Online scheduling with rejection and reordering: exact algorithms for unit size jobs

Abstract

We study an online scheduling problem with rejection on $m\ge 2$ identical machines, in which we deal with unit size jobs. Each arriving job has a rejection value (a rejection cost or penalty for minimization problems, and a rejection profit for maximization problems) associated with it. A buffer of size $K$ is available to store $K$ jobs. A job which is not stored in the buffer must be either assigned to a machine or rejected. Upon the arrival of a new job, the job can be stored in the buffer if there is a free slot (possibly created by evicting other jobs and assigning or rejecting every evicted job). At termination, the buffer must be emptied. We study four variants of the problem, as follows. We study the makespan minimization problem, where the goal is to minimize the sum of the makespan and the penalty of rejected jobs, and the $\ell _p$ norm minimization problem, where the goal is to minimize the sum of the $\ell _p$ norm of the vector of machine completion times and the penalty of rejected jobs. We also study two maximization problems, where the goal in the first version is to maximize the sum of the minimum machine load (the cover value of the machines) and the total rejection profit, and in the second version the goal is to maximize a function of the machine completion times (which measures the balance of machine loads) and the total rejection profit. We show that an optimal solution (an exact solution for the offline problem) can always be obtained in this environment, and determine the required buffer size. Specifically, for all four variants we present optimal algorithms with $K=m-1$ and prove that in each case, using a buffer of size at most $m-2$ does not allow the design of an optimal algorithm, which makes our algorithms optimal in this respect as well. The lower bounds hold even for the special case where the rejection value is equal for all input jobs.

Appendices

Appendix A: Some properties of the considered functions

In this section we prove all technical properties used throughout the paper.

Lemma 10

For any pair of integers $t,y$ such that $0 \le t \le m, y \ge 0$, and $p>1$,

$$\begin{aligned} (t(y+1)^p+(m-t)y^p)^{1/p}-(my^p)^{1/p} \ge t\cdot m^{\frac{1}{p}-1}. \end{aligned}$$

Proof

The inequality holds trivially (with equality) for $t=0$. For $t=m$, we have $(m(y+1)^p+(m-m)y^p)^{1/p}-(my^p)^{1/p}=m^{1/p}((y+1)-y)=m^{1/p}$.

Assume therefore $1\le t \le m-1$. Let $\alpha =\frac{t}{m}$ (and thus $0<\alpha <1$). Since the function $x^p$ is strictly monotonically increasing in $x$, the required inequality is equivalent to $t(y+1)^p+(m-t)y^p \ge (m^{1/p}\cdot y + (t/m) \cdot m^{1/p})^p$, or alternatively, $\frac{t}{m} (y+1)^p+\frac{m-t}{m} \cdot y^p \ge (y + t/m)^p$. Substituting the value of $\alpha$ this is equivalent to $\alpha (y+1)^p+ (1-\alpha ) \cdot y^p \ge (y + \alpha )^p$, which holds due to the convexity of the function $x^p$ for $p>1$. $\square$

Lemma 11

For any pair of integers $t,y$ such that $0 \le t \le m, y \ge 1$, and $p>1$,

$$\begin{aligned} (t(y+1)^p+(m-t)y^p)^{1/p}-(t\cdot y^p+(m-t)(y-1)^p)^{1/p} \le m^{\frac{1}{p}}. \end{aligned}$$

Proof

Let $\bar{v}$ be the vector of length $m$ that has $t$ components equal to $y+1$ and $m-t$ components equal to $y$. The claim is equivalent to the statement that the $\ell _p$ norm of $\bar{v}$ is at most the $\ell _p$ norm of $\bar{v}-(1,1,\ldots ,1)$ plus the $\ell _p$ norm of $(1,1,\ldots ,1)$, which is true for any norm. $\square$

Lemma 12

For any pair of integers $t,y$ such that $0 \le t \le m, y \ge 0$, and $p>1$,

$$\begin{aligned} \left(t(y+1)^{1/p}+(m-t)y^{1/p}\right)^{p}-\left(my^{1/p}\right)^{p} \le t\cdot m^{p-1}. \end{aligned}$$

Proof

The inequality holds trivially (with equality) for $t=0$. For $t=m$, we have $(m(y+1)^{1/p}+(m-m)y^{1/p})^{p}-(my^{1/p})^{p}=m^{p}((y+1)-y)=m^{p}$.

Assume therefore $1\le t \le m-1$. Let $\alpha =\frac{t}{m}$ (and thus $0<\alpha <1$). Since the function $x^{1/p}$ is strictly monotonically increasing in $x$, the inequality is equivalent to $t(y+1)^{1/p}+(m-t)y^{1/p} \le (m^{p}\cdot y + (t/m) \cdot m^{p})^{1/p}$, or alternatively, $\frac{t}{m} (y+1)^{1/p}+\frac{m-t}{m} \cdot y^{1/p} \le (y + t/m)^{1/p}$. Substituting the value of $\alpha$ this is equivalent to $\alpha (y+1)^{1/p}+ (1-\alpha ) \cdot y^{1/p} \le (y + \alpha )^{1/p}$, which holds due to the concavity of the function $x^{1/p}$ for $p>1$.

Lemma 13

For any pair of integers $t,y$ such that $0 \le t \le m, y \ge 1$, and $p>1$,

$$\begin{aligned} \left(t(y+1)^{1/p}+(m-t)y^{1/p}\right)^{p}-\left(t\cdot y^{1/p}+(m-t)(y-1)^{1/p}\right)^{p} \ge m^{p}. \end{aligned}$$

Proof

The inequality holds with equality for $t=0,m$. Assume therefore $1\le t\le m-1$ and let $\alpha =\frac{t}{m}$. The inequality is equivalent to $(\alpha (y+1)^{1/p}+(1-\alpha )y^{1/p})^{p}-(\alpha \cdot y^{1/p}+(1-\alpha )(y-1)^{1/p})^{p} \ge 1$, or alternatively,

$$\begin{aligned} \alpha (y+1)^{1/p}+(1-\alpha )y^{1/p} \ge \left(\left(\alpha \cdot y^{1/p}+(1-\alpha )(y-1)^{1/p}\right)^{p} + 1\right)^{1/p}, \end{aligned}$$

since the function $x^{p}$ is monotonically increasing in $x$. The right hand size is the $\ell _p$ norm of the vector (of two components) $((\alpha \cdot y^{1/p}+(1-\alpha )(y-1)^{1/p},1)$, which is at most the sum of $\ell _p$ norms of the vectors $\alpha (y^{1/p},1)$ and $(1-\alpha )((y-1)^{1/p},1)$, which is equal to $\alpha (y+1)^{1/p}+(1-\alpha )y^{1/p}$. This proves the claim. $\square$

Appendix B: The technical part of the proof of Theorem 1

First, we prove the lower bounds for the minimization problems. Consider an algorithm for one of these problems (${\textsc {MR}}$ or ${\textsc {NR}}$). Let $\alpha$ be a value satisfying $\frac{1}{m}< \alpha < \frac{1}{m-1}$ if the algorithm is for MR, and $\frac{1}{m^{1- \frac{1}{p}}} < \alpha < \frac{1}{(m-1)^{1-\frac{1}{p}}}$ if the algorithm is for the ${\textsc {NR}}$ problem for some $p>1$. We use an input of $m-1$ jobs of rejection penalty $\alpha$. As $K \le m-2$, at least one of these jobs must be either scheduled or rejected. If the algorithm schedules at least one of these jobs, then the input terminates. We find that the cost of a solution which rejects all jobs is $(m-1)\alpha$, and thus ${\textsc {opt}}\le (m-1)\alpha$. Let $1 \le q \le m-1$ denote the number of jobs scheduled by the algorithm (and so $m-1-q \ge 0$ jobs are rejected). For the MRproblem, the makespan is at least 1, and so the cost of the algorithm is at least $1+(m-1-q)\alpha \ge 1 > (m-1)\alpha$, and thus the algorithm is not optimal. For the ${\textsc {NR}}$ problem, the $\ell _p$ norm of the load vector is at least $q^{\frac{1}{p}}$ (corresponding to a schedule where each job is assigned to a different machine, which is minimal by convexity of the function $x^p$ for $p>1$), and the cost of the algorithm is at least $q^{\frac{1}{p}}+(m-1-q)\alpha$. We show $q^{1/p}>q\alpha$. Using $\alpha < (m-1)^{\frac{1}{p}-1}$ we have $q\alpha < q(m-1)^{\frac{1}{p}-1}$. Since $q \le m-1$, we have $(m-1)^{1-1/p} \ge q^{1-1/p}$ (since $p>1$), giving $q(m-1)^{\frac{1}{p}-1} \le q^{{1/p}}$. We get $q^{\frac{1}{p}}+(m-1-q)\alpha > q\alpha + (m-1-q)\alpha = (m-1)\alpha$, and so the algorithm is not optimal.

Now assume that the algorithm rejects at least one job. One final job of the same penalty $\alpha$ arrives. The cost of a solution which schedules all jobs is $1$ for ${\textsc {MR}}$ and $m^{1/p}$ for ${\textsc {NR}}$, and thus this is an upper bound on ${\textsc {opt}}$. Let $0 \le \tilde{q} \le m-1$ denote the number of jobs scheduled by the algorithm (and so $m-\tilde{q} \ge 1$ jobs are rejected). For the ${\textsc {MR}}$ problem, if $\tilde{q} =0$, then the cost is $m\alpha >1 \ge {\textsc {opt}}$, and otherwise the makespan is $1$, and thus the cost is at least $1+\alpha >1$. Therefore the algorithm is not optimal. For the ${\textsc {NR}}$ problem, if $\tilde{q} =0$, then the cost is $m\alpha >m^{1/p} \ge {\textsc {opt}}$. Otherwise, $\tilde{q} >0$, the $\ell _p$ norm of the load vector is at least $\tilde{q} ^{\frac{1}{p}}$, and the cost of the algorithm is at least $\tilde{q} ^{\frac{1}{p}}+(m-\tilde{q} )\alpha$. We show $\tilde{q} ^{1/p} \ge \frac{\tilde{q} m^{1/p}}{m}$. This holds since it is equivalent to $\tilde{q} ^{1-1/p} \le m^{1-1/p}$, which holds since $p>1$ and $\tilde{q} \le m$. We find $\tilde{q} ^{\frac{1}{p}}+(m-\tilde{q} )\alpha > \tilde{q} m^{\frac{1}{p}-1}+(m-\tilde{q} )m^{\frac{1}{p}-1} = m^{1/p}$, and thus the algorithm is not optimal.

Consider an algorithm for one of the maximization problems (${\textsc {CR}}$ or ${\textsc {FR}}$). We use the same basic structure of inputs as follows. There are $m-1$ jobs of rejection profit $\alpha$, such that $(m-1)^{p-1} < \alpha < m^{p-1}$ if the algorithm is for ${\textsc {FR}}$ for some $p>1$, and $0< \alpha < \frac{1}{m}$, if it is for ${\textsc {CR}}$. As $K \le m-2$, at least one of these jobs must be either scheduled or rejected. If the algorithm schedules at least one of these jobs, then the input terminates. The profit of an algorithm which rejects all jobs is $(m-1)\alpha$. For ${\textsc {CR}}$, the profit of the algorithm is at most $(m-2)\alpha$; as at most $m-2$ out of the input $m-1$ jobs can be rejected, and all profit comes from the rejection profit, as the number of jobs is below $m$. For ${\textsc {FR}}$, let $1 \le q \le m-1$ be the number of jobs scheduled by the algorithm. We first show $q^p < q\alpha$. This holds since $q \le m-1$ and thus $q^{p-1} \le (m-1)^{p-1}< \alpha$, and so $q\cdot q^{p-1} < q\alpha$. Thus, the profit of the algorithm is at most $q^{p}+(m-1-q)\alpha < q\alpha +(m-1-q)\alpha =(m-1)\alpha$.

Now assume that the algorithm rejects at least one job. One final job of the same profit $\alpha$ arrives. The profit from scheduling all jobs is $1$ for ${\textsc {CR}}$ and $m^p$ for ${\textsc {FR}}$. For ${\textsc {CR}}$, since out of the $m$ input jobs at least one is rejected, all profit comes from the rejected jobs, and it is at most $(m-1)\alpha <1$. For ${\textsc {FR}}$, let $0 \le q^{\prime } \le m-1$ be the number of scheduled jobs. If $q^{\prime }=0$, then its profit is $m\alpha < m \cdot m^{p-1}=m^p$. Otherwise $q^{\prime }>0$ and $m-q^{\prime }>0$ hold, and the profit is at most $(m-q^{\prime })\alpha +(q^{\prime })^p<(m-q^{\prime })m^{p-1}+(q^{\prime })^p=(m-q^{\prime })m^{p-1}+q^{\prime }\cdot (q^{\prime })^{p-1}<(m-q^{\prime })m^{p-1}+q^{\prime }\cdot (m)^{p-1}=m^p$.