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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 in next lecture Tutorial Questions: Ch 2, 3,6,9,12,16,20 (to p2) (to p38) (to p61)

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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)

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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)

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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)

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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)

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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)

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

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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)

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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)

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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)

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

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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)

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

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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)

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

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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)

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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)

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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)

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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)

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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)

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21 Audience exposure rate (to p20)

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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)

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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)

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

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

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

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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)

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

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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

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

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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)

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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)

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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)

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

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

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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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)

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