(Modeling of Decision Processes) Linear programming (Modeling of Decision Processes)

Linear programming Widely used Mathematical optimization algorithm Model consist of: Objective function System of constrains (linear functions)

General formulation of the model

Terms Optimized variables xj – values whose optimal level is necessary to find solution for decision situation (for example amount of production, transported goods …) Technical coefficients aij – parameters of the model (for example material and energetical needs for production the product)

Terms bi – can be capacity constrains (max production volume, resource constrains, etc.) The variables cj in objective function may be price of the products, variable costs etc.

Equation system constrains ≤ … for existing constrains as in available resources = … for balance of material connections and situations when there is stated ration of the products ≥ …for formulations of the trade needs

General expression of the model max z = cx Ax ≤ b x ≥ 0 Where b … column vector right sides of the constrains A … rectangular matrix (m, n) of technical coefficients c … row vector of variable coefficients in objective function x … column vector of optimized variables

Model formulation Raw material product Available amount A B C S1 0,9 0,3 0,5 75,6 S2 1 0,4 88,2 Price Kč/t 12 600 19 600 15 700 How much of each product should we produce to maximize our utility?

Model formulation Objective function max z = 12 600 x1 + 19 600 x2 + 15 700 x3 Raw material constrains for S1 0,9x1 + 0,3x2 + 0,5x3 ≤ 75,6 for S2 x2 + 0,4x3 ≤ 88,2 min amount of product B <- additional constrain (not in table) x2 ≥ 30

Model formulation It is neccessary to manufacture the products A and B in certain ratio <- additional constrain (not in table) x1 – 0,8x2 = 0 The solution must be greater then 0 <- general constrain, is present in all “linear programs” x1, x2, x3 ≥ 0 Most of the time the ≥ constrain is present in computational package and does not to be expressed explicitly

Using matrixes

Simplex method It is necessary to transform into standard state in witch we transform the non equations on the equations ≤ … we add to left side of new variable known as aditional variable xn+1

Additional variables if the value is not 0 it can be interpreted as amount of the resource bi, which will remain unused. The goal of the computation is to minimize (null) the values of the additional variables Simplex method performs computation in iteration How many variables are we able to compute in the system of linear equations? … Asking because we usually work with more variables then that Minimizing and nulling additional variables is the way to open possibility to compute real variables

Nonequality ≥ We have to add another variable –xn+i

Positives System of equations Any of additional variable is inside of just one equation – so it can be substituted by other variables Easy to compute (even pen and paper computation possible, but who would do such thing?)

Solution Non negative solutions which respect constrains are call basic solution. Each iteration has basic solution Each iteration improves the result of objective function In this case we will get some of xn+i negative witch is unacceptable We solve the problem by adding another variable x’n+i+1 Slides after this one only conclude pen and paper computation

Solution Similarly we add to right sides from original model into

Additional variables Stating of additional variable are very high M, minus if we maximaze and positive if we minimalize The task looks in following way

Additional and helping variables Are easy to express using different variables

First solution We get first solution in standard form when we put xj = 0 for j = 1,2,…,a xn+i = 0 for i = k+p+1,k+p+2,…,k+p+s Max no. Of non-zero variable in basic solution is given by number of constrains m = k + l + s

First solution Values will equal xn+i = bi x’n+i = bi x’n+i+1= bi Chosen m variables will be described as basic, other n-m variables as non-basic

Example max z = 12 600x1 + 19600x2 + 15700x3 + +0x4 + 0x5 – Mx’6 + 0x7 – Mx’8 0,9x1 + 0,3x2 + 0,5x3 + x4 = 75,6 x2 + 0,4x3 + x5 = 88,2 x1 – 0,8x2 +x’6 = 0 x2 – x7 + x’8 = 30 x1,x2,x3,x4,x5,x’6,x7,x8 ≥ 0

First basic solution x4=75,6 x5 = 88,2 x’6 = 0 x’8 = 30 x1,x2,x3,x7 = 0

Using vector model

We divide the elements of model into parts connected to vector of basic variables xB and vector of non-basic variables xR. Similarly vector c we divide into cB and cR And matrix A into B and R

Generally Change of equation system BxB+RxR = b /*B-1 xB + B-1RxR = B-1b xB = B-1b – B-1RxR If we put nonbasic variable equal to 0, xR=0, we get the expression for finding the Solution of linear programming xB = B-1b With value of objective function z = cBxB = cBB-1b

Solution

Solution To get basic solution we only need to find inverse matrix to matrix B And multiply it by vector b