- 267 Views
- Uploaded on
- Presentation posted in: General

Linear Programming: Formulations & Graphical Solution. Introduction To Linear Programming.

Linear Programming: Formulations & Graphical Solution

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Linear Programming: Formulations & Graphical Solution

- After three decades of experimentation and scrutiny, LP has been applied with impressive success to problems ranging from the familiar cases in industry, military, agriculture, economics, transportation, and health systems to the extreme cases in behavioral and social sciences.

- Today many of the resources needed as inputs to operations are in limited supply.
- Operations managers must understand the impact of this situation on meeting their objectives.
- Linear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources.
- A Linear Programming model seeks to maximize or minimize a linear function, subject to a set of linear constraints.

- Linear Programming is a mathematical technique for optimum allocation of limited or scarce resources, such as labour, material, machine, money, energy and so on, to several competing activities such as products, services, jobs and so on, on the basis of a given criteria of optimality.

- The maximization or minimization of some quantity is the objective in all linear programming problems.
- All LP problems have constraints that limit the degree to which the objective can be pursued.
- A feasible solution satisfies all the problem's constraints.
- An optimal solution is a feasible solution that results in the largest possible objective function value when maximizing (or smallest when minimizing).
- A graphical solution method can be used to solve a linear program with two variables.

- If both the objective function and the constraints are linear, the problem is referred to as a linear programming problem.
- Linear functions are functions in which each variable appears in a separate term raised to the first power and is multiplied by a constant (which could be 0).
- Linear constraints are linear functions that are restricted to be "less than or equal to", "equal to", or "greater than or equal to" a constant.
- Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.

- The construction of a mathematical model can be initiated by answering the following three questions:
- What does the model seek to determine? In other words, what are the variables (unknowns) of the problem?
- What constraints must be imposed on the variables to satisfy the limitations of the modeled system?
- What is the objective (goal) that needs to be achieved to determine the optimum (best) solution from among all the feasible values of the variables?
- An effective way to answer these questions is to give a verbal summary of the problem. In terms of the Reddy Mikks example, the situation is described as follows.

- Understand the problem thoroughly.
- Define the decision variables.
- Describe the objective.
- Describe each constraint.
- Write the objective in terms of the decision variables.
- Write the constraints in terms of the decision variables.

- The Reddy Mikks company owns a small paint factory that produces both interior and exterior house paints for wholesale distribution. Two basic raw materials, A and B, are used to manufacture the paints.
- The maximum availability of A is 6 tons a day; that of B is 8 tons a day. The daily requirements of the raw materials per ton of interior and exterior paints are summarized in the following table.

- A market survey has established that the daily demand for the interior paint cannot exceed that of exterior paint by more than 1 ton. The survey also showed that the maximum demand for the interior paint is limited to 2 tons daily.
- The wholesale price per ton is $3000 for exterior paint and $2000 per interior paint. How much interior and exterior paint should the company produce daily to maximize gross income?

- The company seeks to determine the amounts (in tons) of interior and exterior paints to be produced to maximize (increase as much as is feasible) the total gross income (in thousands of dollars) while satisfying the constraints of demand and raw material usage.
- Variables: since we desire to determine the amounts of interior and exterior paints to be produced, the variables of the model can be defined as
- XE = tons produced daily of exterior paint
- XI = tons produced daily of interior paint

- Objective Function: since each ton of exterior paint sells for $3000, the gross income from selling XE tons is 3XE thousand dollars. Similarly, the gross income from XI tons of interior paint is 2XI thousand dollars.
- Under the assumption that the sales of interior and exterior paints are independent, the total gross income becomes the sum of the two revenues.
- If we let Z represents the total gross revenue (in thousands of dollars), the objective function may be written mathematically as
- Z = 3XE +2XI
- The goal is to determine the (feasible) values of XE and XI that will maximize this criterion.

- Constraints: The Reddy Mikks problem imposes restrictions on the usage of raw materials and on demand. The usage restriction may be expressed verbally as
(usage of raw material by both paints) ≤(maximum raw material availability)

- This leads to the following restrictions (see the data for the problem):
XE + 2XI ≤ 6 (raw material A)

2XE + XI ≤ 8 (raw material B)

- The demand restrictions are expressed verbally as
(excess amount of interior over exterior paint) ≤ 1 ton per day

(demand for interior paint) ≤ 2 tons per day

- Mathematically, these are expressed, respectively, as
XI - XE ≤ 1

XI ≤ 2

- An implicit (or "understood-to-be) constraint is that the amount produced of each paint cannot be negative (less than zero). To avoid obtaining such a solution, we impose the nonnegativity restrictions, which are normally written
XI ≥ 0

XE ≥ 0

- The values of the variables XE and XI are said to constitute a feasible solution if they satisfy all the constraints of the model.

- The complete mathematical model for the Reddy Mikks problem may now be summarized as follows:

This is a typical optimization problem.

Any values of x1, x2 that satisfy all the constraints of the model is called a feasible solution. We are interested in finding the optimumfeasible solution that gives the maximum profit while satisfying all the constraints.

More generally, an optimization problem looks as follows:

Determine the decision variablesx1, x2, …, xn so as to optimize an objectivefunctionf (x1, x2, …, xn) satisfying the constraints

gi (x1, x2, …, xn) ≤ bi (i=1, 2, …, m).

An optimization problem is called a Linear Programming Problem (LPP) when the objective function and all the constraints are linear functions of the decision variables, x1, x2, …, xn. We also include the “non-negativity restrictions”, namely xj ≥ 0 for all j=1, 2, …, n.

Thus a typical LPP is of the form:

Optimize (i.e. Maximize or Minimize)

z = c1 x1 + c2 x2+ …+ cn xn

subject to the constraints:

a11 x1 + a12 x2 + … + a1n xn ≤ b1

a21 x1 + a22 x2 + … + a2n xn ≤ b2

. . .

am1 x1 + am2 x2 + … + amn xn ≤ bm

x1, x2, …, xn 0

- When we use LP as an approximate representation of a real-life situation, the following assumptions are inherent:
- Proportionality. - The contribution of each decision variable to the objective or constraint is directly proportional to the value of the decision variable.
- Additivity. - The contribution to the objective function or constraint for any variable is independent of the values of the other decision variables, and the terms can be added together sensibly.
- Divisibility. - The decision variables are continuous and thus can take on fractional values.
- Deterministic (Certainty). - All the parameters (objective function coefficients, right-hand side coefficients, left-hand side, coefficients) are known with certainty.

- Cycle Trends is introducing two new lightweight bicycle frames, the Deluxe and the Professional, to be made from aluminum and steel alloys. The anticipated unit profits are $10 for the Deluxe and $15 for the Professional.
- The number of pounds of each alloy needed per frame is summarized on the next slide. A supplier delivers 100 pounds of the aluminum alloy and 80 pounds of the steel alloy weekly. How many Deluxe and Professional frames should Cycle Trends produce each week?

Aluminum AlloySteel Alloy

Deluxe 2 3

Professional 4 2

Pounds of each alloy needed per frame

- Define the objective
- Maximize total weekly profit

- Define the decision variables
- x1 = number of Deluxe frames produced weekly
- x2 = number of Professional frames produced weekly

- Write the mathematical objective function
- Max Z = 10x1 + 15x2

- LP in Final Form
- Max Z = 10x1 + 15x2
- Subject To
- 2x1 + 4x2 < 100 ( aluminum constraint)
- 3x1 + 2x2 < 80 ( steel constraint)
- x1 , x2 > 0 (non-negativity constraints)

The Burroughs garment company manufactures men's shirts and women’s blouses for Walmark Discount stores. Walmark will accept all the production supplied by Burroughs. The production process includes cutting, sewing and packaging. Burroughs employs 25 workers in the cutting department, 35 in the sewing department and 5 in the packaging department. The factory works one 8-hour shift, 5 days a week. The following table gives the time requirements and the profits per unit for the two garments:

Determine the optimal weekly production schedule for Burroughs.

Assume that Burroughs produces x1 shirts and x2 blouses per week.

Profit got = 8 x1 + 12 x2

Time spent on cutting = 20 x1 + 60 x2 mts

Time spent on sewing = 70 x1 + 60 x2 mts

Time spent on packaging = 12 x1 + 4 x2 mts

The objective is to find x1, x2 so as to

maximize the profit z = 8 x1 + 12 x2

satisfying the constraints:

20 x1 + 60 x2≤ 25 40 60

70 x1 + 60 x2 ≤ 35 40 60

12 x1 + 4 x2 ≤ 5 40 60

x1, x2≥ 0, integers