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Simplex Algorithm: Solving Degenerate Linear Programming Problems - Prof. David Phillips, Study notes of Linux skills

William & Mary Linux skills

Prof. David Phillips

An example of the simplex algorithm applied to a degenerate linear programming problem. The concept of degeneracy in lps, its implications on the simplex algorithm, and how to handle degenerate solutions. The document also includes a lp problem and its solution.

Typology: Study notes

Pre 2010

Uploaded on 03/16/2009

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An example of the simplex algorithm

Original LP

maximize 3x1+x2+ 2x3(1)

subject to

x1+x2+ 3x3≤30 (2)

2x1+ 2x2+ 5x3≤24 (3)

4x1+x2+ 2x3≤36 (4)

x1, x2, x3≥0.(5)

Standard form.

z= 3x1+x2+ 2x3(6)

x4= 30 −x1−x2−3x3(7)

x5= 24 −2x1−2x2−5x3(8)

x6= 36 −4x1−x2−2x3.(9)

Pivot in x1. Remove x6from the basis.

z= 27 + x2

4+x3

2−3x6

4(10)

x1= 9 −x2

4−x3

2−x6

4(11)

x4= 21 −3x2

4−5x3

2+x6

4(12)

x5= 6 −3x2

2−4x3+x6

2.(13)

Pivor in x3. Remove x5.

z=111

4+x2

16 −x5

8−11x6

16 (14)

x1=33

4−x2

16 +x5

8−5x6

16 (15)

x3=3

2−3x2

8−x5

4+x6

8(16)

x4=69

4+3x2

16 +5x5

8−x6

16 .(17)

Pivot in x2. Remove x3.

z= 28 −x3

6−x5

6−2x6

3(18)

x1= 8 + x3

6+x5

6−x6

3(19)

x2= 4 −8x3

3−2x5

3+x6

3(20)

x4= 18 −x3

2+x5

2.(21)

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maximizeAn example of the simplex algorithm^ Original LP 3 x 1 + x 2 + 2 x 3 (1) subject to 2 xx 11 ++ 2 xx 22 ++ 35 xx 33 ≤≤ 3024 (2)(3) 4 x (^1) x 1 +, x (^2) , xx 32 + 2 x 3 ≤≥ (^360). (4)(5) z Standard form.= 3 x 1 + x 2 + 2 x 3 (6) x x x 456 === 302436 −−− 24 xxx 111 −−− 2 xxx 222 −−− 352 xxx (^333). (7)(8)(9)

z Pivot in= 27 x^1 +. Remove x^2 x^6 from the basis. x 1 = 9 − x^442 +−^ xx^2233 −−^3 x^4 x 4 66 (10)(11) x x 45 == 216 −− 33 x (^4) x 2 22 −− 54 x (^2) x 33 ++ xx (^4266). (12)(13)

z Pivor in= 111 x^3. Remove^ x^5. x 1 = 4334 +−^ x^16 x 1622 −+^ xx^8855 −−^1116516 xx^66 (14)(15) x x 34 == 69432 −+ 3316 x (^8) x 22 −+ 5 xx 8455 +− xx 16866. (16)(17)

z Pivot in= 28 x^2 −. Remove x^3 x^3. x 1 = 8 + x^663 −+^ xx^6655 −−^2 x^3 x 3 66 (18)(19) x x 24 == 184 −− 8 x (^3) x 2 33 −+ 2 x (^3) x 2 55 .+ x 36 (20)(21) 1

Iteration 1

s^ z 1 == 8^ − xx^11 −+^ xx^22 +^ x^3 (8)(9)

s 2 = − x 2 + x 3 (10)

variable and^ Note that one of the basic variables is 0. s 1 as the leaving variable.^ We choose^ x^1 as the entering

x^ z 1 = 8= 8^ − x 2 +^ x^3 −−^ ss^11 (11)(12)

s 2 = x 2 − x 3 (13)

increase the objective function value from^ Note again that one of the basic variables is 0. 0 to 8 though.^ The previous pivot did

Iteration 2

x^ z 1 = 8= 8^ − x 2 +^ x^3 −−^ ss^11 (14)(15)

s 2 = x 2 − x 3 (16)

These were our only choices.^ We now choose^ x^3 as the entering variable, and^ s^2 as the leaving variable.

x^ z 1 = 8= 8^ +−^ xx^22 −−^ ss^11 −^ s^2 (17)(18)

x 3 = x 2 − s 2 (19)

of^ Note that the objective function did not increase. degeneracy.^ This occurs because

CSCI 628 – Introduction to Linear ProgrammingCycling

max s.t. z = 3414 xx 11 −− (^15060) xx 22 +− 502511 xx 33 −+9 6 xx (^44) +x 5 = 0 (^12) x 1 − 90 x 2 − 501 x x 33 +3x 4 +x (^6) +x 7 == (^01) x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 ≥ 0

x x 56 == 00 −−^1412 xx 11 +60+90xx 22 ++ 255011 xx 33 −− 93 xx (^44) x z 7 == (^10) + 34 x 1 − 150 x 2 + 50 − 1 xx (^33) − 6 x 4 (2) x x 16 == 00 +240− 30 xx 22 +− 255043 xx 33 −+15 36 xx 44 −+2 4 xx (^55) x z 7 == (^10 30) x 2 + 50 − 7 xx (^33) − 33 x 4 − 3 x 5 (3) x x 12 == (^00) −− 500 2518 xx 33 +84+ 12 xx 44 +12+ 151 xx (^55) −− 3018 xx (^66) x z 7 == (^10) + − 252 xx (^33) − 18 x 4 −x 5 −x 6 (4) 1

x x 32 == 00 − 1602518 xx 11 +−^525 2401 xx (^44) −+ 1207512 xx 55 −+ 25 601 xx (^66) x z 7 == 10 +− 25814 xx 11 −+3^5252 xx 44 −+2^752 xx 55 +25− 3 xx 66 (5)

x x 34 == 00 1252 14 xx 11 − (^10500) − 40 xx 22 −− 5013 xx 55 +150+ 23 xx 66 x z 7 == 10 −^1252 12 xx 11 +10500− 120 xx 22 +50+xx 55 − (^150) −xx 66 (6)

x x 54 == (^00) −^5416 xx 11 −+30 210 xx (^22) +− 150 5011 xx (^33) −+3 13 xx (^66) x z 7 == (^10 74) x 1 − 330 x 2 − − 501 xx (^33) +2x 6 (7)

x x 56 == 00 −−^1412 xx 11 +60+90xx 22 ++ 255011 xx 33 −− 93 xx (^44) x z 7 == (^10) + 34 x 1 − 150 x 2 + 50 − 1 xx (^33) − 6 x 4 (8) 2