## 4 Computational Results
### A Real Case Study  
The MILP model and the proposed algorithm are applied to a vertical machining center (SMTCL VMC 1600P) from a machinery manufacturing company shown in Figure 8. In this real-life instance, the company receives some orders of processing rectangular parts with three product models for continuous casting machines of the steel plants. Figure 9 is the picture of the rectangular part, and the complex cavity surfaces including the planes, camber surfaces, and spherical surfaces are to be processed on the VMC. The power consumption per hour required by the VMC is related to the material of the job, cutting depth, feed speed, and so on. To obtain the average power consumption rate of the machine, a power measurement is performed using a power meter (UNI-T UT232), and the measurement data is presented by Table 2. In addition, the order quantity of the three models is 15, 35, and 10 parts, respectively. Let us temporarily put aside the above real-life instance, and talk about solving a hypothetical instance given in Tables 3 and 4 based on the above data. Specifically, it is assumed that there are twelve parts which need to be processed within two days. That is, the number of periods is ten (i.e., \(m=10\)). According to this example, we will demonstrate the MILP model and the main computational process of GIH-F.  
First, the twelve jobs are sorted in non-increasing order according to their power consumption rates, that is, job 7, job 8, job 9, job 10, job 11, job 12, job 5, job 6, job 1, job 2, job 3, and job 4. Second, the remaining idle time of all the periods are initialized. Obviously, \(t_{7}\leq\max_{k\in A}\{I_{k}\}+I_{k+1}\) and \(t_{7}\leq I_{4}\)(i.e., \(k=4\)), hence job 7 can be inserted into the low-price layer 1 and is placed in the position corresponding to the Condition 1. The same applies to jobs 8, 9, and 10. In this way, the off-peak periods are fully utilized by the jobs with high power consumption rates, resulting in lower total electricity costs. At this stage, the remaining idle time of each period is as follows: \(I_{1}=3.5\), \(I_{2}=7\), \(I_{3}=4.5\), \(I_{4}=1.8\), \(I_{5}=1\), \(I_{6}=3.5\), \(I_{7}=7\), \(I_{8}=4.5\), \(I_{9}=1.8\), \(I_{10}=1\). An illustration is given in Figure 9(a). Now, \(t_{11}>\max_{k\in A}\{I_{k}\}+I_{k+1}\) and \(\exists k_{\prime}\in B,t_{11}\leq I_{k}\), (i.e., \(k=4\) and \(k^{\prime}=2\)), hence job 11 is assigned to medium-price layer 2, and is placed in the position corresponding to the Condition 3 because \(\max_{k\in A}\{I_{k}\}>0\) and \(d_{k+2}=3.5>0\). Moreover, \(x_{i,k}\times(c_{B}-c_{A})>x_{i,k+2}\times(c_{\rm T}-c_{B})\) where \(k=4\) and \(i=11\), thus, job 11 is to be inserted into the position across periods 4-6. At this stage, the remaining idle time of each period is as follows: \(I_{1}=3.5\), \(I_{2}=7\), \(I_{3}=4.5\), \(I_{4}=0\), \(I_{5}=0\), \(I_{6}=3.2\), \(I_{7}=7\), \(I_{8}=4.5\), \(I_{9}=1.8\), \(I_{10}=1\). An illustration is given in Figure 9(b). Let us continue, and it is not hard to check that the job 12 is to be inserted into the position corresponding to Condition 4. As mentioned earlier, there will be five positions waiting for selection at this moment. The insertion costs for these five positions are 12.8, 10.7, 10.7, Inf, and 13.9, respectively. Therefore, job 12 is to be inserted into the Position 2 which crosses periods 8 and 9. Note that \(I_{k}{}^{sm}=0\) (i.e., \(I_{4}=0\)), hence job 12 cannot be processed across periods \(k^{sm}\), \(k^{sm}+1\), and \(k^{sm}+2\), and the corresponding insertion cost is infinity. At this stage, the remaining idle time of each period is as follows: \(I_{1}=3.5\), \(I_{2}=7\), \(I_{3}=4.5\), \(I_{4}=0\), \(I_{5}=0\), \(I_{6}=3.2\), \(I_{7}=7\), \(I_{8}=4.2\), \(I_{9}=0\), \(I_{10}=0\). An illustration is given in Figure 9(c). Next, let us explain the insertion process of other jobs concisely. Jobs 5 and 6 are assigned to layer 2, and they satisfy Condition 5. Therefore, these two jobs are to be inserted into mid-peak period 2.