Design Optimization

(:title Scheduling Optimization:)

Schedule optimization maximizes productivity, minimizes costs and reduces delays. It is widely used in manufacturing, logistics, transportation, construction, project management, and healthcare. By optimizing schedules, organizations can allocate resources effectively, improve delivery times, and meet customer demands efficiently, leading to increased competitiveness and profitability. Manufacturing facilities employ expert schedulers and tools to help visualize and plan for production cycles, scheduled downtime, transitions, etc. This example is a comparison of three methods for scheduling problems:

The order of the sequence is different than Johnson's rule but the total makespan time (121) is the same. The optimizer found the equally optimal solution as the second row in the sorted exhaustive search list.

Schedule optimization maximizes productivity, minimizes costs and reduces delays. It is widely used in manufacturing, logistics, transportation, construction, project management, and healthcare. By optimizing schedules, organizations can allocate resources effectively, improve delivery times, and meet customer demands efficiently, leading to increased competitiveness and profitability.

Manufacturing facilities employ expert schedulers and tools to help visualize and plan for production cycles, scheduled downtime, transitions, etc. This example is a comparison of three methods for scheduling problems:

{$ty_i = \sum_{j=0}^4 Y_j x_{i,j} = 1 \quad \forall \; i=0\ldots4$}

{$tz_i = \sum_{j=0}^4 Z_j x_{i,j} = 1 \quad \forall \; i=0\ldots4$}

The order is ''P4'', ''P0'', ''P1'', ''P3'', ''P2'' with total delay of 26 on machine Z and total makespan time of 121 to complete all processing.

The order of the sequence is different than Johnson's rule but the total makespan time (121) is the same. The optimizer found an equally optimal solution.

Y = { 1: 10, 2: 20, 3: 15, 4: 40, 5: 8 }
Z = { 1: 20, 2: 30, 3: 10, 4: 25, 5: 18 }

   m.Equation(ty[i] == m.sum([x[i,j]*Y[j+1] for j in range(5)]))

   m.Equation(tz[i] == m.sum([x[i,j]*Z[j+1] for j in range(5)]))

m = GEKKO(remote=False)

m.Equation(d[0]==tz[0])

The problem is solved with the APOPT solver in 0.1 sec with a minimized objective (time delay) of 11. The order is ''P4'', ''P2'', ''P0'', ''P1'', ''P3'' with total delay of 11 on machine Z and total makespan time of 121 to complete all processing.

<td><b>Time (131):</b></td>

(:source lang=python:)
Order of processing, rows=order, columns=product
 [[[0.0] [0.0] [0.0] [0.0] [1.0]]
  [[0.0] [0.0] [1.0] [0.0] [0.0]]
  [[1.0] [0.0] [0.0] [0.0] [0.0]]
  [[0.0] [1.0] [0.0] [0.0] [0.0]]
  [[0.0] [0.0] [0.0] [1.0] [0.0]]]
time on machine Y
 [[8.0] [15.0] [10.0] [20.0] [40.0]]
time on machine Z
 [[18.0] [10.0] [20.0] [30.0] [25.0]]
delay
 [[8.0] [3.0] [0.0] [0.0] [0.0]]
(:sourceend:)

<td><b>Time (121):</b></td>

 [[[0.0] [0.0] [1.0] [0.0] [0.0]]
  [[0.0] [0.0] [0.0] [0.0] [1.0]]

 [[15.0] [8.0] [10.0] [20.0] [40.0]]

 [[10.0] [18.0] [20.0] [30.0] [25.0]]

 [[10.0] [2.0] [8.0] [0.0] [0.0]]

The order is ''P4'', ''P0'', ''P1'', ''P3'', ''P2'' with total delay of 26 on machine Z and total time of 121 for all processing.

<td><b>Makespan (121):</b></td>

<td><b>Time (121):</b></td>

The order is ''P4'', ''P0'', ''P1'', ''P3'', ''P2''.

<td><b>Delay:</b></td>

print('time on machine T')

print('time on machine K')

 ---------------------------------------------------
 Solver        :  APOPT (v1.0)
 Solution time  :  0.1003 sec
 Objective      :  20.
 Successful solution
 ---------------------------------------------------
 Order of processing, rows=order, columns=product
  [[[0.0] [0.0] [1.0] [0.0] [0.0]]
   [[0.0] [0.0] [0.0] [0.0] [1.0]]
   [[1.0] [0.0] [0.0] [0.0] [0.0]]
   [[0.0] [1.0] [0.0] [0.0] [0.0]]
   [[0.0] [0.0] [0.0] [1.0] [0.0]]]
 time on machine T
  [[15.0] [8.0] [10.0] [20.0] [40.0]]
 time on machine K
  [[10.0] [18.0] [20.0] [30.0] [25.0]]
 delay
  [[10.0] [2.0] [8.0] [0.0] [0.0]]

Solve the problem with an integer programming solver ([[https://apopt.com|APOPT]]) by setting up each decision as a binary variable (0 or 1). A zero (0) is a decision to not place the item in the schedule while a one (1) is a decision to include it. There are {`5`} products with processing times {`y_1, \ldots, y_5`} and {`z_1, \ldots, z_5`}.  The decision variables are a matrix of 25 values that can be 0 or 1:

The time {`(t)`} to process on machines Y {`(ty)`} and Z {`(tz)`} are:

 

Order of processing, rows=order, columns=product
[[[0.0] [0.0] [1.0] [0.0] [0.0]]
 [[0.0] [0.0] [0.0] [0.0] [1.0]]
 [[1.0] [0.0] [0.0] [0.0] [0.0]]
 [[0.0] [1.0] [0.0] [0.0] [0.0]]
 [[0.0] [0.0] [0.0] [1.0] [0.0]]]
time on machine T
[[15.0] [8.0] [10.0] [20.0] [40.0]]
time on machine K
[[10.0] [18.0] [20.0] [30.0] [25.0]]
delay
[[10.0] [2.0] [8.0] [0.0] [0.0]]
(:sourceend:)

* The lowest time is ''P5'' and it is assigned to be first because the lowest time is for machine Y.
* Then next lowest is ''P3'' and it is assigned to be last because the lowest time is for machine Z.

The order is ''P5'', ''P1'', ''P2'', ''P4'', ''P3''.

<table border=0 align="left">

<td align="center" bgcolor="#abc76d">8</td>

<td align="center" bgcolor="#abc76d">18</td>

<table border=0 align="center">

<td></td><td><b>Product</b></td><td><b>P0</b></td><td><b>P1</b></td><td><b>P2</b></td><td><b>P3</b></td><td><b>P4</b></td>

<td><b>Machine Y</b></td>

<td><b>Machine Z</b></td>

The problem is solved with the APOPT solver in 0.1 sec with a minimized objective (time delay) of 20. The selected order is 3, 5, 1, 2, 4.