1. 1

2. 2

3. 3 Special Variables: 1- Slack Variable (s1): represents the amount by which the available amount of the resources exceeds its usage by the activities.  To find the slack for a ≤ constraint algebraically, we add a slack variable to the constraint and convert it into an equality.  Ex: In Reddy Mikks model The constraint 6x 1 + 4x 2  24 associated with the usage of raw material M1 is equivalent to 6x 1 + 4x 2 + s 1 = 24 provided that s 1 ≥ 0 The slack variable s 1 =24 - 6x 1 - 4x 2 represents the unused amount of raw material M1

4. 4 2- Surplus Variable (S1) : represents the excess of the left hand side over the minimum requirement.  To find the surplus for a ≥ constraint, we subtract a surplus variable from the left-hand side to make it an equality.  Ex: In Diet Model The constraint representing minimum feed requirement, x 1 + x 2 ≥ 800 is equivalent to x 1 + x 2 - S 1 = 800 provided that S 1 ≥ 0 A positive value of S 1 signifies that a surplus amount of feed ( over the minium requirement of 800Ib) will be produced. 3- Unrestricted Variable : In both Reddy Mikks and Diet model, the nature of variables requires them to assume nonnegative values. But there are other situations where variable can assume any real value.

5. 5 Example: McBurger fast – food restaurant sells quarter-pounder and cheeseburger. A quarter-pounder uses a quarter of pound of meat, and cheeseburger uses only 0.2 Ib. The restaurant starts the day with 200 Ib of meat but may order more at an additional cost of 25 cents per pound to cover the delivery cost. Any surplus meat at the end of the day is donated at HotSoup Charity. McBurger’s profits are 20 cents from a quarter pounder and 15 cents for a cheeseburger. All in all, McBurger doesn’t expect to sell more than 900 sandwich in any one day. How many of each sandwich should McBurger make?

6. 6 Solution: Assume x 1 The daily number of quarter-pounders made x 2 The daily number of cheeseburger made The constraints: 0.25x1+ 0.2x 2  200 In first case (Slack) 0.25x 1 + 0.2x 2 ≥ 200 In second case (Surplus) We consider x 3 as unrestricted that allows the variable to play the roles of both a slack and a surplus as desired 0.25x 1 + 0.2x 2 + x 3 = 200

7. 7 The objective function: Maximize the total profit, less any additional cost that may be incurred as a result of ordering special delivery of additional pounds of meat The additional cost is incurred only if x 3 < 0 (role of surplus)  Dealing with an unrestricted variable is cumbersome. So we use standard substitution that converts the unrestricted variable into two nonnegative variables X 3 = x 3 + - x 3 - x 3 +, x 3 - ≥ 0 If x 3 + >0 and x 3 - =0 x 3 + plays a role of slack If x 3 - >0 and x 3 + =0 x 3 - plays a role of surplus The constraint can be expressed as: 0.25x 1 + 0.2x 2 + x 3 + - x 3 - = 200 The objective function is expressed as: Maximize z =0.2 x 1 + 0.15 x 2 - 0.25 x 3 -

8. 8 Sensitivty Analysis: Studying the effect of making changes in the model parameters on a given optimum LP solution. 1) Changes in the objective function coefficient The objective function (Two variable- maximize or minimize )can be expressed as: z = c 1 x 1 + c 2 x 2 The change in the coefficients c 1 and c 2 will change the slope of z which lead to changing the optimum solution to a different corner point of the solution space. So we need to determine the range of optimality for the ratio or that will keep the optimum solution unchanged. Corner point: A point that lies at the intersection of two (or possibly more) constraint lines on the boundary of the feasible region. No interior points in the feasible region need be considered because at least one corner point is better than any interior point

9. 9 Example: In Reddy Mikks problem.

10. 10 The objective function was Z = 5x 1 + 4x 2 If we change the objective function to Z = c 1 x 1 + c 2 x 2 The solution at C point will remain optimum. As the slope of z lies between the slopes of two line intersecting C--- namely 6x 1 + 4x 2  24 raw material M1 x 1 + 2x 2  6 raw material M2 We can express this relationship algebraically: or The line cannot to be horizontal The line cannot to be vertical

11. 11 If c 2 remain equal to 4, then Similarly, If we specify c 1 equal to 5, then

12. 12 2) Unit Worth of a Resource ( Changes in in the right hand – side of the constraints)  Quantifies the rate of change in the optimum value in the objective function as a result of making changes in the availability of the resource.

13. 13 Example: In Reddy Mikks model: The first constraints represent the limitations on the usage of raw material, M1 and M2, respectively. Determine the worth per unit for each resource. Solution:  The worth per unit for raw material M1

14. 14  Point C is the current optimum solution which is the intersection of lines associated with raw materials.  Changing In M1 (increase or decrease around current level of 24 tons), the optimum solution point C will slide along the line segment DG.  Any change in M1 outside the range of line segment DG will make the intersection point C infeasible.  End points D=(2.2) and G=(6.0) delineate the feasibility range for M1  The amount of M1 associated with point D is computed as 6 x 1 + 4 x 2 =20 tons.  Similarly, The amount of M1 associated with point G is computed as 6 x 1 + 4 x 2 =36 tons. Thus the range of feasibility for M1 is  Equivalently, if M1=24+ D1, (D1 is the change in the raw material above or below the current level f 24 tons, then 20  24+ D1  36 or -4  D1  12

15. 15  End points D=(2.2) and G=(6.0) delineate the feasibility range for M1  The amount of M1 associated with point D is computed as 6 x 1 + 4 x 2 =20 tons.  Similarly, The amount of M1 associated with point G is computed as 6 x 1 + 4 x 2 =36 tons. Thus the range of feasibility for M1 is  Let y 1 represent the worth per unit of raw material M1  Given D= (2.2) and G= (6.0) then, z at D = 5*2 +4*2 = 18 (thousand dollars) z at G = 5*6 +4*0 = 30 (thousand dollars)  As a result, the 1-ton change in M1 in range 20  M1  36 will change optimum value of z by 750$

16. 16  The worth per unit for raw material M2

17. 17  End points B=(4,0) and H=(8/3, 2) delineate the feasibility range for M1  The amount of M2 associated with point B is computed as x 1 + 2 x 2 =4 tons.  Similarly, The amount of M2 associated with point H is computed as x 1 + 2 x 2 =20/3 tons. Thus the range of feasibility for M1 is  Let y 2 represent the worth per unit of raw material M2 Given B= (4,0) and H= (8/3, 2) then, z at B = 5*4 +4*0 = 20(thousand dollars) z at H = 5*8/3 +4*2 = 64/3 (thousand dollars)  As a result, the 1-ton change in M1 in range 20  M1  36 will change optimum value of z by 500$