}(\alpha )= & {} (\alpha -1)(H_{m-2}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil \alpha /(m-1). \end{aligned}$$

We obtain \(f_{m-1}(\alpha ) - f_m(\alpha ) = -(\alpha -1)/(m-1) + \lceil (1-1/\alpha )m\rceil \alpha /(m(m-1)) \ge -(\alpha -1)/(m-1) + (\alpha -1)/(m-1) =0\) and thus \(f_{m-1}(\alpha ) \ge f_m(\alpha )\).

If \(\lceil (1-1/\alpha )(m-1)\rceil < \lceil (1-1/\alpha )m\rceil \), then \(\lceil (1-1/\alpha )(m-1)\rceil = \lceil (1-1/\alpha )m\rceil -1\) and

$$\begin{aligned} f_m(\alpha )= & {} (\alpha -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m\rceil -1}) + \lceil (1-1/\alpha )m\rceil \alpha /m\\ f_{m-1}(\alpha )= & {} (\alpha -1)(H_{m-2}-H_{\lceil (1-1/\alpha )m\rceil -2}) + (\lceil (1-1/\alpha )m\rceil -1) \alpha /(m-1). \end{aligned}$$

Since \(\alpha >1\) there holds \(\lceil (1-1/\alpha )(m-1)\rceil \ge 1\). Hence in our case \(\lceil (1-1/\alpha )m\rceil \ge 2\) and \(\lceil (1-1/\alpha )m\rceil -1 >0\). We obtain

$$\begin{aligned} \textstyle {f_{m-1}(\alpha ) - f_m(\alpha ) = -{\alpha -1\over m-1} + {\alpha -1\over \lceil (1-1/\alpha )m\rceil -1} + \lceil (1-1/\alpha )m\rceil {\alpha \over m(m-1)} - {\alpha \over m-1}.} \end{aligned}$$

Choose x, with \(0\le x <1\), such that \(\lceil (1-1/\alpha )m\rceil = (1-1/\alpha )m +x\). Then

$$\begin{aligned} \textstyle {f_{m-1}(\alpha ) - f_m(\alpha )}= & {} \textstyle { -{\alpha -1\over m-1} + {\alpha -1\over (1-1/\alpha )m +x-1} + (1-1/\alpha )m {\alpha \over m(m-1)} + {\alpha x\over m(m-1)}- {\alpha \over m-1}}\\= & {} \textstyle {{\alpha -1\over (1-1/\alpha )m +x-1} + {\alpha x\over m(m-1)}- {\alpha \over m-1}} \end{aligned}$$

In order to establish \(f_{m-1}(\alpha ) - f_m(\alpha )\ge 0\) is suffices to show

$$\begin{aligned} \textstyle {{\alpha -1\over (1-1/\alpha )m +x-1} \ge {\alpha (m-x)\over m(m-1)}.} \end{aligned}$$

This is equivalent to \((\alpha -1)m(m-1) \ge (m-x)((\alpha -1)m +\alpha x-\alpha )\). Standard algebraic manipulation yield that this is equivalent to \(m \ge mx - \alpha x^2+\alpha x\). Let \(g(x) = mx - \alpha x^2+\alpha x\), for any real number x. This function is increasing for any \(x < (m+\alpha )/(2\alpha )\). Since \(\alpha \le m\), the function is increasing for any \(x <1\). As \(g(0) = 0\) and \(g(1) = m\), it follows that \(m \ge mx - \alpha x^2+\alpha x\) holds for all \(0\le x <1\). We conclude \(f_{m-1}(\alpha ) - f_m(\alpha )\ge 0\).

It is easy to verify that \(f_2(4/3)=1\). We show that \(\lim _{m\rightarrow \infty } \alpha _m\) is upper bounded by \(W_{-1}(-1/e^2)/(1+ W_{-1}(-1/e^2))\). Cesáro [5] proved

$$\begin{aligned} 0< H_m - \frac{1}{2} \ln \left( m(m+1) \right) -\gamma < \frac{1}{6m(m+1)}, \end{aligned}$$
(1)

where \(\gamma \approx 0.577\) is the Euler-Mascheroni constant. Using this inequality we find, for any c with \(0< c\le 1\) and \(\lceil cm\rceil -2 >0\),

$$\begin{aligned} H_{m-1} - H_{\lceil cm \rceil -2 }> & {} \frac{1}{2} \ln ((m-1)m) + \gamma - \frac{1}{2} \ln ((\lceil cm \rceil -2)(\lceil cm\rceil -1))\\&-\gamma - \frac{1}{6(\lceil cm \rceil -2)(\lceil cm \rceil -1)}\\\ge & {} \frac{1}{2} \left( \ln (m-1)+ \ln m - \ln (cm -1) - \ln ( cm) \right) - \frac{1}{2(\lceil cm \rceil -1)} \\= & {} \frac{1}{2} \left( \ln (m-1)+ \ln m - \ln (c(m-1/c)) - \ln (cm) \right) - \frac{1}{2(\lceil cm \rceil -1)} \\= & {} \frac{1}{2} \left( \ln (m-1) - \ln (m-1/c) -2 \ln (c) \right) - \frac{1}{2(\lceil cm \rceil -1)} \\\ge & {} \frac{1}{2} \left( 2 \ln (1/c) \right) - \frac{1}{2(\lceil cm \rceil -1)} \\\ge & {} \ln (1/c) - \frac{1}{2(cm-1)}, \end{aligned}$$

where the second to last inequality holds since \(\ln (m-1/c) \le \ln (m-1)\). for \(0<c\le 1\) and sufficiently large m. We obtain

$$\begin{aligned} f_m(\alpha )= & {} (\alpha -1)(H_{m-1}-H_{\lceil (1-1/\alpha )m \rceil -1}) + \left( \lceil (1-1/\alpha )m \rceil \right) \frac{\alpha }{m} \\> & {} (\alpha -1)\left( \ln (\frac{\alpha }{\alpha -1}) - \frac{1}{2((1-1/\alpha )m -1)} - \frac{1}{\lceil (1-1/\alpha )m \rceil -1} \right) \\&+ \left( \lceil (1-1/\alpha )m \rceil \right) \frac{\alpha }{m} \\\ge & {} (\alpha -1)\left( \ln \left( \frac{\alpha }{\alpha -1}\right) - \frac{1}{(1-1/\alpha )m -1} \right) + \alpha -1 =: F(m). \end{aligned}$$

Obviously, \(\lim _{m\rightarrow \infty } F(m) = (\alpha -1) \ln (\frac{\alpha }{\alpha -1}) +\alpha -1\). We show that \((\alpha -1) \ln (\frac{\alpha }{\alpha -1}) +\alpha -1 = 1\), for \(\alpha =\frac{1}{1-\delta }\), where \(\delta = -1/W_{-1}(-1/e^2)\).

Equation \((\alpha -1) \ln (\frac{\alpha }{\alpha -1}) + \alpha -1 = 1\) is equivalent to \(\ln (\frac{\alpha }{\alpha -1})+1 = \frac{1}{\alpha -1}\), which in turn is equivalent to

$$\begin{aligned} \frac{\alpha }{\alpha -1} \cdot e = e^\frac{1}{\alpha -1}. \end{aligned}$$

Substituting \(x=1/(\alpha -1)\), which is equivalent to \(\alpha =1/x+1\), we find that the above is equivalent to \(xe+e = e^x\). Applying the Lambert W function we find that \(x=-W_{-1}(-1/e^2)-1\) is a solution of the former equality. Substituting we conclude that in fact \(\alpha = W_{-1}(-1/e^2)/(1+ W_{-1}(-1/e^2))\) satisfies the equality. Using the same techniques we can show that \(\lim _{m\rightarrow \infty } \alpha _m\) is lower bounded by \(W_{-1}(-1/e^2)/(1+ W_{-1}(-1/e^2))\). In the calculations, (1) yields that \(H_{m-1} - H_{\lceil cm \rceil } < \ln (1/c) + 1/(2m)\). \(\square \)

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Albers, S., Hellwig, M. On the Value of Job Migration in Online Makespan Minimization. Algorithmica 79, 598–623 (2017). https://doi.org/10.1007/s00453-016-0209-9

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Keywords

  • Scheduling
  • Makespan minimization
  • Online algorithm
  • Competitive analysis
  • Job migration