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.

Time Period Covered Minimum # of Agents Shift Time Period needed 1 2 3 4 5 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 170 160 175 180 195

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.

: # 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.

Min s.t. all

Total Personal Cost = $30,610

The Simplex Method

Maximize Subject to and

From a geometric viewpoint : CPF solutions (Corner-Point Feasible) : Corner-point infeasible solutions 8 6 4 Feasible region 2 0 2 4 6 8 10

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 Optimal Solution? Stop Yes No Iteration

1 2 Feasible region

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.

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.

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

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.

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 (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)

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)

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

The most negative coefficient 4 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

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

(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

(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.

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

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

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

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

(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

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

(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.)

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

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

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

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

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

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