# Paint Production

## Contents

## Problem description

As a part of its weekly production a paint company produces five batches of paints, always the same, for some big clients who have a stable demand. Every paint batch is produced in a single production process, all in the same blender that needs to be cleaned between two batches. The durations of blending paint batches 1 to 5 are respectively 40, 35, 45, 32, and 50 minutes. The cleaning times depend on the colors and the paint types. For example, a long cleaning period is required if an oil-based paint is produced after a water-based paint, or to produce white paint after a dark color. The times are given in minutes in the following table CLEAN where CLEANij denotes the cleaning time between batch i and batch j.

Matrix of cleaning times

+-+--+--+--+--+--+ | | 1| 2| 3| 4| 5| +-+--+--+--+--+--+ |1| 0|11| 7|13|11| |2| 5| 0|13|15|15| |3|13|15| 0|23|11| |4| 9|13| 5| 0| 3| |5| 3| 7| 7| 7| 0| +-+--+--+--+--+--+

Since the company also has other activities, it wishes to deal with this weekly production in the shortest possible time (blending and cleaning). Which is the corresponding order of paint batches? The order will be applied every week, so the cleaning time between the last batch of one week and the first of the following week needs to be counted for the total duration of cleaning.

## Variables

cleantimes Times to clean from batch i to j prodtimes Production times per batch

## Reference

Applications of optimization... Gueret, Prins, Seveaux

% Marcus Edvall, Tomlab Optimization Inc, E-mail: tomlab@tomopt.com % Copyright (c) 2005-2009 by Tomlab Optimization Inc., $Release: 7.2.0$ % Written Oct 7, 2005. Last modified Apr 8, 2009.

## Problem setup

cleantimes = [ 0 11 7 13 11;... 5 0 13 15 15;... 13 15 0 23 11;... 9 13 5 0 3;... 3 7 7 7 0]; prodtimes = [40;35;45;32;50]; n = size(cleantimes,1); succ = tom('succ',n,n,'int'); y = tom('y',n,1); % All slots are integers bnds = {0 <= succ <= 1, y >= 0, succ((1:n+1:n^2)) == 0}; % Only one transition at a given time con1 = {sum(succ,1) == 1}; % Only one transition to a given batch con2 = {sum(succ,2) == 1}; % Sub-cycle constraint con3 = cell(n*(n-1),1); for i=1:n for j=2:n con3{(i-1)*(n-1)+j-1} = {y(j) >= y(i)+1-n*(1-succ(i,j))}; end end % Objective objective = sum(sum((repmat(prodtimes,1,n) + cleantimes).*succ)); constraints = {bnds, con1, con2, con3}; options = struct; options.solver = 'cplex'; options.name = 'Paint Production'; sol = ezsolve(objective,constraints,[],options); PriLev = 1; if PriLev > 0 temp1 = [sol.succ, sol.y]; link = []; % connections for i = 1:n for j = 1:n if temp1(i,j) == 1 link = [[i j ]; link ]; % finding connections end end end first = link(1:1); % start batch next = link(1,2); % next batch order = first; % ordered batches for k = 1:n order = [order next]; % adding next next = link(find(link(:,1)==next),2); % finding new next end disp(['one best order: ' num2str(order)]) % display solution end % MODIFICATION LOG % % 051010 med Created. % 060111 per Added documentation. % 060126 per Moved disp to end % 090308 med Converted to tomSym

Problem type appears to be: mip ===== * * * =================================================================== * * * TOMLAB - Tomlab Optimization Inc. Development license 999001. Valid to 2010-02-05 ===================================================================================== Problem: --- 1: Paint Production f_k 243.000000000000000000 f(x_0) 0.000000000000000000 Solver: CPLEX. EXIT=0. INFORM=101. CPLEX Branch-and-Cut MIP solver Optimal integer solution found FuncEv 36 one best order: 5 2 1 4 3 5