1. Linear Programming (LP) Problems MAX (or MIN): c 1 X 1 + c 2 X 2 + … + c n X n Subject to:a 11 X 1 + a 12 X 2 + … + a 1 n X n <= b 1 : a k 1 X 1 + a k 2 X 2 + … + a kn X n >=b k : a m 1 X 1 + a m 2 X 2 + … + a mn X n = b m © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2. An Example LP Problem Blue Ridge Hot Tubs produces two types of hot tubs: Aqua-Spas & Hydro-Luxes. There are 200 pumps, 1566 hours of labor, and 2880 feet of tubing available. Aqua-SpaHydro-Lux Pumps11 Labor 9 hours6 hours Tubing12 feet16 feet Unit Profit$350$300 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3. 5 Steps In Formulating LP Models: X 1 =number of Aqua-Spas to produce X 2 =number of Hydro-Luxes to produce. MAX: 350X 1 + 300X 2 1X 1 + 1X 2 <= 200} pumps 9X 1 + 6X 2 <= 1566} labor 12X 1 + 16X 2 <= 2880} tubing X 1 >= 0 X 2 >= 0 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4. Solving LP Problems: An Intuitive Approach  Idea: Each Aqua-Spa (X 1 ) generates the highest unit profit ($350), so let’s make as many of them as possible!  How many would that be? –Let X 2 = 0  1st constraint:1X 1 <= 200  2nd constraint:9X 1 <=1566 or X 1 <=174  3rd constraint:12X 1 <= 2880 or X 1 <= 240  If X 2 =0, the maximum value of X 1 is 174 and the total profit is $350*174 + $300*0 = $60,900  This solution is feasible, but is it optimal?  No! © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

5. Solving LP Problems: A Graphical Approach  The constraints of an LP problem defines its feasible region.  The best point in the feasible region is the optimal solution to the problem.  For LP problems with 2 variables, it is easy to plot the feasible region and find the optimal solution. © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6. X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 200) (200, 0) boundary line of pump constraint X 1 + X 2 = 200 Plotting the First Constraint © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

7. X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 261) (174, 0) boundary line of labor constraint 9X 1 + 6X 2 = 1566 Plotting the Second Constraint © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8. X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 180) (240, 0) boundary line of tubing constraint 12X 1 + 16X 2 = 2880 Feasible Region Plotting the Third Constraint © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

9. X2X2 Plotting A Level Curve of the Objective Function X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 116.67) (100, 0) objective function 350X 1 + 300X 2 = 35000 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10. A Second Level Curve of the Objective Function X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 (0, 175) (150, 0) objective function 350X 1 + 300X 2 = 35000 objective function 350X 1 + 300X 2 = 52500 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

11. Using A Level Curve to Locate the Optimal Solution X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 objective function 350X 1 + 300X 2 = 35000 objective function 350X 1 + 300X 2 = 52500 optimal solution © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

12. Calculating the Optimal Solution  The optimal solution occurs where the “pumps” and “labor” constraints intersect.  This occurs where: X 1 + X 2 = 200 (1) and 9X 1 + 6X 2 = 1566(2)  From (1) we have, X 2 = 200 -X 1 (3)  Substituting (3) for X 2 in (2) we have, 9X 1 + 6 (200 -X 1 ) = 1566 which reduces to X 1 = 122  So the optimal solution is, X 1 =122, X 2 =200-X 1 =78 Total Profit = $350*122 + $300*78 = $66,100 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

13. Understanding How Things Change See file Fig2-8.xlsmFig2-8.xlsm © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

14. Special Conditions in LP Models  A number of anomalies can occur in LP problems: –Alternate Optimal Solutions –Redundant Constraints –Unbounded Solutions –Infeasibility © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

15. Example of Alternate Optimal Solutions X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 450X 1 + 300X 2 = 78300 objective function level curve alternate optimal solutions © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

16. Example of a Redundant Constraint X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 boundary line of tubing constraint Feasible Region boundary line of pump constraint X1+X2 <= 220 boundary line of labor constraint © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

17. Example of an Unbounded Solution X2X2 X1X1 1000 800 600 400 200 0 0 400 600 8001000 X 1 + X 2 = 400 X 1 + X 2 = 600 objective function X 1 + X 2 = 800 objective function -X 1 + 2X 2 = 400 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

18. Example of Infeasibility X2X2 X1X1 250 200 150 100 50 0 0 100 150 200250 X 1 + X 2 = 200 X 1 + X 2 = 150 feasible region for second constraint feasible region for first constraint © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.