N WORK ON A MACHINE
- Determine the optimal sequence of processing n jobs on one machine. All sequences have the same makespan. Minimizing the mean flow time is the criterion to satisfy. Let us represent the processing times of jobs i as pi (i = 1, n).
- The sequence that minimizes the criterion is the one in which the jobs are ordered from the shortest time to the longest. This sequence also minimizes the average waiting time and the average delay (mean lateness). When the jobs have different priority or weight, the objective It can be to minimize the weighted average flow time. The higher the index value, the more important the work.
The optimal sequence would be by ordering the jobs from lowest pi / wi to highest. Minimize the weighted average of the flow time.
The optimal sequence is (2,5,3,6,1,4).
N WORKS ON 2 MACHINES
- The n jobs are processed on 2 machines with the same order. The criterion is to minimize the makespan. The procedure to use is Johnson's. If pij is the process time of job i on machine j, select the minimum and if it corresponds to machine 1, assign it to the first position of the sequence. If it corresponds to machine 2, the job is assigned to the last position in the sequence. Delete the assigned job from the set and repeat the procedure with the unassigned jobs.
Johnson's Procedure
- Determine the process sequence that minimizes the makespan
The sequence is (2,4,5,3,1).
n Works with Different Route on 2 Machines
- Use Jackson's algorithm. Form 4 sets of jobs; {A} = Those processed only on machine 1. {B} = Those processed on machine 2 only. {AB} = Those processed first on machine 1 and then on machine 2. {BA} = Those processed first in machine 2 and then in machine 1.Sequence the jobs of {AB} and {BA}, separately, with Johnson's algorithm. Define arbitrary sequences for jobs {A} and {B}. Combine the sequences as follows: Machine 1: {AB} before {A} before {BA}. Machine 2: {BA} before {B} before {AB}.
n Works on 3 Machines
- All jobs have the same process sequence. It can be solved with Johnson's algorithm if: min {pi1}> max {pi2}, omin {pi3}> max {pi2}. Machine 2 is completely dominated by machine 1 or 3. The procedure is applied forming 2 dummy machines, 1´ and 2´, with process times: pi1´ = pi1 + pi2 and pi2´ = pi2 + pi3. The procedure provides a feasible and “good” sequence even when they are not met. the conditions
Sequence the following works.
Optimal sequence
is {2,1,4,3}.
N WORK ON MACHINES
- There is no efficient method that provides an exact solution. Heuristic methods such as Dispatch Rules are used. These are rules that determine which job to process by making it available sequentially over time, rather than assuming that all jobs are available. The concept of priority in jobs is handled.
Dynamic Sequencing of Works
- Jobs are randomly processed over a time interval. Their sequence is determined by using dispatch rules that prioritize them. Rules are derived through waiting line analysis, experimentation, and simulation. The most important sequencing and dispatching is the shortest processing time (SPT). Other rules are derived from the SPT, as well as the size of the waiting lines and the promised date to the clients.
Other Dispatch Rules
- Based on process time. Least Remaining Job (LWKR): Considers the sum of the process times for all operations to be performed on the job. Total Job (TWK): Considers the sum of the process times of all job operations. Lowest Number of Operations to Be Performed (FOPR): Considers the number of operations to be performed on the job. Based on Delivery Dates: Promised Date (DDATE): The priority is assigned based on the promised date. Slack Time (SLACK): The priority is assigned based on the time remaining for the promised date minus the missing processing time. Slack / Missing Operation (S / ROP): The priority is determined by the quotient between the SLACK and the number of missing operations.
BALANCE OF ASSEMBLY LINES APPLICATION FOR INDUSTRIAL ENGINEERING
- Assembly lines are characterized by the movement of a workpiece from one workstation to another. The tasks required to complete a product are divided and assigned to the workstations such that each station performs the same operation on each product. The part remains at each station for a period of time called cycle time, which depends on demand. It consists of assigning the tasks to workstations so that a certain performance indicator is optimized. The criterion for selecting a task assignment determined can be the total leisure time. This is determined by: I = kc - pi Where k is the number of workstations, c represents the cycle time and pi corresponds to the total operating time. The purpose is to have I = 0.This would occur if the assignment of tasks can be done to an entire number of stations. Two heuristic methods are provided by Kilbridge & Wester and Helgeson & Birnie.
KIBRIDGE & WESTER METHOD
- Consider precedence constraints between activities, seeking to minimize the number of stations for a given cycle time.
The method is illustrated with the following example: Define the cycle time, c, required to satisfy the demand and start the assignment of tasks to stations respecting the precedents and seeking to minimize the leisure in each station. Considering a cycle of 16, it is estimated that the minimum number of stations would be 48/16 = 3. Observing the total time of I and analyzing the tasks of II, we can see that task 4 could be reassigned to I. When reassigning task 4 to station I, the cycle time.We repeat the process with station II. We can see that task 5, which is located in station III, can be reassigned to station II. The reassignment satisfies the cycle time. We repeat the process and we observe that the rest of the tasks can be reassigned to station III.The line was balanced optimizing the number of stations and with zero leisure.
HELGESON & BIRNIE METHOD
- It consists of estimating the positional weight of each task as the sum of its time plus those of those that follow it. The tasks are assigned to the stations according to the positional weight, taking care not to exceed the cycle time and violate the precedents. It would then form tasks 1, 2 and 4 with weights of 45, 37 and 34. The total time is 16 and no precedence was violated. The following assignment corresponds to tasks 3 and 5 with weights of 25 and 19. Total at Station II is 16. The last assignment includes Tasks 6, 7, 8, and 9, with weights of 16, 9, 5, and 3 respectively. Total time at Station III is 16.
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