E – 1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. Linear Programming E...

E – 1Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Linear ProgrammingLinear ProgrammingE

For For Operations Management, 9eOperations Management, 9e by by Krajewski/Ritzman/Malhotra Krajewski/Ritzman/Malhotra © 2010 Pearson Education© 2010 Pearson Education

PowerPoint Slides PowerPoint Slides by Jeff Heylby Jeff Heyl


Basic ConceptsBasic Concepts

Linear programming is an optimization process A single objective function states mathematically what

is being maximized or minimized Decision variables represent choices that the decision

maker can control Constraints are limitations that restrict the decision

variables. One of three types: ≤, ≥, = The feasible region Parameter or a coefficient Linear objective function and constraints Nonnegativity


Formulating a ProblemFormulating a Problem

Step 1: Define the decision variables

Step 2: Write out the objective function

Step 3: Write out the constraints

As a consistency check, make sure the same unit of measure is being used on both sides of each constraint and the objective function


Formulating a LP ModelFormulating a LP Model

EXAMPLE E.1

The Stratton Company produces two 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. The following data represent next week’s situation. All data are expressed in units of 100 feet of pipe.

Product

Resource Type 1 Type 2 Resource Availability

Extrusion 4 hr 6 hr 48 hr

Packaging 2 hr 2 hr 18 hr

Additive 2 lb 1 lb 16 lb

The contribution to profits and overhead per 100 feet of pipe is $34 for type 1 and $40 for type 2. Formulate a linear programming model to determine how much of each type of pipe should be produced to maximize contribution to profits and to overhead.



SOLUTION

Step 1: To define the decision variables that determine product mix, we let

x1 = amount of type 1 pipe to be produced and sold next week, measured in 100-foot increments (e.g., x1 = 2 means 200 feet of type 1 pipe)

and

x2 = amount of type 2 pipe to be produced and sold next week, measured in 100-foot increments



Step 2: Next, we define the objective function. The goal is to maximize the total contribution that the two products make to profits and overhead. Each unit of x1 yields $34, and each unit of x2 yields $40. For specific values of and x1 and x2, we find the total profit by multiplying the number of units of each product produced by the profit per unit and adding them. Thus, our objective function becomes

Maximize: $34x1 + $40x2 = Z



Step 3: The final step is to formulate the constraints. Each unit of x1 and x2 produced consumes some of the critical resources. In the extrusion department, a unit of x1 requires 4 hours and a unit of x2 requires 6 hours. The total must not exceed the 48 hours of capacity available, so we use the ≤ sign. Thus, the first constraint is

4x1 + 6x2 ≤ 48

Similarly, we can formulate constraints for packaging and raw materials:

2x1 + 2x2 ≤ 18 (packaging)

2x1 + x2 ≤ 16 (additive mix)



These three constraints restrict our choice of values for the decision variable because the values we choose for x1 and x2 must satisfy all of the constraints. Negative values do not make sense, so we add nonnegativity restrictions to the model:

x1 ≥ 0 and x2 ≥ 0 (nonnegativity restrictions)

We can now state the entire model, made complete with the definitions of variables.

Maximize:

Subject to: 4x1 + 6x2 ≤ 48

2x1 + 2x2 ≤ 18

2x1 + x2 ≤ 16

x1 ≥ 0 and x2 ≥ 0

$34x1 + $40x2 = Z

where




Application E.1Application E.1

The Crandon Manufacturing Company produces two principal product lines, a portable circular saw and a precision table saw. There are two crucial operations: fabrication and assembly. 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.

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.

Product

Resource Circular Saw Table Saw Maximum Capacity

Fabrication 2 hrs/month 1 hrs/month 4,000 hrs/month

Assembly 1 hrs/month 2 hrs/month 5,000 hrs/month



SOLUTION

Definition of Decision Variables

x1 = number of circular saws produced and sold per month

x2 = number of table saws produced and sold per month

Formulation

2x1 + 1x2 ≤ 4,000 (Fabrication)

1x1 + 2x2 ≤ 5,000 (Assembly)

1x1 + 1x2 ≤ 3,500 (Demand)

x1, x2 ≥ 0 (Nonnegativity)

Maximize:

Subject to:

900x1 + 600x2 = Z


Graphic AnalysisGraphic Analysis

Five basic steps1. Plot the constraints

2. Identify the feasible region

3. Plot an objective function line

4. Find the visual solution

5. Find the algebraic solution


Plot the ConstraintsPlot the Constraints

Disregard the inequality portion of the constraints; plot the equations

Find the axis intercepts by setting one variable equal to zero and solve for the second variable and repeat to get both intercepts

Once both of the axis intercepts are found, draw a line connecting the two points to get the constraint equation



For

4x1 + 6x2 = 48

At the x1 axis intercept, x2 = 0, so

4x1 + 6(0) = 48

x1 = 12

To find the x2 axis intercept, set x1 = 0 and solve for x2

4(0) + 6x2 = 48

x2 = 8

We connect points (0, 8) and (12, 0) with a straight line, as shown in Figure E.1



4x1 + 6x2 ≤ 48 (extrusion)

18 –

16 –

14 –

12 –

10 –

8 –

6 –

4 –

2 –

0 – | | | | | | | | |

2 4 6 8 10 12 14 16 18x1

x2

Figure E.1 – Graph of the Extrusion Constraint

(0, 8)

(12, 0)



EXAMPLE E.2

For the Stratton Company problem, plot the other constraints: one constraint for packaging and one constraint for the additive mix

SOLUTION

The equation for the packaging process’s line is 2x1 + 2x2 = 18. To find the x1 intercept, set x2 = 0:

2x1 + 2(0) = 18x1 = 9

2(0) + 2x2 = 18x2 = 9

2x1 + 0 = 16x1 = 8

2(0) + x2 = 16x2 = 16

For the packaging constraint

For the additive constraint






18 –

16 –

14 –

12 –

10 –

8 –

6 –

4 –

2 –

0 – | | | | | | | | |

2 4 6 8 10 12 14 16 18x1

x2Figure E.2 – Graph of the Three Constraints

(0, 9)

(9, 0)

(0, 16)

(8, 0)


Identify the Feasible RegionIdentify the Feasible Region

The feasible region is the area on the graph that contains the solutions which satisfy all of the constraints simultaneously, including the nonnegativity restrictions

Locate the area that satisfies all of the constraints using three rules: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 and/or to the left are feasible solutions

3. For the ≥ constraint, the points on the line and the points above and/or to the right are feasible solutions



When one or more of the parameters on the left-hand side of a constraint are negative, we draw the constraint line and test a point on one side of it

2x1 + x2 ≥ 102x1 + 3x2 ≥ 18

x1 ≤ 7 x2 ≤ 5

–6x1 + 5x2 ≤ 5 x1, x2 ≥ 0

12 –

11 –

10 –

9 –

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 – | | | | | | | | | | | |

1 2 3 4 5 6 7 8 9 10 11 12

Feasible region

– 6x1 + 5x2 ≤ 52x1 + x2 ≥ 10

2x1 + 3x2 ≥ 18

x1 ≤ 7

x2 ≤ 5

x2

x1

Test point

Figure E.3 – Identifying the Feasible Region



EXAMPLE E.3

Identify the feasible region for the Stratton Company problem

SOLUTION

Because the problem contains only ≤ constraints, and the parameters on the left-hand side of each constraint are not negative, the feasible portions are to the left of and below each constraint. The feasible region, shaded in Figure E.4, satisfies all three constraints simultaneously.




18 –

16 –

14 –

12 –

10 –

8 –

6 –

4 –

2 –

0 –| | | | | | | | |

2 4 6 8 10 12 14 16 18x1

x2

B

D

C

E

A

Figure E.4 – The Feasible Region

Feasible region



Plot the constraint equations for Crandon Manufacturing and shade feasible region.


1x1 + 2x2 ≤ 5,000 (Assembly)

1x1 + 1x2 ≤ 3,500 (Demand)

x1, x2 ≥ 0 (Nonnegativity)

Point 1 Point 2

Constraint x1 x2 x1 x2

1 0 4,000 2,000 0

2 0 2,500 5,000 0

3 0 3,500 3,500 0

SOLUTION



| | | | | | | | | |

500 1000 1500 2000 2500 3000 3500 4000 4500 5000

4000 –

3500 –

3000 –

2500 –

2000 –

1500 –

1000 –

500 –

0 –

2x1 + x2 ≤ 4,000 (Fabrication)

x1 + x2 ≤ 3,500 (Demand)

x1 + 2x2 ≤ 5,000 (Assembly)

(2,000, 0) (5,000, 0)(3,500, 0)

(0, 4,000)

(0, 3,500)

(0, 2,500)


Plot an Objective Function LinePlot an Objective Function Line

Limit search for solution to the corner points

A corner point lies at the boundary of the feasible region

Interior points need not be considered

Other points on the boundary of the feasible region may be ignored

If the objective function is profits, each line is called an iso-profit line

If Z measures cost, the line is called an iso-cost line


In Figure E.4 we choose corner point B (0, 8)

Substitute these values into the objective function


34x1 + 40x2 = Z

34(0) + 40(8) = 320

At corner point E (8, 0) the objective function is

34(8) + 40(0) = 272

Solving for the other axis intercept

34(0) + 40(x2) = 272

x2 = 6.8



Figure E.5 – Passing an Iso-Profit Line Through (8, 0)




18 –

16 –

14 –

12 –

10 –

8 –

6 –

4 –

2 –

0 – | | | | | | | | |

2 4 6 8 10 12 14 16 18x1

x2

B

D

C

EA

Optimal solution (3, 6)

34x1 + 40x2 = 272





18 –

16 –

14 –

12 –

10 –

8 –

6 –

4 –

2 –

0 – | | | | | | | | |

2 4 6 8 10 12 14 16 18x1

x2

B

D

C

EA

Optimal solution (3, 6)

34x1 + 40x2 = 272

Find the Visual SolutionFind the Visual Solution

Figure E.6 – Drawing the Second Iso-Profit Line



Plot iso-profit lines and identify the visual solution for Crandon Manufacturing

Let Z = $2,000,000 (arbitrary choice)

Plot $900x1 + $600x2 = $2,000,000

Point 1 Point 2

Profit x1 x2 x1 x2

$2,000,000 0 3333.33 2222.22 0

SOLUTION



| | | | | | | | | |

500 1000 1500 2000 2500 3000 3500 4000 4500 5000

4000 –

3500 –

3000 –

2500 –

2000 –

1500 –

1000 –

500 –

0 –

2x1 + x2 ≤ 4,000 (Fabrication)

x1 + x2 ≤ 3,500 (Demand)

x1 + 2x2 ≤ 5,000 (Assembly)

(2,222, 0)

(0, 3,333)

Visual solution is approximately (1,100, 2,100)


Step 1. Develop an equation with just one unknown by multiplying both sides of one equation by a constant so that the coefficient for one of the two decision variables is identical in both equations. Then subtract one equation from the other and solve the resulting equation for its single unknown variable.

Step 2. Insert this decision variable’s value into either one of the original constraints and solve for the other decision variable.

Find the Algebraic SolutionFind the Algebraic Solution


EXAMPLE E.4

Find the optimal solution algebraically for the Stratton Company problem. What is the value of Z when the decision variables have optimal values?

Finding the Optimal Solution Finding the Optimal Solution AlgebraicallyAlgebraically

SOLUTION

Step 1: Figure E.6 showed that the optimal corner point lies at the intersection of the extrusion and packaging constraints. Listing the constraints as equalities, we have

4x1 + 4x2 = 48 (extrusion)

2x1 + 2x2 = 18 (packaging)


We multiply each term in the packaging constraint by 2. The packaging constraint now is 4x1 + 4x2 ≤ 36. Next, we subtract the packaging constraint from the extrusion constraint. The result will be an equation from which has x1 dropped out. (Alternatively, we could multiply the second equation by 3 so that x2 drops out after the subtraction.) Thus,


4x1 + 4x2 = 48

–(4x1 + 4x2 = 36)

2x2 = 12x2 = 6


Step 2: Substituting the value of x2 into the extrusion equation, we get


4x1 + 6(6) = 48

4x1 = 12

x1 = 3

Thus, the optimal point is (3, 6)

This solution gives a total profit of 34(3) + 40(6) = $342



Solve Crandon Manufacturing algebraically with two equations and two unknowns

SOLUTION

From our earlier visual analysis, we know that the optimal solution is at the interesction of the Fabrication and Assembly contraints.


1x1 + 2x2 ≤ 5,000 (Assembly)

Optimal Z: $900(1,000) + $600(2,000) = $2,100,000


Slack and Surplus VariablesSlack and Surplus Variables

A binding constraint is a resource which is completely exhausted when the optimal solution is used because it limits the ability to improve the objective function.

Insert the optimal solution into a constraint equation and solve it. If the number on the left-hand side and the number on the right-hand side are equal, then the constraint is binding.

Relaxing a constraint means increasing the right-hand side for a ≤ constraint and decreasing the right-hand side for a ≥ constraint.

Relaxing a binding constraint means a better solution is possible.


Slack and Surplus VariablesSlack and Surplus Variables

The additive mix constraint, 2x1 + x2 ≤ 16, can be rewritten by adding slack variable s1:

2x1 + x2 + s1 = 16

We then find the slack at the optimal solution (3, 6):

2(3) + 6 + s1 = 16s1 = 4

For a ≥ constraint we subtract a surplus variable from the left-hand side



Find the slack at the optimal solution for Crandon Manufacturing

SOLUTION

Slack in fabrication at (1000, 2000)

2x1 + 1x2 ≤ 4,000

2x1 + 1x2 + s1 = 4,000

2(1000) + (2000) + s1 = 4,000

s1 = 0



Slack in assembly at (1000, 2000)

x1 + 2x2 ≤ 5,000

x1 + 2x2 + s2 = 5,000

(1000) + 2(2000) + s2 = 5,000

s2 = 0

Slack in demand at (1000, 2000)

x1 + x2 ≤ 3,500

x1 + x2 + s3 = 3,500

1000 + 2000 + s3 = 3,500

s3 = 500


Sensitivity AnalysisSensitivity Analysis

Rarely are the parameters in the objective function and constraints known with certainty. Usually parameters are just estimates which don’t reflect

uncertainties such as absenteeism or personal transfers in the Stratton Company problem.

After solving the problem using these estimated values, the analysts can determine how much the optimal values of the decision variables and the objective function value Z would be affected if certain parameters had different values. This type of post solution analysis for answering “what-if” questions is called sensitivity analysis.


Sensitivity AnalysisSensitivity Analysis

TABLE E.1 | SENSITIVITY ANALYSIS INFORMATION PROVIDED BY| LINEAR PROGRAMMING

Term Definition

reducedcost

How much the objective function coefficient of a decision variable must improve (increase for maximum or decrease for minimization) before the optimal solution changes and the decision variable “enters” the solution with some positive number

shadow price

The marginal improvement in Z (increase for maximization and decrease for minimization) caused by relaxing the constraint by one unit

range of optimality

The interval (lower and upper bounds) of an objective function coefficient over which the optimal values of the decision variables remain unchanged

range of feasibility

The interval (lower and upper bounds) over which the right-hand-side parameter can vary while its shadow price remains valid


Simplex method An iterative algebraic procedure Graphic analysis gives insight into the logic The initial feasible solution starts at a corner

point Subsequent iterations result in improved

intermediate solutions In general, a corner point has no more than m

variables greater than 0, where m is the number of constraints

When no further improvement is possible, the optimal solution has been found and the algorithm stops

Computer SolutionComputer Solution


Most real-world linear programming problems are solved on a computer, which can dramatically reduce the amount of time required to solve linear programming problems

POM for Windows in myomlab can handle small- to medium-sized linear programming problems

Microsoft’s Excel Solver offers a second option for similar problem sizes

Computer OutputComputer Output



Figure E.7(a) – Inputs Worksheet



Figure E.7(b) – Inputs Worksheet



Figure E.8 – Results Screen

Figure E.9 – Ranging Screen


Tips on Interpreting OutputTips on Interpreting Output

Reduced cost The sensitivity number is relevant only for a decision

variable that is 0 in the optimal solution It reports how much the objective function coefficient

must improve before it would enter the optimal solution at some positive level

Shadow prices The number is relevant only for binding constraints The shadow price as either positive or negative

The number of variables in the optimal solution > 0 never exceeds the number of constraints

Degeneracy occurs when the number of variables ≠ 0 in the optimal solution is less than the number of constraints


Using Shadow PricesUsing Shadow Prices

EXAMPLE E.5

The Stratton Company needs answers to three important questions: Would increasing capacities in the extrusion or packaging area pay if it cost an extra $8 per hour over and above the normal costs already reflected in the objective function coefficients? Would increasing packaging capacity pay if it cost an additional $6 per hour? Would buying more raw materials pay?

SOLUTION

Expanding extrusion capacity would cost a premium of $8 per hour, but the shadow price for that capacity is only $3 per hour. However, expanding packaging hours would cost only $6 per hour more than the price reflected in the objective function, and the shadow price is $11 per hour. Finally, buying more raw materials would not pay because a surplus of 4 pounds already exists; the shadow price is $0 for that resource.


The Transportation MethodThe Transportation Method

A special case of linear programming Represented as a standard table, sometimes

called a tableau Rows of the table are linear constraints that

impose capacity limitations Columns are linear constraints that require a

certain demand level to be met Each cell in the tableau is a decision variable,

and a per-unit cost is shown in each cell The focus in this section is on the setup and

interpretation of the problem


Transportation Method for LocationTransportation Method for Location

Solve multiple-facility location problems Allocation that minimizes the cost of shipping from two or

more plants, or sources of supply, to two or more warehouses, or destinations

Does not solve all facets of the multiple-facility location problem

Basic steps in setting up the initial tableau Create a row for each plant and a column for each warehouse Add a column for plant capacities and a row for warehouse

demands Each cell not in the requirements row or capacity column

represents a shipping route from a plant to a warehouse. The sum of the shipments in a row must equal the

corresponding plant’s capacity and the sum of shipments in a column must equal the corresponding warehouse’s demand



PlantWarehouse

CapacitySan Antonio, TX (1)

Hot Spring, AR (2)

Sioux Falls, SD (3)

Phoenix5.00 6.00 5.40

400

Atlanta7.00 4.60 6.60

500

Requirements 200 400 300900

900

Figure E.10 – Initial Tableau



Dummy plants or warehouses The sum of capacities must equal the sum of demands If capacity exceeds requirements we add an extra column

(a dummy warehouse) If requirements exceed capacity we add an extra row (a

dummy plant) Assign shipping costs to equal the stockout costs of the

new cells

Finding a solution The goal is to find the least-cost allocation pattern that

satisfies all demands and exhausts all capacities


Interpreting the Optimal SolutionInterpreting the Optimal Solution

EXAMPLE E.6

The optimal solution for the Sunbelt Pool Company, found with POM for Windows, is shown in Figure E.11. Figure E.11(a) displays the data inputs, with the cells showing the unit costs, the bottom row showing the demands, and the last column showing the supply capacities.

Figure E.11(a) – POM for Windows Screens for Sunbelt Pool Company



SOLUTION

Figure E.11(b) shows how the existing network of plants supplies the three warehouses to minimize costs for a total of $4,580. Verify that each plant’s capacity is exhausted and that each warehouse’s demand is filled. Phoenix ships 200 units to warehouse 1 in San Antonio, Texas, and 200 units to warehouse 3 in Sioux Falls, South Dakota, exhausting its 400-unit capacity. Atlanta ships 400 units of its 500-unit capacity to warehouse 2 in Hot Springs, Arkansas, and the remaining 100 units to warehouse 3 in Sioux Falls. All warehouse demand is satisfied: Warehouse 1 in San Antonio is fully supplied by Phoenix and warehouse 2 in Hot Springs is fully supplied by Atlanta. Warehouse 3 in Sioux Falls receives 200 units from Phoenix and 100 units from Atlanta, satisfying its 300-unit demand.



Figure E.11(c) shows the total cost of each shipment. The total optimal cost reported in the upper-left corner of Figure E.11(b) is $4,580, or 200($5.00) + 200($5.40) + 400($4.60) + 100($6.60) = $4,580.

Figure E.11(c) – POM for Windows Screens for Sunbelt Pool Company

Figure E.11(b) – POM for Windows Screens for Sunbelt Pool Company


Transportation Method for Transportation Method for Production PlanningProduction Planning

Making sure that demand and supply are in balance is central to sales and operations planning (SOP)

Helpful in determining anticipation inventoriesTransportation method for production planning is

based on the assumptions Demand forecast and workforce adjustment plan is

available for each period Capacity limits on overtime and the use of

subcontractors also are required All costs are linearly related to the amount of goods

produced


1. Obtain the demand forecasts for each period to be covered by the SOP, identify initial inventory levels

2. Select a candidate workforce adjustment plan and specify capacity limits of each production alternative for each period

3. Estimate the cost of holding inventory and the cost of possible production alternatives and any cost of undertime

4. Input the information gathered in steps 1-3 into a computer routine that solves the transportation problem and use the output to calculate the anticipation inventory levels and identify high-cost elements

5. Repeat the process with other plans until you find the solution that best balances cost and qualitative considerations

Developing a SOPDeveloping a SOP


Preparing a Production PlanPreparing a Production Plan

EXAMPLE E.7

The Tru-Rainbow Company produces a variety of paint products

The demand for paint is highly seasonal

Initial inventory is 250,000 gallons, and ending inventory should be 300,000 gallons

Manufacturing manager wants to determine the best production plan

Regular-time cost is $1.00 per unit, overtime cost is $1.50 per unit, subcontracting cost is $1.90 per unit, and inventory holding cost is $0.30 per unit per quarter

Undertime is paid and the cost is $0.50 per unit



The following constraints apply:

a. The maximum allowable overtime in any quarter is 20 percent of the regular-time capacity in that quarter.

b. The subcontractor can supply a maximum of 200,000 gallons in any quarter. Production can be subcontracted in one period and the excess held in inventory for a future period to avoid a stockout.

c. No backorders or stockouts are permitted.



Demand Regular-time Capacity

Overtime Capacity

Subcontracting Capacity

Quarter 1 300 450 90 200

Quarter 2 850 450 90 200

Quarter 3 1,500 750 150 200

Quarter 4 350 450 90 200

Totals 3,000 2,100 420 800



Figure E.12(a) – POM for Windows Screens for Tru-Rainbow Company



Figure E.12(b) – POM for Windows Screens for Tru-Rainbow Company



SOLUTION

Figure E.12(a) shows the POMS for Windows screen for data input. It looks much like the table shown above, but with one exception. The demand for quarter 4 is shown to be 650,000 gallons rather than the demand forecast of only 350,000. The larger number reflects the desire of the manager to have an ending inventory in quarter 4 of 300,000 gallons.

Figure E.12(b) shows a second screen regarding how the input data gets translated into the transportation tableau. Some points to note include the following:



1. There is a row for each supply alternative (instead of the “sources” or plants in Figure E.10) on a quarter-by-quarter basis. The last column in each row indicates the maximum amount that can be used to meet demand.

2. A column indicates each future quarter of demand and the last row gives its demand forecast. The Excess Capacity column shows the cost of unused capacity.

3. The numbers in the other cells (excluding the cells in the last row or last column) show the cost of producing a unit in one period and, in some cases, carrying the unit in inventory for sale in a future period.



4. The cells in the bottom left portion of the tableau with a cost of $9,999 are associated with backorders (or producing in a period to satisfy demand in a period after it was needed). If backorder costs are so large, the transportation method will try to avoid backorders because it seeks a solution that minimizes total cost.

5. The least expensive alternatives are those in which the output is produced and sold in the same period. We may not always be able to avoid alternatives that create inventory because of capacity restrictions.

6. The per-unit holding cost for the beginning inventory in period 1 is 0 because it is a function of previous production-planning decisions.



Figure E.13 – Solution Screen for Prospective Tru-Rainbow Company Production Plan



Figure E.13 shows the POMS for Windows screen that displays the optimal solution for this particular workforce adjustment plan. Any shortfalls are unused capacity, given in the “Excess Capacity” column.

Similarly, the sum of the allocations down each column must equal the total demand for the quarter. Summed together, they equal the forecasted demand of 300 units.

To further interpret the solution, we can convert Figure E.13 into the following table.



Quarter

Regular-time

ProductionOvertime

Production SubcontractingTotal

SupplyAnticipation

Inventory

1 450 90 20 560 250 + 560 – 300 = 510

2 450 90 200 740 510 + 740 – 850 = 400

3 750 150 200 1,100 400 + 1,100 – 1,500 = 0

4 450 90 110 650 0 + 650 – 350 = 360

Totals 2,100 420 530 3,050

Note: Anticipation inventory is the amount at the end of each quarter, where

Beginning inventory + Total production – Actual Demand= Ending inventory



The breakdown of costs can be found by multiplying the allocation in each cell of Figure E.13 by the cost per unit in that cell in Figure E.12 (b). Computing the cost column by column (it can also be down on a row-by-row basis) yields a total cost of $4,010,000, or $4,010 x 1,000.

Cost Calculations by Column

Quarter 1 230($0) + 50($1.00) + 20($1.90) = $88

Quarter 2 400($1.30) + 450($1.00) = $970

Quarter 320($0.60) + 90($2.10) + 90($1.80) + 200($2.20)

+ 750($1.00)+ 150($1.50) + 200($1.90)= $2,158

Quarter 4 450($1.00) + 90($1.50) + 110($1.90) = $794

Total = $4,010



The Bull Grin Company makes an animal-feed supplement. Sales are seasonal, but Bull Grin's customers refuse to stockpile the supplement and won’t accept backorders. The reactive alternatives that they use, in addition to work-force variation, are regular time, overtime, subcontracting, and anticipation inventory. Backorders are not allowed.

Complete the tableau below by entering the cost per pound produced with each production alternative to meet demand in each period. Bull Grin employs workers who produce 1,000 pounds of supplement for $830 on regular time and $910 on over-time. Holding 1000 pounds of feed supplement in inventory per quarter costs $100. There is no cost for unused regular-time, overtime or subcontracting capacity.


Ap

plicatio

n E

.6A

pp

lication

E.6

Quarter Unused Total

Alternatives 1 2 3 4 Capacity Capacity

Beginning $0 $100 $200 $300

Inventory -

Regular $830 $930 $1,030 $1,130

Time -

1 Overtime $910 $1,010 $1,110 $1,210

-

Subcontract $1,000 $1,100 $1,200 $1,300

-

Regular $99,999

Time -

2 Overtime $99,999

-

Subcontract $99,999

-

Regular $99,999

Time -

3 Overtime $99,999

-

Subcontract $99,999

-

Regular $99,999

Time -

4 Overtime $99,999

-

Subcontract $99,999

-

Demand

$830 $930 $1,030

$910 $1,010 $1,110

$1,000 $1,100 $1,200

$99,999 $830 $930

$99,999 $910 $1,010

$99,999 $1,000 $1,100

$99,999 $99,999 $830

$99,999 $99,999 $910

$99,999 $99,999 $1,000



Now enter data for the capacity column of the tableau as thousands of pounds. The work-force plan being investigating now would provide regular-time capacities of 390 in quarter 1, 400 in quarter 2, 460 in quarter 3, and 380 in quarter 4. Overtime is limited to production of a total of 20,000 pounds per quarter, and subcontractor capacity to only 30,000 pounds per quarter. The current inventory level is 40,000 pounds.

Next enter the data for the demand row. The demand forecasts are 130 in quarter 1, 400 in quarter 2, 800 in quarter 3, and 470 in quarter 4. Management wants 40,000 pounds available at the end the year.


Ap

plicatio

n E

.6A

pp

lication

E.6



Beginning $0 $100 $200 $300

Inventory

Regular $830 $930 $1,030 $1,130

Time -

1 Overtime $910 $1,010 $1,110 $1,210

-

Subcontract $1,000 $1,100 $1,200 $1,300

-

Regular $99,999

Time -

2 Overtime $99,999

-

Subcontract $99,999

-

Regular $99,999

Time -

3 Overtime $99,999

-

Subcontract $99,999

-

Regular $99,999

Time - 380

4 Overtime $99,999

- 20

Subcontract $99,999

- 30

Demand 130 30 1,870

40

390

20

30

400

20

30

460

20

30

400 800 510

$830 $930 $1,030

$910 $1,010 $1,110

$1,000 $1,100 $1,200

$99,999 $830 $930

$99,999 $910 $1,010

$99,999 $1,000 $1,100

$99,999 $99,999 $830

$99,999 $99,999 $910

$99,999 $99,999 $1,000


Ap

plicatio

n E

.6A

pp

lication

E.6



Beginning $0 $100 $200 $300

Inventory 40 0

Regular $830 $930 $1,030 $1,130

Time 90 220 - 80 -

1 Overtime $910 $1,010 $1,110 $1,210

- - 20 - -

Subcontract $1,000 $1,100 $1,200 $1,300

- - - - 30

Regular $99,999

Time -

2 Overtime $99,999

-

Subcontract $99,999

-

Regular $99,999

Time -

3 Overtime $99,999

-

Subcontract $99,999

-

Regular $99,999

Time - 380

4 Overtime $99,999

- 20

Subcontract $99,999

- 30

Demand 130 30 1,870

40

390

20

30

400

20

30

460

20

30

400 800 510

180 220

20

30

460

20

30

380

20

30

400 800 510

$830 $930 $1,030

$910 $1,010 $1,110

$1,000 $1,100 $1,200

$99,999 $830 $930

$99,999 $910 $1,010

$99,999 $1,000 $1,100

$99,999 $99,999 $830

$99,999 $99,999 $910

$99,999 $99,999 $1,000




Quarter 1 Quarter 2 Quarter 3 Quarter 4 Totals

Production

Regular-time 1,630

Overtime 80

Subcontract 90

Total Supply 1,800

Shipments 1,800

Anticipation Inventory 810

What production levels, shipments, and anticipation inventories are called for by the optimal solution in the tableau above?



Quarter 1 Quarter 2 Quarter 3 Quarter 4 Totals

Production

Regular-time 1,630

Overtime 80

Subcontract 90

Total Supply 1,800

Shipments 1,800

Anticipation Inventory 810

What production levels, shipments, and anticipation inventories are called for by the optimal solution in the tableau above?

390 400 460 380

20 20 20 20

0 30 30 30

410 450 570 430

130 400 800 470

320 370 80 40



Quarter 1: = $ 74,700

Quarter 2: = $ 354,000

Quarter 3: = $ 710,000

Quarter 4: = $ 454,000

Total = $1,592,700

What is the total cost of the optimal solution, except for the cost of hiring and layoffs?



Quarter 1: = $ 74,700

Quarter 2: = $ 354,000

Quarter 3: = $ 710,000

Quarter 4: = $ 454,000

Total = $1,592,700

What is the total cost of the optimal solution, except for the cost of hiring and layoffs?

40($0) + 90($830)

220($930) + 180($830)

20($1,110) + 220($930) + 20($1,010) + 30($1,110)+ 460($830) + 20($910) + 30($1,000)

80($1,300) + 380($830) + 20($910) + 30($1,000)


ApplicationsApplications

Examples in operations management and other functional areas

Constraint management Shipping assignments Inventory control Shift scheduling


Solved Problem 1Solved Problem 1

O’Connel Airlines is considering air service from its hub of operations in Cicely, Alaska, to Rome, Wisconsin, and Seattle, Washington. O’Connel has one gate at the Cicely Airport, which operates 12 hours per day. Each flight requires 1 hour of gate time. Each flight to Rome consumes 15 hours of pilot crew time and is expected to produce a profit of $2,500. Serving Seattle uses 10 hours of pilot crew time per flight and will result in a profit of $2,000 per flight. Pilot crew labor is limited to 150 hours per day. The market for service to Rome is limited to nine flights per day.

a. Use the graphic method to maximize profits.

b. Identify slack and surplus constraints, if any.



SOLUTION

a. The objective function is to maximize profits, Z:

Maximize: $2,500x1+ $2,000x2 = Z

where

x1 = number of flights per day to Rome, Wisconsinx2 = number of flights per day to Seattle, Washington

The constraints are

x1 + x2 ≤ 12 (gate capacity)15x1 + 10x2 ≤ 150 (labor)

x1 ≤ 9 (market)x1, x2 ≥ 0



A careful drawing of iso-profit lines parallel to the one shown in Figure E.14 will indicate that point D is the optimal solution. It is at the intersection of the labor and gate capacity constraints. Solving algebraically, we get

The maximum profit results from making six flights to Rome and six flights to Seattle:

$2,500(6)+ $2,000(6) = $27,000

15x1 + 10x2 = 150 (labor)–10x1 – 10x2 = –120 (gate –10)

5x1 + 0x2 = 30x1 = 6

6 + x2 = 12 (gate)x2 = 6



| | |

5 10 15

15 –

10 –

5 –

0 –x1

x2

2,500x1 + 2,000x2 = $20,000(iso-profit line)

x1 + x2 ≤ 12 (gate)

x1 ≤ 9 (market)

15x1 + 10x2 ≤ 150 (labor)

BC

D

E

A

Figure E.14 – Graphic Solution for O’Connel Airlines



b. The market constraint has three units of slack, so the demand for flights to Rome is not fully met:

x1 ≤ 9x1 + s3 = 96 + s3 = 9

s3 = 3



The Arid Company makes canoe paddles to serve distribution centers in Worchester, Rochester, and Dorchester from existing plants in Battle Creek and Cherry Creek. Annual demand is expected to increase as projected in the bottom row of the tableau shown in Figure E.15. Arid is considering locating a plant near the headwaters of Dee Creek. Annual capacity for each plant is shown in the right-hand column of the tableau. Transportation costs per paddle are shown in the tableau in the small boxes. For example, the cost to ship one paddle from Battle Creak to Worchester is $4.37. The optimal allocations are also shown. For example, Battle Creek ships 12,000 units to Rochester. What are the estimated transportation costs associated with this allocation pattern?



SourceDestination

CapacityWorchester Rochester Dorchester

Battle Creek$4.37 $4.25 $4.89

12,000

Cherry Creek$4.00 $5.00 $5.27

10,000

Dee Creek$4.13 $4.50 $3.75

18,000

Demand 6,000 22,000 12,000 40,000

Figure E.15 – Optimal Solution for Arid Company



SourceDestination

CapacityWorchester Rochester Dorchester

Battle Creek$4.37 $4.25 $4.89

12,000

Cherry Creek$4.00 $5.00 $5.27

10,000

Dee Creek$4.13 $4.50 $3.75

18,000

Demand 6,000 22,000 12,000 40,000

Figure E.15 – Optimal Solution for Arid Company

12,000

6,000 4,000

6,000 12,000



SOLUTION

The total cost is $167,000

Ship 12,000 units from Battle Creek to Rochester @ $4.25 Cost = $51,000

Ship 6,000 units from Cheery Creek to Worchester @ $4.00 Cost = $24,000

Ship 6,000 units from Cheery Creek to Rochester @ $5.00 Cost = $20,000

Ship 4,000 units from Dee Creek to Rochester @ $4.50 Cost = $27,000

Ship 12,000 units from Dee Creek to Dorchester @ $3.75 Cost = $45,000

Total = $167,000



The Arctic Air Company produces residential air conditioners. The manufacturing manager wants to develop a sales and operations plan for the next year based on the following demand and capacity data (in hundreds of product units):

DemandRegular-time

CapacityOvertime Capacity

Subcontractor Capacity

Jan-Feb (1) 50 65 13 10

Mar-Apr (2) 60 65 13 10

May-Jun (3) 90 65 13 10

Jul-Aug (4) 120 80 16 10

Sep-Oct (5) 70 80 16 10

Nov-Dec (6) 40 65 13 10

Totals 430 420 84 60



Undertime is unpaid, and no cost is associated with unused overtime or subcontractor capacity. Producing one air conditioning unit on regular time costs $1,000, including $300 for labor. Producing a unit on overtime costs $1,150. A subcontractor can produce a unit to Arctic Air specifications for $1,250. Holding an air conditioner in stock costs $60 for each 2-month period, and 200 air conditioners are currently in stock. The plan calls for 400 units to be in stock at the end of period 6. No backorders are allowed. Use the transportation method to develop a plan that minimizes costs.



SOLUTION

The following tables identify the optimal production and inventory plans. Figure E.16 shows the tableau that corresponds to this solution. An arbitrarily large cost ($99,999 per period) was used for backorders, which effectively ruled them out. Again, all production quantities are in hundreds of units. Note that demand in period 6 is 4,400. That amount is the period 6 demand plus the desired ending inventory of 400. The anticipation inventory is measured as the amount at the end of each period. Cost calculations are based on the assumption that workers are not paid for undertime or are productively put to work elsewhere in the organization whenever they are not needed for this work.



Figure E.16 – Tableau for Optimal Production and Inventory Plans



One initially puzzling aspect of this solution is that it allocates the initial inventory of 200 units to meet demand in period 4 rather than in period 1. The explanation is that multiple optimal solutions exist and this solution is only one of them. However, all solutions result in the same production and anticipation inventory plans derived below:

Production Plan

PeriodRegular-time Production

Overtime Production Subcontracting Total

1 6,500 — — 6,500

2 6,500 400 — 6,900

3 6,500 1,300 — 7,800

4 8,000 1,600 1,000 10,600

5 7,000 — — 7,000

6 4,400 — — 4,400



Anticipation Inventory

PeriodBeginning Inventory Plus

Total Production Minus DemandAnticipation

(Ending) Inventory

1 200 + 6,500 – 5,000 1,700

2 1,700 + 6,900 – 6,000 2,600

3 2,600 + 7,800 – 9,000 1,400

4 1,400 + 10,600 – 12,000 0

5 0 + 7,000 – 7,000 0

6 0 + 4,400 – 4,000 400

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