Difference between revisions of "OR-Tools/C2/Linear-Programming-with-two-index-variables/English"

Welcome

Objective

• Solve a Linear Programming problem having variables with two indices in Scilab.
• Use Karmarkar function.
• Use Transportation Problem as an example of Linear Programming.

Prerequisites

• You should have gone through the spoken tutorial on 'Optimization Using Karmarkar Function'.

• This tutorial is available on the Spoken Tutorial website.

• You should be familiar with Linear Programming.

• And have Scilab installed on your system.

Terminology

• Linear Programming as LP and
• Transportation Problem as TP.

What is two-index variable?

• If a variable has two indices in LP, then it is called a two-index variable.

• For example, x one comma two. Here x has two indices 1 and 2.

• An example of LP with two-index variables is the TP.

What is Transportation Problem?

This network has three supply nodes - plant P1, plant P2 and plant P3, two demand nodes - site S1 and site S2.

We can transport goods from P1 node to S1 and S2 with transporting cost c1,1 and c1,2.

Similarly, goods from P2 to S1 and S2 with transporting cost c2,1 and c2,2.

Goods from P3 to S1 and S2 with transporting cost c3,1 and c3,2, respectively.

• The aim is to supply goods from the plants to the sites, with total minimum transportation cost.

• Supply nodes cannot transport more than their capacity.

• Note that all the demand at the sites must be met.

• Supply capacities, demands and transportation costs are known.

Transportation Problem Example

• A cement company transports cement from plants 1, 2 and 3 to construction sites 1 and 2.

• The available supply at each plant is:
  • plant-1: 45 tons, plant-2: 60 tons and
  • plant-3: 35 tons.

• The demands of sites are:
  • site-1: 50 tons and
  • site-2: 60 tons.

Transportation Problem Example

• The cost of transporting 1 ton of cement from each plant to each site, is given in the table:

• Transporting cost from plant one to site one and site two is three and two respectively.

• Similarly, other plants have transporting cost.

• Each plant has Available Supply.

• Each site has Demand.

• We will formulate it as LP problem and find its optimal solution.

Decision Variable

• Two-index decision variable: xi,j is the number of tons transported from plant i to site j.

Formulation

  * Objective function:
    Minimize
    3 x1,1 plus 2x1,2 plus x2,1 plus 5x2,2 plus 5x3,1 plus 4x3,2.

  * Constraints:
    Since plant-1 can not supply more than its capacity, therefore we have
    x1,1 plus x1,2  is less than or equal to  forty five
    Similary for plant-2, we have
    x2,1 plus x2,2  is less than or equal to  sixty
    Similary for plant-3, we have
    x3,1 plus x3,2 is less than or equal to  thirty five

  * Demand at the site-1 must be met, therefore we have
    x1,1 plus x2,1 plus x3,1  is greater than or equal (>=) to  fifty. 
    It is equivalent to minus x1,1 minus x2,1 minus x3,1 is less than or     equal (<=)to minus fifty(-50). 
    Both constraints are equivalent to each other. 
    We multiplied with -1 to make inequality constraints in 'less than or equal to' form for Karmarkar function.
    Similarly at site-2, we have
    x1,2  plus x2,2  plus x3,2  is greater than or equal to (>=) sixty. It is equivalent to minus x1,2  minus x2,2  minus x32 is less than or       equal (<=) to minus sixty(-60). 

 * xi,j is greater than or equal to zero (This is because transported quantity can not be negative). 
   i is 1, 2, 3 because we have 3 supply nodes in this example.
   And j is 1, 2 because we have two demand nodes.

Karmarkar Function in Scilab

• Karmarkar function is LP solver in Scilab.

• Karmarkar function solves the LP problem in the following form-

Karmarkar Function in Scilab

  Minimize c transpose X.
     subject to AeqX is equal to beq
     which means constraints have equality sign.
     AX less than or equal to b
     which means constraints with greater than or equal to sign must be by multiplied by -1.
     lb less than or equal to X less than or equal to ub.
     Where, lb stands for lower bounds and ub stands for upper bounds on the variables. 

  c transpose,  Aeq, beq, A, b,  lb and ub are the known parameters.

  x is the vector of decision variables.

Input Parameters

    The coefficient vector of objective function is c transpose 3 2 1 5 5 4.
    This vector size will change according to the example.In our example, it contains 6 numbers. 

Input Parameters

    The coefficient matrix of constraints is capital A . 

    In the first row, we have 1 1 and rest of the values are zero.
    This is because in our 1st constraint only variables x1,1 and x1,2 are present, and rest of the variables have zero coefficient.

    Row 2 is 0 0 1 1 0 0 because 2nd constraint  has variables x2,1 and x2,2.

    And Row 3 is 0 0 0 0 1 1 because 3rd constraint  has variables x3,1 and x3,2.

    In fourth row, we have -1 0 -1 0 -1 0. 
    This because in our 4th constraint only variables are x1,1, x2,1 and x3,1 are present.

    5th row is 0 -1 0 -1 0 -1 because 5th constraint has variables x1,2, x2,2 and x3,2.

Input Parameters

• The vector of right hand side of constraints is b 45 60 35 -50 -60.

• lb is equal to zero for all variables as all xi,j are greater than or equal to zero.

• In our example, we do not require Aeq, beq and ub.

Steps to download the required code

• Pause this tutorial and click on the Code Files link below the player.

• This will download “opt.sce” file on your machine.

• Locate this file and open it in the Scilab console.

• Now resume the tutorial.

c = [3;2;1;5;5;4];

In the formulation of TP,

c is the coefficient vector of the objective function.

  A = [
       1 1 0 0 0 0
       0 0 1 1 0 0
       0 0 0 0 1 1
       -1 0 -1 0 -1 0
       0 -1 0 -1 0 -1
       ];

b=[45;60;35;-50;-60];

lb=[0;0;0;0;0;0];

lb is the vector of the lower bound of the variables.

ub =[];

Upper bound is empty vector because we do not require it in this example.

[xopt,fopt,exitflag,iter,yopt]= karmarkar([],[],c,[],[],[],[],[],A,b,lb,ub)

Now we can call the Karmarkar function to solve TP. Each parameter in this function is explained in the spoken tutorial mentioned earlier. In TP, four parameters c, A, b, lb and ub will always be present, and rest will be empty.

disp(xopt,'Optimal solution')

This line will display xopt value, which is the optimal solution of TP.

disp(fopt, 'Optimal value')

   0.0000691  
   44.999793  
   50.000138  
   0.0002763  
   0.0000691  
    15.

  First value is x1,1, 
  second value is x1,2, 
  third value is x2,1, 
  fourth value is x2,2,  
  fifth value is x3,1 
  and sixth value is x3,2.

  Note x1,1 can be taken as zero since the decimal value after zero, is floating point error. 
  x1,2 can be rounded upto forty five.
  x2,1 can be rounded to fifty.
  x2,2 and x3,1 can be rounded to zero.

    200.00166 

Result

In the optimal solution, we can transport,

  45 tons from plant-1 to site-2 with transportation cost 2 per ton,  
  50 tons from plant-2 to site-1 with transportation cost 1 per ton, 
  15 tons from plant-3 to site-2 with transportation cost 4 per ton. 

Other values of x are zero. So we do not transport goods on those paths.

Thus, the minimum transportation cost to transport the goods is two hundred.

Assignment

  * Here we have 2 plants and 2 sites. Transportation cost is given in the table.

  * Hint: Formulate this problem as explained in the previous example.

  * Please solve it using Karmarkar function and verify. 

  * The optimal value should be 520 and

  * The optimal solution  should have x1,2=60, x2,1= 100 , and all others have zero.

Summary

In this tutorial, we have learnt:

  * About variables with two indices.

  * Example of LP with two-index variables and its input parameters.

  * We have also learnt to apply Karmarkar function to solve Linear  Transportation problem in Scilab.

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