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WORKSHEET 3 (GRAPHICAL SOLUTION: THEORY AND EXAMPLES) The set of all points that satisfy all the constraints of the model is called the feasible region. Of the feasible points some are: 1. interior points, that satisfies all constraints but none with equality 2. boundary points, that satisfies all constrains, at least one with equality, 3. extreme points, that satisfies all constraints, two with equality (these are the “corner points” of the feasible region, occurring at the intersection of two of the boundary constraints. The coordinates for extreme points can be found by solving the two equations in two unknowns that determine the point) Associated with each feasible point, there is slack on the ≤ constraints, which is found by subtracting the value of the left-hand side of the constraint from the constant on the right-hand side. Slack can be thought of as the amount of resource that is left over at this point. If the point satisfies the “≤” constraint with equality, the slack is zero; otherwise it is positive. Similarly, for each feasible point, surplus is associated with “≥” constraints, which is found by subtracting the constant on the right-hand side from the value of the left-hand size. The surplus indicates the extent to which some minimum restriction is exceeded at a given point. If the point satisfies the “≥” constraint with equality, the surplus is zero; otherwise it is positive. In general, no feasible point satisfies all the constraints with equality; that is there is slack or surplus on at least some of the constraints. Those points lying outside the feasible region (those that violate one or more of the constraints) are infeasible points. If a linear programing problem has an optimal solution, an extreme (corner) point is optimal, ie., the optimal solution lies only on one (or more) of the corner points of the feasible area. 1 Graphical Solution Procedure for Two-Variable Linear Programs. 1st Method: Complete Enumeration of the Corner Points. 1. Graph the constraints to find the feasible region, ie. Plot equality Convert the inequality to an equality and plot the straight line that represents this equation. - first set x2=0 and solve for x1 and plot this point on the horizontal (x1) axis; then set x1=0 and solve for x2 and plot this point on the vertical (x2) axis. - connect these two points by a straight line Determine the direction of the feasible area for a ≤ or ≥ constraint, - in order to determine the direction of the feasible area for a ≤ or ≥ constraint, choose a trial point (x1 and x2 coordinates) that is not on the line (either below the line or above it). Substitute the trial point into the expression on the left-hand side of the inequality. - determine if the trial point satisfies the inequality - if the trial point satisfies the original inequality, then the direction of the inequality is this side of the line. - if the trial point does not satisfy the original inequality, then the direction of the inequality is on the opposite side of the line. 2. Find the coordinates (x1 and x2) of each of the corner points, by solving the two equations that intersect at this point. 3. Insert the coordinates of each of the corner points in the objective function and calculate the objective function value of each of these points. 4. For a maximization problem, the point(s) that has (have) the maximum objective function value will be the optimal solution. For a minimization problem, the point(s) that has (have) the minimum objective function value will be the optimal solution. 2nd Method: Isoprofit (isocost) Line Method 1. Graph the constraints to find the feasible region by following the method described above 2. Set the objective function equal to an arbitrary value (Plot the objective function by following the method you used to plot the constraints) 3. Move the objective function line paralel to itself until it touches the last point of the feasible region. 4. Solve for x1 and x2 by solving the two equations that intersect, to determine the coordinates of this point. 5. Substitute these values into the objective function to determine its optimal value. 2 EXAMPLE 1 Galaxy Industries is an emerging toy manufacturing company that produces two water guns that are marketed nationwide. Although many parents object to the potentially violent implications of these products, the products have proven very popular and are in such demand that Galaxy has had no problem selling all the items it manufactures. The two models, Space Rays and Zappers are produced in lots of one-dozen each and are made exclusively from a special plastic compound. Two of the limiting resources are the 1200 pounds of the plastic compound and the 40 hours of production time that are available each week. Galaxy’s marketing department is more concerned with building a strong customer demand base for the company’s products than with meeting high production quotas. Two of its recommendations, which Galaxy’s management has already accepted, are to limit total weekly production to at most 800 dozen units and to prevent weekly production of Space Rays from exceeding that of Zappers by more than 450 dozen. The relevant data are summarized below: Product Profit per dozen Plastic req. (lb) per dozen Prod. time (min) req. per dozen Space Ray $8 2 3 Zapper $5 1 4 Galaxy has reasoned that since the $8 profit per dozen Space Rays exceeds the $5 profit per dozen Zappers by 60 %, the company could maximize its profit by producing as many Space Rays as possible, while remaining within the marketing guidelines, and using any remaining resources to produce Zappers. As a result, Galaxy has been producing 550 dozen Space Rays and 100 dozen Zappers weekly, earning a profit of $4900. Management is interested in determining whether a different production schedule could improve company profits. Recall that a mathematical model consists of three parts: 1. a well defined set of decision variables 2. an overall objective to be maximized or minimized 3. a set of constraints From the problem statement for Galaxy Industries, we can make a number of observations: production is to be in terms of dozens and scheduled on a weekly basis the overall objective is to maximize weekly profit production must be scheduled so that the weekly supply of plastic and the availability of production time are not exceeded. the two marketing department quidelines concerning maximum total production and the product mix must be met. 3 The decision maker can control the production levels of Space Rays and Zappers. The production units are in terms of dozens and production is done on a weekly basis, so the appropriate decision variables are: x1 : x2: Then the model can be expressed as: Max. Z = 8 x1 + 5x2 s.t. ………………………………………………………………………………...plastic constraint ………………………………………………………………………production time constraint ………………………………………………………………...maximum total production limit ………………………………………………...the limit on the mix of Space Rays and Zappers …………………………………………………………………………nonnegativity constraint x2 x1 4 5 EXAMPLE 2 The department of Navy has been downsizing and is looking for cost savings. One suggestion under consideration is to change the makeup of the content of Navy sea ration, the canned food supplies containing certain minimum quantities of Vitamin A, Vitamin D, iron and other nutrients, which soldiers carry. According to Texfoods, the current supplier of sea rations for the Navy, each two ounce portion of its product supplies 20% of the required amount of Vitamin A, 25% of the required amount of Vitamin D, and 50% of the required amount of iron. Each portion costs the Navy $0.60. Because all minimum standards must be met in each serving, the current sea ration container must contain 10 ounces of Texfoods product (to meet the minimum amount of Vitamin A). This costs the Navy $3 (=5x$0.60) per serving. The Navy is considering switching to another product from a different supplier, Calration. The two-ounce portion of Calration costs $0.50 and provides 50% of the minimum requirement of Vitamin A and 25% of the requirement of Vitamin D, but only 10% of the requirement for iron. Substituting the Calration product, the sea ration container would have to contain 20 ounces in order to meet the minimum iron requirement, costing the Navy $5(=10x$0.50) per serving. One bright soldier has suggessed that a mixture of the two products might meet the overall standards at a lower cost than the current $3 per serving. The Navy has never worried about the taste of sea rations; hence mixing them, either by combining them or by packing a portion of each, is an acceptable alternative. The soldier has been given permission to evaluate the idea. The following is a brief summary of the problem posed by the soldier: Determine the amount of two-ounce portions of each sea ration product in the mix Minimize the total cost of the sea rations Meet the minimum requirements for Vitamin A, Vitamin B and iron The decision variables are: x1 : x2 : Min. Z.= s.t. .................................................................................................minimum Vitamin A requirement ...............................................................................…………..minimum Vitamin D requirement ..........................................................................................………….minimum iron requirement ............................................................................................……………nonnegativity constraint 6 x2 x1 7 EXAMPLE 3 Max.Z= 5x1+ 4x2 s.t. 6x1+ 4x2 ≤ 24 x1+ 2x2 ≤ 6 -x1 + x2 ≤ 1 x2 ≤ 2 x1, x2 ≥ 0 x2 x1 8 EXAMPLE 4 Min.Z= x1+ 2x2 s.t. -3x1+ 2x2 ≤ 6 x1+ x2 ≤ 10.5 -x1 + 3 x2 ≥ 6 x1, x2 ≥ 0 x2 x1 9 EXAMPLE 5 Min. Z= 3x1+ 9x2 s.t. x1+ 2x2 ≥ 8 5x1+ 3x2 ≥ 0 x1 -2 x2 ≤ 2 x1 -2 x2 ≥ 0 x1 ≤ 9 x1, x2 ≥ 0 x2 x1 10 EXAMPLE 6 Max. Z= 5000x1+ 4000x2 s.t. EXAMPLE 7 Max. Z= 5000x1+ 10000x2 (use the same constraints of eg.6) x1+ x2 ≥ 4 x1- 3x2 ≤ 0 10x1 +15x2 ≤ 150 20x1 +10x2 ≤ 160 30x1 +10x2 ≤ 135 x1, x2 ≥ 0 x2 x1 11 EXAMPLE 8 Max. Z= 5000x1+ 4000x2 s.t. x1+ x2 ≥ 4 x1- 3x2 ≤ 0 10x1 +15x2 ≤ 150 20x1 +10x2 ≤ 160 30x1 +10x2 ≥ 210 x1, x2 ≥ 0 x2 x1 12