The Longest Processing Time rule for identical parallel machines revisited

Abstract

We consider the $P_m || C_{\max }$ scheduling problem where the goal is to schedule n jobs on m identical parallel machines $(m < n)$ to minimize makespan. We revisit the famous Longest Processing Time (LPT) rule proposed by Graham in 1969. LPT requires to sort jobs in non-ascending order of processing times and then to assign one job at a time to the machine whose load is smallest so far. We provide new insights into LPT and discuss the approximation ratio of a modification of LPT that improves Graham’s bound from $\left( \frac{4}{3} - \frac{1}{3m} \right)$ to $\left( \frac{4}{3} - \frac{1}{3(m-1)} \right)$ for $m \ge 3$ and from $\frac{7}{6}$ to $\frac{9}{8}$ for $m=2$. We use linear programming to analyze the approximation ratio of our approach. This performance analysis can be seen as a valid alternative to formal proofs based on analytical derivation. Also, we derive from the proposed approach an $O(n \log n)$ time complexity heuristic. The heuristic splits the sorted job set in tuples of m consecutive jobs ($1,\dots ,m; m+1,\dots ,2m;$ etc.) and sorts the tuples in non-increasing order of the difference (slack) between largest job and smallest job in the tuple. Then, given this new ordering of the job set, list scheduling is applied. This approach strongly outperforms LPT on benchmark literature instances and is competitive with more involved approaches such as COMBINE and LDM.

Appendix: Proof of Proposition 6

Proof

The relevant cases involve instances with $2m + 2 \le n \le 3m$, where LPT schedules the critical job in third position on a machine. Also, notice that each non-critical machine must process at most three jobs to contradict the claim; otherwise, the results of Lemma 1 hold for $k \ge 4$. In addition, an optimal solution must assign at most three jobs to each machine. Otherwise, we would have $C_{m}^*({\mathcal {J}}) \ge \sum \nolimits _{j=n-3}^{n} p_j \implies p_n \le \frac{C_{m}^*({\mathcal {J}})}{4}$ which induces a $\left( \frac{5}{4} - \frac{1}{4m}\right)$ performance guarantee according to expression (3) (with $j^\prime = n$).

Considering the above requirements for both LPT schedule and the optimal solution, we introduce different LP formulations which consider appropriate bounds on the completion times of the machines as well as on the optimal makespan according to the number of jobs and machines involved. We analyze two macro-cases defined by the two different LP modelings.

• Case a): $n = 3m ~(m=3 ~ n=9; m=4 ~ n=12);$

• Case b): $2m+2 \le n \le 3m - 1 ~(m=3 ~ n=8; m=4 ~ n=10,11).$

1.1 Case a)

Since $n=3m$, an optimal solution must process exactly three jobs on each machine to contradict the claim. This implies a lower bound on the optimum equal to $p_1 + p_{(3m-1)} + p_{3m}$ as well as that condition $p_1 \le 2p_{3m}$ must hold (otherwise $p_{n} \le \frac{C_{m}^*({\mathcal {J}})}{4}$). Given the last condition, LPT couples jobs $1,2,\dots ,m$, respectively, with jobs $2m,(2m-1),\dots ,(m+1)$ on the machines. A valid upper bound on $C_{m}^{LPT}({\mathcal {J}})$ is hence given by $p_1 + p_{(m+1)} + p_{3m}$. In order to evaluate the worst-case LPT performance, we introduce a simple LP formulation with nonnegative variables $p_j$ ($j=1,\dots ,n$) related to job processing times and variable opt which again represents $C_{m}^*({\mathcal {J}})$. As in model (8)–(16), we minimize the value of $C_{m}^*({\mathcal {J}})$ after setting w.l.o.g. $C_{m}^{LPT}({\mathcal {J}})=1$. The following LP model is implied:

$$\begin{aligned} \text {minimize}\quad&opt \end{aligned}$$

(64)

$$\begin{aligned} \text {subject to}\quad&\sum \limits _{j=1}^{3m} p_j \le m\cdot opt \end{aligned}$$

(65)

$$\begin{aligned}&p_1 + p_{(3m -1)} + p_{3m} - opt \le 0 \end{aligned}$$

(66)

$$\begin{aligned}&p_1 - 2p_{3m} \le 0 \end{aligned}$$

(67)

$$\begin{aligned}&p_1 + p_{(m + 1)} + p_{3m} \ge 1 \end{aligned}$$

(68)

$$\begin{aligned}&p_{(j + 1)} - p_j \le 0 \quad j=1, \dots , 3m - 1; \end{aligned}$$

(69)

$$\begin{aligned}&p_j \ge 0 \quad j=1, \dots , 3m; \end{aligned}$$

(70)

Constraints (65)–(70) represent the above-specified conditions together with the sorting of the jobs by decreasing processing time (constraint (69)). Solving model (64)–(70) by means of an LP solver (e.g., CPLEX) provides the optimal solution value of opt for any value of m. More precisely, we have $opt=0.8571\dots =\frac{6}{7}$ for $m=3$ and $opt=0.8421\dots =\frac{16}{19}$ for $m=4$. The claim is showed by the corresponding upper bounds $\frac{1}{opt}$ on the ratio $\frac{C_{m}^{LPT}({\mathcal {J}})}{C_{m}^*({\mathcal {J}})}$ which are equal to $\frac{7}{6}$ for $m=3$ and to $\frac{19}{16} (< \frac{11}{9})$ for $m=4$.

1.2 Case b)

As said above, LPT assigns three jobs to the critical machine. The possible values of n and m also imply that there is a non-critical machine in the LPT solution which schedules three jobs. As in Case a), we introduce an LP formulation which reconsiders model (8)–(16). For the target non-critical machine loading three jobs, we distinguish the processing time of the last job assigned to the machine, represented by a nonnegative variable $p^{\prime }$, from the contribution of the other two jobs, represented by nonnegative variable $t^{\prime }$. Variables $t_{c},t^{\prime \prime }, opt$ have the same meaning as in model (8)–(16).

First notice that $p^{\prime } \ge p_{(n-1)}$ holds as the critical job is n and that neither job $n-1$ nor job n can contribute to the value of $t^{\prime }$. Also, in any LPT schedule both $t_{c}$ and $t^{\prime }$ are given by the sum of two jobs where the first job is between job 1 and job m. The following relations straightforwardly hold:

$$\begin{aligned} t_{c} \le p_1 + p_{(m+1)}; ~ t^{\prime } \ge p_m + p_{(n-2)} \end{aligned}$$

Considering these conditions and the reasoning applied to model (8)–(16), we get the following LP model:

$$\begin{aligned} \text {minimize}\quad&opt \end{aligned}$$

(71)

$$\begin{aligned} \text {subject to}\quad&\sum \limits _{j=1}^{n} p_j - m\cdot opt \le 0 \end{aligned}$$

(72)

$$\begin{aligned}&(t_{c} + p_n) + (t^{\prime } + p^{\prime }) + t^{\prime \prime } - \sum \limits _{j=1}^{n} p_j = 0 \end{aligned}$$

(73)

$$\begin{aligned}&t_{c} - (t^{\prime } + p^{\prime }) \le 0 \end{aligned}$$

(74)

$$\begin{aligned}&(m-2)t_{c} - t^{\prime \prime } \le 0 \end{aligned}$$

(75)

$$\begin{aligned}&t_{c} + p_n = 1 \end{aligned}$$

(76)

$$\begin{aligned}&p_{(n-1)} - p^{\prime } \le 0 \end{aligned}$$

(77)

$$\begin{aligned}&p_m + p_{(n-2)} - t^{\prime } \le 0 \end{aligned}$$

(78)

$$\begin{aligned}&t_{c} - (p_1 + p_{(m+1)}) \le 0 \end{aligned}$$

(79)

$$\begin{aligned}&p_{(j + 1)} - p_j \le 0 \quad j=1, \dots , n - 1; \end{aligned}$$

(80)

$$\begin{aligned}&p_j \ge 0 \quad j=1, \dots , n; \end{aligned}$$

(81)

$$\begin{aligned}&t_{c}, t^{\prime }, t^{\prime \prime }, p^{\prime }, opt \ge 0 \end{aligned}$$

(82)

Model (71)–(82) constitutes a backbone LP formulation for all subcases analyzed in the following.

1.2.1 $P_m || C_{\max }$ instances with $m=4$, $n=11$

To contradict the claim, an optimal solution must assign three jobs to three machines and two jobs to one machine. Assume first that the optimal solution processes the first three jobs on two machines. Since the optimal solution must schedule at least five jobs on two machines, a valid lower bound on $C_{m}^*({\mathcal {J}})$ is equal to one half of the sum $(p_1 + p_2 + p_3 + p_{10} + p_{11})$. Thus, we can add constraint

$$\begin{aligned} opt \ge \frac{p_1 + p_2 + p_3 + p_{10} + p_{11}}{2} \end{aligned}$$

to model (71)–(82). The corresponding LP optimal solution gives an upper bound on ratio $\frac{C_{m}^{LPT}({\mathcal {J}})}{C_{m}^*({\mathcal {J}})}$ equal to $\frac{1}{opt} = \frac{11}{9}$.

Likewise, should the optimal solution schedule the first three jobs on three different machines, a valid lower bound on $C_{m}^*({\mathcal {J}})$ would correspond to the following constraint

$$\begin{aligned} opt \ge \frac{p_1 + p_2 + p_3 + p_7 + p_8 + p_9 + p_{10} + p_{11}}{3} \end{aligned}$$

since the optimal solution has to process at least eight jobs on three machines. If we add the last constraint to model (71)–(82), the LP optimal solution yields again a value of $\frac{1}{opt}$ equal to $\frac{11}{9}$.

1.2.2 $P_m || C_{\max }$ instances with $m=4$, $n=10$

An optimal solution assigns either three jobs to three machines and one job to the other machine, or three jobs to two machines and two jobs to the others. We again distinguish whether the optimal solution processes the first three jobs either on two or three machines. In the first case, since two machines must process at least four jobs in an optimal solution, a valid lower bound on $C_{m}^*({\mathcal {J}})$ is equal to one half of the sum $(p_1 + p_2 + p_3 + p_{10})$. The optimal objective value of model (71)–(82) after adding constraint

$$\begin{aligned} opt \ge \frac{p_1 + p_2 + p_3 + p_{10}}{2} \end{aligned}$$

provides an approximation ratio equal to $\frac{1}{opt} = \frac{11}{9}$. In the second case, as three machines must necessarily process at least seven jobs, we can add to the LP model the following constraint which represents a valid lower bound on $C_{m}^*({\mathcal {J}})$:

$$\begin{aligned} opt \ge \frac{p_1 + p_2 + p_3 + p_7 + p_8 + p_9 + p_{10}}{3} \end{aligned}$$

From the corresponding LP optimal solution, we get again $\frac{1}{opt}=\frac{11}{9}$.

1.2.3 $P_m || C_{\max }$ instances with $m=3$, $n=8$

We have to address the case where an optimal solution assigns two jobs to a machine and three jobs to the other two machines. This implies relations $C_{m}^*({\mathcal {J}}) \ge p_1 + p_8$ and $C_{m}^*({\mathcal {J}}) \ge \frac{p_1 + p_2 + p_6 + p_7 + p_8}{2}$ and the corresponding constraints to be added to model (71)–(82):

$$\begin{aligned} opt \ge p_1 + p_8; ~ opt \ge \frac{p_1 + p_2 + p_6 + p_7 + p_8}{2} \end{aligned}$$

For the solution provided by LPT, we now analyze all possible assignments of the jobs which can give $t^{\prime }$. LPT separately schedules the first three jobs and job 4 with job 3 on the third machine. It immediately follows that if the target non-critical machine is the first one, then it must necessarily process job 6 as second job, i.e., $t^{\prime } = p_1 + p_6$. Instead, if the non-critical machine is the second machine, then $t^{\prime }$ must be contributed by $p_2$ and $p_5$, i.e., $t^{\prime } = p_2 + p_5$. To show this, first notice that the assignment $t^{\prime } = p_2 + p_6$, which also implies $p^{\prime } = p_7$ as well as having job 5 with job 3 and 4, cannot occur, as it would prevent the critical machine from scheduling three jobs. Also, job 7 cannot be scheduled as second job on such non-critical machine. Eventually, a third possibility is to have $t^{\prime } = p_3 + p_4$. Solving model (71)–(82) with the constraints related to the three possible assignments of $t^{\prime }$ provides the following upper bounds on ratio $\frac{C_{m}^{LPT}({\mathcal {J}})}{C_{m}^*({\mathcal {J}})}$:

$$\begin{aligned}&t^{\prime } = p_1 + p_6 \implies \frac{1}{opt} = \frac{15}{13} \left( < \frac{7}{6} \right) ; \\&t^{\prime } = p_2 + p_5 \implies \frac{1}{opt} = \frac{7}{6}; \\&t^{\prime } = p_3 + p_4 \implies \frac{1}{opt} = \frac{7}{6}. \end{aligned}$$

Putting together all the stated results, we get the claim. $\square$