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Linear Programming

Linear Programming. We are to learn two topics today: LP formulation determining LP solutions Graphical approach Simplex method – will be taken up in next lecture Tutorial Questions: Ch 2, 3,6,9,12,16,20. (to p2). (to p38). (to p61). LP formulation.

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Linear Programming

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  1. Linear Programming We are to learn two topics today: • LP formulation • determining LP solutions • Graphical approach • Simplex method – will be taken up in next lecture Tutorial Questions: Ch 2, 3,6,9,12,16,20 (to p2) (to p38) (to p61)

  2. LP formulation • Note that Chapter 4 has presented 8 LP formulations • Here, we pick up the following three examples for discussions • Example 1 (Investment) • Example 2 (Marketing) • Example 3 (Transportation) (to p3) (to p17) (to p30) (to p1)

  3. Example 1 (Investment) • Please refer to the handouts(p112) • Spend 5 mins to read through them • Question asked was to formulate its LP problem! • Solution: (to p59) (to p4)

  4. How to achieve it? • From lecture 1, we know LP formulation involve the following steps: • Step 1: define decision variables • Step 2: define the objective function • Step 3: state all the resource constraints • Step 4: define non-negativity constraints • Overall LP formulation (to p5) (to p6) (to p8) (to p15) (to p16) (to p2)

  5. Step 1: define decision variables Question: how much to invest in four alternative choices: Let, x1 = amount invested in municipal bonds ($) x2 = amount invested in certificates of deposit ($) x3 = amount invested in treasury bills ($) x4 = amount invested in growth stock fund($) (to p4)

  6. Step 2: define the objective function Expected profits are: 8.5% for x1 = 0.085x1, 5% for x2 = 0.05x2 6.5% for x3 = 0.065x3 13% for x4 = 0.13x4 Thus, objective function is (to p7)

  7. Objective function maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4 (to p4)

  8. Step 3: state all the resource constraints • Four constraints • Not more than 20%for x1 • X2 should not exceed total of other investments • At least 30% spent for x2 and x3 • X2 and x3 should be greater than x1 and x4, by ratio of at least 1.2 to 1 Note: total investment is $70,000 All constraints are: (to p9) (to p10) (to p11) (to p12) (to p13) (to p14) (to p4)

  9. Not more than 20%for x1 • Given that total investment is $70,000 • 20% of it = $14,000 • Thus this constraint could be expressed as x1  14,000 (to p8)

  10. X2 should not exceed total of other investments • Total 4 investments, x1, x2, x3 and x4 • Thus, other investments would be x1, x3 and x4 • This constraint is then expressed as: x2  x1 + x3 + x4 Or x2 - x1 - x3- x4  0 (to p8)

  11. At least 30% spent for x2 and x3 • That is, • 30% of $70,000 = $21,000 Or x2 + x3  21,000 (to p8)

  12. X2 and x3 should be greater than x1 and x4, by ratio of at least 1.2 to 1 That is x2 + x3  1.2(x1 + x4) or -1.2x1 + x2 + x3 - 1.2 x4  0 (to p8)

  13. total investment is $70,000 That is, x1 + x2 + x3 + x4 = 70,000 (to p8)

  14. Over constraints x1  14,000 x2 - x1 - x3- x4  0 x2 + x3  21,000 -1.2x1 + x2 + x3 - 1.2 x4  0 x1 + x2 + x3 + x4 = 70,000 (to p8)

  15. Step 4: define non-negativity constraints That is, x1, x2, x3, x4  0 (to p4)

  16. Overall LP formulation maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4 subject to x1  14,000 x2 - x1 - x3- x4  0 x2 + x3  21,000 -1.2x1 + x2 + x3 - 1.2 x4  0 x1 + x2 + x3 + x4 = 70,000 x1, x2, x3, x4  0 (at least 20% on x1) (x2 spent not more than total investment) (at least 20% on x2 and x3) (x2+x3 has at least 1.2 ratio to x1+x4 (total investment is $70,000) (to p2) Note: it is up to you if you like to arrange in this format or let it be the way in according to how constraints were stated!

  17. Example 2 (Marketing) • Please read the handout (118) • Read it for 5 mins and then try to formulate it! • How to start? • Again we following the four steps (to p59) (to p18)

  18. 4 procedural steps Step 1: define decision variables Step 2: define the objective function Step 3: state all the resource constraints Step 4: define non-negativity constraints Overall LP formulation (to p19) (to p20) (to p22) (to p28) (to p29)

  19. Step 1: define decision variables Question: how to achieve a max exposure to three of possible media Let x1 = number of television commercials x2 = number of radio commercials x3 = number of newspaper aids (to p18)

  20. Step 2: define the objective function • Exposure are: (refer to the attached table) 20,000 for x1 12,000 for x2 9,000 for x3 Thus the objective function is maximize Z = 20,000x1 + 12,000x2 + 9,000x3 (to p21) (to p18)

  21. Audience exposure rate (to p20)

  22. Step 3: state all the resource constraints • We have the following resources: • Budget limit $100,000 • Television time for four commercials • Radio time for 10 commercials • Newspaper space for 7 ads • Resources for no more than 15 commercials and/or ads. (to p) (to p24) (to p25) (to p26) (to p27) (to p18)

  23. Budget limit $100,000 • All associated cost to each variables has a budget of $100,000 • Thus, $15,000x1 + 6,000x 2+ 4,000x3  100,000 (to p22)

  24. Television time for four commercials • That is x1  4 (max of 4 TV ads) (to p22)

  25. Radio time for 10 commercials • That is, x2  10 (max of 10 radio ads) (to p22)

  26. Newspaper space for 7 ads • That is, x3  7 (max of 7 newspaper ads) (to p22)

  27. Resources for no more than 15 commercials and/or ads • That is x1 + x2 + x3  15 (max of 15 ads can be subscribed) (to p22)

  28. Step 4: define non-negativity constraints • That is, x1, x2, x3  0 (to p18)

  29. Overall LP formulation maximize Z = 20,000x1 + 12,000x2 + 9,000x3 subject to $15,000x1 + 6,000x 2+ 4,000x3  100,000 x1  4 x2  10 x3  7 x1 + x2 + x3  15 x1, x2, x3  0 (note it is advisable to include explanation here!) (to p2)

  30. Example 3 (Transportation) • Please read the handout (121) • Read it for 5 mins and then try to formulate it! • How to start? • Again we following the four steps (to p60) (to p31)

  31. 4 procedural steps Step 1: define decision variables Step 2: define the objective function Step 3: state all the resource constraints Step 4: define non-negativity constraints Overall LP formulation (to p32) (to p33) (to p34) (to p36) (to p37)

  32. Step 1: define decision variables • We are to determine all combination of shipping from warehouse i to store j, i=1,2,3;j=A,B,C • That is, xij, where i is warehouse, j is store or X1A= no of TV shipped from warehouse I to A store X1B = no of TV shipped from warehouse 1 to B store X1C = no of TV shipped from warehouse 1 to C store X2A = no of TV shipped from warehouse 2 to A store X2B = no of TV shipped from warehouse 2 to B store X2C = no of TV shipped from warehouse 2 to C store X3A = no of TV shipped from warehouse 3 to A store X3B = no of TV shipped from warehouse 3to B store X3C = no of TV shipped from warehouse 3 to C store OR Xij = no of TV shipped from warehouse i to j store, i=1,2,3;j=A,B,C (to p31)

  33. Step 2: define the objective function • Here, we attempt to min the shipping cost, i.e. Cost Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C (to p31)

  34. Step 3: state all the resource constraints • Our resources are: Warehouse supply of televisions sets: Retail store demand for television sets: 1- Cincinnati 300 A - New York 150 2- Atlanta 200 B - Dallas 250 3- Pittsburgh 200 C - Detroit 200 total 700 total 600 So, the resource constraints are: x1A + x1B+ x1  300 x2A+ x2B + x2C 200 x3A+ x3B + x3C 200 x1A + x2A + x3A = 150 x1B + x2B + x3B = 250 x1C + x2C + x3C= 200 How to get these? Note, how this Problem really looks Like? (to p31) (to p35)

  35. Transportation Problem Total supply 300 200 200 700 150 250 200 Total demand 600 Note: total supply is greater than total demand. Thus, we set demand constraint to = and supply to <= (to p34)

  36. Step 4: define non-negativity constraints That is, xij  0, i=1,2,3; j=A,B,C (to p31)

  37. Overall LP formulation Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C Subject to x1A + x1B+ x1C  300 x2A+ x2B + x2C  200 x3A+ x3B + x3C  200 x1A + x2A + x3A = 150 x1B + x2B + x3B = 250 x1C + x2C + x3C = 200 xij  0, i=1,2,3; j=A,B,C (to p2)

  38. LP - Graphical approach • Here, we now try to solve an LP formulation using a graphical approach, that is • Plot all equations onto a graph and determine Z values of decision variables that satisfying all resource constraints • How it works? • Types of LP solutions (to p39) (to p53)

  39. LP - Graphical approach Consider a simple LP problem: maximize Z=$40x1 + 50x2 subject to 1x1 + 2x2  40 ………(e1) 4x2 + 3x2  120 ……..(e2) x1  0 ……...(e3) x2  0 ………(e4) Objective function Resource Constraints Solution steps of graphical approach (to p40)

  40. Solution steps of graphical approach • Step 1: take one resource constraint and treat it as an equation and draw them onto a graph • Step 2: determine the region that satisfying the inequality equation of step 1 • Step 3: repeat steps 1-2 for all resource constraints • Step 4: super imposing objective function to determine the solution set for decision variables (to p41)

  41. Step 1: • Consider 1x1 + 2x2  40 ……… (1) And teating it as 1x1 + 2x2 = 40 ……… (1) How to draw them ? (to p42)

  42. 1x1 + 2x2= 40 How to plot? Consider any points on x and y axis in turn, then we have : x1=0, x2=20 and x2=0, x1=40 Step2: Where is 1x1 + 2x2 40 ? (to p43)

  43. Determine 1x1 + 2x2  40 This region Is known as feasible region! We take a point at (0,0) to check which side satisfying this constraint! (0,0) At (0,0), 1*0 + 0*0  40 ? Yes, then all points on the side of (0,0) are feasible region Repeat these steps for equation e2 (to p44)

  44. Constraint e2 (to p45) When combine this to the one we obtained before, then we have ….

  45. Constraints e1 to e2 Interception point (x1,x2)=(24,8) (0,20) (30,0) (to p46) When add e3 and e4, we have ……

  46. Common region of e1 to e4 Now, we proceed to Step 4, that is impose Objective function Z onto this graph ………. How to plot it? (to p47)

  47. maximize Z=$40x1 + 50x2 To plot this equation Z, we can simply assign an aribitrary number for Z, Say, Z = 800 Then 800 =$40x1 + 50x2 When x2 = 0, then x1= 20 ….. A plotting point (20,0) When x1 = 0, then x2 = 16 ….a plotting point (0,16) When plotting it on the graph, we would have …. (to p48) Where is the optimal solution?

  48. Optimal solution • The optimal solution is at the corner point of the feasible region! • To obtain the optimal point, we use a ruler move toward the right hand side and the last corner point • Solution (to p49) For max problem if min then it is on the left hand side

  49. Optimal point This is the solution (x1,x2)=(24,8) (to p50) Let check to see if it is the optimal point/solution ….

  50. Checking! Optimal solution X1=0 X2-0 Z=0 (to p51) More exercise! ……..

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