We break the barrier of $3/2$ for the problem of online load balancing with known makespan, also known as bin stretching. In this problem, $m$ identical machines and the optimal makespan are given. The load of a machine is the total size of all the jobs assigned to it and the makespan is the maximum load of all the machines. Jobs arrive online and the goal is to assign each job to a machine while staying within a small factor (the competitive ratio) of the optimal makespan. We present an algorithm that maintains a competitive ratio of $139/93<1.495$ for sufficiently large values of $m$, improving the previous bound of $3/2$. The value 3/2 represents a natural bound for this problem: as long as the online bins are of size at least $3/2$ of the offline bin, all items that fit at least two times in an offline bin have two nice properties. They fit three times in an online bin and a single such item can be packed together with an item of any size in an online bin. These properties are now both lost, which means that putting even one job on a wrong machine can leave some job unassigned at the end. It also makes it harder to determine good thresholds for the item types. This was one of the main technical issues in getting below $3/2$. The analysis consists of an intricate mixture of size and weight arguments.

翻译：我们突破了已知完工时间在线负载均衡问题（亦称箱体拉伸问题）的$3/2$障碍。该问题给定$m$台相同机器及最优完工时间。机器负载为其分配的所有作业尺寸之和，完工时间指所有机器中的最大负载。作业在线到达，目标是在将每个作业分配到机器的同时，使完工时间保持在最优完工时间的一个较小倍数（竞争比）内。我们提出一种算法，在$m$足够大时保持$139/93<1.495$的竞争比，改进了原有的$3/2$上界。数值3/2代表该问题的一个自然界限：只要在线箱体尺寸不小于离线箱体的$3/2$，所有能至少两次装入离线箱体的物品均具有两个优良特性——它们能在在线箱体中装入三次，且单个此类物品可与任意尺寸物品共同装入在线箱体。如今这两个特性均已丧失，这意味着即使将单个作业分配到错误机器，也可能导致最终有作业无法分配。这也使得确定物品类型的合适阈值变得更加困难，这正是突破$3/2$界限的主要技术难点之一。我们的分析融合了尺寸论证与权重论证的复杂技术体系。