1. Linear Programming (LP)

2. [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.

3. Time Period Covered Minimum # of Agents Shift Time Period needed
6:00 am to 8:00 am
8:00 am to10:00 am
10:00 am to noon
Noon to 2:00 pm
2:00 pm to 4:00 pm
4:00 pm to 6:00 pm
6:00 pm to 8:00 pm
8:00 pm to 10:00 pm
10:00 pm to midnight
Midnight to 6:00 am *
* *
* * * 48 79 65 87 64 73 82 43 52 15
Daily cost per agent

4. 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.

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

6. Min
s.t.
all

7. Total Personal Cost = $30,610

8. The Simplex Method

9. Maximize
Subject to
and

10. From a geometric viewpoint
: CPF solutions
(Corner-Point Feasible)
: Corner-point infeasible
solutions 8 6 4
Feasible
region 2

11. 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.

12. Initialization
Optimal
Solution?
Stop
Yes No
Iteration

13. 1 2
Feasible
region

14. The Key Solution Concepts
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.

15. 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.

16. 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.

17. 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.

18. Simplex Method
To convert the functional inequality constraints to equivalent equality constraints, we need to incorporate slack variables.

19. Original Form
of Model
Augmented Form
of the Model
Slack variables
Max
Max
s.t.
s.t.
and
and
for

20. Properties of BF Solution
A basic solution is an augmented corner-point solution.
A Basic Feasible (BF) solution is an augmented CPF solution.
Properties of BF Solution
1. Each variable is designated as either a nonbasic
variable or a basic variable.
2. # of nonbasic variables = # of functional
constraints.

21. 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.

22. Simplex in Tabular Form
(a) Algebraic Form
(0)
(1)
(2)
(3)
(b) Tabular Form
Coefficient of:
Basic
Variable
Right
Side
Eq. Z
(0) 1 -3 1 3 -5 2 1 1 1 4 12 18
(1)
(2)
(3)

23. The most negative coefficient
minimum
(b) Tabular Form
Coefficient of:
Right
Side
Eq. BV Z
(0) 1 -3 1 3 -5 2 1 1 1 4 12 18
(1)
(2)
(3)

24. The most negative coefficient
minimum 12 18
The most negative coefficient
(b) Tabular Form
Coefficient of:
Right
Side
Iteration
Eq. BV Z
(0)
(1)
(2)
(3) 1 -3 1 3 -5 2 1 1 1 4 12 18

25. The most negative coefficient4
minimum 6
The most negative coefficient
(b) Tabular Form
Coefficient of:
Right
Side
Iteration
Eq. BV Z
(0)
(1)
(2)
(3) 1 -3 1 3 1 1 -1 1 30 4 6 1

26. None of the coefficient is negative.
The optimal solution
None of the coefficient is negative.
(b) Tabular Form
Coefficient of:
Right
Side
Iteration
Eq. BV Z
(0)
(1)
(2)
(3) 1 1 1 1 1 36 2 6 2

27. (a) Optimality Test: (a) Optimality Test (b) Minimum Ratio 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

28. (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.

29. 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.

30. Algebraic Form
(0)
(1)
(2)
(3)
Coefficient of:
Right
Side
Solution
Optimal?
Eq. BV Z
(0)
(1)
(2)
(3) 1 -3 1 3 -2 2 1 1 1 4 12 18 No

31. 1 Coefficient of: Right Side Solution Optimal? Eq. BV Z (0) (1) (2)
(3) 1 1 -2 2 3 1 -3 1 1 12 4 6 1 No

32. 2 Coefficient of: Right Side Solution Optimal? Eq. BV Z (0) (1) (2)
(3) 1 1 1 1 3 1 1 -1 18 4 6 3 2
Yes

33. Extra Coefficient of: Right Side Solution Optimal? Eq. BV Z (0) (1)
(2)
(3) 1 1 1 1 1 18 2 6
Extra
Yes

34. (c) Unbounded Solution
If An entering variable has zero in these coefficients
in its pivoting column, then its solution can be
increased indefinitely.
Basic
Variable
Coefficient of:
Right
side
Eq. Z
Ratio
(0) 1 -3 1 -5 1 4
(1)
None

35. Other Model Forms (a) Big M Method
(b) Variables - Allowed to be Negative

36. (a) Big M Method Original Problem Artificial Problem Max Max s.t. s.t.
and
and
for
: Slack Variables
: Artificial Variable
(M: a large positive
number.)

37. Max:
(0)
s.t.
(3)
Eq (3) can be changed to
(4)
Put (4) into (0), then
Max:
or Max:
or Max:

38. Max
s.t.
(1)
(2)
(3)
Coefficient of:
Right
Side
Eq. BV Z
(0)
(1)
(2)
(3) 1
-3M-3 1 3
-2M-5 2 1 1 1
-18M 4 12 18

39. 1 Coefficient of: Right Side Eq. BV Z (0) (1) (2) (3) 1 -2M-5 1 3 3M+3
-2M-5 1 3
3M+3 2 1 1 1
-6M+12 4 12 18 1

40. 2 Coefficient of: Right Side Eq. BV Z (0) (1) (2) (3) 1 1 1 1 3 1 -11 1 1 3 1 -1 27 4 6 3 2

41. Extra Coefficient of: Right Side Eq. BV Z (0) (1) (2) (3) 1 1 1 1 27 41 1 1 27 4 6 3
Extra

42. (b) Variables with a Negative Value
Min
Min
s.t.
s.t.