Minimize Makespan in Parallel Processing

Problem Setup

This example has 11 processors and 40 tasks. The time for each processor to process each task is given in the array processingTime. For this example, generate random processing times.

rng default % for reproducibility
numberOfProcessors = 11;
numberOfTasks = 40;
processingTime = [10;7;2;5;3;4;7;6;4;3;1] .* rand(numberOfProcessors,numberOfTasks);

processingTime(i,j) represents the amount of time that processor i takes to process task j.

To solve the problem using binary integer programming, create process as a binary optimization variable array, where process(i,j) = 1 means processor i processes task j.

process = optimvar('process',size(processingTime),'Type','integer','LowerBound',0,'UpperBound',1);

Each task must be assigned to exactly one processor.

assigneachtask = sum(process,1) == 1;

To represent the objective, define a nonnegative optimization variable named makespan.

makespan = optimvar('makespan','LowerBound',0);

Compute the time that each processor requires to process its tasks.

computetime = sum(process.*processingTime,2);

Relate the compute times to the makespan. The makespan is greater than or equal to each compute time.

makespanbound = makespan >= computetime;

Create an optimization problem whose objective is to minimize the makespan, and include the problem constraints.

scheduleproblem = optimproblem('Objective',makespan);
scheduleproblem.Constraints.assigneachtask = assigneachtask;
scheduleproblem.Constraints.makespanbound = makespanbound;

Solve Problem and View Solution

Solve the problem, suppressing the usual display.

options = optimoptions(scheduleproblem,'Display',"off");
[sol,fval,exitflag] = solve(scheduleproblem,'Options',options)

sol = struct with fields:
    makespan: 1.3634
     process: [11x40 double]

fval = 
1.3634

exitflag = 
    OptimalSolution

The returned exitflag indicates that the solver found an optimal solution, meaning the returned solution has minimal makespan.

The returned makespan is 1.3634. Is this an efficient schedule? To find out, view the resulting schedule as a stacked bar chart. First, create a schedule matrix where row i represents the tasks done by processor i. Then, find the processing time for each entry in the schedule.

processval = round(sol.process);
maxlen = max(sum(processval,2)); % Required width of the matrix
% Now calculate the schedule matrix
optimalSchedule = zeros(numberOfProcessors,maxlen);
ptime = optimalSchedule;
for i = 1:numberOfProcessors
    schedi = find(processval(i,:));
    optimalSchedule(i,1:length(schedi)) = schedi;
    ptime(i,1:length(schedi)) = processingTime(i,schedi);
end
optimalSchedule

optimalSchedule = 11×10

    25    38     0     0     0     0     0     0     0     0
     4    12    23    32     0     0     0     0     0     0
     7    13    14    18    31    37     0     0     0     0
    35     0     0     0     0     0     0     0     0     0
     6    22    39     0     0     0     0     0     0     0
    10    26    28    30     0     0     0     0     0     0
    20     0     0     0     0     0     0     0     0     0
    21    24    27     0     0     0     0     0     0     0
     8    16    33     0     0     0     0     0     0     0
     3    17    34     0     0     0     0     0     0     0
      ⋮

Display the schedule matrix as a stacked bar chart. Label the top of each bar with the task number.

figure
bar(ptime,'stacked')
xlabel('Processor Number')
ylabel('Processing Time')
title('Task Assignments to Processors')
for i=1:size(optimalSchedule,1)
    for j=1:size(optimalSchedule,2)
        if optimalSchedule(i,j) > 0
            processText = num2str(optimalSchedule(i,j),"%d");
            hText = text(i,sum(ptime(i,1:j),2),processText);
            set(hText,"VerticalAlignment","top","HorizontalAlignment","center","FontSize",10,"Color","w");
        end
    end
end

Find the minimum height of the stacked bars, which represents the earliest time a processor stops working. Then, find the processor corresponding to the maximum height.

minlength = min(sum(ptime,2))

minlength = 
1.0652

[~,processormaxlength] = max(sum(ptime,2))

processormaxlength = 
7

All processors are busy until time minlength = 1.0652. From the stacked bar chart, you can see that processor 8 stops working at that time. Processor processormaxlength = 7 is the last processor to stop working, which occurs at time makespan = 1.3634.