Linear Programming (LP)

[8] Case Study - Personal Scheduling

UNION AIRWAYS needs to hire additional customer service agents.

Management recognizes the need for cost control while also consistently providing a satisfactory level of service to customers.

Based on the new schedule of flights, an analysis has been made of the minimum number of customer service agents that need to be on duty at different times of the day to provide a satisfactory level of service.

** ** ** * * * * * * * * * * * *

ShiftTime Period Covered Minimum #

of Agents needed

Time Period

6:00 am to 8:00 am8:00 am to10:00 am10:00 am to noon Noon to 2:00 pm2:00 pm to 4:00 pm4:00 pm to 6:00 pm6:00 pm to 8:00 pm8:00 pm to 10:00 pm10:00 pm to midnightMidnight to 6:00 am

1 2 3 4 548796587647382435215

170 160 175 180 195Daily cost per agent

The problem is to determine how many agents should be assigned to the respective shifts each day to minimize the total personnel cost for agents, while meeting (or surpassing) the service requirements.

Activities correspond to shifts, where the level of each activity is the number of agents assigned to that shift.

This problem involves finding the best mix of shift sizes.

1x2x3x

4x5x

: # of agents for shift 1 (6AM - 2PM)

: # of agents for shift 2 (8AM - 4PM)

: # of agents for shift 3 (Noon - 8PM)

: # of agents for shift 4 (4PM - Midnight)

: # of agents for shift 5 (10PM - 6AM)

The objective is to minimize the total cost of the agents assigned to the five shifts.

Min

s.t.54321 195180175160170 xxxxx

0ix )5~1( iall 15

52

43

82

73

64

87

65

79

48

5

54

4

43

43

32

321

21

21

1

x

xx

x

xx

xx

xx

xxx

xx

xx

x

15

52

43

82

64

87

79

48

5

54

4

43

32

321

21

1

x

xx

x

xx

xx

xxx

xx

x

)15,43,39,31,48(),,,,( 54321 xxxxx

Total Personal Cost = $30,610

The Simplex Method

,x5x3Z 21

1x

0,0

1823

21

21

xx

xx

122 2 x4

and

Maximize

Subject to

From a geometric viewpoint

: CPF solutions (Corner-Point Feasible): Corner-point infeasible solutions

1823 21 xx

1x0 2 4 6 8 10

2x

2

4

6

8

Feasibleregion

122 2 x

0x2

01 x

)6,4(

4x1

Optimality test:

There is at least one optimal solution.

If a CPF solution has no adjacent CPF solutions

that are better (as measured by Z) than itself,

then it must be an optimal solution.

Initialization

Iteration

Optimal Solution?

No

YesStop

30Z

1x

2x)6,2(

)3,4(

)0,4()0,0(

)6,0( 36Z

27Z

12Z

0Z

Feasibleregion

1 2

0

The Key Solution Concepts

Solution concept 1:

The simplex method focuses solely on CPF

solutions.

For any problem with at least one optimal

solution, finding one requires only finding a best

CPF solution.

Solution concept 2:

The simplex method is an iterative algorithm ( a systematic solution procedure that keeps repeating a fixed series of steps, called an iteration).

Solution concept 3:

The initialization of the simplex method

chooses the origin to be the initial CPF

solution.

Solution concept 4:

Given a CPF solution, it is much quicker

computationally to gather information about its

adjacent CPF solutions than other CPF

solutions.

Therefore, each time the simplex method

performs an iteration to move from the current

CPF solution to a better one, it always chooses a

CPF solution that is adjacent to the current one.

Solution concept 5:

After the current CPF solution is identified, the simplex method identifies the rate of improvement in Z that would be obtained by moving along edge.

Solution concept 6:

The optimality test consists simply of checking whether any of the edges give a positive rate of improvement in Z. If no improvement is identified, then the current CPF solution is optimal.

Simplex Method

To convert the functional inequality

constraints to equivalent equality constraints,

we need to incorporate slack variables.

,x5x3Z 21

1x

1823 21 xx

122 2 x

4

0,0 21 xx

,x5x3Z 21

1x

21 23 xx

12

4

and

Maxs.t.

,0jx

3x

18

4x

5x22x

.5,4,3,2,1jfor

Original Form of Model

Augmented Form of the Model

Slack variables

and

Maxs.t.

A basic solution is an augmented corner-point solution.

A Basic Feasible (BF) solution is an augmented CPF solution.

Properties of BF Solution1. Each variable is designated as either a nonbasic

variable or a basic variable.

2. # of nonbasic variables = # of functional

constraints.

3. The nonbasic variables are set equal to zero.

4. The values of the basic variables are obtained

from the simultaneous equations.

5. If the basic variables satisfy the

nonnegativity constraints, the basic solution

is a BF solution.

Simplex in Tabular Form

1x 2x 3x 4x 5xZ

3x4x5x

(0)(1)(2)(3)

21 53 xxZ 1x 3x

4x21 23 xx 5x 18

12

4

0

(0)

(1)(2)(3)

(a) Algebraic Form

(b) Tabular FormCoefficient of: Right

SideBasic

Variable ZEq.1000

-3103

-5022

0100

0010

0001

04

1218

2x2

1x 2x 3x 4x 5xZ

3x4x5x

(0)(1)(2)(3)

(b) Tabular FormCoefficient of: Right

SideBV ZEq.1000

-3103

-5022

0100

0010

0001

04

1218

62

12

92

18

minimumThe most negative coefficient

1x 2x 3x 4x 5xZ

3x

4x

5x

(0)

(1)

(2)

(3)

(b) Tabular FormCoefficient of: Right

SideBV ZEq.1

0

0

0

-3

1

0

3

-5

0

2

2

0

1

0

0

0

0

1

0

0

0

0

1

0

4

12

18

minimum

The most negative coefficient

Iteration

0

12

18

62

12

92

18

1x 2x 3x 4x 5xZ

3x

2x

5x

(0)

(1)

(2)

(3)

(b) Tabular FormCoefficient of: Right

SideBV ZEq.1

0

0

0

-3

1

0

3

0

0

1

0

0

1

0

0

0

-1

0

0

0

1

30

4

6

6

minimum

The most negative coefficient

Iteration

1

4

6

41

4

23

6

25

21

1x 2x 3x 4x 5x

Z3x

2x

1x

(0)

(1)

(2)

(3)

(b) Tabular FormCoefficient of: Right

SideBV ZEq.

1

0

0

0

0

0

0

1

0

0

1

0

0

1

0

0

1

0

36

2

6

2

None of the coefficient is negative.

Iteration

22

3

21

31

31 3

1

31

The optimal solution

6,2 21 xx

(a) Optimality Test:

The current BF solution is optimal

if and only if every coefficient in row 0 is

nonnegative .

Pivot Column:

A column with the most negative coefficient

)0(

(a) Optimality Test

(b) Minimum Ratio Test:

(b) Minimum Ratio Test:1. Pick out each coefficient in the pivot column

that is strictly positive (>0).

2. Divide each of these coefficients into

the right side entry for the same row.

3. Identify the row that has the smallest of

these ratios.

4. The basic variable for that row is the leaving

basic variable, so replace that variable by the

entering basic variable in the basic variable

column of the next simplex tableau.

Breaking in Simplex Method

(a) Tie for Entering Basic Variable

Several nonbasic variable have largest and

same negative coefficients.

(b) Degeneracy

Multiple Optimal Solution occur if a non

BF solution has zero or at its coefficient

at row 0.

1x 2x 3x 4x 5x

Z3x

4x

5x

(0)

(1)

(2)

(3)

Coefficient of: RightSideBV ZEq.

1

0

0

0

-3

1

0

3

-2

0

2

2

0

1

0

0

0

0

0

1

0

4

12

18

0

0

0

1

0

SolutionOptimal?

No

21 23 xxZ 1x 3x

22x 4x21 23 xx 5x 18

12

4

0

(0)

(1)(2)(3)

Algebraic Form

1x 2x 3x 4x 5x

Z1x

4x

5x

(0)

(1)

(2)

(3)

Coefficient of: RightSideBV ZEq.

1

0

0

0

0

1

0

0

-2

0

2

2

3

1

0

-3

0

0

0

1

12

4

12

6

1

0

0

1

0

SolutionOptimal?

No

1x 2x 3x 4x 5x

Z1x

4x

2x

(0)

(1)

(2)

(3)

Coefficient of: RightSideBV ZEq.

1

0

0

0

0

1

0

0

0

0

0

1

0

1

3

1

0

-1

18

4

6

3

2

0

0

1

0

SolutionOptimal?

Yes

23 2

1

1x 2x 3x 4x 5x

Z1x

3x

2x

(0)

(1)

(2)

(3)

Coefficient of: RightSideBV ZEq.

1

0

0

0

0

1

0

0

0

0

0

1

0

0

1

0

1

0

18

2

2

6

Extra

0

SolutionOptimal?

Yes

21

31

31

31

31

(c) Unbounded Solution

If An entering variable has zero in these coefficients

in its pivoting column, then its solution can be

increased indefinitely.

Z3x

3x2x1x(0)

(1)

Coefficient of: Rightside Ratio

BasicVariable ZEq.

None

1

0

-3

1

-5

0

0

1

0

4

Other Model Forms

(a) Big M Method

(b) Variables - Allowed to be Negative

(M: a large positive number.)

1823 21 xx

(a) Big M Method

21 53 xxZ

1x122 2 x4

and

Maxs.t.

0,0 21 xx

521 53 xMxxZ

1x

21 23 xx 124

and

Maxs.t.

,0jx

3x

184x

5x22x

.5,4,3,2,1jfor

Original Problem Artificial Problem

: Artificial Variable43 , xx

5x: Slack Variables

521 53 xMxx

21 23 xx 185x (3)

(0)

)2318(53 2121 xxMxx

(4)215 2318 xxx

MxMxM 18)52()33( 21

MxMxMZ 18)52()33( 21

Max:

or Max:

or Max:

Max:

s.t.

Eq (3) can be changed to

Put (4) into (0), then

1x 2x 3x 4x 5x

Z3x

4x

5x

(0)

(1)

(2)

(3)

Coefficient of: RightSideBV ZEq.

1

0

0

0

-3M-3

1

3

3

-2M-5

0

2

2

0

1

0

0

0

0

0

1

-18M

4

12

18

0

0

0

1

0

MxMxMZ 18)52()33( 21 Max

1x

21 23 xx 124s.t.

3x

184x

5x22x

(1)

(2)

(3)

1x 2x 3x 4x 5x

Z1x

4x

5x

(0)

(1)

(2)

(3)

Coefficient of: RightSideBV ZEq.

1

0

0

0

-2M-5

1

0

3

3M+3

0

2

2

0

1

0

0

0

0

0

1

-6M+12

4

12

18

1

0

0

1

0

1x 2x 3x 4x 5x

Z1x

4x

2x

(0)

(1)

(2)

(3)

Coefficient of: RightSideBV ZEq.

1

0

0

0

0

1

0

0

0

0

0

1

1

3

0

-1

27

4

6

3

2

0

0

1

0 21

25M2

9

23

1x 2x 3x 4x 5x

Z1x

3x

2x

(0)

(1)

(2)

(3)

Coefficient of: RightSideBV ZEq.

1

0

0

0

0

1

0

0

0

0

0

1

0

0

1

0 0

27

4

6

3

Extra 31

1M

31

31

31

21

23

(b) Variables with a Negative Value jjj xxx 0,0,

jj xx

21 53 xx Mins.t.

0,: 21 xURSx

4122 2 x

1823 21 xx

1x

211 533 xxx Mins.t.

00,,0 211 xxx

4122 2 x

18233 211 xxx

11 xx