CHAPTER3 Linear Programming 3.1 INTRODUCTION Linear programming (LP) is a quantitative method for solving problems concerning resource allocation. This study is about managing the resources – how the management should allocate the resources efficiently so that maximum benefits can be obtained from the limited resources available. Examples of resources are raw materials, employees, labor hours, machine hours, financial budgets, etc. Since the resources are limited, the management must utilize and allocate them efficiently so that the firm could obtain maximum profit. 3.2 REQUIREMENTS OF A LINEAR PROGRAMMING PROBLEM The linear programming problems have four important properties in common. 1. All Linear Programming problems in practice seek to maximize or to minimize certain quantity, usually profit or cost. We refer to this property as the objective function of a Linear Programming problem. 2. All problems have certain restrictions, or constraints that limit the degree to which we can achieve our objective. Therefore, we want to maximize or minimize the objective function, subject to some limited resources available (constraints). 3. There must be some alternative courses of action to choose from in the solution. If there is no alternative solution to choose from, then we would not need LP. 4. The objective and constraints in Linear Programming problems must be expressed in terms of linear equations or inequalities. The linear mathematical relationship requires all terms used in the objective functions and the constraints involved to be of the first degree. 54 3.3 FORMULATING LINEAR PROGRAMMING PROBLEMS Three components of Linear Programming are the variables, objective and constraints. The variables are continuous, controllable, and non-negative. X1 represents the first variable, X2 represents the second variable, and so on; X1, X 2, …X n… > 0. The objective of the Linear Programming problem is a mathematical representation of the goals in terms of a measurable objective such as the amount of profit, cost, revenue, or quantity. It must be in a linear function and is represented by one of these two forms: i) Maximize or ii) Minimize C1X1 + C 2X 2 +….. + C nX n, C 1X 1 + C 2X 2 +….. + C nX n. The constraints are a set of mathematical expressions representing the restrictions imposed on the controllable variables and thereby limiting the possible values of those variables. The mathematical expressions of the constraints can be given either as linear equations (=) or as linear inequalities (<, >, >, <). A linear programming model will be structured to find the values of controllable variables that maximize or minimize a defined linear objective function, subject to a set of defined linear constraints and subject to the non-negativity restrictions. Example 3.1 The owner of a small company that manufactures clocks must decide how many clocks of each type to produce daily in order to maximize profit. The company manufactures two types of clocks, that is regular clocks and alarm clocks. The resources required to produce the clocks are labor and processing machine. The limited resources available every day are as follows: Labor hours Machine hours = 1,600 hours = 1,800 hours The requirement for resources for each type of clocks is as follows: 55 To produce one unit of regular clock requires 2 hours of labor, and 6 hours of machine time. While to produce one unit of alarm clock requires 4 hours of labor, and 2 hours of machine time. The profit for each unit of regular clock sold is RM 3.00, while the profit for each unit of alarm clock sold is RM 8.00. Determine how many units of each type of clock should be manufactured every day so that the profit is maximum? Solution: First of all we need to define the decision variables involved: Let Let X1 = number of regular clocks to manufacture per day, X2 = number of alarm clocks to manufacture per day. Now we should construct a table to present the total resources available, the required resources for each clock, and the profit obtained if each unit of the respective clocks is sold. Refer to Table 3.1. The manager must now determine the values of X1 and X2. Table 3.1: The resources, the requirements, and the profits Resources X1 X2 Total resources required (regular clock) (alarm clock) available Labor hours Machine hours Profit per unit 2 6 3 4 2 8 1600 1800 The objective function is to maximize the profit: ⇒ Maximize profit 3X1 + 8X2 Subject to the constraints: Labor hours constraint : 2X1 + 4X2 < 1600 Machine hours constraint non-negativity conditions : 6X1 + 2X2 < 1800 : X1 > 0, X2 > 0. This is a two-variable Linear Programming problem. The manager’s job is to determine how many regular clocks (X1) and how many alarm clocks (X2) to produce in order to maximize the profit. We can solve this problem using a graphical method. 56 Example 3.2 A manufacturer produces two types of cotton cloth: Denim and Corduroy. Corduroy is a heavier grade cotton cloth and, as such, requires 3.5kg of raw cotton per meter whereas Denim requires 2.5kg of raw cotton per meter. A meter of Corduroy requires 3.2 hours of processing time, and a meter of Denim requires 3 hours. Although the demand for Denim is practically unlimited, the maximum demand for Corduroy is 510 meters per month. The manufacturer has 3250kg of cotton and 3000 hours of processing time available each month. The manufacturer makes a profit of RM4.50 per meter from Denim and RM6 per meter from Corduroy. The manufacturer wants to know how many meters of each type of cloth to produce to maximize profit. (a) Formulate a linear programming model for the above problem. Solution: Let X1 be the quantity of denim cloth type to be produced Let X2 be the quantity of corduroy cloth type to be produced Resources required X1 (Denim) 2.5 3.0 4.5 Maximize Profit = 4.50X1 + 6.00X2 Subject to constraints: 2.5X1 + 3.5X2 < 3250 (cotton cloth constraint) 3.0X1 + 3.2X2 < 3000 (processing time constraint) X1 > 0 0 < X2 < 510 (the production of courdroy should not exceed the maximum demand of 510 meters) X2 (Corduroy) 3.5 3.2 6.0 Total resources available 3250 3000 Cotton cloth Processing time Profit 57 Profit is made at RM 3 for each batch of cherry fizz and RM 4 for each batch of lemon pop. They decided to set up a soft drink stall in Kuala Lumpur. and X2 = number of batches of lemon pop to produce. There are 30 liters of ingredient A and 55 liters of ingredient B available for use. The resources required to produce the drinks are ingredient A and B. the organization has decided not to produce more than 4 batches of lemon pop at any one time. X2 < 4 (not more than 4 batches of lemon pop to be produced) 58 . X1 > 0. Resources required Ingredient A Ingredient B Profit Cherry fizz (X1) 5 5 3 Lemon pop (X2) 3 11 4 Resources Available 30 55 From the table we can write the mathematical inequalities as follows: Constraints: 5X1 + 3X2 < 30. How many batches of each type should be produced every time so that the profit is maximum? Solution: Define the decision variables: Let X1 = number of batches of cherry fizz to produce. 5X1 + 11X1 < 55. Due to some marketing strategy. while to make one batch of lemon pop requires 3 liters of ingredient A and 11 liters of ingredient B.Example 3.3 A non-profit organization is raising fund for AIDS victims. They plan to sell two types of soft drinks namely cherry fizz and lemon pop. To produce one batch of cherry fizz requires 5 liters of ingredient A and 5 liters of ingredient B. However.Objective function: The total profit made is 3X1 + 4X2. 5X1+ 11X2 < 55. this technique is not commonly employed since most LP problems involve more than two variables. In the graphical solution method. 59 . 3.4 GRAPHICAL SOLUTION TO A LINEAR PROGRAMMING PROBLEM If a linear programming problem involves only two decision variables (X1 and X2).1. X1 > 0. the solution may be found using the graphical technique. and this is to be maximized Thus. what you need to do is: a) b) c) d) Obtain the linear equation for each constraint in the problem Plot the linear equations of the constraints on the graph Determine the feasible region based on the inequalities in (a) Obtain the feasible solution points based on the feasible region in (c) e) Obtain an optimal solution for the problem by applying each feasible solution point into the objective function. the technique is illustrated using the problem in Examples 3. the linear programming model may be written as: Maximize 3X1 + 4X2 subject to: 5X1 + 3X2 < 30. In practice. X2 < 4. the optimal solution occurs at the feasible solution that yields the maximum value. input each coordinate of the feasible solution into the objective function and determine its value.4. the optimal solution occurs at the feasible solution that yields the minimum value. The coordinates that fall in the feasible region are the feasible solutions for the problem at hand. that is the common region which belongs to the inequalities. ≤ etc) of the constraints. For the maximization objective function. x2 > 0. Secondly. And for the minimization objective function. : 2x1 + 4x2 < 1600 -----(1) : 6x1 + 2x2 < 1800 -----(2) : x1. Finally. 60 . plot all linear equations of the constraints on the graph. determine the feasible region. Find the optimal solution to the problem by using the graphical method. The objective is to maximize the profit: ⇒ Maximize 3x1 + 8x2.1.3.1 THE FEASIBLE REGION AND CORNER POINTS METHOD Example 3. shade the area covered by the respective inequality (≥ . Thirdly. Solve all intersections between constraints and also determine their coordinates.4 Consider the problem in Example 3. Find all coordinates (X1 and X2) that fall in the feasible region. The feasible region is the common area or the area that overlapps all constraints. Subject to: The Constraints: labor hours machine hours non-negativity conditions Solution: Firstly. then we have 2X1 + 4(0) = 1600 ↔ 2X1 = 1600 or X1 = 800 Then (800. In order to draw any equation. Now. Constraint Equation (1) 2X1 + 4X2 = 1600 Let X2 = 0. Constraint Equation (2) 6X1 + 2X2 = 1800 Let X2 = 0. 400) will be the second coordinate on the graph.400) 400 (300. 900) will be the second coordinate on the graph. 0) will be the first coordinate on the graph. then we have 6(0) + 2X2 = 1800 ↔ hence 2X2 = 1800 or X2 = 900 Then (0. X2 (0. we need to obtain at least two coordinates from the respective equation.4. Now. you can draw the linear equation 2X1 + 4X2 = 1600 by connecting the two coordinates. then we have 6X1 + 2(0) = 1800 ↔ 6X1 = 1800 or X1 = 300 Then (300. Let X1 = 0. you can draw the linear equation 6X1 + 2X2 = 1800 by connecting the two coordinates. Let X1 = 0.0) 0 X1 61 .0) (800. 0) will be the first coordinate on the graph. The first thing to do is to draw both equations (Equation 1 and Equation 2) on the same graph.Applying to Example 3. then we have 2(0) + 4X2 = 1600 ↔ 4X2 = 1600 or X2 = 400 Then (0.900) 900 (0. we will find the two equations intercept with one another. 300) 62 . substitute X1 = 200 into equation (2): 200 + 2X2 2X2 2X2 Therefore = 800 = 800 – 200 = 600 = 1000 X2 = 300 So the interception point beween the two equations occurs at (200.Once we have plotted the two constraints. How to estimate the coordinate of interceptions between the two equations: The equation of constraints: Machine Hours Labor Hours Non negativity conditions : 6X1 + 2X2 = 1800 : 2X1 + 4X2 = 1600 : X1 > 0. we obtain: 6X1 + 2X2 = 1800 ----(1) (-) X1 + 2X2 = 800 ----(2) 5X1 Therefore X1 = 200 Then. we have to solve the two equations simultaneously to determine the coordinate where the two equations intercept. X2 > 0 Divide the second equation (labor hours) by 2. In this case. we have: 1/2 [2X1 + 4X2 = 1600] » Machine Hours Labor Hours X1 + 2X2 = 800 : 6X1 + 2X2 = 1800 ----(1) : X1 + 2X2 = 800 ----(2) Subtract the two equations (1) – (2). 200 * = RM900 At (200. 300).4 X2 (0.000 At (300.200 occurs at (0. Applying the objective function ⇒ Maximize 3X1 + 8X2 on the feasible solution.900) 900 (0. and (300. 400). The maximum profit RM3. the feasible solutions are (0.300) (0. 300) the profit is 3(200) + 8(300) = RM3.0) 300 (800.1. the feasible solution only occurs at the coordinates in the feasible region.0) X1 800 From Figure 3. 400). 0) the profit is 3(300) + 8(0) The objective of the above linear programming problem is to maximize the profit. The feasible region is the region covered by all constraints requirement in the problem (in this case there are two constraints). we have: At (0.1: Feasible Region as defined by all constraints for Example 3. the company should decide to produce X1 = 0 (regular clock) and X2 = 400 (alarm clock) in order to maximize the profit based on the current constraints. 0). (200. Hence. 63 .Figure 3.400) 400 (200. For your information.0) 0 (300. 400) the profit is 3(0) + 8(400) = RM3. Chemical AA costs RM2 per liter and chemical BB costs RM3 per liter to manufacture. at least 6 liter of AA and 2 liter of BB must be produced. Each liter of AA requires 3gm of this material while each liter of BB requires 5gm. One of the raw materials needed to make each chemical is in short supply and only 30gm are available.2 SOLVING THE MINIMIZATION PROBLEMS Sometimes the problem is stated in terms of the cost of materials used in production instead of the profit contribution for each product.4. In this case the objective function should be changed to minimizing the cost. x1 > 6 (at least 6 litres of AA should be produced) x2 > 2 (at least 2 litres of BB should be produced) 3x1+ 5x2 < 30. Minimize total cost = 2x1+ 3x2 Define the constraints.3. x2 > 0 (the non-negativity constraints) 64 . Example 3.5 A firm produces 2 types of chemicals namely AA and BB. the problem becomes: Minimize total cost = 2x1 + 3x2 Subject to constraints: x1 > 6 (at least 6 liters of x1 to be produced) x2 > 2 (at least 2 liters of x2 to be produced) 3x1+ 5x2 < 30 (each x1 requires 3gm while each x2 requires 5gm and the supply is limited to 30gm in total) x1. How many liters of AA and BB should the firm produce in order to minimize the total cost? Solution: Define decision variables: Let x1 = number of liters of chemical AA to be produced Let x2 = number of liters of chemical BB to be produced Define the objective function. Thus. Over the week. we have: At (6. Figure 3. 2) : 2(20/3) + 3(2) = RM19. 2). 2) Applying the objective function 2X1 + 3X2. (6.2 = RM18. we should plot the equation of constraints. obtain the feasible region.3 Hence the optimal solution occurs at X1 = 6. 2) : 2(6) + 3(2) At (20/3. and finally apply the objective function on the feasible solution points to determine the optimal solution. X2 = 2 x1 = number of liters of chemical X to be produced x2 = number of liters of chemical Y to be produced 65 .Again.2: Feasible Region for Example 3.0 (minimum cost) = RM19. 12/5).3).5 From the graph (Figure 3. the feasible solution points. 12/5) : 2(6) + 3(12/5) At (6. the feasible solution points occur at (6. and (20/3. and a non-slip handle bar. X2 be the number of girls’ bikes produced per day. 66 . ASC plant’s overall production capacity is 400 bicycles per day. Assume the company does not retrench workers since the company believes its stable work force is one of its biggest assets. However realizing that the demand for boys’ bikes are more than girls’ bikes.6 AGM Super-Bike Company (ASC) has the latest product on the upscale toy market. ASC is confident that it can sell all the bicycles produced. girls’ bike at RM 350 and toddlers’ bike at RM 150. boys’. chromeplated chairs. we need to define the decision variables as follows: Let: X1 be the number of boys’ bikes produced per day. each girls’ bike requires 2. RM 35 per girls’ bike and RM 20 per toddlers’ bike. the marketing manager insists that ASC must produce not more than 200 units of girls’ bikes. brackets and valves. Solution: First of all. who work 8-hours per day. 40% of the price ASC receives from the girls’ model and 30% from the toddlers’ model. Each boys’ bikes requires 3 labor hours. girls’ and toddlers’ bike in bright fashion colors. but excluding painting and packaging are RM 45 per boys’ bike. Production costs other than labor. ASC currently employs 120 workers.4 hours and each toddlers’ bike requires 2 hours to complete. and not more than 100 units of toddlers’.Example 3. Due to the best sellers market for high quality toys. boys’ bike at RM 420. X3 be the number of toddlers’ bikes produced per day. (a) Formulate the above Linear Programming problem. Painting and packaging cost is RM 25 per bike regardless of the model. a strong padded frame. at the following prices. The company’s accountant has determined that direct labor costs will be 45% of the price ASC receives from the boys’ model. 0 Variables X2 X2 2. X3 > 0 (non-negativity constraints) 67 .0 Resources 400 200 100 960 hours 120 workers x 8 hours Maximize profit: 161X1 + 150X2 + 60X3 Subject to (constraints) : X1 + X2 + X3 < 400 (capacity constraints) X2 < 200 (production constraint for X2 girl’s bike) X3 < 100 (production constraint for X3 toddler’s) 3 X1 + 2. X2.Variables Selling price (RM) Labor Cost Production Cost Painting Cost Net Profit X1 420 -189 -45 -25 161 X2 350 -140 -35 -25 150 X3 150 -45 -20 -25 60 Requirements X1 X2 X3 X2 X3 Labor Objective function is: X1 3.4 X3 X3 2.4 X2 + 2 X3 < 960 (labor hours constraint) X1. 7 Toy Race Car Company manufactures three different race cars called Racer_1. X2 be the number of Racer_2 to be produced. and RM4 respectively. Solution: Let X1 be the number of Racer_1 to be produced. X3 be the number of Racer_3 to be produced. 600 hours of labor. The above informations are summarized in the following table: Resources Technical Labor Administration Profit X1 1 10 2 10 X2 2 4 2 6 X3 3 5 6 4 Total Resources 100 600 300 The objective function: Maximize profit 10X1 + 6X2 + 4X3 Subject to (constraints): 1X1 + 2X2 + 3X3 < 100 (Technical services constraints) 68 . RM6. and administration (300 hours). labor (600 hours). The profits for each unit sold for Racer_1. Product Racer 1 Racer 2 Racer 3 Technical Services (hour) 1 2 3 Labor (hour) 10 4 5 Administration (hour) 2 2 6 Unit profit (RM) 10 6 4 The resources available for each week are 100 hours of technical services. and Racer_3 are RM10. The resources required to make the products are technical services. Racer_2. labor and administration. and Racer_3. (a) Formulate the problem as a linear programming problem. The limited resources available are technical services (100 hours). Racer_2. Table below provides the requirements for the production.Example 3. and 300 hours of administration. Case 4: Alternate Optimal Solutions Two or more feasible solutions that give the same optimal value.3 SPECIAL CASES IN LINEAR PROGRAMMING (WHEN USING THE GRAPHICAL APPROACH) Case 1: Infeasibility A condition that arises when there is no solutions to a linear programming that satisfies all of the constraints. if the linear 69 . graphically. Case 2: Unboundedness A linear programming problem will not have a finite solution. 3. we saw how a linear programming problem with only two variables (X1 and X2) was solved by using the graphical method. no feasible region exists. the feasible region is open-ended. graphically.5 THE SIMPLEX METHOD In the previous section. simply one that does not affect the feasible solution region. However. Case 3: Redundancy The presence of redundant constraints.4. when the slope of the objective function is parallel to one of the constraints (infinitely many solutions). X2. graphically. X3 > 0 (Non-negativity constraints) 3.10X1 + 4X2 + 5X3 < 600 (Labor constraints) 2X1 + 2X2 + 6X3 <300 (Administration constraints) X1. In addition only Mathematicians and Statisticians have the capacity and patience to discuss such process. and this is precisely what the simplex method does. the graphical method is impossible to apply since normal graphs only possess two dimensions.. In this section. we have seen that it is only necessary to examine the feasible solution points obtained in the feasible region. For us non-Mathematicians and nonStatisticians. The manager only needs to know how to intepret the results in the final table for their decision making.5. In particular. it is relevant for us to be more interested in how to set-up the actual problem into the initial simplex table. it is essential to write the problems in their standard form: 70 . called the simplex method. X2. 3.Xn). we will not discuss the process in details since its complexity requires a longer medium of instruction. Essentially the method repeatedly searches for a better solution point until the objective function cannot be improved further even if we move to another feasible solution point. an algebraic method is developed to solve Linear Programming (LP) problems with more than two variables. by means of computational routines. ….programming problem has more than two variables (X1. for a two-variable problem. In this section. assuming the process would be carried out by a computer package called “QM for Windows” (manual computation is very tedious). The computer package is available in the market and is being used by managers to analyze their data.1 HOW TO SET UP THE INITIAL SIMPLEX SOLUTION Before setting up the initial simplex table. is an iterative process that begins with the initial solution and. and the interpretation of the results in a final table in term of management problem. moves to an improved solution until the optimal or final solution is reached. The technique. 7 Objective function: Maximize profit 10X1 + 6X2 + 4X3 Subject to (constraints): 1X1 + 2X2 + 3X3 < 100 (Technical services constraints) 10X1 + 4X2 + 5X3 < 600 (Labor constraints) 2X1 + 2X2 + 6X3 < 300 (Administration constraints) X1. S2. Maximize profit 10X1 + 6X2 + 4X3 + 0S1 + 0S2 + 0S3 1X1 + 2X2 + 3X3 + 1S1 + 0S2 + 0S3 10X1 + 4X2 + 5X3 + 0S1 + 1S2 + 0S3 2X1 + 2X2 + 6X3 + 0S1 + 0S2 + 1S3 = 100 = 600 = 300 Profit contribution for each variable (X1.Xn) must be non-negative (> 0). X2. and resources 3 (administration hours) respectively. d) All constraints must be in the form of mathematical equations. Example 3. and S3 to represent the unused resources 1 (technical services hours). X2. we write the above equation using the standard form of simplex method. S2. Now.. S3) 10 6 4 0 0 0 X1 X2 X3 S1 S2 S3 1 2 3 1 0 0 10 4 5 0 1 0 Quantity 100 600 71 . We need to introduce the slack variables to represent the “unused resources”. we need to introduce three slack variables namely S1. * Slack variables will become an Identity Matrix in the initial solution.a) b) c) e) All decision variables (X1. resources 2 (labor hours).8 (APR 2001) Obtain the initial simplex table for Example 3. X3 > 0 (Non-negativity constraints) Solution: Since we have 3 resources in each of the equations. X3. X2. The objective function is either to maximize (for profit) or to minimize (for cost) The Right-Hand-Side of each constraint is a non-negative (quantity of resources). S1. …. To prepare the problem for the initial simplex table.Z j 10 x1 1 10 2 0 10 6 x2 2 4 2 0 6 4 x3 3 5 6 0 4 0 S1 1 0 0 0 0 0 S2 0 1 0 0 0 0 S3 0 0 1 0 0 Resources Quantity 100 600 300 0 0 0 Example 3. S3 > 0. x2 > 0. Maximize (profit): Subject to 3x1 + 8x2 2x1 + 4x2 < 1600 (labor hours constraint) 6x1 + 2x2 < 1800 (machine hours constraint) x2 < 350 (materials constraint) x1.2 2 6 0 0 1 300 The initial simplex table for Example 3. S2. the model should be written as: Maximize (profit): 3x1 + 8x2 + 0S1 + 0S2 + 0S3 Subject to 2x1 + 4x2 + 1S1 + 0S2 + 0S3 = 1600 6x1 + 2x2 + 0S1 + 1S2 + 0S3 = 1800 0x1 + 1x2 + 0S1 + 0S2 + 1S3 = 350 x1. S2. Where S1.2. S1. Table 3.9 Solve the following problem using the simplex method.2: Initial Simplex Table for Example 3.9 Cj profit Solution Mix S1 S2 S3 Zj 3 x1 2 6 0 0 8 x2 4 2 1 0 0 S1 1 0 0 0 0 S2 0 1 0 0 0 S3 0 0 1 0 Resources Quantity 1600 1800 350 0 72 0 0 0 . S3 are called Slack Variables (representing unused resources) The above equations are now expressed in Table 3. x2.8 Cj profit Solution Mix S1 S2 S3 Zj Cj . Cj . the Cj . In this case the profit is 0(1600) + 0(1800) + 0(350) = 0 Meanwhile. we can intepret the initial solution as follows: S1 = 1600. if there are variables that have positive signs in row Cj. 3. select the highest number – and the variable which corresponds to this is called 73 . x2 = 0 The total profit = 0. From the above table. S2 = 1800. S3 = 350 ⇒ All resources remain in original quantity (unused) since the production has not started The solution for decision variables x1 = 0. To compute these Cj . For the maximization problem. For example.2 SIMPLEX SOLUTION PROCEDURES The Simplex Algorithm 1. and S3 respectively.Zj under x1 is 3 – 0 = 3.Z j 3 8 0 0 0 Always remember that the optimal solution occurs at the identity matrix. The Value of Zj and Cj .Zj values. The same steps are used to compute the values for S1. the Cj . simply deduct the Zj value obtained for each decision variable from the Cj value at the top of every variable.Zj value in each column represents the net profit for each variable. The the Cj . S2.Zj row: The Zj value in each column in the table is the sum of each number in the C j column multiplied by the corresponding number in the decision variable column.5. For example: The Zj value under variable x1 is the summation of 0(2) + 0(6) + 0(0) = 0 The Zj value under variable x2 is the summation of 0(4) + 0(2) + 0(1) = 0 The Zj value for the quantity column provides the profit obtained in the current solution.Zj under x2 is 8 – 0 = 8.Zj. 2. 3. When this is done. This variable is called the leaving variable. Select the smallest ratio. The variable which corresponds to this smallest ratio. it should be firstly noted that the column of a variable in the current solution mix are all zeros except for a 1 in the row corresponding to this variable.5 -1. The pivot number is the number at the intersection of the column of the entering variable and the row of the leaving variable. an optimal solution has been reached).Z j 0 S1 0. return to (1). will be the variable to be omitted from the solution mix. Divide every number in the pivot row by the pivot number and this row (R) is then written in the next table in the same row position as it was in the previous table.5 -3 0 1. For the problem at hand. this algorithm should be applied twice to obtain a final (third) simplex table. (If all numbers in the row Cj . The final simplex table (using simplex algorithm) is shown in Table 3. This is obtained by adding and subtracting suitable multiples of row R from the rows of the previous table.Zj are either 0 or negative values. This condition must now be obtained for the variable that is entered in (1).9 Cj 3 0 8 Solution Mix 3 x1 1 0 0 3 0 8 x2 0 0 1 8 0 x1 S2 x2 Zj Cj .the entering variable. 4.3 Table 3. Find the ratios of the resources quantity to the numbers in the column of the entering variable.3 Final Simplex Table for Example 3.5 0 S2 0 1 0 0 0 0 S3 -2 10 1 2 -2 Resources Quantity 100 500 350 3100 74 . To obtain the remaining rows in the new table. this can be recognized when there is a zero value. S3 = 0. S1 = 0. since this knowledge is very useful when performing a sensitivity analysis on the solution.3).3 INTERPRETATION OF SIMPLEX SOLUTION From the final table (Table 3.e. the management should allocate only 1300 hours of processing time. 75 . an arithmetic error would occur when solving an LP problem manually. This optimal solution (as expected) agrees with those found by graphical method. It is clear that with the number of arithmetic calculations involved in the simplex method. there may be more than one set of values for the decision variables that gives the optimal value of the objective function. it is important to understand how the simplex algorithm works. (i. then the problem is said to have no feasible solution.5. S2 = 500. the optimal solution for this production is as follows: ⇒ X1 = 100. Case 2 : Alternative optimal solution For some LP problems. with total profit = $ 3100. Nevertheless.4 SPECIAL CASES IN LINEAR PROGRAMMING FIVE special cases may arise when solving an LP problem using the simplex method.3. For the next production. mean the labor hours and assemblies hours are all used (balance = 0). For this reason. i. Case 1 : No feasible solution (infeasibility) If there are no values of the decision variables that satisfy all of the constraints. we will be solving the LP problems with the aid of a computer package.e. 3. In the simplex method. only 1800-500 = 1300 hours of processing are used). This case can be recognized in the simplex method when one or more artificial variables appear with a positive value in the final table. To maximize the total profit.5. X2 = 350. the company should produce 100 units of X1 and 350 units of X2. S2 = 500 means 500 hours of processing time left unused at the end of production process. S1 = 0 and S3 = 0. This ‘degeneracy’ in LP problem may cause a situation. Case 3 : Unbounded solution In some LP problems. At the next iteration. in which the simplex iteration enter a loop and repeat the same sequence of iterations without reaching the optimal solution. A solution in which one or more of the variables in the solution mix has a value of zero is said to be degenerate. the objective function has a finite optimal value.In row Cj .Zj of the final table under a variable that is not in the final solution mix. it was stated that if there was a tie for minimum ratio to decide the variable leaving the solution mix then the tie should be broken arbitrarily. An alternative optimal solution may be found by entering the said variable into a solution mix and performing a further iteration on the table. at any iteration for a maximization (minimization) problem there is a positive (negative) number in row Cj .e. there is a positive (negative) in row Cj . 3.Zj in that column for a maximization (minimization) problem. such LP problems are said to have an alternative optimal solution. there are zero and/or negative numbers. the value of the objective function may be made as large (maximization) or as small (minimization) as we please.Zj) of any table. The solution will be unbounded only if in addition to this condition. the values of those variables in the tie that did not leave the solution mix will be zero. fortunately this does not happen very often in practice. An unbounded feasible region can be recognized in LP problem when in any column (excluding row Cj . Such problems are said to have an unbounded solution. They can be recognized in the simplex method when.6 THE DUAL IN LINEAR PROGRAMMING 76 .Zj and zero and/or negative number in the column underneath. Case 4 : Unbounded feasible region but bounded optimal solution Although the feasible region is unbounded. i. the optimal solution is bounded. Case 5 : Degenerate solution In Step (2) of the simplex algorithm. however. 6. Looking at the “dual” perspective is useful for two reasons: 1) There is less computation required to solve the dual problem compared to solving the primal problem 2) The solution of the dual problem has a meaningful economic interpretation. One problem is called “primal” while the other is called “dual”. the dual of the dual problem would be the primal problem. the dual is simply looking at the same problem but from the “opposite perspective”. The primal problem is maximization. The coefficient of objective function in the primal would become the quantity of the constraint in the dual 6. The quantity of resources of the constraints in primal problem would become the coefficient of the objective function in the dual problem. 3. For example. For example.1 THE RELATIONSHIP BETWEEN THE PRIMAL AND DUAL PROBLEMS 1. 5.10 Primal problem Maximize 3X1 + 4X2 77 . Example 3. Each constraint of the primal would become a variable in the dual problem. In an easier term. 2. 3. The maximization of primal problem has < (less or equal) constraints while the minimization of dual problem has > (greater or equal constraints). From a symmetry.Every linear programming problem has a twin problem associated with it. we can have the actual prices that should be paid for each unit of extra resources if the company wants to continue with the production (shadow price). while the dual problem is minimization. in the dual solution. 4. the primal is looking at maximizing profit but the dual is looking at minimizing resources. 4X2 + X3 < 10 : Y2 3. X2 > 0. X3 > 0. * The optimal values of the primal and dual objective functions are equal. : Y3 Dual problem Minimize 30Y1 + 55Y2 + 4Y3 subject to 5Y1 + 5Y2 >3 3Y1 + 11Y2 + Y3 > 4 Y1.6. The corresponding dual problem is Minimize subject to 40Y1 + 10Y2 2Y1 + 3Y2 > 3 Y1 – 4Y2 > 2 5Y1 + Y2 > 4 Y1. X2.2 THE OPTIMAL SOLUTION FOR DUAL VARIABLES 78 . Y2 > 0. Y2.subject to 5X1 + 3X2 < 30 : Y1 5X1 + 11X2 < 55 : Y2 X2 < 4 X1. 3X1 . Y3 > 0 Example 3. then the minimization of dual problem will have ‘>’ constraints.11 Maximize subject to 3X1 + 2X2 + 4X3 2X1 + X2 + 5X3 < 40 : Y1 X1. * If the maximization of primal problem has ‘<’ constraints. Zj. S2.. Example 3. S2. S3 > 0.. X2.Sn respectively in the row Cj . …. …..Yn) can be obtained from the final simplex table of primal problem under the resources columns (S1. Since the objective function of dual is to minimize resources.. …. Y2.4: Initial Simplex Table for Example 3.9 (problem in the primal form) Cj Solution 0 0 0 Resources 3 8 X1 X2 S1 S2 S3 Mix Quantity 0 0 0 S1 S2 S3 Zj Cj . Hence.Yn can be found under S1.Zj (ignore the negative sign) of the final simplex table (primal). the optimal solutions for Y1. …. Y2. Table 3. S1. the optimal value of Y1 can be obtained under column S1 in row Cj . 79 .Sn).. The dual variable Y2 corresponds to the second primal constraint that contains the slack variable S2 (unused resources 2).The optimal solution for the dual variables (Y1. S2.12 Using Example 3..9: Maximize (profit): 3X1 + 8X 2 + 0S1 + 0S2 + 0S3 Subject to 2X1 + 4X2 + 1S1 + 0S2 + 0S3 = 1600 : Y1 6X1 + 2X2 + 0S1 + 1S2 + 0S3 = 1800 : Y2 0X1 + 1X2 + 0S1 + 0S2 + 1S3 = 350 : Y3 X1.Z j 2 6 0 0 3 4 2 1 0 8 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 1600 1800 350 0 The dual variable Y1 corresponds to the first primal constraint that contains the slack variable S1 (unused resources 1). The dual variable Y3 corresponds to the third primal constraint that contains the slack variable S3 (unused resources 3). 5).9 (problem in the primal form) Cj 3 0 8 Solution Mix X1 S2 X2 Zj Cj .4).00 (Maximum) 3.5 (Y1) S2 (0) 0 1 0 0 0 (Y2) S3 (0) -2 10 1 2 -2 (Y3) Resources Quantity 100 500 350 3100 * Y1.5 -1.12 is: Minimize (resources): 1600Y1 + 1800Y2 + 350Y3 Subject to 2Y1 + 6Y2 Y1. Y2.5.That is.6.Zj under S1. we get Y1= 1. Y2 = 0 and Y3 = 2. X2 = 350 units S1 = 0 (all resource 1 is used) S2 = 500 (500 units of resource 2 is unused) S3 = 0 (all resource 3 is being used) Maximum profit = 3 (100) + 8 (350) = RM 3. Table 3. Y1 = 0.5 -3 0 1. Y3 are located in row Cj . Y3 = 0 in the initial simplex table (Table 3. Y2.Z j X1 (3) 1 0 0 3 0 X2 (8) 0 0 1 8 0 S1 (0) 0. S2. From Table 3. >3 4Y1 + 2Y2 + Y3 > 8 80 .3 INTERPRETATION OF THE DUAL VARIABLES The corresponding dual problem for Example 3. we can intepret the solution as follows: X1 = 100 units. 2. Y2. Y3 are the shadow prices for resources 1. Y2 = 0. Y3 > 0 The optimal values for dual variables Y1.100.5: Final Simplex Table for Example 3. S3 columns respectively in the final simplex table (Table 3. and 3 respectively From the final simplex table of the primal.5. Y3: ⇒ Each additional unit of the first resource (labor hours) would contribute RM1. 2. That is. ⇒ Each additional unit of the second resource (processing hours) contributes RM0. Y2 = 0 and Y3 = 2. hence all negative values are changed to positive values. RM1. The optimal value of a dual variable which is associated with a primal constraint gives an indication of how much the objective function (profit) increases for every unit increase in the resources provided that the current solution remains feasible and optimal. Y2 = 0 and Y3 = 2 into the objective function: Minimize (resources): 1600Y1 + 1800Y 2 + 350Y3 . These values indicate the actual worth of each resources if the management wants to obtain additional resources to continue production. The dual variable Yi may be interpreted as the per-unit contribution of the ith resource (Si) to the value of the objective function. This is because of the current production which 81 .12. In Example 3. In other words.50 is the shadow price for labor hours. the value of the objective function is increased by Yi unit as long as the current solution mix is feasible. This value (RM3100) is equivalent to the maximum profit under the primal form. Y3 > 0). and 3 respectively.50 per hour.5. Y2.5) + 1800 (0) + 350 (2) = RM 3100 (Minimum). In other words. the solution must always be greater than zero (Y1. if the resources of a primal constraint is increased by one unit. Remember. at the moment any increase in processing hour does not have any monetary value or does not contribute anything to the profit. Y2.If we input the values of Y1= 1. we will get : The total resources used = 1600 (1.5.50 to the profit in the primal objective function. the optimal values of the dual variables are Y1 = 1. Hence. Y3) are called “shadow prices” for resources 1. The interpretation of Y1. if the management wants to obtain additional labor (overtime hours). Y2. the cost should not be higher than RM1. the optimal values of the dual variable (Y 1. So. 00. In other words.00 per unit since RM2. this additional resource should not cost more than RM2. labour and administration. Product Racer 1 Racer 2 Racer 3 Technical Services (hour) 1 2 3 Labor (hour) 10 4 5 Administration (hour) 2 2 6 Unit profit (RM) 10 6 4 The resources available for each week are 100 hours of technical services. 82 . Any increase of this resources is unnecessary. (a) Formulate the problem as a linear programming problem. EXERCISES 1 (APR 2001) The Toy Race Car Company manufactures three different race-cars called Racer 1. and 300 hours of administration. Racer 2. ⇒ Each additional unit of the third resource (alarm assemblies) would contribute RM2.has resulted in the unused resource 2 (S2 = 500). The resources required to make the products are technical services. 600 hours of labor. and Racer 3.00 is the shadow price for resource 3. The table below gives the requirements for the production. 2 (SEPT 2002) a) Summit Events Management plans to rent out sales booths during the Ramadan Food and Fashion Fair. There should be a maximum of 100 booths. Cj 6 10 0 Solution Mix X2 X1 S3 Zj Cj . obtain the solution of the dual problem. how much is the spare capacity? (iii) Write down the dual of the problem. and administration hours respectively. of man-hours Type 1 RM 400 1 Type 2 RM 200 2 Type 3 RM 300 1 83 . labor. What is the total profit? (ii)Which resource is not fully utilized? If so. S2.(b) The optimal simplex table for the above problem is given below. Type 2. The cost per unit and the number of hours to set up each booth are as follows: Cost per unit No. comprising three different types: Type 1. However. The rental will be at a flat rate of RM500 each.32 (i) Specify the optimal daily production levels for the company for the three products. The S1.Zj 10 X1 0 1 0 10 0 6 X2 1 0 0 6 0 4 X3 5/6 1/6 4 20/3 -8/3 0 S1 5/3 -2/3 -2 10/3 -10/3 0 S2 -1/6 1/6 0 2/3 -2/3 0 S3 0 0 1 0 0 Resources Quantity 66.33 100. The booths are to be set up at a popular shopping mall. (iv) Based on the table.67 33. the cost of setting up the booth varies depending on the type of booth.00 733. S3 are the slack variables for technical services. and Type 3. v.4 RM0 S3 0 Resources 32 84 . A standard shipment can be packaged in a class A container. or a class C container. iii. What would be the effect on the profit if 5 more man-hour is made available? Puncak Distribution packages distributes industrial supplies. ii. How much of the budget will be used to reach the optimal solution? Formulate the dual and determine the optimal solution. The final simplex table of the problem is given below: Cj 9 Solution Mix X1 RM9 X1 1 RM7 X2 0 RM8 X3 0. Each shipment requires a certain amount of packing materials.8 RM0 S1 0. Formulate a linear programming problem to determine the number of each type of booth to maximize profit: Let: X1 = Quantity of Type 1 booth X2 = Quantity of Type 2 booth X3 = Quantity of Type 3 booth S1=slack variable for the number of booths S2=slack variable for the budget S3=slack variable for man-hour b) The following is the final simplex table for the above problem.Summit has a RM32. and one unit of class C container yields a profit of RM8.6 RM0 S2 -0. Cj 200 0 300 Solution Mix X3 S2 X2 Zj Cj-Zj i. material I and material II ( in meters). One unit of class A container yields a profit of RM9. iv. (c) 100 X1 1 10 0 200 -200 300 X2 0 0 1 300 0 200 X3 1 0 0 200 0 0 S1 2 -40 -1 100 -100 0 S2 0 1 0 0 0 0 S3 -1 10 1 100 -100 Resources 40 8000 60 State the optimal solution and interpret the values of all variables.000 budget to cover the cost. and packing time (in hour). There are 4 workers available to set up the booths. a class B container. one unit of class B container yields a profit of RM7. Determine the maximum profit. Each of them works 8 hours per day for 5 days. 3 (MAR 2004) Muarlite Company produces three types of desk lamps: Lamp A. (b) The optimal solution to the above problem is given by the following simplex table. There are 500 units of fiberglass. Lamp B requires 2 units of fiberglass and 8 units of wood.2 0. (a) Formulate a linear programming model for this problem with the objective of maximizing the weekly profit.5 0 S3 -0. Lamp C requires 1 unit of fiberglass and 2 units of plastic.2 2. S2 and S3 are the slack variables representing the unused amount of packing materials (I and II) and packing time respectively. class B. respectively. The lamps require three limited resources: fiberglass. The resource requirements for each lamp are as follows: Lamp A requires 1 unit of fiberglass.6 -2.4 -0.7 0 X2 S3 Zj Cj .6 -0. Should the firm do so? (iii) What is the optimal solution of the dual problem? Interpret.8 -0. 460 units of plastic and 840 units of wood available each week. Lamp B and Lamp C are RM30.6 -0. plastic and wood.8 -4. and class C containers respectively. (i) (ii) What is the optimal number of each class of containers and the total profits? Puncak is considering getting an extra packing material I at a cost of RM2.25 Resources 60 85 .00 per meter.Z j 0 0 9 0 1 0 7 0 0. Cj 0 Solution Mix S1 30 X1 -1 10 X2 0 40 X3 0 0 S1 1 0 S2 -0. RM10 and RM40. S1. Lamp B and Lamp C. 3 units of plastic and 2 units of wood.6 -0. X2 and X3 represent the number of class A.6 12.6 0 1 0 0 12 6 372 X1.8 -0. Profits from the sales of Lamp A. 2 5.32 -0.32 -0. Obtain the optimal solution of the dual and briefly explain the values.8 -2. S2 and S3 are the respective amounts of slacks of fiberglass. plastic and wood. Lamp B and Lamp C.00 per unit? Explain your answer.25 -1.12 0.2 Quantity 12 8 30 Where S1. Cj 40 0 50 Solution Mix X2 S2 X1 Zj Cj .5 0.125 1. S1. Formulate the dual of the problem. iv) Are there any changes in the total profit when additional units of Resource 2 are obtained? Explain your answer. 86 . 4 (OCT 2004) The objective of a linear programming problem is to maximize profit. (i) Obtain the quantity of each type of desk lamps to produce per week and determine the total weekly profit. 2 and 3 respectively.2 -5. Explain why this solution is already optimal. Obtain the optimal solution and determine the total profit.5 0 20 -20 0 0.Zj 50 X1 0 0 1 50 0 40 X2 1 0 0 40 0 0 S1 0.12 0.Z j 1. The optimal solution to the problem is given by the following simplex table.25 230 105 X1.40 10 X3 X2 Zj Cj .25 62.5 0 1 10 0 1 0 40 0 0 0 0 0 0. X2 and X3 are the respective productions of Lamp A. S2 and S3 are the slacks for Resources 1.2 2. i) ii) iii) Write down the objective function for the above problem. v) Is it worthwhile to obtain Resource 3 at RM5. (ii) (iii) (iv) State the resource(s) that is or are not fully utilized in the optimal production.5 -32.8 0 S2 0 1 0 0 0 0 S3 -0. 5 (SEPT 2001) A manufacturer produces two types of cotton cloth: Denim and Corduroy. The manufacturer wants to know how many meters of each type of cloth to produce to maximize profit.50 per meter from Denim and RM6 per meter from Corduroy.5 X1 0 1 0 4.83 -1.83 0.5kg of raw cotton per meter.2 hours of processing time. processing time and demand for corduroy respectively. Corduroy is a heavier grade cotton cloth and. (i) Determine the optimal production level and calculate the maximum profit obtained. (a) Formulate a linear programming model for the above problem.5 6 Solution Mix S1 X1 X2 Zj Cj – Zj 4. (b) The final simplex table of the LP problem is shown below: Cj 0 4. X2 = number of meters of denim produced. S1.07 1 1. as such. The manufacturer makes a profit of RM4. The manufacturer has 3250kg of cotton and 3000 hours of processing time available each month.5 0 6 X2 0 0 1 6 0 0 S1 1 0 0 0 0 0 S2 -0.2 -1. and a meter of Denim requires 3 hours. S2 and S3 are the slack variables for cotton usage. requires 3. the maximum demand for Corduroy is 510 meters per month.33 0 1.5 0 S3 -0.5kg of raw cotton per meter whereas Denim requires 2. Although the demand of Denim is practically unlimited.2 Quantity 325 456 510 X1 = number of meters of corduroy produced. A meter of Corduroy requires 3.5 -1. (ii) How many cotton (S1) and processing time (S2) are left over the optimal solution? (iii) What is the maximum amount the manufacturer is willing to spend for an additional processing time? 87 . and Z must be sold to maximize the total sales? Formulate the problem as a linear programming model and develop an initial simplex table for the problem (Do not solve the problem) b. Design 2. a sales person is only allowed delivery expenses of not more than RM75 and selling time is expected not to exceed 30 working hours. The production manager of the formulated the equation given below. The goods are left with the customer at the time of sale.20.80 to deliver it to the customer. 6 (MAR 2005) a. and Design 3.(iv) What will be the effect on the optimal solution if the manufacturer could only obtain 3000 kg of cotton per month? Justify your answer. Y. Z is sold at RM12 each unit. Y is sold at RM5 each. and it takes a sales person 15 minutes to sell one unit. A marketing company is doing research in order to maximize its profit of selling products X. how many units of X. is in the process of introducing new design of vase namely Design 1. Indah Pottery Sdn Bhd. and Z are respectively RM2.80. a manufacturer of products made from clay. RM1. During the week. If the unit costs of product X. These products will be sold from door to door. Y. The products will be sold to the local market and to be exported to neighboring countries. X is sold at RM7 per unit. X1 = Quantity of Design 1 vase X2 = Quantity of Design 2 vase X3 = Quantity of Design 3 vase Maximize Z = 12X1 + 18X2 + 10X3 Subject to: 2X1 + 3X2 + 4X3 < 50 X1 + X2 + X3 < 0 X2 + 1. and RM4. and it costs RM0. Y.25. and Z. It takes a sales person 12 minutes to sell one unit of Z. It takes a sales person 10 minutes to sell one unit of X and it costs RM1 to deliver the goods to the customer.5X3 < 0 88 . 6 0. X2. the company already has orders for 500 units of component A that must be delivered.7 0 S1 0.00 respectively. B and C are RM8. while 89 . In fact.00.2 1.Zj 12 X1 1 0 0 18 X2 0 1 0 10 X3 0. 7 (NOV 2005) A company supplies three components to automotive manufacturers. 8 (APR 2006) a) C-Bizz Mart is a retail catalog store specializing in cosmetics. and X3 > 0 State the dual for this problem ii) The following is the final simplex table of the primal problem.2 0. Phone orders are taken each day by a pool of computer operators. Formulate the above Linear Programming problem. but up to 1000 units of component B can be sold.All X1. A permanent operator can process an average of 75 orders per day. The duration in minutes required on each machine are as follows: Phase Component A B C Moulding 16 5 8 Polishing 12 6 8 The moulding machine is available for 120 hours and the polishing machine is available for 110 hours. The profit per unit for component A. some of them are permanent while some are temporary staff. The components are processed on 2 phases: Moulding and Polishing.2 2. Cj 12 18 0 Solution Mix X1 X2 S3 Zj Cj .00 and RM9.2 0 S2 -0. No more than 200 units of component C can be sold.4 0. RM6. Complete the table by finding all values in the Zj and Cj – Zj rows.4 0 S3 0 0 1 Quantity 10 10 10 iii) Give the optimal solution of the primal and briefly interpret it.2 0. while a temporary operator averages 4.5 30 Y2 1 0 0 0 S1 0. The company receives an average of at least 600 orders per day.5 0 S2 0 1 0 0 S3 0 0 1 Quantity 500 200 600 90 . The company wants to know the number of permanent and temporary operators to hire in order to minimize costs. Cj 30 0 0 Solution Mix Y2 S2 S3 Zj Cj .25 0. X3 > 0 i) Write the dual for the above linear programming problem.75 -0. Formulate a linear programming model for the above problem.25 -0. 9 (OCT 2006) 20 Y1 1.Zj ii) Complete the above simplex table iii) State the optimal solution (including the minimum cost) for the dual.25 -0. A permanent operator is paid RM60 per day including benefits and a temporary operator is paid RM40 per day. The store wants to limit errors to 24 per day.3 orders with errors each day. The following is an incomplete final simplex table for the dual.a temporary operator can process an average of 52 orders per day.2 orders per day with errors. iv) State the optimal solution (including the maximum profit) for the primal. A permanent operator will process about 1. X2. b) Consider the following linear programming problem: Minimize Z = 2000X1 + 700X2 + 1600X3 Subject to: 5X1 + 2X2 + 2X3 > 20 4X1 + X2 + 2X3 > 30 X1. (APR 2007) a) ACE Sdn Bhd manufactures three products X1. The products must go through two production processes: Machining and Assembly.5 0. and X3 are RM150.0 2. The capacity in each production process is limited by the number of labor hours available as shown in the following table. and X3.5 Available X3 1. It makes a profit of RM50 on each GE25 and RM75 on each GE29. RM130. and RM250 respectively. Determine the production levels for GE25 and GE29 that would optimize the profit per shift. Each GE25 requires 2 man-hours in the production area and 1 man-hour in the assembly area while each GE29 requires 2 man-hours in the production area and 3 man-hours in the assembly area. During each shift.a) Xlent Electronics manufactures several products including 25-inch GE25 and 29inch GE29 televisions. X2.5 Profits for the products X1. X2.5 0. Formulate the above linear programming problem Solve the problem by using the graphical method Complete the following initial simplex table: Cj Solution Mix S1 S2 A1 Zj Cj-Zj X1 (3) 6 5 1 X2 (5) 3 7 1 S1 (0) 1 0 0 S2 (0) 0 1 0 A1 (-M) 0 0 1 Quantity Bi 300 200 50 ) 10. Labor hour for each product X2 0. i) Formulate the above problem as a linear programming model 91 . Xlent allocates up to 300 man-hours in its production area and 240 man-hours in its assembly area to manufacture the televisions.0 hours 700 600 Process Machining Assembly X1 0. 00 143. X2.75 243.75 375.25 0 87.25 1 -0.ii) iii) Write the dual of the problem Set-up an initial simplex table for the problem b) Given the following simplex table of a manufacturing firm: Cj 0 150 250 0 130 Solution mix S3 X1 X3 S4 X2 Zj Cj .5 -87.5 0 S2 0.5 -67.5 -0. and X3) that should be produced? Determine the total profit.00 i) What is the quantity of each product (X1.5 0 S3 1 0 0 0 0 0 0 0 S4 0 0 0 1 0 0 0 0 S5 -0.25 250.Zj 150 X1 0 1 0 0 0 150 0 130 X2 0 0 0 0 1 130 0 250 X3 0 0 1 0 0 250 0 0 S1 0.5 0 125 -125 Quantity 56.25 -0.25 1 67.5 0 0. (OCT 2007) Write down the standard form and construct the initial simplex table for the following linear programming problem: Maximize Z = 3X1 + 2X2 + X3 Subject to X1 + 2X2 + 2X3 < 30 2X1 + 3X2 + X3 < 50 92 . Calculate the minimum resources used.25 -0. Intepret the shadow price for Resource 5 State the resources that are not fully utilized. i) ii) v) v) 11. Obtain the solution for the dual problem.25 0 -0. X1 – 2X2 > 0 X1. X2. X3 > 0 12. using the corner point method. What is the optimal solution for the primal? Interpret. Explain why this table gives the optimal solution iii. Minimize Cost = X1 + 2X2 Subject to 1X1 + 3X2 > 900 8X1 + 2X2 > 1600 3X1 + 2X2 > 1200 X2 < 700 X1 X2 > 0 b) The following is the incomplete final simplex table for the dual. What is the total cost? 93 .Zj i. 10 Y1 1 0 0 24 Y2 0 1 0 50 Y3 0 0 1 0 S1 4 0 0 0 S2 -2 2 0 0 S3 -1 -1 1 Quantity 4 3 2 Complete the above simplex table ii. S2 and S3 are the slack variables for the dual Cj 10 24 50 Solution mix Y1 Y2 Y3 Zj Cj . (OCT 2007) a) Solve the following linear programming problem graphically. where S1. 5 1 Quantity 12 16 4 Where. A detached house costs RM60.000 to build and requires 300 m2 of land.000 and RM6. (APR 2008) a) The optimal simplex table for the linear programming problem is given in the following table: Cj Solution Mix S1 X2 X1 ZJ CJ-ZJ 12 X1 0 0 1 15 X2 0 1 0 14 X3 -1 1 0 0 S1 1 0 0 0 S2 2 1 -1 0 S3 -3 -0. From past experience.50 per hour in Department C.13. 94 . iii) The firm is considering hiring an extra labor on a part time basis at a rate of RM3. i) Formulate a linear programming (LP) model for this problem. the developer estimates the profits of a detached house and a semi-detached house to be about RM9.000 to build and requires 650 m2 of land and a semi-detached house costs RM40. ii) Write down the dual for this LP primal problem. detached and semi-detached within an overall budget of RM3 million. X1= number of units of Product 1 produced per week X2= number of units of Product 2 produced per week X3= number of units of Product 3 produced per week S1= Slack variable for labor hours in Department A S2= Slack variable for labor hours in Department B S3= Slack variable for labor hours in Department C i) Complete the optimal simplex table ii) Determine the optimal production level and the maximum profit obtained.000 m2 of land on which he is planning to build two types of houses.000 respectively. (OCT 2008) a) A housing developer has purchased 20. What would be the daily profit obtained by the firm? (Assume the firm operates 8 hours daily) 14. S2 and S3 are the slack variables of the three resources A.67 -1.22 0.Zj i) Complete the above simplex table ii) State the optimal solution.b) The following is the optimal simplex table for the linear programming problem of a company manufacturing two products X1 and X2. B and C respectively involved in the manufacturing process.00 per unit? Why? 40 X1 0 0 1 30 X2 1 0 0 0 S1 3. S1 . Solution Cj Mix X2 S2 X1 Zj Cj. iii) Is it worthwhile to purchase additional units of resource A at a cost of RM35.67 0 S2 0 1 0 0 S3 -2.78 Quantity 20 1 25 95 . including the total profit.33 -0.44 2.