Maximize: 5 x - PowerPoint Presentation

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Maximize: 5 x1 + 4 x2 + 3 x3subject to: x1 , x2 , x3 ≥ 0 and2 x1 + 3 x2 + 1 x3 ≤ 54 x1 + 1 x2 + 2 x3 ≤ 113 x1 + 4 x2 + 2 x3 ≤ 8After one pivot: HNT= [4 2 3]X1 = 2.5 -1.5 X2 -0.5 X3 -0.5 X4 X5 = 1.0 +5.0 X2 +0.0 X3 +2.0 X4 X6 = 0.5 +0.5 X2 -0.5 X3 +1.5 X4 What are HBT, CBT, CHT, AB, AH, xB and xH for this dictionary?2. Solve for z: (a) Solve ABT y = cB for y. (b) z= yT b + [cNT - yT AN ] xN

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2Announcements:The deadline for grad lecture notes was extended to Friday Nov. 9 at 3:30pm.Assignment #4 is posted: Due Wed. Nov. 14.Nov. 12-14 is reading break.I plan on holding a Test #2 tutorial on Wed. Nov. 14 at 3:30pm, room TBA.Test #2 is on Thursday Nov. 15, in class.Mandatory attendance starts Mon. Nov. 19.A handout for the integer programming material is coming soon.Slide3

HBT =[1 5 6]AB =[2 0 0][4 1 0][3 0 1] cB = [5 0 0]TcN= [0 4 3]TxB= [x1 x5 x6]TxN= [x4 x2 x3]TCurrent solution:xB= [2.5 1 0.5]THNT =[4 2 3]AN =[1 3 1][0 1 2][0 4 2]Z= 12.53Slide4

2. Solve for z:(a) Solve ABT y = cB for y. ABT =[2 4 3] [y1] [5][0 1 0] [y2] = [0][0 0 1] [y3] [0]y= [2.5] [ 0 ] [ 0 ]4Slide5

(b) z= yT b + (cNT - yT AN ) xNz= + ([0 4 3]- z = 12.5 – 2.5 x4 -3.5 x2 + 0.5 x3 5Slide6

The Revised Simplex Method uses more work to determine the z row coefficients, the column that should enter, and the pivot row number (exiting variable) but then it takes less work to pivot. 6Slide7

The Revised Simplex Algorithm Step 1: Determine pivot column. Solve ABT y = cB for y.compute [cNT - yT AN] * xNto get coefficients of non-basic variables.Look for a positive coefficient, say r corresponding to non-basic xj.7Slide8

Step 2: Determine the leaving variable. Solve for entering column d in current dictionary: d= AB-1 a where a is the entering column taken from the initial problem.Or equivalently, solve for d: AB d = aIf tightest constraint corresponds to xleaving= v – t * xenteringthen the new value of xentering will be s= v/t.8Slide9

Step 3: Update variables (xj enters, xk leaves). Update basic variables headers HB by replacing k with j. Update HN by replacing j with k. Set xj = s in the new solution. Plug this value for xj into the other equations to update the values of the other basic variables. The leaving variable will be 0.Recall r = the coefficient of xj in the z row:Set z = z + r s.Update AB, AN, cB, cN, xB, xN to match basis headers.9Slide10

After the first iteration:HBT =[1 5 6]AB =[2 0 0][4 1 0][3 0 1] cB = [5 0 0]TcN= [0 4 3]TxB= [x1 x5 x6]TxN= [x4 x2 x3]TCurrent solution:[5/2 1 1/2]HNT =[4 2 3]AN =[1 3 1][0 1 2][0 4 2]Z= 25/210Slide11

Step 1: Determine pivot column. Solve ABT y = cB for y. [2 4 3] [y1] [5][0 1 0] [y2] = [0][0 0 1] [y3] [0]y1= 2.5, y2=0, y3=011Slide12

Compute [cNT - yT AN] xN to get coefficients of non-basic variables. Look for a positive coefficient, say r corresponding to non-basic xj. [0 4 3]-[ 2.5 0 0][1 3 1] [0 1 2] [0 4 2]= [ -2.5 -3.5 0.5]The third variable should enter. Looking at xN, this is x3: xN= [x4 x2 x3]Tr = 1/2 12Slide13

Step 2: Find pivot row. Solve AB d = a where a is column in A for entering variable xj. x3 is entering.[2 0 0] [d1] [1][4 1 0] [d2] = [2][3 0 1] [d3] [2]d = [0.5] [ 0 ] [0.5]13Slide14

d = [0.5] [ 0 ] [0.5]Find the tightest constraint: Current solution is:[5/2 1 1/2]Tx1 ≤ 5/2 - 0.5 x3 ⟹ x3 ≤ 5x5 ≤ 1 - 0 x3 NO CONSTRAINTx6 ≤ 1/2 - 0.5 x3 ⟹ x3 ≤ 1 (*) s=1So the third basis variable exits. HBT= [1 5 6], x6 exits.14Slide15

Pivot:The new basis value variables:x1 = 5/2 - 0.5 * 1 = 2x5 = 1 - 0 * 1 = 1x6 = 1/2 - 0.5 * 1 = 0  x3=1 z= 12.5 + r *s = 12.5 + 0.5*1= 1315Slide16

After the second iteration:HBT =[1 5 3]AB =[2 0 1][4 1 2][3 0 2] cB = [5 0 3]TcN= [0 4 0]TxB= [x1 x5 x3]TxN= [x4 x2 x6]TCurrent solution:[2 1 1]THNT =[4 2 6]AN =[1 3 0][0 1 0][0 4 1]Z= 1316Slide17

Iteration 3: Step 1: Determine pivot column. Solve ABT y = cB for y. [2 4 3] [y1] [5][0 1 0] [y2] = [0][1 2 2] [y3] [3]y1= 1, y2=0, y3=117Slide18

Compute [cNT - yT AN] xN to get coefficients of non-basic variables.y1= 1, y2=0, y3=1[0 4 0]- [1 0 1] = [ -1 -3 -1] All the coefficients are negative so the current solution is OPTIMAL 18Slide19

After 1 pivot:X1 = 2.5- 1.5 X2 - 0.5 X3 - 0.5 X4 X5 = 1.0+ 5.0 X2 + 0.0 X3 + 2.0 X4 X6 = 0.5+ 0.5 X2 - 0.5 X3 + 1.5 X4 ------------------------------------z = 12.5- 3.5 X2 + 0.5 X3 - 2.5 X4 X3 enters. X6 leaves. After 2 pivots:X1 = 2.0- 2.0 X2 - 2.0 X4 + 1.0 X6 X5 = 1.0+ 5.0 X2 + 2.0 X4 + 0.0 X6 X3 = 1.0+ 1.0 X2 + 3.0 X4 - 2.0 X6 -------------------------------------z = 13.0- 3.0 X2 - 1.0 X4 - 1.0 X6yi’s at each step: -(coeff. of xn+i) 19