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MATH2070 Week 3-4

PART I Linear programming problem setup

PART 1.1 Standard form of linear programming

Maximise  Z  = c1x1  + c2x2  +... + cnxn

Subject to  a11x1  + a12x2  +... + a1nxn  b1

a21x1  + a22x2  +... + a2nxn  b2

 

am1x1  + am2x2  +... + amnxn  bm

with      x1, x2,  ... , xn  ≥  0

Definitions

a) Objective function

b) Decision variables

c) Cost coefficients

d) Constraint matrix

e) Resource vector

f) Feasible solution

g) Feasible region

h) Optimal solution

Exercise 1: the following is the final simplex tableau for linear programming problems:

x       x

2        0

−  1   0

1        1

x

1

1

1

x

0

1

0

x

3

 1

1

RHS

6

2

2

a) What are the non-basic variables?

b) What is the optimal value of the objective functions?

c) Identify the solution vertex (x1, x2, x3 )

PART 1.2 The simplex algorithm

Algebraic representation of corner points

a) At the corner point, basic variables are non-zero and non-basic variables are zero.

b) Two corner points are adjacent if they differ in precisely one non-basic variable (or equivalently, in precisely one basic variable)

The simplex algorithm

Aim: move from one FCP to an adjacent FCP with largest potential increase in the objective function

Z

.

a) Initialization: start on a feasible corner point (FCP).

i.e by setting decision variables to zero and slack variables to resource constraints b) Iteration steps: move to an adjacent FCP with best potential increase of Z.

1)   Which non-basic variable enters the basis?

Largest coefficient rule: increasing the non-basic variable Xl to Xl  >  0 with the largest cost coefficient cl might yield the largest increase of Z

2)   Which basic variables leave the basis?

Choose X  such that    bl     is minimal

c) Stopping rule: stop at FCP if its Z is greater than the Z-value of all its adjacent FCPs. Stop when all

modified cost coefficient c  ≥  0.

Exercise 2: Use the simplex algorithm to solve the system by hand

Max Z  = −  3x1  +  2x2  +  3x3

Subject to   2x1  + x2  + x3  ≤  12

3x1  x2  +  3x3   8

with x1, x2, x3  ≥  0

Exercise 3:

John Smith instals and demonstrates burglar alarms. There are two burglar alarms that he instals, type A and B. Type A requires 1 hour to install and  hour to demonstrate. Type B requires 2 hours to      install and   hour to demonstrate. Union rules require Smith to work a minimum of 20 hours per      week as an installer and a maximum of 20 hours per week as a demonstrator. If he gets paid $13 per

hour for installing and $10 per hour as a demonstrator, how many alarms of each type should Smith install and demonstrate each week to maximise his earnings?

Exercise 4: The following is the final simplex tableau for a linear programming problem:

 

a) What are the basic and non-basic variables?

b) What is the optimal value of the objective function?

c) What is the optimal solution(s)?

Exercise 5: The following is the final simplex tableau for a linear programming problem:

 

Exercise 6: Use the Simplex method to solve the system by hand:

Max  Z  =  2x1  +  2x2

Subject to  x1  +  2x2  ≤  11

2x1  + x2  ≤  10

x1  + x2  ≤  6

with x1, x2  ≥  0

PART 1.3 Potential problems of the simplex algorithm

Tie-breaking rules

a) Largest coefficient rule ambiguity

Basis

Z

X

1

X

2

X

3

X

4

X

5

b

Z

1

0

-3

-3

0

0

20

Solution: choosing either will work.

b) Smallest non-negative ration ambiguity

Solution: choosing either will work.

Smallest index rule: to break a tie, always choose the variable with smallest index 

No variable leaving the basis

Degeneracy

The equations in a tableau do not permit any increment of the selected non-basic variables, and it may be impossible to increase the objective function in a single pivot step.

Multiple optimal solutions

If contours of the objective function are parallel to a boundary, there may be a family of solutions.

PART 1.4 Efficiency of the simplex algorithm

Klee-Minty cube

The set of feasible solutions is a deformed n-dimensional cube, constructed such that the   simplex algorithm passes through all its vertices: 2n  −  1 iterations are needed with a fixed pivoting rule (worst case).


PART II Non-standard Linear Programming

PART 2.1 Minimising the objective function

To minimise Z  =    cixi, define Z = − Z, then minimising Z is equivalent to maximising Z :

i=1

min Z = max Z

PART 2.2 Negative resource elements

Suppose that for some constraint a1x1  + a2x2  + a3x3  +... + anxn  ≤  b  <  0, this is equivalent to −  a1x1  a2x2  a3x3  − ... − anxn  ≥  b  >  0

PART 2.3 Greater-than or equal-to constraints

For a constraint a1x1  + a2x2  + a3x3  +... + anxn  ≥  b  ≥  0, introduce a surplus variable xn+1 such that

a1x1  + a2x2  + a3x3  +... + anxn  xn+1  =  b

PART 2.4 Negative decision variables

If x   ≤  0, introduce a variable x   = − x  ,  then x   ≥  0.

PART 2.5 Unrestricted decision variables

If x   is unrestricted in sign (x   ∈  R), introduce two new variables  x   ≥  0 and  x   ≥  0, then

k            k                k

PART 2.6 Finding an initial FCP

a) Artificial variables: added into each constraint without a slack variable.

b) Decision variables and surplus variables: non-basic.

Slack and artificial variables: basic.

c) A feasible solution of the new LP problem is a feasible solution of the original LP problem if and only if the artificial variables are zero.


PART III Two-phase Simplex Algorithm

A hospital wants to design a dinner menu containing two items M and N. Each gm of M provides 1   unit of vitamin A and 2 units of vitamin B. Each gm of N provides 1 unit of vitamin A and 1 unit of   vitamin B. The two dishes must provide at least 7 units of vitamin A and at least 10 units of vitamin  B. If each gm of M costs 8 cents and each gm of N costs 12 cents, how many gm of each item should the hospital serve to minimise its cost?