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在安排建筑群体工程施工总进度计划时,如何确定各单位工程之间的先后施工顺序,使完成各工程的总工期最短? 象这一类的施工顺序选择问题,在数学上称之为排序问题。排序问题,当n=2 (即两个有序作业的工序) 时,数学方法处理比较成功,计算也很简单。但n>2时,要找出最优排序方案计算工作十分繁琐,而且仅仅是n=3,这对于解决施工顺序安排问题是不够的。由于n>2的排序问题,在方法上没有得到解决,常常在处理上,都用n=2的方法的近似解,有时接近于最优解(按枚举法),有时误差比较
When arranging the construction schedule for the construction of a group of buildings, how to determine the order of construction between each unit of the project so that the total duration of each project is the shortest? The question of the order of construction like this is called the sorting problem in mathematics. . For the ordering problem, when n=2 (ie, the operation of two sequential jobs), the mathematical method is more successful and the calculation is very simple. But when n>2, it is very tedious to find out the optimal sorting plan calculation work, and it is only n=3. This is not enough to solve the construction scheduling problem. Due to the ordering problem of n>2, there is no solution in the method. Often, the approximate solution of the n=2 method is used in the processing, sometimes it is close to the optimal solution (by enumeration), and sometimes the error is compared.