Operational Research Linear Programming With Simplex Method

1. Operational ResearchLinear Programming With Simplex Method Minggu 2 Part 1

2. Introduction • George Dantzig (1947) • This is an iterative procedure that leads to the optimal solution in a finite number of steps. Begin with a basic feasible solution and then moves from one basic solution to the next until an optimal basic feasible solution is found.

3. Definisi • Metode simplex adalahmetodeoptimasipemrograman linear dengancaraevaluasisederetantitik-titikekstrimsehingganilaiobjektifdarisuatutitikekstrimlebihbaikatausamadengannilaiobjektifsuatutitikekstrim yang dievaluasisebelumnya

4. Contohsoal • Suatuperusahaan skateboard akanmemproduksi 2 jenisprodukyaitu skateboard deluxe dan professional. Prosesproduksiterdiridari 2 tahapyaituprosesperakitandanprosespenyesuaian. Waktu yang tersediauntukprosesperakitanadl 50 jam sedangkanutkprosespenyesuaianadl 60 jam. Jumlah unit rakitanroda yang tersediahanya 1200 unit. Setiapproduk deluxe membutuhkan 1 unit rakitanroda, 2 menitprosesperakitan, dan 1 menitprosespenyesuaian. Setiapproduk professional membutuhkan 1 unit rakitanroda, 3 menitprosesperakitan, dan 4 menitprosespenyesuaian. Jikadijual, keuntungansetiapproduk deluxe dan professional berturut-turutadl $3 dan $4. Tentukankombinasiproduksi yang optimal!

5. Case x1 = number of deluxe product x2 = number of professional product Maximize Z = 3x1 + 4x2 (profit) subject to x1 + x2 1200 2x1 + 3x2 3000 x1 + 4x2 3600 with x1, x2 0

6. The Initial Simplex Tableau • Each constraint must be converted to an equation • In every equation there must be a variable that is basic in that equation. • The Right Hand Side (RHS) of every equation must be nonnegative constant.

7. Converting constraints into equations Maximize Z = 3x1 + 4x2 + 0S1+ 0S2 + 0S3 subject to x1 + x2 + S1 = 1200 2x1 + 3x2 + S2 = 3000 x1 + 4x2 + S3 = 3600 with x1, x2, S1, S2, S3 0

8. ..contd cj = objective function coefficient for variable j bi = right-hand-side value for constraint i aij = coefficient of variable j in constant i c row : a row of objective function coefficients b column : a column of RHS values of the constraint equations A matrix : a matrix with m rows and n columns of the coefficients of the variables in the constraint equations.

9. ..contd

11. In every equation there must be a variable that is basic in that equation If S1, S2, and S3 are basic, x1 and x2 must be nonbasic. Therefore, the constraints are simply (1)(0) + (1) (0) + (1) S1 + (0)S2 + (0)S3 = 1200 (2)(0) + (3) (0) + (0) S1 + (1)S2 + (0)S3 = 3000 (1)(0) + (4) (0) + (0) S1 + (0)S2 + (1)S3 = 3600 Or S1 = 1200 S2 = 3000 S3 = 3600

12. Choosing the pivot column • Rule : for maximization problem, the nonbasic variable with the largest cj-Zj value for all cj-Zj 0 is the pivot variable. • The variable that is nonbasic and becomes a basic variable is often called the entering variable.

13. Choosing the pivot row • If the jth the pivot column, compute all ratios bi/aij, where aij > 0. Select the variable basic in the row with the minimum ratio to leave the basis. The row with the minimum ratio is the pivot row

15. The pivot operation and the optimal solution • Suppose the jthis the pivot column and the kthis the pivot row, then the element akjis the pivot element. The pivot operation consists of m elementary row operations organized as follows: • Divide the pivot row (ak , bk) by the pivot element akj. Call the result (ak , bk). • For every other row (ai , bi), replace that row by (a , b) + (-aij)(ak , bk). In other words, multiply the revised pivot row by the negative of the aijth element and add it to the row under consideration.

16. …pivot operation • Step 1 : (1/4 4/4 0/4 0/4, 3600/4) • Step 2* (modify row 2 by multiplying the revised pivot row by -3 adding it to row 2) (-0.75 -3 0 0 -0.75, -2700) + ( 2 3 0 1 0 , 3000) ( 1.25 0 0 1 -0.75, 300)

17. Step 2* (modify row 1 by multiplying the revised pivot row by -1 adding it to row 1) (-0.25 -1 0 0 -0.25, -900) + ( 1 1 1 0 0 , 1200) ( 0.75 0 1 0 -0.25, 300)

19. ..where Z1 = (0)(0.75) + (0)(1.25) + (4)(0.25) = 1 Z2 = (0)(0) + (0)(0) + (4)(1) = 4 Z3 = (0)(1) + (0)(0) + (4)(0) = 0 Z4 = (0)(0) + (0)(1) + (4)(0) = 0 Z5 = (0)(-0.25)+ (0)(-0.75)+ (4)(0.25) = 1 and the objective value is (0)(300) + (0)(300) + 4(900) = 3600

23. Optimal Condition for Linear Programming • The optimal solution to a linear programming problem with a maximization objective has been found when cj– Zj 0 for all variable columns in the simplex tableau.

24. Kesimpulan • Solusi optimal didapatkandengannilai skateboard deluxe (X1)= 600; skateboard professional (X2)=600 dankeuntungan yang didapatkanadalah $4200

25. Review metode Simplex • Mengubahbentukbatasan model pertidaksamaanmenjadipersamaan. • Membentuktabelawaluntuksolusifisibeldasarpadatitik origin danmenghitungnilai-nilaibariszjdancj-zj • Menentukan pivot column dengancaramemilihkolom yang memilikinilaipositiftertinggipadabariscj-zj • Menentukanpivot row dengancaramembaginilai-nilaipadakolomsolusidengannilai-nilaipadapivot column danmemilihbarisdenganhasilbagi non negatifterkecil.

26. Review… • Menghitungnilai pivot row yang barumenggunakan formula: nilai pivot row tabel lama dibagidengan pivot elemen. • Menghitungnilai pivot yang lain menggunakan formula: nilaibaristabel lama – (koef.pivot column yang berhubungandikalidengannilai pivot row yang berhubungan) • Menghitungbaris-bariszjdancj-zj yang baru. • Lakukaniterasisampainilaicj-zjadalahnolataunegatif. Diperolehlahsolusi optimal.

27. Contohsoal: • Selesaikan model program linear berikutinimenggunakanmetode simplex! • Maksimumkan Z= 4x1+5x2 • Constrains: • x1+2x2 ≤10 • 6x1+6x2 ≤36 • x1 ≤4 • X1,x2 ≥0

28. Thank You…