# LP EXAMPLES: ANOTHER MAX AND A MIN Dr. Ron Lembke.

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LP EXAMPLES: ANOTHER MAX AND A MIN Dr. Ron Lembke

Example 2 mp3- 4 min electronics - 2 min assembly DVD- 3 min electronics - 1 min assembly Min available: 240 (elect) 100 (assy) Profit / unit: mp3 \$7, DVD \$5 X 1 = number of mp3 players to make X 2 = number of DVD players to make

Standard Form Max7x 1 +5x 2 s.t. 4x 1 +3x 2 <=240 2x 1 +1x 2 <=100 x 1 >=0 x 2 >=0 electronics assembly

Graphical Solution 020406080 80 20 40 60 0 100 DVD players mp3 X2X2 X1X1

Graphical Solution 020406080 80 20 40 60 0 100 DVD players mp3 X 1 = 0, X 2 = 80 X 1 = 60, X 2 = 0 Electronics Constraint X2X2 X1X1 4x 1 + 3x 2 <=240 x 1 =0, x 2 =80 x 2 =0, x 1 =60

Graphical Solution 020406080 80 20 40 60 0 100 DVD players mp3 X 1 = 0, X 2 = 100 X 1 = 50, X 2 = 0 Assembly Constraint X2X2 X1X1 2x 1 + 1x 2 <=100 x 1 =0, x 2 =100 x 2 =0, x 1 =50

Graphical Solution 020406080 80 20 40 60 0 100 DVD players mp3 Assembly Constraint Electronics Constraint Feasible Region – Satisfies all constraints X2X2 X1X1

020406080 80 20 40 60 0 100 DVD players mp3 Isoprofit Line: \$7X 1 + \$5X 2 = \$210 (0, 42) (30,0) Isoprofit Lnes X2X2 X1X1

Isoprofit Lines 020406080 80 20 40 60 0 100 DVD players mp3 \$210 \$280 X2X2 X1X1

Isoprofit Lines 020406080 80 20 40 60 0 100 DVD players mp3 \$210 \$280 \$350 X2X2 X1X1

Isoprofit Lines 020406080 80 20 40 60 0 100 DVD players mp3 (0, 82) (58.6, 0) \$7X 1 + \$5X 2 = \$410 X2X2 X1X1

Mathematical Solution Obviously, graphical solution is slow We can prove that an optimal solution always exists at the intersection of constraints. Why not just go directly to the places where the constraints intersect?

Constraint Intersections 020406080 80 20 40 60 0 100 DVD players mp3 X 1 = 0 and 4X 1 + 3X 2 <= 240 So X 2 = 80 X2X2 X1X1 4X 1 + 3X 2 <= 240 (0, 0) (0, 80)

Constraint Intersections 020406080 80 20 40 60 0 100 DVD players mp3 X 2 = 0 and 2X 1 + 1X 2 <= 100 So X 1 = 50 X2X2 X1X1 (0, 0) (0, 80) (50, 0)

Constraint Intersections 020406080 80 20 40 60 0 100 DVD players mp3 4X 1 + 3X 2 = 240 2X 1 + 1X 2 = 100 – multiply by -2 X2X2 X1X1 (0, 0) (0, 80) (50, 0) 4X 1 + 3X 2 = 240 -4X 1 -2X 2 = -200 add rows together 0X 1 + 1X 2 = 40 X 2 = 40 substitute into #2 2 X 1 + 40 = 100 So X 1 = 30

Constraint Intersections 020406080 80 20 40 60 0 100 DVD players mp3 X2X2 X1X1 (0, 0) \$0 (0, 80) \$400 (50, 0) \$350 (30,40) \$410 Find profits of each point. Substitute into \$7X 1 + \$5X 2

Do we have to do this? Obviously, this is not much fun: slow and tedious Yes, you have to know how to do this to solve a two-variable problem. We won’t solve every problem this way.

Constraint Intersections 020406080 80 20 40 60 0 100 DVD players mp3 X2X2 X1X1 Start at (0,0), or some other easy feasible point. 1.Find a profitable direction to go along an edge 2.Go until you hit a corner, find profits of point. 3.If new is better, repeat, otherwise, stop. Good news: Excel can do this for us. Using the Simplex Algorithm

Minimization Example Min8x 1 +12x 2 s.t. 5x 1 +2x 2 ≥20 4x 1 +3x 2 ≥ 24 x 2 ≥ 2 x 1, x 2 ≥ 0

Minimization Example Min8x 1 +12x 2 s.t. 5x 1 +2x 2 ≥20 4x 1 +3x 2 ≥ 24 x 2 ≥ 2 x 1, x 2 ≥ 0 5x 1 + 2x 2 =20 If x 1 =0, 2x 2 =20, x 2 =10 (0,10) If x 2 =0, 5x 1 =20, x 1 =4 (4,0) 4x 1 + 3x 2 =24 If x 1 =0, 3x 2 =24, x 2 =8 (0,8) If x 2 =0, 4x 1 =24, x 1 =6 (6,0) x 2 = 2 If x 1 =0, x 2 =2 No matter what x 1 is, x 2 =2

Graphical Solution 0246802468 8 2 4 6 0 10 5x 1 +2x 2 =20 X2X2 X1X1 4x 1 +3x 2 =24 x 2 =2

0246802468 8 2 4 6 0 10 5x 1 +2x 2 =20 X2X2 X1X1 4x 1 +3x 2 =24 x 2 =2 (0,10) [5x 1 +2x 2 =20]*3 [4x 1 +3x 2 =24]*2 15x 1 +6x 2 = 60 8x 1 +6x 2 = 48 - 7x 1 = 12 x 1 = 12/7= 1.71 5x 1 +2x 2 =20 5*1.71 + 2x 2 =20 2x 2 = 11.45 x 2 = 5.725 (1.71,5.73)

0246802468 8 2 4 6 0 10 5x 1 +2x 2 =20 X2X2 X1X1 4x 1 +3x 2 =24 x 2 =2 (0,10) (1.71,5.73) 4x 1 +3x 2 =24 x 2 =2 4x 1 +3*2 =24 4x 1 =18 x 1 =18/4 = 4.5 (4.5,2)

0246802468 8 2 4 6 0 10 5x 1 +2x 2 =20 X2X2 X1X1 4x 1 +3x 2 =24 x 2 =2 (0,10) (1.71,5.73) Z=8x 1 +12x 2 8*0 + 12*10 = 120 (4.5,2) Z=8x 1 +12x 2 8*1.71 + 12*5.73 = 82.44 Z=8x 1 +12x 2 8*4.5+ 12*2 = 60 Lowest Cost

IsoCost Lines 0246810 12 8 2 4 6 0 10 5x 1 +2x 2 =20 X2X2 X1X1 4x 1 +3x 2 =24 x 2 =2 Z=8x 1 +12x 2 Try 8*12 = 96 x 1 =0 12x 2 =96, x 2 =8 x 2 =0 8x 1 =96, x 1 =12

Summary Method for solving a two-variable problem graphically 1. Find end points of each constraint 2. Draw constraints 3. Figure out which intersections are interesting 4. Use algebra to solve for intersection pts 5. Find profits (or costs) of intersections 6. Choose the best one Iso-profit (or Iso-Cost) lines can help find the most interesting points

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