Bin stretching with migration on two hierarchical machines

Abstract

In this paper, we consider semi-online scheduling with migration on two hierarchical machines, with the purpose of minimizing the makespan. The meaning of two hierarchical machines is that one of the machines can run any job, while the other machine can only run specific jobs. Every instance also has a fixed parameter $M \ge 0$, known as the migration factor. Jobs are presented one by one. Each new job has to be assigned to a machine when it arrives, and at the same time it is possible to modify the assignment of previously assigned jobs, such that the moved jobs have a total size not exceeding M times the size of the new job. The semi-online variant studied here is called bin stretching. In this problem, the optimal offline makespan is provided to the scheduler in advance. This is still a non-trivial variant for any migration factor $M > 0$. We prove tight bounds on the competitive ratio for any migration factor M. The design and analysis is split into several cases, based on the value of M, and on the resulting competitive ratio. Unlike the online variant with migration for two hierarchical machines, this case allows an online fully polynomial time approximation scheme.

Examples for the action of all algorithms

We provide a large number of examples for our algorithms. The examples cover a multitude of cases occurring in the execution.

1.1 Algorithm A, where $\varvec{M \ge 2.5}$

In this section, we provide three examples for Algorithm A. Two examples are for the migration factor $M=3$. In both of them step 4 is applied once. In the first example, after applying step 4, only step 2 will be applied, and in the second example, both step 3 and step 2 are applied. The third example is for $M=20$, and in that example, step 4 is performed several times.

Fig. 4

The schedules produced by Algorithm A with migration factor $M = 3$ for $I_1$ (see Sect. A.1.1) after three, four, and six jobs, respectively have been presented, and an optimal solution (with makespan 1) for that input. The final makespan for the algorithm is 1.12

1.1.1 Examples for Algorithm A with $\varvec{M=3}$

In the two examples of this section, we show the action of Algorithm A with a migration factor of $M = 3$. For this value of M, we have $\mu =\frac{2}{9} \approx 0.2222$, the value used by the algorithm in step 2 is $1-\mu =\frac{7}{9} \approx 0.7778$, and the value used in step 3, which is also the competitive ratio for this algorithm, is $1+\mu =\frac{11}{9}\approx 1.2222$.

Input $I_1$ is defined as follows:

$$\begin{aligned} J_1= & {} (0.19,2),\ J_2=(0.38,2),\ J_3=(0.39,1),\ J_4=(0.68,2),\\ J_5= & {} (0.32,2),\ J_6=(0.03,1). \end{aligned}$$

The algorithm schedules the first two jobs on the second machine in step 3, and it schedules the third job on the first machine in step 2. We get $x_3=0.39$, $y_3=0.57$, and $z_3=0$. When the algorithm receives the fourth job, it reaches step 4 because $1.25=y_3+p_4>1+\mu$ holds. In step 4, the algorithm updates W to be the set of the two jobs $J_1$ and $J_4$, and it schedules these jobs on the second machine. The algorithm schedules the other jobs with GoS 2 that are not in W on the first machine (this set consists of a single job, which is $J_2$), and we get $x_4=0.39,\ y_4=0.87,\ z_4=0.38$. Afterwards, the algorithm schedules the $J_5$ on the first machine in step 2, because the load of the second machine is already not smaller than $1-\mu$, i.e. $y_4 \ge \frac{7}{9}$. Job $J_6$ is assigned by step 2 as well, because both $g_6=1$ and $y_5 \ge \frac{7}{9}$ hold. Consequently, the final load of the first machine is 1.12, and the load of the second machine is 0.87. See Fig. 4 for an illustration of the process of execution of the algorithm, and an optimal solution.

Fig. 5

The schedules produced by Algorithm A with migration factor $M = 3$ for $I_2$ (see Sect. A.1.1), and an optimal solution (with makespan 1) for that input. The final makespan for the algorithm is 1.05

In the second example, we use input $I_2$, defined as follows:

$$\begin{aligned} J_1=(0.6,2),\ J_2=(0.65,2),\ J_3=(0.4,2),\ J_4=(0.35,1). \end{aligned}$$

The algorithm schedules the first job on the second machine in step 3, and we get $x_1=0,\ y_1=0.6,\ z_1=0$. When the algorithm receives the second job, it reaches step 4 because $1.25=y_1+p_2>1+\mu$ holds. In step 4, the algorithm updates W to be the set $\{J_2\}$, schedules it on the second machine, and migrates the first job to the first machine. We get $x_2=0,\ y_2=0.65,\ z_2=0.6$. The algorithm schedules the third job on the second machine in step 3 because $y_2+p_3 \le 1+\mu$ holds, and the fourth job is scheduled on the first machine in step 2 because $g_4=1$ (and additionally, at this time, it holds that $y_3 \ge 1-\mu$). Consequently, the load of the first machine is 0.95, and the load of the second machine is 1.05. See Fig. 5 for an illustration of the process of execution of the algorithm, and an optimal solution.

1.1.2 An example for Algorithm A with $\varvec{M=20}$

In this example, we show an input where step 4 is applied three times. We define input $I_3$ for Algorithm A with a migration factor of $M=20$. For this value of M, we have $\mu =\frac{2}{43} \approx 0.04651$, the value used by the algorithm in step 2 is $1-\mu =\frac{41}{43} \approx 0.95349$, and the value used in step 3, which is also the competitive ratio for this algorithm, is $1+\mu =\frac{45}{43}\approx 1.04651$.

Fig. 6

The schedules produced by Algorithm A with migration factor $M = 20$, after three, four, five, six, seven, and eight jobs were presented, for input $I_3$ of Sect. A.1.2. There are migrations of $J_1$ (when $J_5$ arrives), of $J_5$ (when $J_6$ arrives), and of three jobs when $J_7$ arrives. The final makespan for the algorithm is 1, and the obtained solution is optimal

Input $I_3$ is defined as follows:

$$\begin{aligned} J_1= & {} (0.25,2),\ J_2=(0.28,2),\ J_3=(0.29,2),\ J_4=(0.1,1),\\ J_5= & {} (0.26,2),\ J_6=(0.27,2),\ J_7=(0.49,2),\ J_8=(0.06,1). \end{aligned}$$

The algorithm schedules the first three jobs on the second machine in step 3, and it schedules $J_4$ on the first machine in step 2. We get $x_4=0.1,\ y_4=0.82,\ z_4=0$. When the algorithm receives the fifth job, it reaches step 4 because $1.08=y_4+p_5>1+\mu$ holds. In step 4, the algorithm updates W to be $\{J_2, J_3, J_5\}$ (the subset of $Z\cup Y\cup \{J_5\}$ of maximum total size not exceeding 1), it migrates $J_1$ to the first machine, and it schedules W on the second machine. We get: $x_5=0.1,\ y_5=0.83,\ z_5=0.25$. Afterwards, the algorithm repeats the same process for job $J_6$ (because $1.1=y_5+p_6>1+\mu$ holds), it updates W to be $\{J_2, J_3, J_6\}$, it migrates $J_5$ to the first machine and schedules W on the second machine, and we get: $x_6=0.1,\ y_6=0.84,\ z_6=0.51$. The algorithm migrates $J_1$ and $J_5$ a second time when it receives job $J_7$, since $1.33=y_6+p_7>1+\mu$ holds. Here the algorithm updates W to be $\{J_1, J_5, J_7\}$, where $w_7=1$, it migrates all jobs in Y to the first machine, and it schedules W on the second machine. We get: $x_7=0.1,\ y_7=1,\ z_7=0.84$. Job $J_8$ is assigned to the first machine by step 2 because both $g_8=1$ holds (and additionally, $y_7\ge 1-\mu$), and we get: $x_8=0.16,\ y_8=1,\ z_8=0.84$. Consequently, the loads of both machines are equal to 1. Thus, in this specific case, the obtained solution is optimal. See Fig. 6 for an illustration of the process of execution of the algorithm.

1.2 Algorithm B, where $\varvec{0.75 \le M < 2.5}$

We provide four examples for Algorithm B. In the first one, we show the action of this algorithm when it receives input $I_1$ from Sect. A.1.1. In the second one, we show the action of this algorithm in step 4 by using input $I_4$ (which we define). In the third example and in the fourth example, we show the action of the algorithm in step 5, when it receives inputs $I_5$ and $I_6$, which we define, where the actual assignment is performed by step 5.1 and step 5.2, respectively, for these two inputs.

Recall that the competitive ratio for this algorithm is 1.25, and the values used in steps 2 and 3 are based on this value.

1.2.1 An example for Algorithm B, such that there is no migration

Algorithm B schedules input $I_1$ (see Sect. A.1.1) only using steps 2 and 3, i.e. it does not migrate any jobs. This holds because scheduling the fourth job on the second machine would not exceed the threshold of 1.25, but the load becomes greater than 0.75, i.e. $0.75 \le y_4 =1.25$, which keeps the machines balanced with respect to loads. Consequently, $x_6=0.42,\ y_6=1.25,\ z_6=0.32$, the load of the first machine is 0.74, and the load of the second machine is 1.25. See Fig. 7 for an illustration and additional details.

Fig. 7

The schedules produced by Algorithm B for input $I_1$ of Sect. A.1.1: (1) The algorithm schedules the first two jobs on the second machine in step 3, and the third job is scheduled on the first machine in step 2. (2) The algorithm schedules the fourth job on the second machine in step 3 and we get $0.75 \le y_4$. (3) All other jobs will be scheduled on the first machine in step 2, and we have $c_{B}(I_1)=1.25$ and $c^*(I_1)=1$ (which holds by assigning $J_4$ and $J_5$ to the second machine, and the other jobs to the first machine)

1.2.2 An example for Algorithm B, such that step 4 is applied

In this example, we define input $I_4$ and in particular, we exhibit the action of the algorithm in step 4 when the fifth job of $I_4$ arrives. The input is:

$$\begin{aligned} J_1= & {} (0.01,2),\ J_2=(0.36,1),\ J_3=(0.2,2),\ J_4=(0.36,2),\\ J_5= & {} (0.79,2),\ J_6=(0.28,2). \end{aligned}$$

The first four jobs are scheduled in both steps 2 (job $J_2$) and 3 (the other three jobs), so we get: $x_4=0.36,\ y_4=0.57$, and $z_4=0$. In the process of assignment of the fifth job, the algorithm reaches step 4 because $y_4+p_5>1.25$ and $p_5\ge 0.75$ hold. We have $0.75\cdot p_5=0.5925$. In step 4, the algorithm updates W be the set of the three jobs $J_1$, $J_3$, and $J_4$ (whose total size is 0.57), it migrates W to the first machine, and it schedules $J_5$ on the second machine, because $y_4-w_5+p_5 = p_5 \le 1.25$. We get: $x_5=0.36,\ y_5=0.79$, and $z_5=0.57$. The algorithm schedules the last job on the first machine in step 2 because $y_5 \ge 0.75$ holds, and it updates the variables as follows: $x_6=0.36,\ y_6=0.79,\ z_6=0.85$. Consequently, the load of the first machine is 1.21, and the load of the second machine is 0.79. See Fig. 8.

Fig. 8

The schedules produced by Algorithm B for input $I_4$ of Sect. A.2.2 (from the moment when the fourth job had arrived), and an optimal solution for that input. We have $c_{B}(I_4)=1.21$ and $c^*(I_4)=1$

1.2.3 An example for Algorithm B, such that step 5.1 is applied

In this example, we use a new input $I_5$ to show the action of Algorithm B. In particular, we exhibit the action of the algorithm in step 5 when the fourth job arrives. Input $I_5$ is:

$$\begin{aligned} J_1= & {} (0.2,2),\ J_2=(0.36,2),\ J_3=(0.07,2),\\ J_4= & {} (0.73,2),\ J_5=(0.28,1),\ J_6=(0.36,2). \end{aligned}$$

The first three jobs are scheduled on the second machine in step 3, and we get: $x_3=0,\ y_3=0.63,\ z_3=0$. In the process of assignment of the fourth job, the algorithm reaches step 5, because $1.36=y_3+p_4>1.25$ and $0.73=p_4<0.75$ hold. The largest job of the current set Y is $J_2$, and $p_4^{\max Y}=p_2=0.36$, and together with the fourth job, it holds that $p_4+p_4^{\max Y}=1.09<1.25$, so W will be computed. Since $p_4^{\max Y}=p_2=0.36 \ge \frac{y_3}{2}=0.315$ holds, the algorithm defines W in step 5.1, and it is defined to be the set of the two jobs $J_1$ and $J_3$. The algorithm migrates W to the first machine and schedules $J_4$ on the second machine, and we get: $x_4=0,\ y_4=1.09$, and $z_4=0.27$. The algorithm schedules the last two jobs on the first machine in step 2 (for both jobs it holds that the load of the second machine is sufficiently large, and for the fifth job the GoS is 1), and we get: $x_6=0.28,\ y_6=1.09,\ z_6=0.63$. As a result, the load of the first machine is 0.91, the load of the second machine is 1.09. See Fig. 9.

Fig. 9

The schedules produced by Algorithm B for input $I_5$ of Sect. A.2.3 (from the moment when the third job had arrived), and an optimal solution for that input. We have $c_{B}(I_5)=1.09$ and $c^*(I_5)=1$

1.2.4 An example for Algorithm B, such that step 5.2 is applied

In this example, we use a new input $I_6$ to show the action of Algorithm B. We exhibit the action of the algorithm in step 5 when the fourth job arrives. Input $I_6$ is:

$$\begin{aligned} J_1= & {} (0.17,2),\ J_2=(0.2,2),\ J_3=(0.26,2),\ J_4=(0.7,2),\\ J_5= & {} (0.1,1),\ J_6=(0.57,2). \end{aligned}$$

The first three jobs are scheduled on the second machine in step 3, and we get: $x_3=0,\ y_3=0.63,\ z_3=0$. For the fourth job, the algorithm reaches step 5 because $1.33=y_3+p_4>1.25$ and $0.7=p_4<0.75$ hold. The largest job of the current set Y is $J_3$, and $p_4^{\max Y}=p_3=0.26$, and together with the fourth job, it holds that $p_4+p_4^{\max Y}=0.96<1.25$, so W will be computed. Since $p_4^{\max Y}=p_3=0.26$ while $y_3=0.63$ holds, the algorithm defines W in step 5.2. The set W is defined to be the set $\{J_3\}$. Specifically, step 5.2 is applied because $0.25 \le p_4^{\max Y} < \frac{y_3}{2}=0.315$ holds. The algorithm migrates W to the first machine, and it schedules $J_4$ on the second machine. We get: $x_4=0,\ y_4=1.07$, and $z_4=0.26$. Similarly to the previous examples, the algorithm schedules the last two jobs on the first machine in step 2, and it defines its new values as follows: $x_6=0.1,\ y_6=1.07,\ z_6=0.83$. As a result, the load of the first machine is 0.93, the load of the second machine is 1.07. See Fig. 10.

Fig. 10

The schedules produced by Algorithm B for input $I_6$ of Sect. A.2.4 (from the moment when the third job had arrived), and an optimal solution for that input. We have $c_{B}(I_6)=1.07$ and $c^*(I_6)=1$

1.3 Algorithm C, where $\varvec{0.5 \le M < \frac{2}{3} }$

Here, we provide three examples for Algorithm C. In the first example, we show the action of the algorithm when it receives input $I_1$ of Sect. A.1.1. In the second one, we show the action of the algorithm in step 4 by using a new input $I_7$. In the third example, we show the action of the algorithm in step 5 when it receives input $I_8$, which we define.

In all three examples for this algorithm, we use $M=0.6$, and thus the value used in step 2 is 0.6, and the value used in step 3 (and the competitive ratio) is 1.4.

1.3.1 An example Algorithm C, using $\varvec{I_1}$

Algorithm C with migration factor $M = 0.6$ schedules $I_1$ of Sect. A.1.1 only using the two steps 2 and 3, i.e. it does not migrate any jobs. This is because the load of the second machine after scheduling the fourth job on this machine does not exceed the upper bound (competitive ratio) $2-M= 1.4$, and this keeps loads relatively balanced. Consequently, $x_6=0.42,\ y_6=1.25,\ z_6=0.32$, the load of the first machine is 0.74, and the load of the second machine is 1.25 (see the output of Fig. 7, which belongs to the run of another algorithm and the same input, though the output is identical).

1.3.2 An example for Algorithm C, such that step $\varvec{4}$ is applied

In this example, we exhibit the action of the algorithm in step 4 when the third job of our new input arrives. Input $I_7$ is:

$$\begin{aligned} J_1=(0.5,2),\ J_2=(0.09,2),\ J_3=(0.82,2),\ J_4=(0.18,2),\ J_5=(0.41,1). \end{aligned}$$

Recall that $M=0.6$. The first two jobs are scheduled on the second machine in step 3, and we get: $x_2=0,\ y_2=0.59,\ z_2=0$. When the third job arrives, the algorithm reaches step 4 because $p_3^{\max Y}=p_1=0.5 > M\cdot p_3=0.492$ holds. In step 4, the algorithm schedules $J_3$ on the first machine, and we get: $x_3=0,\ y_3=0.59$, and $z_3=0.82$. The algorithm schedules the fourth job on the second machine in step 3 because $0.77=y_3+p_4 \le 1.4=2-M$ holds, and the last job on the first machine in step 2 because $g_6=1$ holds. As a result, the load of the first machine is 1.23, the load of the second machine is 0.77. See Fig. 11.

Fig. 11

The schedules produced by Algorithm C with migration factor $M =0.6$ for input $I_7$ of Sect. A.3.2 (from the moment when the second job had arrived), and an optimal solution for that input. We have $c_{C}(I_7)=1.23$ and $c^*(I_7)=1$

1.3.3 An example for Algorithm C, such that step 5 is applied

In this example, we exhibit the action of the algorithm in step 5 when the fifth job of the input arrives. Input $I_8$ is:

$$\begin{aligned} J_1= & {} (0.35,2),\ J_2=(0.11,2),\ J_3=(0.13,2),\ J_4=(0.41,1),\\ J_5= & {} (0.82,2),\ J_6=(0.18,2). \end{aligned}$$

Recall that $M=0.6$. The first three jobs are scheduled on the second machine in step 3, and the fourth job is scheduled on the first machine in step 2. At this time, we get: $x_4=0.41,\ y_4=0.59,\ z_4=0$. When the fifth job arrives, the algorithm reaches step 5, because $p_5^{\max Y}=p_1=0.35 \le M \cdot p_5=0.492$. It updates W to be $\{J_1\}$ (the minimum length prefix with total size at least $p_5+y_4-(2-M)=0.01$), it migrates W to the first machine, and it schedules $J_5$ on the second machine. We get: $x_5=0.41,\ y_5=1.06$, and $z_5=0.35$. The algorithm schedules the last job on the first machine in step 2 and updates: $x_6=0.41,\ y_6=1.06,\ z_6=0.35$. As a result, the load of the first machine is 0.94, the load of the second machine is 1.06. See Fig. 12.

Fig. 12

The schedules produced by Algorithm C with migration factor $M =0.6$ for input $I_8$ of Sect. A.3.3 (from the moment when the fourth job had arrived), and an optimal solution for that input. We have $c_{C}(I_8)=1.06$ and $c^*(I_8)=1$

1.4 Algorithm D, where $\varvec{\frac{2}{3} \le M < 0.75}$

In this part we provide six examples for Algorithm D. In the first example we show the action of the algorithm when it receives input $I_1$ of Sect. A.1.1. In the second and the third examples we show the action of the algorithm in step 4, by using inputs $I_9$, $I_{10}$, which we define. In the last three examples, we show the action of the algorithm in step 5 for three inputs $I_{11}$, $I_{12}$ and $I_{13}$, which we also define here.

In all six examples, we use $M=0.7$. As a result, the value used by the algorithm in step 2 is $M=0.7$, and the value used in step 3, which is also the competitive ratio for this algorithm, is $2-M=1.3$.

1.4.1 An example for Algorithm D, using $\varvec{I_1}$

Recall that $M=0.7$. Algorithm D schedules input $I_1$ of Sect. A.1.1 only using the two steps 2 and 3, i.e. it does not migrate any jobs. This holds because after scheduling the fourth job on the second machine we get: $x_4=0.31,\ y_4=1.25,\ z_4=0$, i.e. the load of the second machine is between the two bounds $M=0.7$ and $2-M=1.3$, so any job that arrives after $J_4$ will be scheduled on the first machine. Consequently, $x_6=0.42,\ y_6=1.25,\ z_6=0.32$, and the makespan is 1.25 (see Fig. 7 again).

Fig. 13

The schedules produced by Algorithm D with migration factor $M =0.7$ for input $I_9$ of Sect. A.4.2 (from the moment when the third job had arrived), and an optimal solution for that input. We have $c_{D}(I_9)=1.08$ and $c^*(I_9)=1$

1.4.2 An example for Algorithm D, such that step 4.2 is applied

In this example, we exhibit the action of the algorithm in step 4 where $y_{j-1}-w_j+p_j \le 2-M$ holds, which is tested when the fourth job in input $I_9$ arrives. Input $I_9$ is:

$$\begin{aligned} J_1=(0.54,2),\ J_2=(0.14,2),\ J_3=(0.32,1),\ J_4=(0.78,2),\ J_5=(0.22,2). \end{aligned}$$

Recall that $M=0.7$. The first two jobs are scheduled on the first machine in step 3, and the third job on the first machine in step 2. We get: $x_3=0.32,\ y_3=0.68$, and $z_3=0$. When the fourth job arrives, the algorithm reaches step 4, because $1.46=y_3+p_4 >2-M=1.3$ and $p_4=0.78\ge M=0.7$ hold. In step 4, the algorithm updates W to be $\{J_1\}$ (the maximum length prefix with total size at most $M \cdot p_4=0.7 \cdot 0.78=0.546$, where $M\cdot p_4 \ge p_1=0.54$, but $p_1+p_2=0.68>0.546$). We have $y_3-w_4+p_4 = 0.68-0.54+0.78 = 0.92 \le 1.3=2-M$. Thus, the algorithm continues to step 4.2, and it migrates W to the first machine. The algorithm schedules $J_4$ on the second machine, and we get: $x_4=0.32,\ y_4=0.92$, and $z_4=0.54$. The algorithm schedules the fifth job on the first machine in step 2 and it has $x_5=0.32,\ y_5=0.92$, and $z_5=0.76$. As a result, the load of the first machine is 1.08, the load of the second machine is 0.92. See Fig. 13.

1.4.3 An example for Algorithm D, such that step 4.1 is applied

In this example, we exhibit the action of the algorithm in step 4 where $y_{j-1}-w_j+p_j > 2-M$ holds, which holds for the fourth job arrives. Input $I_{10}$ which is used for this example is:

$$\begin{aligned} J_1=(0.54,2),\ J_2=(0.14,2),\ J_3=(0.32,1),\ J_4=(0.71,2),\ J_5=(0.29,2). \end{aligned}$$

Recall that $M=0.7$. The first two jobs are scheduled on the second machine in step 3, and the third job is scheduled on the first machine in step 2. At this time we have $x_3=0.32,\ y_3=0.68$, and $z_3=0$. When the fourth job arrives, the algorithm reaches step 4, because $p_4=0.71\ge M=0.7$ holds, and it did not apply earlier steps because $1.39=y_3+p_4 >2-M=1.3$ holds while $y_3<M=0.7$. In step 4, the algorithm updates W to be an empty set (the maximum length prefix with total size at most $M \cdot p_4=0.7 \cdot 0.71=0.497 < p_1=0.54$), and we get: $y_3-w_4+p_4 = 0.68-0+0.71 = 1.39 > 1.3=2-M$, so the algorithm schedules $J_4$ on the first machine, and we get: $x_4=0.32,\ y_4=0.68$, and $z_4=0.71$. The algorithm schedules the fifth job on the second machine in step 3 and it has $x_5=0.32,\ y_5=0.97$, and $z_5=0.71$. As a result, the load of the first machine is 1.03, the load of the second machine is 0.97. See Fig. 14.

Fig. 14

The schedules produced by Algorithm D with migration factor $M =0.7$ for input $I_{10}$ of Sect. A.4.3 (from the moment when the third job had arrived), and an optimal solution for that input. We have $c_{D}(I_{10})=1.03$ and $c^*(I_{10})=1$

1.4.4 An example for Algorithm D, such that both steps 5.1 and 5.3 are applied

In this example, we exhibit the action of the algorithm in step 5 where $w_j > \min \{\frac{2\,M}{3}, M\cdot p_j\}$ and $y_{i-1}-w_j+p_j \le 2-M$ hold, which happens when the fourth input job arrives. Input $I_{11}$ which is used for this example is:

$$\begin{aligned} J_1=(0.54,2),\ J_2=(0.14,2),\ J_3=(0.32,1),\ J_4=(0.69,2),\ J_5=(0.31,2). \end{aligned}$$

Recall that $M=0.7$. The first two jobs are scheduled on the second machine in step 3, and the third job is scheduled on the first machine in step 2. At this time, we get: $x_3=0.32,\ y_3=0.68,\ z_3=0$. When the fourth job arrives, the algorithm reaches step 5 because $1.37=y_3+p_4 >2-M=1.3$ and $p_4=0.69 < M=0.7$ hold. In step 5 the algorithm updates W twice. First, W is defined as $\{J_1\}$ (the minimum length prefix with total size at least $\frac{M}{3}=\frac{7}{30}\approx 0.2333 \le p_1=w_4=0.54$). At this time, we have: $w_4 > \min \{\frac{2\,M}{3},\ M\cdot p_4 \} = \min \{\frac{7}{15}\approx 0.46667,\ 0.483\}$. The algorithm updates W in step 5.1 again to be $Y \setminus W = \{J_2\}$. In step 5.3, the algorithm migrates W to the first machine, and it schedules $J_4$ on the first machine. Step 5.2 is not applied because $y_3 - w_4+p_4 = 1.23 \le 1.3 = 2-M$ holds. We get: $x_4=0.32,\ y_4=1.23$, and $z_4=0.14$. The algorithm schedules the last job on the first machine in step 2 and we get: $x_5=0.32,\ y_5=1.23$, and $z_5=0.45$. As a result, the load of the first machine is 0.77, and the load of the second machine is 1.23. See Fig. 15.

Fig. 15

The schedules produced by Algorithm D with migration factor $M =0.7$ for input $I_{11}$ of Sect. A.4.4 (from the moment when the third job had arrived), and an optimal solution for that input. We have $c_{D}(I_{11})=1.23$ and $c^*(I_{11})=1$

1.4.5 An example for Algorithm D, such that step 5.3 is applied

In this example, we exhibit the action of the algorithm in step 5 where $w_j \le \min \{\frac{2\,M}{3}, M\cdot p_j\}$ and $y_{i-1}-w_j+p_j \le 2-M$ hold, and this happens when the second input job arrives. Input $I_{12}$ which is used for this example is:

$$\begin{aligned} J_1=(0.45,2),\ J_2=(0.24,2),\ J_3=(0.65,2),\ J_4=(0.31,1),\ J_5=(0.35,2). \end{aligned}$$

Recall that $M=0.7$. The first two jobs are scheduled on the second machine in step 3, and we get: $x_2=0,\ y_2=0.69,\ z_2=0$. When the third job arrives, the algorithm reaches step 5 because $1.34=y_2+p_3 >2-M$ and $p_3=0.65 < M=0.7$ hold. In step 5, the algorithm defines W to be $\{J_1\}$ (the minimum length prefix with total size at least $\frac{M}{3}=\frac{7}{30}\approx 0.2333 \le p_1=w_3=0.45$), here we get: $w_3 \le \min \{\frac{2\,M}{3},\ M\cdot p_3 \} = \min \{\frac{7}{15},\ 0.455\}$. In this case $0.89=y_2 - w_3+p_3 \le 2-M=1.3$ holds, so the algorithm only applies step 5.3, and it migrates W to the first machine while scheduling $J_3$ on the second machine. We get: $x_3=0,\ y_3=0.89$, and $z_3=0.45$. The algorithm schedules the last two jobs on the first machine in step 2 and it updates its variables as follows: $x_5=0.31,\ y_5=0.89,\ z_5=0.8$. As a result, the load of the first machine is 1.11, the load of the second machine is 0.89. See Fig. 16.

Fig. 16

The schedules produced by Algorithm D with migration factor $M =0.7$ for input $I_{12}$ of Sect. A.4.5 (from the moment when the second job had arrived), and an optimal solution for that input. We have $c_{D}(I_{12})=1.11$ and $c^*(I_{12})=1$

Fig. 17

The schedules produced by Algorithm D with migration factor $M =0.7$ for input $I_{13}$ of Sect. A.4.6, and an optimal solution for that input. We have $c_{D}(I_{13})=1.07,\ c^*(I_{13})=1$

1.4.6 An example for Algorithm D, such that both steps 5.1 and 5.2 are applied

In this example, we exhibit the action of the algorithm in step 5 where $w_j > \min \{\frac{2\,M}{3}, M\cdot p_j\}$ and $y_{i-1}-w_j+p_j > 2-M$ hold, and this happens when the second input job arrives. Input $I_{13}$ which is used for this example is:

$$\begin{aligned} J_1=(0.69,2),\ J_2=(0.62,2),\ J_3=(0.38,2),\ J_4=(0.31,1). \end{aligned}$$

Recall that $M=0.7$. The first job is scheduled on the second machine in step 3, and we get: $x_1=0,\ y_1=0.69,\ z_1=0$. When the second job arrives, the algorithm reaches step 5 because $1.31=y_1+p_2 >2-M=1.3$ and $p_2=0.62 < M=0.7$ hold. In step 5, the algorithm defines W to be $\{J_1\}$ (the minimum length prefix with total size at least $\frac{M}{3}=\frac{7}{30}\approx 0.2333 \le p_1=w_2=0.69$). Here, we get: $w_2 > \min \{\frac{2\,M}{3},\ M\cdot p_2 \} = \min \{\frac{7}{15},\ 0.434\}$, and in this case the algorithm updates W again to be $Y \setminus W = \{\}$, it schedules $J_2$ on the first machine, because $y_1 - w_2+p_2 = 1.31 > 1.3 = 2-M$ holds, and we get: $x_2=0,\ y_2=0.69,\ z_2=0.62$. The algorithm schedules the third job on the second machine in step 3, and it schedules the last job on the first machine in step 2. We get: $x_4=0.31,\ y_4=1.07$, and $z_4=0.62$. As a result, the load of the first machine is 0.93, the load of the second machine is 1.07. See Fig. 17.