“AT-MOST” TYPE OF CONSTRAINTS
Q. Solve the following problem using simplex method (NDV 93 – equality constraint)
This type of problem is resolved broadly in the same manner as the problem with same type of constraints. The methodology differs in writing the standard form. When there is an equality constraint, LHS is already equal to RHS. Hence neither a slack variable nor a surplus variable is required. When the standard form is written, every constraint needs a slack variable, real or artificial. Hence the equality constraint is written in the standard form with an artificial variable. This is explained with an example below:
Case2-1
Q. Solve the following problem using simplex method (NDV 93 – equality constraint)
Objective function: Maximize Z =2X1+ 4X2, Subject to the constraints
2 X1+ X2 18 3X1+ 2X2≥ 30 X1+ 2X2= 26 X1,X2 ≥ 0
The above problem is already formulated and X1 & X2 are decision variables. Hence steps 1&2 are already done. Now standard form is to be written
3. Standard form:
2 X1+ X2+ S1+0S2+0A1+0A2= 18 3 X1+ 2X2+0S1– S2+ A1+0A2= 30 X1+ 2X2+0S1+0S2+0A1+A2= 26 X1,,X2, S1,S2, A1, A2 ≥ 0
S1is a slack variable; S2is a surplus variable and artificial variables A1&
A2serve as artificial slack variables.
4. Starting solution:
Put all non-slack variables equal to zero.X1= 0,,X2= 0, S2=0, hence, S1= 18, A1= 30, A2= 26
5. Structuring of Simplex table
From the information collected, following simplex table ST-1 is constructed:
Simplex table ST-1
Contributions >>> 2 4 0 0 -M -M C
The above table doesn’t give optimal solution as NER values are not non positive. As per the methodology first iteration is performed and next table ST2is written.
Simplex table ST2
Contributions >>> 2 4 0 0 -M -M C
The above table doesn’t give optimal solution as NER values are not non positive. As per the methodology second iteration is performed and next table ST3is written.
Simplex table ST3
Contributions >>> 2 4 0 0 -M
In the above ST3columns of A1& A2are ignored as these artificial variables are driven out of the system.
The above solution is optimal as all NER values are non-positive.
The values of the basic variables which form the optimal solution are given in solution values column. The unknown values of the decision variables are substituted in the objective function to calculate the maximum profit as per the optimal solution.
Objective function: Maximize Z =2X1+ 4X2 , Z =2 (2) + 4 (2) = 12,
Maximum value of Z as per the above optimal solution is 12 in appropriate units.
4.3 MINIMIZATION CASE (PROBLEM) - “BIG M”
METHOD
The methodology for solving minimization problems follows the same flow as in maximization case but certain steps are different.
Minimization problems deal with parameters like cost. The objective function is minimization type with constraints of at-least or at-most type.
Minimization problems also come with mixed constraints. The formulation is done in the same manner as in maximization. Hence these cases are not discussed separately. For minimization type of problems, study of the business case and mathematical formulation are done in the same way as in the case of a maximization problem. Construction of objective function in standard form and application of optimality test are different as compared to maximization problems. This chapter covers two methods for solving minimization problem. The methods are “Big M”
method and “two phases” method. The Big M method is taken up first.
The concept of big M as a finite large quantity is already understood and applied while solving maximization problem with mixed constraints.
“Two phases” method will be taken up later.
1. Study of the business case: done similarly as in the case of maximization
2. Formulation: done similarly as in the case of maximization. But nature of the objective function is now minimization of cost.
3. Standard form:
a) Objective function: objective being minimization type in order to drive out the artificial variable from the system, the artificial variable should be made very high. For this purpose artificial variable is associated with a positive high value M.
b) Constraints: done similarly as in the case of maximization.
4. Starting solution: done similarly as in the case of maximization.
5. Structuring of Simplex table: done similarly as in the case of maximization.
6. Calculation of Z row values: done similarly as in the case of maximization.
7. Optimality test: when all the values in NER, index row or ∆ row are non-negative (when calculated as = C – Z) the solution for a minimization problem is optimal. If solution is optimal, the process stops and solution offered by the simplex table which can be read in bi column is put to use. If the solution is not optimal, the simplex method improves the solution as described below.
8. Simplex Iterations: iterations are carried out as in the case of maximization to improve the solution until optimality is reached.
Minimization case (problem) with all “at-least” type of constraints A business problem is taken through simplex methodology discussed earlier.
Case3 is considered below. All the steps leading to the construction of first simplex table are discussed earlier. The nature of the constraints is “at least” type. A definite quantity of resources should be used while running the process. Hence surplus variables and artificial variables enter the standard form of formulation.
Following is a business case taken up for step by step discussion
Case3: Food F1contains 20 units of vitamin A and 40 units of vitamin B per gramme; Food X2 contains 30 units each of vitamin A and B per gramme. The minimum daily requirements for an individual are 900 units of A and 1200 units of B. How many grammes of each food must be consumed to satisfy daily vitamin requirements at minimum cost? If X1 costs 60 paise per gram and X2costs 80 paise per gram, find the optimal solution to minimize the costs using simplex method.
1.Study of the business case: Given Case3 is analyzed as per the guidelines and following facts are established:
This is a case of cost minimization. An optimal solution minimizing costs is contemplated.
Company is considering producing food products F1& F2. Quantities of food products are decision variables.
Expected cost contribution by a unit of F1is Rs.0.60/-, by a unit of F2
is
Rs.0.80/- Availability of each vitamin in each food product in units, decision variables and minimum requirement of each vitamin in appropriate units are tabulated below:
Vitamin
2 .Formulation: Let unknown quantity X1 units of F1, unknown quantity X2 units of F2 be produced. The model will finally give the values of decision variables X1 & X2 for minimum cost at optimum level. Cost contribution per unit for A & B are Rs 0.60/- & Rs 0.80/-. The cost contributions are written in terms of paise for simplicity of calculations.
Decision variables: Unknown production quantities X1& X2are decision variables.
Objective function: 60X1 + 80X2 = G (G is the total cost per day which is to be minimized)
Constraints:
20X1+30X2≥ 900 ……….(1) Vitamin A Constraint (at-least) 40X1+30X2≥ 1200..…….(2) Vitamin B Constraint (at-least) X1, X2 ≥ 0………Non-negativity constraint
3. Standard form: The mathematical formulation suitable for simplex method is written in terms of decision variables, surplus variables, artificial variables and their coefficients. While X1& X2 are decision variables, S1 & S2are surplus variables introduced to established equality between LHS & RHS, and A1& A2 are artificial variable required as per simplex method. A1& A2serve as artificial slack variables. M is a finite high value introduced with a positive sign as a coefficient to A1& A2 in objective function to drive out artificial variable A1.
Objective function: 60X1+ 80X2+ 0S1+0 S2+ MA1+MA2= G (G is the total cost per day which is to be minimized)
Constraints:
20X1+ 30X2- S1+ 0S2+A1+0A2 = 900 ……….(1) Vitamin A Constraint (at-least)
40X1+ 30X2+ 0S1- S2+ 0A1+ A2= 1200..…….(2) Vitamin B Constraint (at-least)
X1, X2, S1, S2, A1, A2≥ 0………Non negativity constraint
4. Starting feasible solution: Put all non-slack variables equal to zero, then
X1=0, X2=0, S1=0, S2=0; A1=900, A2=1200.
5. Structuring of Simplex table: setup the simplex table as per the guide lines already laid down. Put the starting feasible solution in the first simplex table ST1.
Contributions >>> 60 80 0 0 M M C
The optimality test for a minimization problem is applied to the solution in ST1. As all NER values are not non-negative the starting solution is not optimal. Hence the first iteration is carried out to improve the solution in ST1. The key column, key row and key element are identified from the table STI. This being minimization problem, the NER value selected is most negative unlike in the maximization case. The rule for selection of replacement ratio continues to be the same as in maximization case.
It can be seen that column of X1is the key column, row of A2is the key row and the value 40 (cell A2-X1) is the key element. Using the information from ST1following table ST2is written.
ST2
Contributions >>> 60 80 0 0 M C
The optimality test for a minimization problem is applied to the solution in ST2. As all NER values are not non-negative the starting solution is not optimal. Hence the second iteration is carried out to improve the solution in ST2. The key information from the table ST2 is collected. This being minimization problem, the NER value selected is most negative unlike in the maximization case. The rule for selection of replacement ratio continues to be the same as in maximization case.
It can be seen that column of X2is the key column, row of A1 is the key row and the value 15 (cell A1-X2) is the key element. Using the information from ST2 following table ST3 is written.
ST3
Contributions >>> 60 80 0 0 C
The solution in ST3 is optimal as the optimality rule for minimization problem is that the NER values should be non-negative.
When the values of the decision variables of the optimal solution are put in the objective function the total minimum cost is obtained
60X1+ 80X2= G
60 X (15) + 80 (20) = 2500
Rs 25.00/- is the minimum cost as cost contributions were written in terms of paise.
4.4 MINIMIZATION CASE (PROBLEM) - “TWO PHASES” METHOD
The two phase method, under simplex methodology, is the other method to be studied for minimization problem. Study of the business case and formulation are done as per earlier methodology.
Standard form: the constraints are written but while writing the objective function artificial variables are not associated with M. the coefficients are only 1 each.
As the name suggests there are two phases to this method. The objective function is divided into two parts.
Phase I: the objective function is maximized to remove artificial variables