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Linear Programming
Supplement E
¾ Linear programming: A technique that is useful for allocating scarce resources among competing
demands.
¾ Objective function: An expression in linear
programming models that states mathematically what is being maximized (e.g., profit or present value) or minimized (e.g., cost or scrap).
¾ Decision variables: The variables that represent choices the decision maker can control.
¾ Constraints: The limitations that restrict the permissible choices for the decision variables.
Linear Programming
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¾ Feasible region:
A region that represents all permissible combinations of the decision variables in a linearprogramming model.
¾ Parameter:
A value that the decision maker cannot control and that does not change when the solution is implemented.¾ Certainty:
The word that is used to describe that a fact is known without doubt.
¾ Linearity:
A characteristic of linear programming models that implies proportionality and additivity – there can be no products or powers of decision variables.¾ Nonnegativity:
An assumption that the decision variables must be positive or zero.
Linear Programming
Formulating a Problem
¾ Step 1. Define the Decision Variables.
¾ Step 2.Write Out the Objective Function.
¾ Step 3. Write Out the Constraints.
¾ Product-mix problem: A one-period type of planning problem, the solution of which yields optimal output quantities (or product mix) of a group of services or products
subject to resource capacity and market
demand constraints.
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¾ The Stratton Company produces 2 basic types of plastic pipe. Three resources are crucial to the output of pipe: extrusion hours, packaging hours, and a special additive to the plastic raw material.
¾ Below is next week’s situation.
Formulating a Problem
Example E.1
16 lb 1 lb
2 lb Additive mix
18 hr 2 hr
2 hr Packaging
48 hr 6 hr
4 hr Extrusion
Resource Availability Type 2
Type 1 Resource
Product
x
1= amount of type 1 pipe produced and sold next week, 100-foot increments
x
2= amount of type 2 pipe produced and sold next week, 100-foot increments
Step 1 – Define the decision variables
Formulating a Problem
Example E.1 continued
16 lb 1 lb
2 lb Additive mix
18 hr 2 hr
2 hr Packaging
48 hr 6 hr
4 hr Extrusion
Resource Availability Type 2
Type 1 Resource
Product
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Step 2 – Define the objective function
16 lb 1 lb
2 lb Additive mix
18 hr 2 hr
2 hr Packaging
48 hr 6 hr
4 hr Extrusion
Resource Availability Type 2
Type 1 Resource
Product
Formulating a Problem
Example E.1 continued
Max Z = $34 x
1+ $40 x
2Objective is to
maximize profits (Z)
Each unit of x
1yields $34, and each unit of x
2yields $40.
Step 3 – Formulate the constraints
Formulating a Problem
Example E.1 continued
16 lb 1 lb
2 lb Additive mix
18 hr 2 hr
2 hr Packaging
48 hr 6 hr
4 hr Extrusion
Resource Availability Type 2
Type 1 Resource
Product
4 x
1+ 6 x
2≤ 48 2 x
1+ 2 x
2≤ 18 2 x
1+ x
2≤ 16
Extrusion Packaging Additive mix
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¾ Typically the constraining resources have upper or lower limits.
¾ e.g., for the Stratton Company, the total extrusion time must not exceed the 48 hours of capacity available, so we use the ≤ sign.
¾ Negative values for constraints x
1and x
2do not
make sense, so we add nonnegativity restrictions to the model:
x
1≥ 0 and x
2≥ 0 (nonnegativity restrictions)
¾ Other problem might have constraining resources requiring > , > , = , or < restrictions.
Formulating a Problem
with Inequalities
Formulating a Problem
Application E.1
¾ The Crandon Manufacturing Company produces two
principal product lines. One is a portable circular saw, and the other is a precision table saw. Two basic operations are crucial to the output of these saws: fabrication and
assembly. The maximum fabrication capacity is 4000 hours per month; each circular saw requires 2 hours, and each table saw requires 1 hour. The maximum assembly
capacity is 5000 hours per month; each circular saw
requires 1 hour, and each table saw requires 2 hours. The marketing department estimates that the maximum market demand next year is 3500 saws per month for both
products. The average contribution to profits and overhead is $900 for each circular saw and $600 for each table saw.
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Application E.1
¾ Management wants to determine the best product mix for the next year so as to maximize contribution to profits and overhead. Also, it is interested in the payoff of expanding capacity or increasing market share.
Maximize: 900x
1+ 600x
2= Z
Subject to: 2x
1+ 1x
2≤ 4,000 (Fabrication) 1x
1+ 2x
2≤ 5,000 (Assembly)
1x
1+ 1x
2≤ 3,500 (Demand)
x
1, x
2≥ 0 (Nonnegativity)
Graphic Analysis
¾ Most linear programming problems are solved with a computer.
¾ However, insight into the meaning of the computer output, and linear programming concepts in general, can be gained by analyzing a simple two-variable problem graphically.
¾ Graphic method of linear programming: A type of graphic analysis that involves the following five
steps:
¾ plotting the constraints
¾ identifying the feasible region
¾ plotting an objective function line
¾ finding a visual solution
¾ finding the algebraic solution
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18 — 16 — 14 — 12 — 10 — 8 — 6 — 4 — 2 —
0 | | | | | | | | |
2 4 6 8 10 12 14 16 18 x1 x2
4x
1+ 6x
2≤ 48 (extrusion) 2x
1+ 2x
2≤ 18 (packaging) 2x
1+ x
2≤ 16 (additive mix)
Graphic Analysis
Example E.2
We begin by plotting the constraint equations, disregarding the inequality portion of the constraints (< or >). Making each constraint an equality (=)
transforms it into the equation for a straight line.
Graphic Analysis
Example E.3
¾ The feasible region is the area on the graph that contains the solutions that satisfy all the constraints simultaneously.
¾ To find the feasible region, first locate the feasible points for each constraint and then the area that satisfies all constraints.
¾ Generally, the following three rules identify the feasible points for a given constraint:
1. For the = constraint, only the points on the line are feasible solutions.
2. For the ≤ constraint, the points on the line and the points below or to the left of the line are feasible.
3. For the ≥ constraint, the points on the line and the points above or to the right of the line are feasible.
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Graphic Analysis
Identify the feasible region
18 — 16 — 14 — 12 — 10 — 8 — 6 — 4 — 2 —
0 | | | | | | | | |
2 4 6 8 10 12 14 16 18 x1 x2
4x
1+ 6x
2≤ 48 (extrusion) 2x
1+ 2x
2≤ 18 (packaging) 2x
1+ x
2≤ 16 (additive mix)
Feasible region
A A
B B C C
D D
E
Plotting Crandon Mfg.
Constraints
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¾ Now we want to find the solution that optimizes the objective function.
¾ Even though all the points in the feasible region represent possible solutions, we can limit our search to the corner points.
¾ Corner point: A point that lies at the intersection of two (or possibly more) constraint lines on the
boundary of the feasible region.
¾ No interior points in the feasible region need be
considered because at least one corner point is better than any interior point.
¾ The best approach is to plot the objective function on the graph of the feasible region for some
arbitrary Z values.
Graphic Analysis
Plotting an Objective Function Line
18 — 16 — 14 — 12 — 10 — 8 — 6 — 4 — 2 —
0 | | | | | | | | |
x x2
4x
1+ 6x
2≤ 48 (extrusion) 2x
1+ 2x
2≤ 18 (packaging) 2x
1+ x
2≤ 16 (additive mix)
Feasible region
B C
D E
34x
1+ 40x
2= $272
Graphic Analysis
Plotting an Objective Function Line
For
Example E.3, t
he equation for an arbitrary objective function line passing through E is 34x1 + 40x2 = 272© 2007 Pearson Education
18 — 16 — 14 — 12 — 10 — 8 — 6 — 4 — 2 —
0 | | | | | | | | |
2 4 6 8 10 12 14 16 18 x1 x2
4 4 x x
11+ 6 + 6 x x
22≤ ≤ 48 (extrusion) 48 (extrusion) 2 2 x x
11+ 2 + 2 x x
22≤ ≤ 18 (packaging) 18 (packaging) 2x 2 x
11+ + x x
22≤ ≤ 16 (additive mix) 16 (additive mix)
Feasible region
A A
B B C C
D D E
A series of dashed lines can be drawn parallel to this first line. Each would have its own Z value.
Lines above the first line would have higher Z
values. Lines below it would have lower Z values.
Graphic Analysis
Plotting an Objective Function Line
18 — 16 — 14 — 12 — 10 — 8 — 6 — 4 — 2 —
0 | | | | | | | | |
x x2
4
x1+ 6
x2≤ 48 (extrusion) 2
x1+ 2
x2≤ 18 (packaging) 2
x1+
x2≤ 16 (additive mix)
Feasible region
B C
D E
Our goal is to maximize profits, so the best solution is a point on the iso-profit line farthest
from the origin but still touching the feasible region.
Graphic Analysis
Identifying the Visual Solution
Optimal solution (3,6)
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Application E.3
Iso-profit Line
and Visual Solution for Crandon Mfg.
¾ Step 1: Develop an equation with just one unknown.
¾ Start by multiplying both sides by a constant so that the coefficient for one of the two decision variables is identical in both equations.
¾ Then subtract one equation from the other and solve the resulting equation for its single
unknown variable.
¾ Step 2: Insert this decision variable’s value into either one of the original constraints and solve for the other decision variable.
Finding the
Algebraic Solution
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Application E.4
Algebraic Solution for Crandon Mfg.
Solve algebraically, with two equations and two unknowns
Slack & Surplus Variables
¾ Binding constraint: A constraint that helps form the optimal corner point; it limits the ability to
improve the objective function.
¾ Slack: The amount by which the left-hand side falls short of the right-hand side.
¾To find the slack for a ≤ constraint algebraically, we add a slack variable to the constraint and convert it to an equality.
¾ Surplus: The amount by which the left-hand side exceeds the right-hand side.
¾To find the surplus for a ≥ constraint, we subtract a surplus variable from the left-hand side to make it an equality.
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Slack Variables for Crandon Mfg.
Application E.5
Sensitivity Analysis
¾ Coefficient sensitivity
: How much the objective function coefficient of a decision variable must improve (increase for maximization or decrease for minimization) before the optimal solution changes and the decision variable becomes some positive number.¾ Range of feasibility
: The interval over which the right- hand-side parameter can vary while its shadow price remains valid.¾ Range of optimality
: The lower and upper limits over which the optimal values of the decision variables remain unchanged.¾ Shadow price
: The marginal improvement in Z (increase for maximization and decrease for minimization) caused by relaxing the constraint by one unit.© 2007 Pearson Education
Computer Solutions
¾ Computer programs dramatically reduce the time required to solve linear programming problems.
¾ Special-purpose programs can be developed for applications that must be repeated frequently.
¾ Such programs simplify data input and generate the objective function and constraints for the problem. In addition, they can prepare customized managerial reports.
¾ Simplex method: An iterative algebraic procedure for solving linear programming problems.
¾ Most real-world linear programming problems are solved on a computer. The solution procedure in computer codes is some form of the simplex method.
Computer Solution
Output from OM Explorer
for the Stratton Company
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Computer Solution
Output from OM Explorer for the Stratton Company
Results Worksheet
Computer Solution
The coefficient sensitivities
provide no new insight because they are always 0 when
decision variables have positive values in the optimal solution.
Optimal solution is to make 300 ft of type 1 pipe and 600 ft of type 2 pipe. Thus the product mix is x1 and x2.
Maximum profit
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© 2007 Pearson Education
4 lbs of additive were unused (surplus) so its shadow price is zero.
An additional hour of extrusion time would contribute $3 to profits.
Computer Solution
An additional hour of packaging time is worth
$11.
All of the extrusion and packaging time was used.
Computer Solution
If either objective
function coefficient goes above or below its
sensitivity range, the product mix will change.
If the availability of one of the constraints goes above or below its
sensitivity range, the product mix will change.
Increased additive has no limit because there was a 4 lb surplus.
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© 2007 Pearson Education
APPLICATIONS OF LINEAR PROGRAMMING
¾ Product Mix: Find the best mix of products to produce.
¾ Shipping: Find the optimal shipping assignments.
¾ Stock Control: Determine the optimal mix of products to hold in inventory.
¾ Supplier Selection: Find the optimal combination of suppliers to minimize unwanted inventory.
¾ Plants or Warehouses: Determine optimal location of a plant or warehouse.
¾ Stock Cutting: Find the cutting pattern that minimizes the amount of scrap.
¾ Production: Find the minimum-cost production schedule.
¾ Staffing: Find the optimal staffing levels.
¾ Blends: Find the optimal proportions of various ingredients used to make products.
¾ Shifts: Determine the minimum-cost assignment of workers to shifts.
¾ Vehicles: Assign vehicles to products or customers.
¾ Routing: Find the optimal routing of a service or product through several sequential processes.
Product Mix Problem
Application E.6
The Trim-Look Company makes several lines of skirts, dresses, and sport coats for women. Recently it was
suggested that the company reevaluate its South Islander line and allocate its resources to those products that would maximize contribution to profits and overhead. Each product must pass through the cutting and sewing departments. In addition, each product in the South Islander line requires the same polyester fabric. The following data were collected for the study.
The Cutting department has 100 hours of capacity, sewing has 180 hours, and 60 yards of material are available. Each skirt contributes $5 to profits and overhead; each dress, $17;
and each sport coat, $30.
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Product Mix Problem
Application E.6
Process Design
Application E.7
The plant manager of a plastic pipe manufacturer has the
opportunity to use two different routings for a particular type of plastic pipe: Routing 1 uses extruder A, and routing 2 uses extruder B. Both routings require the same melting process.
The following table shows the time requirements and capacities of these processes.
In addition, each 100 feet of pipe processed on routing 1 uses 5 pounds of raw material, whereas each 100 feet of pipe
processed on routing 2 uses only 4 pounds. This difference results from differing scrap rates of the extruding machines.
Consequently, the profit per 100 feet of pipe processed on routing 1 is $60 and on routing 2, $80. A total of 200 pounds of
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Process Design
Application E.7
Blending Problem
Application E.8
Consider the task facing the procurement manager of a company that manufactures special additives. She must determine the proper amounts of each raw material to purchase for the production of a certain product.
Three raw materials are available. Each gallon of the finished product must have a combustion point of at least 220°F. In addition, the gamma content (which causes hydrocarbon pollution) cannot exceed 6 percent of the volume. The zeta content (which cleans the internal moving parts of engines) must be at least 12 percent by volume. Each raw material has varying degrees of these characteristics.
Raw material A costs $0.60 per gallon; raw material B, $0.40; and raw material C, $0.50. The procurement manager wishes to minimize the cost of raw materials per gallon of product. What are the optimal
proportions of each raw material to use in a gallon of finished product?
Hint: Express your decision variables in terms of fractions of a gallon.
The sum of the fractions must equal 1.00.
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Blending Problem
Application E.8
Portfolio Selection
Application E.9
E-Traders, Inc. invests in various types of securities. The firm has $5 million for immediate investment and wishes to maximize the interest earned over the next year. Risk is not a factor. There are four investment possibilities, as outlined below.
To further structure the portfolio, the board of directors has specified that at least 40 percent of the investment must be placed in corporate bonds and common stock. Furthermore, no more than 20 percent of the investment can be in real estate.
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Portfolio Selection
Application E.9
Shift Scheduling
Application E.10
NYNEX has a scheduling problem. Operators work eight-hour shifts and can begin work at either midnight, 4 A.M., 8 A.M., noon, 4 P.M., or 8 P.M. Operators are needed according to the following demand pattern.
Hint: Let xj equal the number of operators beginning work (an eight-hour shift) in time period j, where j = 1, 2, . . . , 6.
Formulate the model to cover the demand requirements with the minimum number of operators.
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Shift Scheduling
Application E.10
Production Planning
Application E.11
Bull Grin employs manual, unskilled labor, who require little or no
training. Producing 1000 pounds of supplement costs $810 on regular time and $900 on overtime. These figures include materials, which
account for over 80 percent of the cost. Overtime is limited to production of 30,000 pounds per quarter. In addition, subcontractors can be hired at
$1100 per thousand pounds, but only 10,000 pounds per quarter can be produced this way.
The current level of inventory is 40,000 pounds, and management wants to end the year at that level. Holding 1000 pounds of feed supplement in inventory per quarter costs $110. The latest annual forecast follows.
The firm currently has 180 workers, a figure that management wants to keep in quarter 4. Each worker can produce 2000 pounds per quarter, so that regular-time production costs $1620 per worker. Idle workers must be paid at that same rate. Hiring one worker costs $1000, and laying off
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Production Planning
Application E.11
Production Planning
Application E.11
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Solved Problem 1
¾ O’Connel Airlines is considering air service from its hub of operations in Cicely, Alaska to Rome, Wisconsin, and Seattle.
¾ They have one gate at the Cicely Airport, which operates 12 hours per day. Each flight requires 1 hour of gate time.
¾ Each flight to Rome consumes 15 hours of pilot crew time and is expected to produce a profit of $2,500.
¾ Serving Seattle uses 10 hours of pilot crew time per flight and will result in a profit of $2,000 per flight.
¾ Pilot crew labor is limited to 150 hours per day.
¾ The market for service to Rome is limited to 9 flights per day.
1. Use the graphic method to maximize profits.
2. Identify slack and surplus constraints, if any.
The objective function is to maximize profits (Z) Maximize Z = $2,500
x
1 + $2,000x
2where
x1 = number of flights per day to Rome, Wisconsin x2 = number of flights per day to Seattle, Washington
The constraints are
x1 + x2 ≤ 12 (gate capacity) 15 x1 + 10 x2 ≤ 150 (labor)
x1 ≤ 9 (market) x1 ≥ 0 and x2 ≥ 0
Solved Problem 1
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Solved Problem 1
15 —15
10 —10
5 5 —
0 0 — | | |
AA 55 1010 1515
xx11 + + xx22 ≤≤ 12 (gate12 (gate))
xx11 ≤≤ 9 (market)9 (market)
BB CC DD
15 x15 x11 + 10 x+ 10 x22 ≤≤ 150 (labor)150 (labor)
2,500
2,500 xx11 +2,000 x+2,000 x22 = $20,000= $20,000 (iso-(iso-profit line)profit line)
EE
xx11 xx22
| | |
A 5 10 15
x1 + x2 ≤ 12 (gate) x1 ≤ 9 (market)
B C D´
15 x1 + 10 x2 ≤ 150 (labor)
E´
x1 x2
15 —
10 —
5 —
0 —
Solved Problem 1
A careful drawing of iso-profit lines parallel to the one shown indicates that point D is the optimal solution.
The maximum profit results from making six flights to Rome and six flights to Seattle:
