Example 2.4: Sky store promotes its products from a big city to different parts of the state. Product A can be sold at a profit of Rs.3 per unit and B with a profit of Rs.8 per unit. Example 2.8: International City Trust (ICT) invests in short-term trade credits, corporate bonds, gold stocks and construction loans.
A wine maker has a stock of three different wines with the following characteristics
General form of LPP
Mathematically speaking, we have m – linear inequalities or equalities in n – variables (m can be greater than, less than or equal to n) of the form. Note: After introducing slack/surplus variables, any given LPP can be expressed as below: Maximize Z = c1x1 + c2x2. Using matrix notations, above LPP can be expressed in canonical form as well as standard form as follows: Canonical form: Maximize (Minimize) Z = cTx subject to Ax ≤ (≥) b, x ≥ 0.
Solving the Linear Programming Problem
Optimal basic feasible solution : A basic feasible solution is said to be optimal if it optimizes the objective function. Unbounded Solution : If the value of the objective function can be infinitely increased or decreased without violating the constraints, the solution is known as an unbounded solution. So, if there is a basic feasible solution for a given LPP, then one of the nodes will give the optimal value to the objective function.
The set of optimal solutions to the LPP is convex
Now consider the point x0 є SF , which is not an extreme point, and let Z0 be the corresponding value of the objective function. Since x0 is not an extreme point, it can be expressed as a convex combination of the extreme points x1, x2, …, xk of the feasible region F, where F is assumed to be a closed and bounded set. Let x1, x2, …, xr (r ≤ k) be the extreme points of the feasible region F at which the objective function assumes the same optimal value.
If there exists a feasible solution to the LPP then there exists a basic feasible solution to a given LPP
Graphical Method of solving LPP
- Extreme point approach
- Iso-profit (cost) function line approach : Follow step 1 and step 2 as stated in 3.6.1
Using results proved in section 2.5, the optimal solution for LPP can be found by evaluating the value of the objective function at each vertex of the feasible region. Theorem 2.2 also states that an optimal solution to LPP will only occur at one of the extreme points. Evaluate the values of the objective function at each extreme point and the extreme point of the feasible region that optimizes (maximizes or minimizes) the objective function value is the required basic feasible solution.
This means that as we move the iso-gain line in the desired direction, the last point beyond which we leave the feasible region is the required optimal solution. The constraint equations can be plotted simply by making one variable equal to zero and solving for the intersection of the axis of the other. It can be shown mathematically that the optimal combination of decision variables is always found at an extreme point (vertex) of the convex polygon.
This involves simultaneously solving the equations of several pairs of intersecting lines and replacing the quantities of the resulting variables in the objective function. The procedure simply involves drawing a straight line parallel to a randomly selected initial iso-profit line so that the iso-profit line is farthest from the origin of the chart. Note that the initial randomly selected iso-gain line is needed to represent the slope of the objective function for the particular problem.
Coming back to the problem, we now evaluate the value of the objective function at all vertices of the feasible region and select that point as the optimal solution that gives the maximum value.
Special cases in LP
- Alternative (or Multiple) Optimal Solution : We try to under the concept of alternative or multiple solution by considering the example
- An Unbounded Solution
- Infeasible Solution
The unfeasible solution occurs when no value of the variables satisfies all constraints at the same time; Thus, the redundant constraint will not have any effect on the optimal value of the objective redundant constraint. Thus, the redundant constraint has no effect at all on the optimal value of the objective function.
Simplex Method
Since simplex method is an iterative method, we assumed that an initial basic feasible solution is available. The columns of A that are included in B are called basic vectors and those that are not in B are called non-basic vectors. The variables vector x chosen corresponding to vectors in basis matrix B are known as basic variables and rest are known as non-basic variables.
The problem is “Which basic variable should be removed from B and which of the non-basic variables should be inserted in its place. Suppose we remove a basis vector br from B and insert a non-basis vector aj є A in its place. Let B* denote the non-basic matrix obtained by substituting aj for br then.
We must first determine the vector aj to be inserted into the new basis and then using (2.13) determine the vector br to be removed from B. While discussing the simplex method, we have assumed an initial feasible basis solution, we have assumed a feasible baseline. solution with a basis matrix B. Therefore, to obtain the initial feasible basis solution, we will assume that a unit matrix is present as a sub-matrix of the coefficient matrix A.
If two or more zj–cj (cj–zj) have the same most negative (positive) value, choose one of the corresponding vectors to get into the basis.
Every basic feasible solution to a LPP corresponds to a vertex of the set of feasible solution
Simplex Algorithm : (Maximization Case)
For choosing the input vector aj in the other basis, choose the one zj – cj ( cj – zj) that is more negative (positive). If this minimum value is assumed for more than two vectors, choose any of the corresponding basis vectors from B. The element at the intersection of the key row and the main column is called the key element.
Note that division by a negative or zero element in the key column is not allowed. If the key element is 1, do not change the row in the following simplex table. If the key element is other than 1, divide the element in that key row by that element including the element in xB column and formulate the new row.
The new values of the elements in the remaining rows for the next iteration can be evaluated by performing elementary operations on all rows so that all elements except the key element in the key column are zero. Solution: If we write the given LPP in standard form, we must add the slack variables s1, s2 and s3 to the constraints. The leading (key) element is 8, which is different from 1, so divide all elements of the key row by 8 and using elementary row transformations so that the entries in the key columns of the first and second rows are zero.
The key (key) element is 7/2 which is different from 1, so divide all the key row elements by 2/7 and use elementary row transformations so that in the key columns the second and third row entries are zero .
Minimization Case
- Two – Phase Method
- Big – M method
In phase I of this method, we will try to minimize the sum of the artificial variables subject to the constraints of the given LPP. If all zj – cj ≥ 0 and at least one artificial variable occurs in the optimal basis and thus max z* < 0, then the LPP has an impossible solution. If all zj – cj ≥ 0, max z* = 0 and at least one artificial variable occurs in the optimal basis, go to phase II.
Assign actual cost to the original variables and '0' to other variables in the objective function. Since zj - cj are all non-negative and no artificial variable occurs in the basis, the optimum basic feasible solution for the objective function is obtained from phase I and goes to phase II. Since zj - cj are all non-negative, an optimum basic feasible solution for the reduced problem is reached, but at the same time artificial variable A1 appears at a positive level in the basis, so given LPP does not possess any feasible solution.
Add slack, excess, and artificial variables to the constraints mentioned in the previous two paragraphs, but assign a very high value. Net valuations zj – cj (j n) are not negative and the artificial variables are not present in the basis. Net evaluations zj – cj(j n) are non-negative and there is at least one artificial variable in the basis and the objective function value z contains M.
Since all NER entries are ≤ 0, the optimality criteria is satisfied, but the Z value in the final table contains the coefficient of the artificial variable.
Duality in LPP
- Duality Theorems
Since this is a maximization problem, the optimality criteria are satisfied since all N.E.R entries viz. all cj - zj entries are ≤ 0. a) Each positive number in the cj - zj row represents the net profit that could be obtained if 1 unit of the variable heading this column was added to the solution. A negative number in the row cj - zj under one of the columns representing time also has a different interpretation. The values under the SA, SF and SP columns in the cj - zj row indicate that removing 1 productive hour from each of the three departments would reduce the total contribution by Rs.
These are of course the same values that we got by looking at the cj-zj row in the final table of the primary problem. The objective function coefficients of the primal problem have become right-hand side constants of the dual. Similarly, right-hand constants of the primal have become the objective function coefficients of the dual.
Transferring the technological (input-output) coefficient matrix of the primal becomes the technological (input-output) coefficient matrix of the dual. Suppose that T, C, B is a feasible solution for the primal (the level of production that can be achieved with current resources) and A, F, P is a feasible solution for the dual. a set of rents that would induce the manager to lease the plant rather than use it himself). This shows us that the value of the time required to produce one unit of T is Rs.
3 Coefficients of the objective function RHS of constraints 4 Input – output matrix A Input – output matrix AT.
The dual of dual is the primal
The value of the objective function z for any feasible solution of the primal is not less than the value of the objective function z * for any feasible solution of the dual
- Find the maximum value of p = x + 2y + 3z
- Find the maximum value of p = 2x - 3y + 5z
- Find the maximum value of p = 2x + 4y + z + w
- Find the maximum value of p = 107x + y + 2z
- Find the maximum value of p = 2x + 4y + 3z
- Find the maximum value of p = 2x + y + 3z
- Find the minimum value of p = 2x + 8y
- Find the maximum value of p = x + 2y + 3z – w
- Find the minimum value of p = x + 2y + 3z
- Find the minimum value of p = 50x + 150y + 100z
- Find the minimum value of p = x – 2y – 3z
Solve graphically: A diet-conscious housewife wants to ensure certain minimum intakes of vitamins A, B and C for the family. The minimum daily (quantity) needs of the vitamins A, B and C for the family are 30, 20 respectively for the family. The minimum daily (quantity) needs of the vitamins A, B and C for the family are 30, 20 and 16 units respectively.
The first respectively provides 7.5.2 units of the three vitamins per gram and the second respectively provides 2.4.8 units of the same three vitamins per gram of food. The problem is how many ounces of each food item the housewife needs to buy each day to keep her food bill as low as possible. Hint: divide the first equation by 3 (coefficient of w) and then treat w as the soft variable].
