Script | Spoken-Tutorial:About

Time	Narration
Slide 1 Welcome	Welcome to the spoken tutorial on solving Linear programming problems with two-index variables in Scilab.
Slide 2 Objective	In this tutorial, we will learn how to: * Solve a Linear Programming problem having variables with two indices in Scilab. * Use Karmarkar function. * Use Transportation Problem as an example of Linear Programming.
Slide 3 Prerequisites	In order to understand this tutorial: * 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.
Slide 4 Terminology	In this tutorial, we will refer to: * Linear Programming as LP. And * Transportation Problem as TP.
Slide 5 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.
Slide 6 What is Transportation Problem?	To understand the transportation problem, we consider the following network as an example 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.
Slide 7 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.
Slide 8 Transportation Problem Example	Point to the table * 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.
Slide 9 Decision Variable	* Two-index decision variable: xi,j is the number of tons transported from plant i to site j.
Slide 10 Formulation	Formulation of this problem is: * 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.
Slide 11 Karmarkar Function in Scilab	* Karmarkar function is LP solver in Scilab. * Karmarkar function solves the LP problem in the following form-
Slide 12 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.
Slide 13 Input Parameters	Coming back to our example. * 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.
Slide 14 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.
Slide 15 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.
Slide 16 Steps to download the required code	I have a working example with me. I will open my file opt.sce in the Scilab console. * 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 ];	Capital A is the coefficient matrix of constraints.
b=[45;60;35;-50;-60];	b is the vector of right hand side of constraints.
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')	This line will display fopt value, which is the optimal value of TP.
Optimal-solution 0.0000691 44.999793 50.000138 0.0002763 0.0000691 15.	The optimal solution and optimal value are given. 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.
Optimal-value 200.00166	Optimal value is two hundred only as decimal places are only floating point errors.
Slide 17 Result	In our example, red edges denote the non zero variables in the optimal solution. 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.
Slide 18 Assignment	Here is an assignment for you. Here we have 2 plants and 2 sites. Transporation 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.
Slide 19 Summary	This brings us to the end of this tutorial. Let us summarise. 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 learn applying Karmarkar function to solve Linear Transportation problem in Scilab.
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Slide 23 Acknowledgements	Spoken Tutorial Project is funded by NMEICT, MHRD, Government of India. * More information on this Mission is available at http://spoken-tutorial.org /NMEICT-Intro Script is created by Suraj Trivedi This is Rahul Joshi from FOSSEE Project, IIT Bombay. Thank you for watching.