1. 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 = X X X4
X5 = X X X4
X6 = X X 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

2. 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 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.

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

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

5. (b) z= yT b + (cNT - yT AN ) xN
([0 4 3] ) 𝑥4 𝑥2 𝑥3
z = – 2.5 x x x3

6. 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.

7. 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.

8. 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.

9. 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.

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

11. Step 1: Determine pivot column.
Solve ABT y = cB for y.
[ ] [y1] [5]
[ ] [y2] = [0]
[ ] [y3] [0]
y1= 2.5, y2=0, y3=0

12. 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]-[ ][ ]
[ ]
[ ]
= [ ]
The third variable should enter. Looking at xN, this is x3:
xN= [x4 x2 x3]T
r = 1/2

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

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

15. Pivot:
The new basis value variables:
x1 = 5/ * 1 = 2
x5 = * 1 = 1
x6 = 1/ * 1 = 0  x3=1
z= r *s = *1= 13

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

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

18. Compute [cNT - yT AN] xN to get coefficients of non-basic variables.
y1= 1, y2=0, y3=1
[0 4 0]- [1 0 1] 𝑥4 𝑥2 𝑥6
= [ ] 𝑥4 𝑥2 𝑥6
All the coefficients are negative so the current solution is OPTIMAL

19. After 1 pivot:
X1 = X X X4
X5 = X X X4
X6 = X X X4
z = X X X4
X3 enters. X6 leaves.
After 2 pivots:
X1 = X X X6
X5 = X X X6
X3 = X X X6
z = X X X6
yi’s at each step: -(coeff. of xn+i)