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Linear Programming For Operations Management, 9e by Krajewski/Ritzman/Malhotra © 2010 Pearson Education PowerPoint Slides by Jeff Heyl Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Basic 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Formulating 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Resource Availability
Formulating 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 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Formulating a LP Model 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Formulating a LP Model 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Formulating a LP Model 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) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Formulating a LP Model 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: \$34x1 + \$40x2 = Z Subject to: 4x1 + 6x2 ≤ 48 2x1 + 2x2 ≤ 18 2x1 + x2 ≤ 16 x1 ≥ 0 and x2 ≥ 0 where x1 = amount of type 1 pipe to be produced and sold next week, measured in 100-foot increments x2 = amount of type 2 pipe to be produced and sold next week, measured in 100-foot increments Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application 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 5,000 hrs/month Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.1 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 Maximize: 900x x2 = Z Subject to: 2x1 + 1x2 ≤ 4,000 (Fabrication) 1x1 + 2x2 ≤ 5,000 (Assembly) 1x1 + 1x2 ≤ 3,500 (Demand) x1, x2 ≥ 0 (Nonnegativity) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Graphic Analysis Five basic steps Plot the constraints Identify the feasible region Plot an objective function line Find the visual solution Find the algebraic solution Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Plot 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Plot the Constraints 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Plot the Constraints 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – | | | | | | | | | x1 x2 Figure E.1 – Graph of the Extrusion Constraint (0, 8) (12, 0) 4x1 + 6x2 ≤ 48 (extrusion) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Plot the Constraints 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: For the packaging constraint For the additive constraint 2x1 + 2(0) = 18 x1 = 9 2(0) + 2x2 = 18 x2 = 9 2x1 + 0 = 16 x1 = 8 2(0) + x2 = 16 x2 = 16 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Plot the Constraints 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – | | | | | | | | | x1 x2 Figure E.2 – Graph of the Three Constraints (0, 16) (8, 0) 4x1 + 6x2 ≤ 48 (extrusion) 2x1 + 2x2 ≤ 18 (packaging) 2x1 + x2 ≤ 16 (additive mix) (0, 9) (9, 0) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Identify 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: For the = constraint, only the points on the line are feasible solutions For the ≤ constraint, the points on the line and the points below and/or to the left are feasible solutions For the ≥ constraint, the points on the line and the points above and/or to the right are feasible solutions Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 6

Identify the Feasible Region
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 12 – 11 – 10 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 – 0 – | | | | | | | | | | | | Feasible region – 6x1 + 5x2 ≤ 5 2x1 + x2 ≥ 10 2x1 + 3x2 ≥ 18 x1 ≤ 7 x2 ≤ 5 x2 x1 2x1 + x2 ≥ 10 2x1 + 3x2 ≥ 18 x1 ≤ 7 x2 ≤ 5 –6x1 + 5x2 ≤ 5 x1, x2 ≥ 0 Test point Figure E.3 – Identifying the Feasible Region Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 14

Identify the Feasible Region
EXAMPLE E.3 Identify the feasible region for the Stratton Company problem 4x1 + 6x2 ≤ 48 (extrusion) 2x1 + 2x2 ≤ 18 (packaging) 2x1 + x2 ≤ 16 (additive mix) 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – | | | | | | | | | x1 x2 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. B D C E A Feasible region Figure E.4 – The Feasible Region Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 42

Application E.2 Plot the constraint equations for Crandon Manufacturing and shade feasible region. 2x1 + 1x2 ≤ 4,000 (Fabrication) 1x1 + 2x2 ≤ 5,000 (Assembly) 1x1 + 1x2 ≤ 3,500 (Demand) x1, x2 ≥ 0 (Nonnegativity) SOLUTION Point 1 Point 2 Constraint x1 x2 1 4,000 2,000 2 2,500 5,000 3 3,500 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.2 | | | | | | | | | | 4000 – 3500 – 3000 – 2500 – 2000 – 1500 – 1000 – 500 – 0 – 2x1 + x2 ≤ 4,000 (Fabrication) x1 + x2 ≤ 3,500 (Demand) x1 + 2x2 ≤ 5,000 (Assembly) (0, 4,000) (0, 3,500) (0, 2,500) (2,000, 0) (3,500, 0) (5,000, 0) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Plot 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Plot an Objective Function 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Plot an Objective Function Line
4x1 + 6x2 ≤ 48 (extrusion) 2x1 + 2x2 ≤ 18 (packaging) 2x1 + x2 ≤ 16 (additive mix) 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – | | | | | | | | | x1 x2 B D C E A Optimal solution (3, 6) Figure E.5 – Passing an Iso-Profit Line Through (8, 0) 34x1 + 40x2 = 272 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Find the Visual Solution
4x1 + 6x2 ≤ 48 (extrusion) 2x1 + 2x2 ≤ 18 (packaging) 2x1 + x2 ≤ 16 (additive mix) 18 – 16 – 14 – 12 – 10 – 8 – 6 – 4 – 2 – 0 – | | | | | | | | | x1 x2 B D C E A Optimal solution (3, 6) 34x1 + 40x2 = 272 Figure E.6 – Drawing the Second Iso-Profit Line Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.3 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 SOLUTION Point 1 Point 2 Profit x1 x2 \$2,000,000 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.3 | | | | | | | | | | 4000 – 3500 – 3000 – 2500 – 2000 – 1500 – 1000 – 500 – 0 – 2x1 + x2 ≤ 4,000 (Fabrication) x1 + x2 ≤ 3,500 (Demand) x1 + 2x2 ≤ 5,000 (Assembly) (0, 3,333) Visual solution is approximately (1,100, 2,100) (2,222, 0) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Find the Algebraic Solution
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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Finding the Optimal Solution Algebraically
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? 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) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Finding the Optimal Solution Algebraically
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 = 12 x2 = 6 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Finding the Optimal Solution Algebraically
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.4 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. 2x1 + 1x2 ≤ 4,000 (Fabrication) 1x1 + 2x2 ≤ 5,000 (Assembly) Optimal Z: \$900(1,000) + \$600(2,000) = \$2,100,000 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Slack 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Slack 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) s1 = 16 s1 = 4 For a ≥ constraint we subtract a surplus variable from the left-hand side Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 36

Application E.5 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.5 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 s3 = 3,500 s3 = 500 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Sensitivity 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 41

Sensitivity Analysis TABLE E.1 | SENSITIVITY ANALYSIS INFORMATION PROVIDED BY | LINEAR PROGRAMMING Term Definition reduced cost 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Computer Solution 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Computer Output 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Computer Output Figure E.7(a) – Inputs Worksheet Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 41

Computer Output Figure E.7(b) – Inputs Worksheet Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 41

Computer Output Figure E.8 – Results Screen Figure E.9 – Ranging Screen Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall. 41

Tips 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

The 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Transportation 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Transportation Method for Location
Plant Warehouse Capacity San Antonio, TX (1) Hot Spring, AR (2) Sioux Falls, SD (3) Phoenix 5.00 6.00 5.40 400 Atlanta 7.00 4.60 6.60 500 Requirements 200 300 900 Figure E.10 – Initial Tableau Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Transportation Method for Location
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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Interpreting 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Interpreting the Optimal 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Interpreting the Optimal Solution
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(b) – POM for Windows Screens for Sunbelt Pool Company Figure E.11(c) – POM for Windows Screens for Sunbelt Pool Company Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Transportation Method for Production Planning
Making sure that demand and supply are in balance is central to sales and operations planning (SOP) Helpful in determining anticipation inventories Transportation 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Developing a SOP Obtain the demand forecasts for each period to be covered by the SOP, identify initial inventory levels Select a candidate workforce adjustment plan and specify capacity limits of each production alternative for each period Estimate the cost of holding inventory and the cost of possible production alternatives and any cost of undertime 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 Repeat the process with other plans until you find the solution that best balances cost and qualitative considerations Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
Demand Regular-time Capacity Overtime Capacity Subcontracting Capacity Quarter 1 300 450 90 200 Quarter 2 850 Quarter 3 1,500 750 150 Quarter 4 350 Totals 3,000 2,100 420 800 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
Figure E.12(a) – POM for Windows Screens for Tru-Rainbow Company Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
Figure E.12(b) – POM for Windows Screens for Tru-Rainbow Company Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
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: Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
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. 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. 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
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. 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. The per-unit holding cost for the beginning inventory in period 1 is 0 because it is a function of previous production-planning decisions. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
Figure E.13 – Solution Screen for Prospective Tru-Rainbow Company Production Plan Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
Quarter Regular-time Production Overtime Subcontracting Total Supply Anticipation Inventory 1 450 90 20 560 – 300 = 510 2 200 740 – 850 = 400 3 750 150 1,100 ,100 – 1,500 = 0 4 110 650 – 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Preparing a Production Plan
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 3 20(\$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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 Quarter Unused Total Alternatives 1 2 3 4 Capacity Beginning \$0 \$100 \$200 \$300 Inventory - Regular \$830 \$930 \$1,030 \$1,130 Time Overtime \$910 \$1,010 \$1,110 \$1,210 Subcontract \$1,000 \$1,100 \$1,200 \$1,300 \$99,999 Demand \$830 \$930 \$1,030 \$910 \$1,010 \$1,110 \$1,000 \$1,100 \$1,200 \$99,999 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 Quarter Unused Total Alternatives 1 2 3 4 Capacity Beginning \$0 \$100 \$200 \$300 Inventory Regular \$830 \$930 \$1,030 \$1,130 Time - Overtime \$910 \$1,010 \$1,110 \$1,210 Subcontract \$1,000 \$1,100 \$1,200 \$1,300 \$99,999 380 20 30 Demand 130 1,870 400 800 510 40 390 20 30 400 460 \$830 \$930 \$1,030 \$910 \$1,010 \$1,110 \$1,000 \$1,100 \$1,200 \$99,999 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 Quarter Unused Total Alternatives 1 2 3 4 Capacity Beginning \$0 \$100 \$200 \$300 Inventory 40 Regular \$830 \$930 \$1,030 \$1,130 Time 90 220 - 80 Overtime \$910 \$1,010 \$1,110 \$1,210 20 Subcontract \$1,000 \$1,100 \$1,200 \$1,300 30 \$99,999 380 Demand 130 1,870 400 800 510 40 390 20 30 400 460 180 220 20 30 460 380 400 800 510 \$830 \$930 \$1,030 \$910 \$1,010 \$1,110 \$1,000 \$1,100 \$1,200 \$99,999 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 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 Anticipation Inventory 810 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 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 Anticipation Inventory 810 390 400 460 380 20 30 410 450 570 430 130 800 470 320 370 80 40 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Application E.6 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 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) Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Applications Examples in operations management and other functional areas Constraint management Shipping assignments Inventory control Shift scheduling Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 1 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, Wisconsin x2 = 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 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 1 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 15x1 + 10x2 = 150 (labor) –10x1 – 10x2 = –120 (gate  –10) 5x1 + 0x2 = 30 x1 = 6 6 + x2 = 12 (gate) x2 = 6 The maximum profit results from making six flights to Rome and six flights to Seattle: \$2,500(6)+ \$2,000(6) = \$27,000 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 1 | | | 15 – 10 – 5 – 0 – x1 x2 x1 + x2 ≤ 12 (gate) x1 ≤ 9 (market) 15x1 + 10x2 ≤ 150 (labor) B C D E A 2,500x1 + 2,000x2 = \$20,000 (iso-profit line) Figure E.14 – Graphic Solution for O’Connel Airlines Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 1 b. The market constraint has three units of slack, so the demand for flights to Rome is not fully met: x1 ≤ 9 x1 + s3 = 9 6 + s3 = 9 s3 = 3 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 2 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? Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 2 Source Destination Capacity Worchester 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 40,000 Figure E.15 – Optimal Solution for Arid Company Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 2 Source Destination Capacity Worchester 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 40,000 12,000 6,000 4,000 Figure E.15 – Optimal Solution for Arid Company Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 2 SOLUTION The total cost is \$167,000 Ship 12,000 units from Battle Creek to \$4.25 Cost = \$51,000 Ship 6,000 units from Cheery Creek to \$4.00 \$24,000 to \$5.00 \$20,000 Ship 4,000 units from Dee Creek to \$4.50 \$27,000 Ship 12,000 units from Dee Creek to \$3.75 \$45,000 Total = \$167,000 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Regular-time Capacity Subcontractor Capacity
Solved Problem 3 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): Demand Regular-time Capacity Overtime Capacity Subcontractor Capacity Jan-Feb (1) 50 65 13 10 Mar-Apr (2) 60 May-Jun (3) 90 Jul-Aug (4) 120 80 16 Sep-Oct (5) 70 Nov-Dec (6) 40 Totals 430 420 84 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 3 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 3 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. Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 3 Figure E.16 – Tableau for Optimal Production and Inventory Plans Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Regular-time Production
Solved Problem 3 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 Period Regular-time Production Overtime Production Subcontracting Total 1 6,500 2 400 6,900 3 1,300 7,800 4 8,000 1,600 1,000 10,600 5 7,000 6 4,400 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Solved Problem 3 Anticipation Inventory Period
Beginning Inventory Plus Total Production Minus Demand Anticipation (Ending) Inventory 1 ,500 – 5,000 1,700 2 1, ,900 – 6,000 2,600 3 2, ,800 – 9,000 1,400 4 1, ,600 – 12,000 5 0 + 7,000 – 7,000 6 0 + 4,400 – 4,000 400 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.