Solving LPs-Simplex Method

HighTech Industries imports electronic components that are used to assemble two different models of personal computers. One model is called the Deskpro, and the other model is called the Portable.

HighTech’s management is currently interested in developing a weekly production schedule for both products. The Deskpro generates a profit contribution of $50 per unit, and the Portable generates a profit contribution of $40 per unit. For next week’s production, a maximum of 150 hours of assembly time can be made available. Each unit of the Deskpro requires 3 hours of assembly time, and each unit of the Portable requires 5 hours of assembly time. In addition, HighTech currently has only 20 Portable display components in inventory; thus, no more than 20 units of the Portable may be

assembled. Finally, only 300 square feet of warehouse space can be made available for new

production. Assembly of each Deskpro requires 8 square feet of warehouse space; similarly, each Portable requires 5 square feet. Formulate a mathematical model of HighTech’s situation that can be used to maximize the companies profit.

Solving LPs-Simplex Method

Example:

Source: Anderson (2012)

Maximize Z = 50X

+ 40X

Subject to:

3X

+ 5X

≤ 150

X

≤ 20 8X

+ 5X

≤ 300 X

≥ 0

X

≥ 0

X₁ = Number of units of Deskpro X₂ = Number of units of Portable

Solving LPs-Simplex Method

Note: The simplex method applied to this problem is for maximization LPs with “≤” constraints. Similar method applies for minimization with few changes

Max Z = 50X

+ 40X

S.t

3X

+ 5X

+ S

= 150 X

+ S

= 20 8X

+ 5X

+ S

= 300 X

X

S

S

S

≥ 𝟎

Solving LPs-Simplex Method

Step 1: Convert LP into standard form

Solving LPs – Simplex Method

Basic solution and basic feasible solutions

Solving LPs – Simplex Method

Picturing basic feasible and non feasible basic solutions

Points (1)-(5) are basic feasible solutions. Points (6)-(9) are basic solutions that are not feasible

Solving LPs – Simplex Method

Graphically, the simplex method

moves from one extreme point to the next, carefully improving upon the objective value on each move.

Picturing the simplex method graphically

Max Z = 50X₁+40X₂ s.t

3X₁ + 5X₂ ≤ 150 X₂ ≤ 20 8X₁ + 5X₂≤ 300

Solving LPs-Simplex Method

Step 2: Determine the initial Basic Feasible Solution

Basic solution:

A basic solution to an LP with n variables and m constraints is obtained by setting (n-m) of the variables to zero and solving the m linear constraint equations for the remaining m variables

For example if we set 𝑥₂ = 0 and 𝑠₁ = 0 the system of constraint equations becomes:

3𝑥₁ = 150 1𝑠₂ = 20

8𝑥₁ +1𝑠₃ = 300

A basic feasible solution (BFS):

This is a basic solution that also satisfies the nonnegativity conditions of the LP problem.

For example if we set 𝑥₁ = 0 and 𝑥₂ = 0 the system of constraint equations becomes:

1𝑠₁ = 150 1𝑠₂ = 20

+1𝑠₃ = 300

Hint: Start by setting the actual decision variables to zero

Solving this yield 𝑠₁=150, 𝑠₂=20, 𝑠₃=300, and Z =0

Solving LPs-Simplex Method

How can the current solution be improved?

Max Z = 50X₁ + 40X₂+ 0S₁+0S₂+0S₃ S.t

3X₁ + 5X₂ + S1 = 150 X2 + S2 = 20 8X₁ + 5X₂ + S₃ = 300 X_1,X_2,S_1, S_2, S₃≥ 𝟎

At S₁ =150, S₂ =20, S₃ =300, Z = 0 . Is there a combination of the variables than will improve upon the current objective value?

When x₁ =1, Z improves by 50, which is a gain. However, setting x₁ =1 affect the values of S_1, S_2, and S_3.

From the constraints, S₁ = 147, S₂ =20, S₃ =292. This results in no loss to Z only because S_1, S_2, and S₃ all have zero coefficient. Total gain is 50

When x₂ =1, Z improves by 40, which is a gain. However, setting x₂ =1 affect the values of S_1, S_2, and S_3.

From the constraints, S₁ = 145, S₂ =19, S₃ =295. This results in no loss to Z only because S_1, S_2, and S₃ all have zero coefficient. Total gain is 40.

Between x₁ and x₂ which one would you select to improve on the objective Z?

Solving LPs-Simplex Method

Step 3: Form the initial simplex tableau

For our example we have:

Solving LPs-Simplex Method

Step 4: Determine Reduced Cost for each variable

Reduced cost is the amount by which an objective function coefficient would have to improve before it would be possible for the corresponding variable to assume a positive value in the optimal solution.

𝑧_𝑗 represents the decrease in the value of the objective function that will result if one unit of the variable corresponding to the j th column of matrix A is brought into the basis Value of the obj.

function

How can the current solution be improved?

Solving LPs-Simplex Method

Step 4: Improve the solution

Determine which variable leaves and which enters the basic feasible solution

For maximization, pick a variable from the current non-basic with the largest 𝑐_𝑗 − 𝑧_𝑗 to enter the BFS. In the case of a tie, select the variable to enter the basis that corresponds to the leftmost of the columns.

Thus, x₁ enter the BFS

Minimum Ratio Test

Pick a variable from the current BFS with the smallest ^𝑏𝑖

𝑎_𝑖𝑗 to leave the BFS. In case of a tie, select the variable that corresponds to the uppermost of the tied rows. Thus, S₃ leaves the BFS

Solving LPs-Simplex Method

Step 4: Calculate the next improved tableau

Convert the column associated with the new basic variable into a unit column; in this way its value will be given by the right-hand- side value of the corresponding row

𝑥₂ and 𝑠₃ are non-basic, Thus, 𝑥₂ = 0 and 𝑠₃ = 0.

If The new basic feasible solution has:

𝑠₁ = ⁷⁵

2 , 𝑠₂ = 20, and 𝑥₁ = ⁷⁵

2 . This yield a new objective value of:

0 ∗ 75

2 + 0 ∗ 20 + 50 ∗ 75

2 = 1875

Solving LPs-Simplex Method

Step 4: Move towards a better solution

Calculate a new 𝑐_𝑗 − 𝑧_𝑗

Pick a variable from the current

non-basic with the largest 𝑐_𝑗 − 𝑧_𝑗 to enter the BFS

Pick a variable from the current BFS with the smallest ^𝑏𝑖

𝑎_𝑖𝑗 to leave the BFS

Optimality Criterion

When all of the entries in the net evaluation row (𝑐_𝑗 − 𝑧_𝑗) are zero or negative, stop. Current basic feasible solution is the optimal solution.

Solving LPs-Simplex Method

Convert column of new basic variable to unit column.

Elements of 𝑐_𝑗 − 𝑧_𝑗 are all zero or negative

Solving LPs-Simplex Method

Interpreting the optimal solution.

The optimal solution has:

𝑥₁ = 30, 𝑥₂ = 12, 𝑠₁ = 0, 𝑠₂ = 8, 𝑠₃= 0 And the value of the objective function is: $1980

If management wants to maximize the total profit contribution, HighTech should produce 30 units of the Deskpro and 12 units of the Portable. When 𝑠₂ = 8, management should note that there will be eight unused Portable display units.

Also, because 𝑠₁ = 0 and 𝑠₃ = 0, no slack is associated with the assembly

time constraint and the warehouse capacity constraint; in other words, these constraints are both binding. Consequently, if it is possible to obtain additional assembly time and/or additional warehouse space, management should consider doing so.

Solving LPs-Simplex Method

Summary of the simplex method