The Price of Fixed Assignments in Stochastic Extensible Bin Packing

Abstract

We consider the stochastic extensible bin packing problem (SEBP) in which n items of stochastic size are packed into m bins of unit capacity. In contrast to the classical bin packing problem, the number of bins is fixed and they can be extended at extra cost. This problem plays an important role in stochastic environments such as in surgery scheduling: Patients must be assigned to operating rooms beforehand, such that the regular capacity is fully utilized while the amount of overtime is as small as possible.

This paper focuses on essential ratios between different classes of policies: First, we consider the price of non-splittability, in which we compare the optimal non-anticipatory policy against the optimal fractional assignment policy. We show that this ratio has a tight upper bound of 2. Moreover, we develop an analysis of a fixed assignment variant of the LEPT rule yielding a tight approximation ratio of $(1+e^{-1}) \approx 1.368$ under a reasonable assumption on the distributions of job durations.

Furthermore, we prove that the price of fixed assignments, related to the benefit of adaptivity, which describes the loss when restricting to fixed assignment policies, is within the same factor. This shows that in some sense, LEPT is the best fixed assignment policy we can hope for.

The research of the first two authors is carried out in the framework of MATHEON supported by Einstein Foundation Berlin.

A Proofs of Intermediate Results

Proof

(of Proposition 2). To prove this result, we examine the change in the objective value of $\varPi$ when we move one job to the machine with highest load in $\varPi$, for a realization ${{\varvec{p}}}$ of the processing times. W.l.o.g. let machine 1 be the one with highest workload in $\varPi ({{\varvec{p}}})$. Consider another machine $i\in \mathcal {M}\setminus \{1\}$ on which at least one job is scheduled. Let k be the last job on machine i, i.e., $C_k^\varPi ({{\varvec{p}}})=W_i^\varPi ({{\varvec{p}}})$. For the sake of simplicity, we define $A:= \{j\in \mathcal {J}|j\xrightarrow {\varPi ({{\varvec{p}}})} i\}\setminus \{k\}$ and $B:= \{j\in \mathcal {J}|j\xrightarrow {\varPi ({{\varvec{p}}})} 1\}$. We consider another schedule $\varPi '({{\varvec{p}}})$ which coincides with $\varPi ({{\varvec{p}}})$ except that job k is scheduled on machine 1 right after all jobs in B. We obtain

$$\begin{aligned}&\!\phi (\varPi ,{{\varvec{p}}})- \phi (\varPi ',{{\varvec{p}}})\\&=\!&\max \Bigl (\sum _{j\in A} p_j + p_k, 1\Bigr )\! +\! \max \Bigl (\sum _{j\in B} p_j, 1\Bigr ) - \Bigl (\max \Bigl (\sum _{j\in A} p_j, 1\Bigr ) + \max \Bigl (\sum _{j\in B} p_j + p_k, 1\Bigr )\Bigr )\\= & {} {\left\{ \begin{array}{ll} 1 + \max \Bigl (\displaystyle {\sum _{j\in B}} p_j, 1\Bigr ) - \Bigl (1 + \max \Bigl (\sum _{j\in B} p_j + p_k, 1\Bigr )\Bigr ) \quad &{}\ \text {if } \displaystyle {\sum _{j\in A}} p_j + p_k\le 1\\ \displaystyle {\sum _{j\in A}} p_j + p_k + \sum _{j\in B} p_j - \Bigl (\max \Bigl (\sum _{j\in A} p_j, 1\Bigr ) + \sum _{j\in B} p_j + p_k\Bigr ) \quad &{}\ \text {otherwise } \end{array}\right. }\\\le & {} \! 0. \end{aligned}$$

Hence, iteratively moving some job k to the fullest machine yields $\phi (\varPi ,{{\varvec{p}}})\le \phi (\varPi _1,{{\varvec{p}}})$. Finally, the result follows by taking the expectation.

Proof

(of Lemma 1). The proof simply works by exploiting the analytical form of Poisson probabilities:

$$\begin{aligned} \frac{1}{\lambda }\mathbb {E}\Big [\max (Y, \lambda )\Big ]&= \frac{1}{\lambda }\sum _{k=0}^{\infty }\max (k, \lambda )\cdot \frac{e^{-\lambda }\lambda ^k}{k!}\\&= \frac{1}{\lambda }\sum _{k=0}^{\infty } k\cdot \frac{e^{-\lambda }\lambda ^k}{k!} + \frac{1}{\lambda }\sum _{k=0}^{\infty } \max (0, \lambda - k)\cdot \frac{e^{-\lambda }\lambda ^k}{k!}\\&= 1 + \sum _{k=0}^{\lambda } \Bigl (1 - \frac{k}{\lambda }\Bigr )\cdot \frac{e^{-\lambda }\lambda ^k}{k!}\\&= 1 + e^{-\lambda }\cdot \Bigl (\sum _{k=0}^{\lambda } \frac{\lambda ^k}{k!} - \sum _{k=1}^{\lambda } \frac{\lambda ^{k-1}}{(k-1)!}\Bigr )\\&= 1 + \frac{e^{-\lambda }\lambda ^\lambda }{\lambda !}, \end{aligned}$$

where the last step follows from the property of a telescoping sum.

Proof

(of Lemma 2).

Let X and Y be random variables with $\mathbb {P}[ 0\le X\le 1]=1.$ Observe that $0\le \mathbb {E}[X]\le 1$. We are going to show that $\mathbb {E}[\max (X+Y,1)]$ can be bounded from above by choosing the two point distribution $X^*\sim {\text {Bernoulli}}(\mathbb {E}[X])$, such that $\mathbb {P}[X^*=0] = (1-\mathbb {E}[X])$ and $\mathbb {P}[X^*=1] = \mathbb {E}[X].$ To do so, we define the function $g:[0,1]\rightarrow \mathbb {R}$, $x\mapsto \mathbb {E}_Y[\max (x+Y,1)].$ This function is convex, since it is the expectation of a pointwise maximum of two affine functions [4]. Therefore, for all $x\in [0,1]$ we have $g(x) \le g(0) + x (g(1) - g(0)).$ Then, by definition of g,

$$\begin{aligned} \mathbb {E}[\max (X+Y,1)] = \mathbb {E}_X[g(X)]&\le g(0) + \mathbb {E}_X[X]\cdot (g(1) - g(0))\\&= \mathbb {E}_{X^*}[g(X^*)]=\mathbb {E}[\max (X^*+Y,1)]. \end{aligned}$$

Using this bound for all $j\in [k]$, we obtain $\mathbb {E}\Bigl [\max \Bigl (\sum _{j=1}^k P_j,1\Bigr )\Bigr ] \le \mathbb {E}\Bigl [\max \Bigl (\sum _{j=1}^k P_j^*,1\Bigr )\Bigr ],$ where $P_j^* \sim {\text {Bernoulli}}(\mathbb {E}[P_j])$. Then, by the law of total expectation, we have:

$$\begin{aligned} \mathbb {E}\Bigl [\max \Bigl (\sum _{j=1}^k P_j^*,1\Bigr )\Bigr ]&= \mathbb {E}\Bigl [\sum _{j=1}^k P_j^* \Big |\sum _{j=1}^k P_j^*\ge 1\Bigr ] \mathbb {P}[\sum _{j=1}^k P_j^*\ge 1] + \mathbb {E}[1]\ \mathbb {P}[\sum _{j=1}^k P_j^*< 1]. \end{aligned}$$

Since the random variable $\sum _{j=1}^k P_j^*$ is a nonnegative integer, it cannot lie in the interval (0, 1), so the first term in the above sum is equal to $\mathbb {E}\Bigl [\sum _{j=1}^k P_j^*\Bigr ]=\sum _{j=1}^k \mathbb {E}\Big [P_j\Big ]$, and the second term is equal to $\mathbb {P}[P_1^*=\ldots =P_k^*=0]=\prod _{j=1}^k (1-\mathbb {E}[P_j])$.

Proof

(of Lemma 3). We set $\ell := \min \{x_i : i\in \mathcal {M}\}$. Then, the first inequality follows immediately. Next, we will show that in each step that $\text {LEPT}_{\mathcal {F}}$ assigns a job to a machine the second inequality is fulfilled. Let j denote the job which is put on machine i in the current step. Furthermore, let $\ell '$ and $\ell$ denote the minimum expected load among all machines before and after the allocation, respectively. Trivially, $\ell '\le \ell$ is true. Moreover, let $x_i'$ and $x_i$ denote the expected workload of i before and after assigning j to it, respectively. Clearly, we have

$$\begin{aligned} x_i=x_i'+\mathbb {E}[P_j]. \end{aligned}$$

Observe, that $\ell ' = x_i'$, because $\text {LEPT}_{\mathcal {F}}$ assigns j to the machine with the smallest expected load. In addition, let $n_i$ denote the number of jobs running on machine i after the insertion of j. Since $\text {LEPT}_{\mathcal {F}}$ sorts jobs in decreasing order of their expected processing times, it holds

$$\begin{aligned} \mathbb {E}[P_j] \le \frac{x_i'}{n_i-1} = \frac{\ell '}{n_i-1}. \end{aligned}$$

Consider a machine other than i. If the inequality of the statement was fulfilled in an earlier step, then by setting the new $\ell$ it still is true. In the beginning, when we have no job at all, the inequality is true, so we only have to take care of machine i.

Finally, we obtain on machine i

$$\begin{aligned}&\frac{x_i}{\ell }&= \frac{x_i'+\mathbb {E}[P_j]}{\ell } \le \frac{x_i'+\mathbb {E}[P_j]}{\ell '} \le 1 + \frac{\mathbb {E}[P_j]}{\ell '} \le 1 + \frac{\ell '}{\ell ' (n_i-1)} = \frac{n_i}{(n_i-1)}. \end{aligned}$$

Proof

(of Lemma 4). First, we argue that $h: y \mapsto (1-y)^{1+\frac{\ell }{y}}$ is convex over [0, 1]. To see this, we compute its second derivative:

$$\begin{aligned} h''(y) = \frac{\ell (1-y)^{\frac{\ell }{y}-1}}{y^4} h_2(y), \end{aligned}$$

where $h_2(y) := y^2 (\ell -y+2)+\ell (y-1)^2 \log ^2(1-y)-2 (\ell +1) (y-1) y \log (1-y).$ Now, we use the fact that $\log (1-y) = -\sum _{k=1}^\infty \frac{y^k}{k}$ for all $y\in [0,1)$. Hence, $\log ^2 (1-y) = \sum _{k=2}^\infty \gamma _k y^k$, where $\gamma _k:=\sum _{i=1}^{k-1} \frac{1}{i(k-i)}$. After some calculus, the terms of order 2 and 3 vanish and we obtain the following series representation of $h_2$ over [0, 1):

$$\begin{aligned} h_2(y) = (\frac{\ell }{4}+\frac{1}{3}) y^4 + \sum _{k=5}^\infty (\frac{2(\ell +1)}{(k-1)(k-2)} + \ell ( \gamma _k + \gamma _{k-2} - 2 \gamma _{k-1})) y^k. \end{aligned}$$

We are going to show that $\gamma _k + \gamma _{k-2} - 2 \gamma _{k-1}\ge 0$ for $k\ge 5$ implying that $h''(y)\ge 0$ for all $y\in [0,1)$. To do so, we rewrite the sums using the partial fraction decomposition. As a consequence, we obtain

The last inequality results from the fact that for all $k\ge 5$ we have $4\sum _{i=1}^{k-3}\frac{1}{i} \ge 6$. Hence, h is convex on [0, 1), and even on [0, 1] by continuity. Now, let $v^*(\rho ,\ell )$ denote the optimal value of the problem

$$\begin{aligned} \underset{{{\varvec{y}}}\in \mathbb {R}^m}{\varvec{{\text {maximize}}}}&\quad \sum _{i\in \mathcal {M}} h(y_i)\end{aligned}$$

(8a)

$$\begin{aligned} s.t.&\quad \sum _{i\in \mathcal {M}} y_i = m (\rho -\ell ) \end{aligned}$$

(8b)

$$\begin{aligned}&\quad 0\le y_i \le 1,\quad (\forall i\in \mathcal {M}). \end{aligned}$$

(8c)

As h is convex, a maximizer of the optimization problem above is an extreme point of the polytope induced by the constraints (8b) and (8c). Let $k:=\lfloor m(\rho -\ell ) \rfloor$ and $u:= m(\rho -\ell ) - k$, where $\lfloor . \rfloor$ denotes the floor function, that is, $\lfloor x \rfloor$ is the largest integer less than or equal to x. By construction, it holds $0\le u \le 1$, and $u+k=m(\rho -\ell )$. At an extreme point, at least $m-1$ inequalities of (8c) must be tight. Hence, one coordinate of ${{\varvec{y}}}$ must be u, k coordinates must be 1 and the remaining $(m-k-1)$ coordinates must be 0.

It follows that $v^*(\rho ,\ell ) = (m-k-1) h(0) + h(u) = (m-k-1) e^{-\ell } + (1-u)^{1+\ell /u}$. Now, we observe that $(1-u)^{\ell /u} \le e^{-\ell }$, so

where the first equivalence is due to the decomposition $m(\rho -\ell )=k+u$.

Finally, the inequality of the proposition follows from the fact that $(1+\ell -\rho )e^{-\ell }$ is a nondecreasing function of $\ell$ over $[0,\rho ]$.