After one pivot: HNT= [4 2 3] Maximize: 5 x1 + 4 x2 + 3 x3 subject to: x1 , x2 , x3 ≥ 0 and 2 x1 + 3 x2 + 1 x3 ≤ 5 4 x1 + 1 x2 + 2 x3 ≤ 11 3 x1 + 4 x2 + 2 x3 ≤ 8 After 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

Announcements: 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.

HBT = [1 5 6] AB = [2 0 0] [4 1 0] [3 0 1] cB = [5 0 0]T cN= [0 4 3]T xB= [x1 x5 x6]T xN= [x4 x2 x3]T Current solution: xB= [2.5 1 0.5]T HNT = [4 2 3] AN = [1 3 1] [0 1 2] [0 4 2] Z= 12.5

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

(b) z= yT b + (cNT - yT AN ) xN ([0 4 3]- 2.5 0 0 1 3 1 0 1 2 0 4 2 ) 𝑥4 𝑥2 𝑥3 z = 12.5 – 2.5 x4 -3.5 x2 + 0.5 x3

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.

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

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 = a If tightest constraint corresponds to xleaving= v – t * xentering then the new value of xentering will be s= v/t.

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.

After the first iteration: HBT = [1 5 6] AB = [2 0 0] [4 1 0] [3 0 1] cB = [5 0 0]T cN= [0 4 3]T xB= [x1 x5 x6]T xN= [x4 x2 x3]T Current solution: [5/2 1 1/2] HNT = [4 2 3] AN = [1 3 1] [0 1 2] [0 4 2] Z= 25/2

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=0

Compute [cNT - yT AN] xN to get coefficients of non-basic variables 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]T r = 1/2

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]

d = [0.5] [ 0 ] [0.5] Find the tightest constraint: Current solution is:[5/2 1 1/2]T x1 ≤ 5/2 - 0.5 x3 ⟹ x3 ≤ 5 x5 ≤ 1 - 0 x3 NO CONSTRAINT x6 ≤ 1/2 - 0.5 x3 ⟹ x3 ≤ 1 (*) s=1 So the third basis variable exits. HBT= [1 5 6], x6 exits.

Pivot: The new basis value variables: x1 = 5/2 - 0.5 * 1 = 2 x5 = 1 - 0 * 1 = 1 x6 = 1/2 - 0.5 * 1 = 0  x3=1 z= 12.5 + r *s = 12.5 + 0.5*1= 13

After the second iteration: HBT = [1 5 3] AB = [2 0 1] [4 1 2] [3 0 2] cB = [5 0 3]T cN= [0 4 0]T xB= [x1 x5 x3]T xN= [x4 x2 x6]T Current solution: [2 1 1]T HNT = [4 2 6] AN = [1 3 0] [0 1 0] [0 4 1] Z= 13

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=1

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 0 0 1 0 0 4 1 𝑥4 𝑥2 𝑥6 = [ -1 -3 -1] 𝑥4 𝑥2 𝑥6 All the coefficients are negative so the current solution is OPTIMAL

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 X6 yi’s at each step: -(coeff. of xn+i)