Optimization problems using excel solver McGraw-Hill/Irwin Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved.

Linear Programming Using the Excel Solver LAB 3 Linear Programming Using the Excel Solver

OBJECTIVES Linear Programming Basics A Maximization Problem A Minimization Problem 2

Linear Programming Essential Conditions Is used in problems where we have limited resources or constrained resources that we wish to allocate The model must have an explicit objective (function) Generally maximizing profit or minimizing costs subject to resource-based, or other, constraints 3

Linear Programming Essential Conditions (Continued) Limited Resources to allocate Linearity is a requirement of the model in both objective function and constraints Homogeneity of products produced (i.e., products must the identical) and all hours of labor used are assumed equally productive Divisibility assumes products and resources divisible (i.e., permit fractional values if need be) 3

Aggregate sales and operations planning Common Applications Aggregate sales and operations planning Service/manufacturing productivity analysis Product planning Product routing Vehicle/crew scheduling Process control Inventory control

2A-7 Objective Function Maximize (or Minimize) Z = C1X1 + C2X2 + ... + CnXn Cj is a constant that describes the rate of contribution to costs or profit of (Xj) units being produced Z is the total cost or profit from the given number of units being produced 5

Bi are the available resource requirements Constraints A11X1 + A12X2 + ... + A1nXnB1 A21X1 + A22X2 + ... + A2nXn B2 : AM1X1 + AM2X2 + ... + AMnXn=BM Aij are resource requirements for each of the related (Xj) decision variables Bi are the available resource requirements Note that the direction of the inequalities can be all or a combination of , , or = linear mathematical expressions 5

Non-Negativity Requirement X1,X2, …, Xn  0 All linear programming model formulations require their decision variables to be non-negative While these non-negativity requirements take the form of a constraint, they are considered a mathematical requirement to complete the formulation of an LP model 5

2A-10 Excel solver The Excel Solver is a tool for solving linear and nonlinear optimization problems. For linear optimization problems: Simplex method, with branch and bound algorithm for integer design variables. For nonlinear optimization problems: Generalized reduced gradient method. Solver is an Add-In for Microsoft Excel which can solve optimization problems, including multiple constraint problems. Launching the Excel Solver: Start the Excel program. Tools > Solver… 7

Installing Excel solver On Excel Menu, choose Tools Add-Ins... Put a Check in the Box Next to ‘Solver Add-in’ 7

On Excel Menu, choose Tools Using Excel solver Solver This brings 2A-12 Using Excel solver On Excel Menu, choose Tools Solver This brings up the Solver Parameters box which will be discussed next. 7

2A-13 Excel solver 7

2A-14 Excel solver 7

An Example of a Maximization Problem LawnGrow Manufacturing Company must determine the unit mix of its commercial riding mower products to be produced next year. The company produces two product lines, the Max and the Multimax. The average profit is $400 for each Max and $800 for each Multimax. Fabrication hours and assembly hours are limited resources. There is a maximum of 5,000 hours of fabrication capacity available per month (each Max requires 3 hours and each Multimax requires 5 hours). There is a maximum of 3,000 hours of assembly capacity available per month (each Max requires 1 hour and each Multimax requires 4 hours). Question: How many units of each riding mower should be produced each month in order to maximize profit? Now let’s formula this problem as an LP model… 6

The Objective Function 2A-16 The Objective Function If we define the Max and Multimax products as the two decision variables X1 and X2, and since we want to maximize profit, we can state the objective function as follows: 8

Given the resource information below from the problem: Constraints Given the resource information below from the problem: We can now state the constraints and non-negativity requirements as: Note that the inequalities are less-than-or-equal since the time resources represent the total available resources for production 9

Produce 715 Max and 571 Multimax per month for a profit of $742,800 Solution Produce 715 Max and 571 Multimax per month for a profit of $742,800 16

The Excel Solver Formulation

An Example of a Minimization Problem HiTech Metal Company is developing a plan for buying scrap metal for its operations. HiTech receives scrap metal from two sources, Hasbeen Industries and Gentro Scrap in daily shipments using large trucks. Each truckload of scrap from Hasbeen yields 1.5 tons of zinc and 1 ton of lead at a cost of $15,000. Each truckload of scrap from Gentro yields 1 ton of zinc and 3 tons of lead at a cost of $18,000. HiTech requires at least 6 tons of zinc and at least 10 tons of lead per day. Question: How many truckloads of scrap should be purchased per day from each source in order to minimize scrap metal costs to HiTech? Now let’s formula this problem as an LP model… 17

The Objective Function 2A-21 The Objective Function If we define the Has been truckloads and the Gentro truckloads as the two decision variables X1 and X2, and since we want to minimize cost, we can state the objective function as follows: Minimize Z = 15,000 X1 + 18,000 X2 Where Z = daily scrap cost X1 = truckloads from Hasbeen X2 = truckloads from Gentro Hasbeen Gentro 18

Given the demand information below from the problem: Constraints Given the demand information below from the problem: We can now state the constraints and non-negativity requirements as: Note that the inequalities are greater-than-or-equal since the demand information represents the minimum necessary for production. 1.5X1 + X2 > 6(Zinc/tons) X1 + 3X2 > 10(Lead/tons) X1, X2 > 0(Non-negativity) 19

Order 2.29 truckloads from Hasbeen and 2.57 Solution Order 2.29 truckloads from Hasbeen and 2.57 truckloads from Gentro for daily delivery. The daily cost will be $80,610. Note: Do you see why in this solution that “integer” linear programming methodologies can have useful applications in industry? 25

The Excel Solver solution 2A-24 The Excel Solver solution

maximize f = 10x1+ 6.5x2+ 6x3+ 5x4 2y1+ 5.5y2+ 5y3+ 4y4 Question Find x1, x2, x3, x4, y1, y2, y3, y4 to maximize f = 10x1+ 6.5x2+ 6x3+ 5x4 2y1+ 5.5y2+ 5y3+ 4y4 subject to x1+ y1 = 25 x2+ y2 = 45 x3+ y3 = 50 x4+ y4 = 60 x1+ x2+ x3+ x4  50 y1+ y2+ y3+ y4  1000 x1, x2, x3, x4, y1, y2, y3, y4  0. 7

Using the Excel Solver: Design variables: Answer Using the Excel Solver: Design variables: A1 = x1, A2 = x2, A3 = x3, A4 = x4 B1 = y1, B2 = y2, B3 = y3, B4 = y4 Objective function: C1 = 10*A1+6.5*A2+6*A3+5*A42*B1+5.5*B2+5*B3+4*B4 Constraints: D1 = A1+B1, D2 = A2+B2, D3 = A3+B3, D4 = A4+B4 D5 = A1+A2+A3+A4, D6 = B1+B2+B3+B4 7

By Changing Cells: A1 to B4 Subject to the Constraints: Add Answer (…continued) Set Target Cell: C1 Equal To: Max By Changing Cells: A1 to B4 Subject to the Constraints: Add D1 = 25 Add D2 = 45 Add D3 = 50 Add D4 = 60 Add D5 <= 50 Add D6 <= 1000 OK Options > Assume Non-Negative > OK Solve Keep Solver Solution > OK 7

Solutions in A1 to A4 and B1 to B4: (x1, x2, x3, x4) = (25, 0, 10, 15) Answer (…continued) Solutions in A1 to A4 and B1 to B4: (x1, x2, x3, x4) = (25, 0, 10, 15) (y1, y2, y3, y4) = (0, 45, 40, 45) Maximum profit in C1 = 1012.5 Note that the solution is not unique. Verify that another solution is given by (x1, x2, x3, x4) = (25, 0, 0, 25) (y1, y2, y3, y4) = (0, 45, 50, 35) 7

2A-29 End