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1 Linear Programming We are to learn two topics today: 1.LP formulation 2.determining LP solutions Graphical approach Simplex method – will be taken up.

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Presentation on theme: "1 Linear Programming We are to learn two topics today: 1.LP formulation 2.determining LP solutions Graphical approach Simplex method – will be taken up."— Presentation transcript:

1 1 Linear Programming We are to learn two topics today: 1.LP formulation 2.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 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 p1) (to p30) (to p17) (to p3)

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

4 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 p2) (to p15) (to p8) (to p6) (to p5) (to p16)

5 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 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 7 Objective function maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4 (to p4)

8 8 Step 3: state all the resource constraints Four constraints 1.Not more than 20%for x1 2.X2 should not exceed total of other investments 3.At least 30% spent for x2 and x3 4.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 p4) (to p9) (to p11) (to p12) (to p13) (to p14) (to p10)

9 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 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 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 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 13 total investment is $70,000 That is, x1 + x2 + x3 + x4 = 70,000 (to p8)

14 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 15 Step 4: define non-negativity constraints That is, x1, x2, x3, x4  0 (to p4)

16 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 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! (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)

17 17 Example 2 (Marketing) Please read the handout (118)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 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 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 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 p18) (to p21)

21 21 Audience exposure rate (to p20)

22 22 Step 3: state all the resource constraints We have the following resources: 1.Budget limit $100,000 2.Television time for four commercials 3.Radio time for 10 commercials 4.Newspaper space for 7 ads 5.Resources for no more than 15 commercials and/or ads. (to p18) (to p) (to p24) (to p25) (to p26) (to p27)

23 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 24 Television time for four commercials That is x1  4 (max of 4 TV ads) (to p22)

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

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

27 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 28 Step 4: define non-negativity constraints That is, x1, x2, x3  0 (to p18)

29 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 30 Example 3 (Transportation) Please read the handout (121)121 Read it for 5 mins and then try to formulate it! How to start? –Again we following the four steps (to p31) (to p60)

31 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 p34) (to p33) (to p32) (to p37) (to p36)

32 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, x ij, where i is warehouse, j is store or X 1A = no of TV shipped from warehouse I to A store X 1B = no of TV shipped from warehouse 1 to B store X 1C = no of TV shipped from warehouse 1 to C store X 2A = no of TV shipped from warehouse 2 to A store X 2B = no of TV shipped from warehouse 2 to B store X 2C = no of TV shipped from warehouse 2 to C store X 3A = no of TV shipped from warehouse 3 to A store X 3B = no of TV shipped from warehouse 3to B store X 3C = no of TV shipped from warehouse 3 to C store OR X ij = no of TV shipped from warehouse i to j store, i=1,2,3;j=A,B,C (to p31)

33 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 34 Step 3: state all the resource constraints Our resources are: Warehouse supply of televisions sets:Retail store demand for television sets: 1- Cincinnati 300A - New York 150 2- Atlanta 200B - Dallas250 3- Pittsburgh 200C - Detroit200 total 700 total600 So, the resource constraints are: x 1A + x 1B + x 1  300 x 2A + x 2B + x 2C  200 x 3A + x 3B + x 3C  200 x 1A + x 2A + x 3A = 150 x 1B + x 2B + x 3B = 250 x 1C + x 2C + x 3C = 200 Note, how this Problem really looks Like? How to get these? (to p31) (to p35)

35 35 Transportation Problem Total supply 300 200 150250200 600 Total demand 700 Note: total supply is greater than total demand. Thus, we set demand constraint to = and supply to <= (to p34)

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

37 37 Overall LP formulation Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C Subject to x1A + x1B+ x 1C  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 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 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 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 41 Step 1: Consider 1x 1 + 2x 2  40 ……… (1) And teating it as 1x 1 + 2x 2 = 40 ……… (1) How to draw them ? (to p42)

42 42 1x 1 + 2x 2 = 40 Step2: Where is 1x 1 + 2x 2  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 (to p43)

43 43 Determine 1x 1 + 2x 2  40 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 This region Is known as feasible region! (to p44)

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

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

46 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 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 …. Where is the optimal solution? (to p48)

48 48 Optimal solution corner pointThe 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 For max problem if min then it is on the left hand side (to p49)

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

50 50 Checking! X 1 =0 X 2 -0 Z=0 Optimal solution More exercise! …….. (to p51)

51 51 Another example Consider Minimize Z = $6x 1 + 3x 2 subject to 2x 1 + 4x 2  16 …………. (e1) 4x 1 + 3x 2  24 …………...(e2) x 1  0 …………..(e3) x 2  0 …………..(e4) Feasible region for constraints e1 to e4 ……… (to p52)

52 52 Resource constraints Feasible region 2x1 + 4x2  16 4x1 + 3x2  24 And also x1  0, x2  0 Imposing inequalities Imposing function Z Note, Min Z is at the most left Hand side of the Z function (to p38)

53 53 Types of LP solutions Two types of LP solution: 1.The solution – the unique solution 2.Multiple solutions Question: Do all LP problems have a solution? (to p55) (to p54) (to p56)

54 54 The solution – unique solution only only oneIt is referred to ….there is only one optimal solution Z has satisfying all LP formulation …. i.e. only one (corner) point from the feasible region (to p53)

55 55 Multiple solutions more than one setIt refers to that we have more than one set of solutions satisfying to the LP formulation. That is, the optimal function Z lies onto one of resource constraints that forms the feasible region …… and that all points (x 1,x 2 ) on that region are optimal solutions to the LP problem For Max Z, all points between here are optimal solutions (to p53)

56 56 LP solutions LP solution does not exit if our LP problems are: 1.Infeasible problems 2.Unbound problems (to p58) (to p1) (to p57)

57 57 An Infeasible Problem Every possible solution violates at least one constraint: maximize Z = 5x 1 + 3x 2 subject to 4x 1 + 2x 2  8 x 1  4 x 2  6 x 1, x 2  0 Here, we cannot find a region that has satisfying all constraints, i.e. no feasible region! (to p56)

58 58 An Unbounded Problem Value of objective function increases indefinetly: maximize Z = 4x 1 + 2x 2 subject to x 1  4 x 2  2 x 1, x 2  0 Note: there is no bounded feasible region, thus it cannot find Max Z. But it has solution if Z is a Min problem! (Why?) (to p56)

59 59 (to p17) (to p3)

60 60 ( to p30)

61 61

62 62

63 63


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