2.  Minimization Problem  First Approach  Introduce the basis variable  To solve minimization problem we simple reverse the rule that is we select the variable with most negative cj-zj to select new basic variable in the next iteration  The stopping rule is also changed ; the iteration is stopped when every value is the cj-zj row is zero or positive.

3. SIMPLEX METHOD Second Approach  Change the minimization problem to maximization problem by multiplying the objective function with -1  Solve the problem using the simplex method  Stop the iteration when cj-zj row is either negative or zero.  Multiply the objective function value to -1 in the last iteration to convert the maximization to original minimization problem

4.  Management want to minimize the cost of producing two products to a demand constraint of product A,a minimum total production quantity requirement and a constraint on available process time  Min 2x 1 +3x 2  S.t  1x 1 >=125 (Demand for Product A)  1x 1 +1x 2 >=350(Total Production)  2x 1 +1x 2 <=600 (Processing Time)  X 1,x 2 >=0

5.  To solve this minimization problem we multiply the objective solution by -1 to convert in to maximization problem  Max -2x1-3x2  S.t  1x 1 >=125 (Demand for Product A)  1x 1 +1x 2 >=350(Total Production)  2x 1 +1x 2 <=600 (Processing Time)  X 1,x 2 >=0  Standardized form ?????

6. Standard form Max -2x 1 -3x 2 +0s 1 +0s 2 +0s 3 -Ma 1 -Ma 2 1x 1 -1s 1 +1a 1 =125 1x 1 +1x 2 -1s 2 +1a 2 =350 2x 1 +1x 2 +1s 3 =600 x 1.,x 2,s 1,s 2,s 3,a 1,a 2 >=0 Initial Simplex Table ????

7.  x1 x2 s1 s2 s3 a1 a2  -2 -3 0 0 0 -M -M  1 0 -1 0 0 1 0 125  1 1 0 -1 0 0 1 350  2 1 0 0 1 0 0 600  What next ????  What are basic variables????

8. Basis CB x1 x2 s1 s2 s3 a1 a2 -2 -3 0 0 0 -M -M a 1 -M 1 0 -1 0 0 1 0 125 a 2 -M 1 1 0 -1 0 0 1 350 s 3 0 2 1 0 0 1 0 0 600  Z j -2M -M M -M 0 -M -M -475  C j -Z j -2+2M -3+M -M -M 0 0 0  Which variable need to go basis ????

9.  X1 need to go basis because -2+2M is largest  A1 need to go nonbasis because 125/1=125;  350/1=350; 600/2=300 hence a1 is smallest  Also a11=1;a21=0;a31=0 as need for basis variable  Row operation : Subtract row 1 from row 2  Multiply row 1 by 2 and subtract from row 1  a 1 is removed as it is now nonbasic and need tobe zero.

10. Basis CB x1 x2 s1 s2 s3 a2 -2 -3 0 0 0 -M x 1 -2 1 0 -1 0 0 0 125 a 2 -M 0 1 1 -1 0 1 225 s 3 0 0 1 2 0 1 0 350  Z j -2 -M 2- M M 0 -M -250-225M  C j -Z j 0 -3+M -2+M -M 0 0  Which variable need to go basis ????

11.  Now S1 will be basis and a2 will be non basis hence removed from tableau continue with two other iterations.

12. Basis CB x1 x2 s1 s2 s3 -2 -3 0 0 0 x 1 -2 1 0 -0 1 1 250 x 2 -3 0 1 0 -2 -1 100 s 1 0 0 0 1 1 1 125  Z j -2 -3 0 4 1 -800  C j -Z j 0 0 0 -4 -1  Optimal solution is +800 becaz it is minimization problem

13.  Occurs when no solution can be found that satisfies all constraints. Identified by positive value of artificial variable in the solution  Max 50x 1 + 40x 2  S.t  3x 1 +5x 2 <=150 (Assembly Time of two product)  1x 2 <=20 (portable display)  8x 1 +5x 2 <=300 (warehouse space)  1x 1 +1x 2 >=50 (Minimum total production)

14.  Basic CB x1 x2 s1 s2 s3 s4 a4  50 40 0 0 0 0 -M  X 2 40 0 1 8 / 25 0 - 3 / 25 0 0 12  S 2 0 0 0 - 8 / 25 1 3 / 25 0 0 8  X 1 50 1 0 - 5 / 25 0 5 / 25 0 0 30  a 4 -M 0 0 - 3 / 25 0 - 2 / 25 -1 1 8  Z j 50 40 (70+3M)/ 25 0 (130+2M)/ 25 M -M 1980-8M  C j -Z j 0 0 (-70-3M)/ 25 0 (-130-2M)/ 25 -M 0  Feasible Solution ???????

15.  Presence of a4 =8 in the solution means it is not feasible although c j -z j is non negative.  X1=30;x2=12 x1+x2=42 <50 hence violates the fourth constraint of at least 50 units, a4 =8 indicates that constraint 4 th is violates by 8 unit  S1 and s3=0 means warehouse space and assembly time constraint are binding as not enough spare house space and time is available hence minimum combined total production of 50 units is lowered by 8 units.

16.  If more time or space is not allotted than management will have to relax total production by 8 units

17. 1. A Maximization Problem is unbounded if it is possible to make the value of optimal solution as large as possible without violating the constraints. 2. A surplus variable can be interpreted as amount of the basis variable over the minimum amount required. 3. If a solution is unbounded then we can increases the minimum amount of a basic variable as much as we want and the objective function will have no upper bound provided the basis variable has positive coefficient in the objective function

18.  In simplex tableau we recognized the unbounded solution is that all a ij are less than or equal to zero in column associated with incoming variables.  Max 20x 1 +10x 2  S.t 1x 1 >=2  1x 2 <=5  X1,x2>=0

19.  Range of optimality for an objective function  Is the range of that coefficient for which the current optimal solution will remain optimal (keeping all other coefficients constant). However the objective function value may change.  The range of optimality for the basic variable defines the objective function coefficient values for which current variable will remain the part of the basic feasible solution. Range of optimality for nonbasic variable defines objective function coefficient values for which that variable remain nonbaisc

20.  Change the objective function coefficient to c k in the c j row.  If x k is basic, then also change the objective function coefficient to ck in the c B column and recalculate the z j row in terms of c k.  Recalculate the c j - z j row in terms of c k.  Determine the range of values for ck that keep all entries in the c j - z j row less than or equal to 0.

21.  Max 50x 1 + 40x 2  S.t  3x 1 +5x 2 + <=150 (Assembly time)  1x 2 + <=20 (Portable display)  8x 1 +5x 2 <=300 (warehouse capacity)  X 1 is number of units of Desktop PC  X 2 is number of the portable display  X1,x2>=0

22.  Basic cb X 1 x 2 s 1 s 2 s 3  50 40 0 0 0  X 2 40 0 1 8 / 25 0 -3 / 25 12  s 2 0 0 0 -8 / 25 1 3 / 25 8  X 1 50 1 0 -5 / 25 0 5 / 25 30  Z j 50 40 14 / 5 0 26 / 5 1980  C j -Z j 0 0 -14 / 5 0 -26 / 5  The range of optimality for an objective function coefficient is determined by those coefficient values that maintain  Cj-zj<=0

23.  Basic cb X 1 x 2 s 1 s 2 s 3  C 1 40 0 0 0  X 2 40 0 1 8 / 25 0 -3 / 25 12  s 2 0 0 0 -8 / 25 1 3 / 25 8  X 1 C 1 1 0 -5 / 25 0 5 / 25 30  Z j c 1 40 (64-c1) / 5 0 (c1-24) / 5 480+30c 1  C j -Z j 0 0 c1-64 / 5 0 24-c1 / 5  Applying cj-zj c 1 -64/5 <=0  C 1 -64<=0 --  C 1 <=64  24-c1<=0--  24<=C1 thus 24<=C1<=64

24.  Suppose increase in material cost reduces profit contribution per unit of desktop to 30  Range of optimality of X1 indicates still the solution x2=12;s1=0;s3=0 would be an optimal solution. The final tableau for x1=30 verifies this.  Basic cb X 1 x 2 s 1 s 2 s 3  30 40 0 0 0  X 2 40 0 1 8 / 25 0 -3 / 25 12  s 2 0 0 0 -8 / 25 1 3 / 25 8  X 1 30 1 0 -5 / 25 0 5 / 25 30  Z j 30 40 34 / 5 0 6 / 5 1380  C j -Z j 0 0 -34 / 5 0 - 6 / 5 HOW???

25.  Current solution is not optimal as can be seen from cj-zj row  S3 is positive means it is not optimal solution.  Basic cb X 1 x 2 s 1 s 2 s 3  20 40 0 0 0  X 2 40 0 1 8 / 25 0 -3 / 25 12  s 2 0 0 0 -8 / 25 1 3 / 25 8  X 1 20 1 0 -5 / 25 0 5 / 25 30  Z j 20 40 44 / 5 0 4 / 5 1080  C j -Z j 0 0 -44 / 5 0 4/ 5

26.  Basic cb X1 x2 s1 s2 s3  50 40 c s1 0 0  X2 40 0 1 8 / 25 0 -3 / 25 12  s2 0 0 0 -8 / 25 1 3 / 25 8  X1 50 1 0 -5 / 25 0 5 / 25 30  Zj 50 40 14 / 5 0 26 / 5 1980  Cj-Zj 0 0 c s1 -14 / 5 0 -26 / 5 C s1 -14/5<=0 cs1<=14/5

27.  Dual Price  Increase in objective per unit increase in RHS of constraint. In Simplex method they are identified in zj row row of final simplex method  Value of slack variables in final  S1=14/5=2.80;s2=0;s3=26/5=5.20  The dual price for assembly constraint mean that one 1 unit increase in assembly hours increases the objective function by 2.80$

28.  Basic cb X 1 x 2 s 1 s 2 s 3  50 40 0 0 0  X 2 40 0 1 8 / 25 0 -3 / 25 12  s 2 0 0 0 -8 / 25 1 3 / 25 8  X 1 50 1 0 -5 / 25 0 5 / 25 30  Z j 50 40 14 / 5 0 26 / 5 1980  C j -Z j 0 0 -14 / 5 0 -26 / 5

29.  Zj corresponding to S2 is 0 means dual price for portable display is zero. Also S2 is basic (slack)variable = 8 which shows there are 8 unused display left so increasing them will not effect objective function  If a slack variable is a basic variable in optimal solution then its dual price will be zero.  Dual price for >= constraint is given by negative of zj entry for surplus variable  Dual Price for = constraint is determined by zj values for corresponding artificial variables

30.  Is the Range of RHS of constraint that does not make current basis solution infeasible.  This means range of elements of b column matrix that does not make the solution infeasible.  Using the dual price concept the increase in one unit of element in b vector appeared in objective function is called dual price. Now we wish to know the range of this b column matrix.

31.  If we want to know the range of say b1 than we need to use the corresponding slack of final tableu. s1. WHY  IT is because the coefficient in this matrix show corresponding decrease in the basic variable or in other word how many units of basic variable will be driven out from solution or alternatively decreasing b column.

32.  B1 a1j 0 B2 a2j 0 B3 + delB* a3j >= 0. BN anj 0  Current solution or l  Last column of the final column corresponding to slack variable  tableu

33.  Example

34.  Is a two different perspective of the same problem. The original linear problem is called as primal and to each primal problem a corresponding optimization problem is associated which is termed as dual problem  Primal Problem  the objective function is a linear combination of n variables and m constraints to maximize the value of the objective function subject to the constraints

35.  Primal problem  A per unit value of each product is given and it is determined how much of each product is produced to maximized the value of total production. the constraint require amount of each resourced used to be less than or equal to amount available.  Dual Problem  Is resource valuation problem. In dual problem availability of each resource is given and we determine the per unit value of each constraint such that total resources used is minimized. In other word we determine what is the value of unit consumption of the available resource.

36.  The number of variables in the dual problem is equal to the number of constraints in the original (primal) problem. The number of constraints in the dual problem is equal to the number of variables in the original problem.  If the original problem is a max model, the dual is a min model; if the original problem is a min model, the dual problem is the max problem.  Convert the problem in to conical form

37.  Conical form for Maximization problem  All constraint should be less than or equal to constraint.  Conical form for Minimization problem  All constraint should be greater than or equal to constraint.

38.  Decision variable in primal becomes constraint in dual problem  the first constraint in dual will be associated with the first decision variable second with second and so.  The RHS of the constraint in the primal becomes the coefficient of objective function in the dual

39.  The objective function coefficient of the primal becomes RHS of the constraint in the dual.  the constraint coefficient ith primal variable becomes coefficient of ith constraint in the dual. In other words the row becomes column or the ‘A’ matrix is transposed.

40.  If the dual problem has an optimal solution, the primal has an optimal solution and vice versa. Both have same value of optimal solution in term of objective function  The optimal values of the primal decision variables in the final simplex tableau are given by zj entries for surplus variables. The optimal value of the slack variable are given by negative of cj-zj entries for uj variables.

41. Primal Problem  Max 50x 1 + 40x 2  S.t  3x 1 +5x 2 <=150 (Assembly Time of two product)  1x 2 <=20 (portable display)  8x 1 +5x 2 <=300 (warehouse space)  Is it in conical Form???

42.  Dual Problem  Min 150u 1 + 20u 2 +300u 3  S.t  3u 1 +8u 3 >=50  5u 1 +u 2 +5u 3 >=40  U 1,u 2,u 3 >=0  u1 is associated with assembly time constraint  u2 is associated with portable display constraint  u3 with warehouse space constraint.

43.  u1 u2 u3 s1 s2 a1 a2  Basic Cb -150 -20 -300 0 0 -M -M why-  A1 -M 3 0 8 -1 0 1 0 50  A2 -M 5 1 5 0 -1 0 1 40  Zj -8M -M -13M M M -M -M -90M  Cj-1 -150+8M -20+M -300+13M –M –M 0 0

44.  u1 u2 u3 s1 s2  Basic Cb -150 -20 -300 0 0  u3 -300 0 - 3 / 25 1 - 5 / 25 3 / 25 26/5  u1 -150 1 8 / 25 0 5 / 25 -8 / 25 14/5  Zj -150 -12 -300 30 12  Cj-1 0 -8 0 -30 -12 -1980  Assembly Time $2.80  Portable Display= 0  Warehouse space = $5.20