LINEAR PROGRAMMING PROBLEM: A GEOMETRIC APPROACH


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1 59 LINEAR PRGRAMMING PRBLEM: A GEMETRIC APPRACH 59.1 INTRDUCTIN Let us consider a simple problem in two variables x and y. Find x and y which satisfy the following equations x + y = 4 3x + 4y = 14 Solving these equations, we get x=2 and y=2. What happens when the number of equations and variables are more? Can we find a unique solution for such system of equations? However, a unique solution for a set of simultaneous equations in n variables can be obtained if there are exactly n relations. What will happen when the number of relations is greater than or less then n? A unique solution will not exist, but a number of trial solutions can be found. Again, if the number of relations are greater than or less than the number of variables involved and the relation are in the form of inequalities. Can we find a solution for such a system? Whenever the analysis of a problem leads to minimizing or maximizing a linear expression in which the variable must obey a collection of linear inequalities, a solution may be obtained using linear programming techniques. ne way to solve linear programming problems that involve only two variables is geometric approach called graphical solution of the linear programming problem.
2 18 :: Mathematics 59.2 BJECTIVES After the study of this lesson you will be able to define feasible solution define optimal solution solve linear programming problems by graphical method SLUTIN F LINEAR PRGRAMMING PRBLEMS In the previous lesson we have seen the problems in which the number of relations are not equal to the number of variables and many of the relations are in the form of inequalities (i.e., or ) to maximize (or minimize) a linear function of the variables subject to such conditions. Now the question is how one can find a solution for such problems? To answer this questions, let us consider the system of equations and inequations (or inequalities). We know that x represents a region lying towards the right of y axis including the y axis. y x x x y Fig Similarly, the region represented by y, lies above the x axis including the x axis. y x y 0 x y Fig. 59.2
3 Linear Programming Problem: A Geometric Approach:: 19 The question arises what region will be represented by x and y simultaneously. bviously, the region given by x, y will consist of those points which are common to both x and y. It is the first quadrant of the plane. y x, y x x y Fig Next, we consider the graph of the equation x + 2y 8. For this, first we draw the line x + 2y = 8 and then find the region satisfying x + 2y 8. Usually we choose x = 0 and calculate the corresponding value of y and choose y = 0 and calculate the corresponding value of x to obtain two sets of values (This method fails, if the line is parallel to either of the axes or passes through the origin. In that case, we choose any arbitrary value for x and choose y so as to satisfy the equation). Plotting the points (0,4) and (8,0) and joining them by a straight line, we obtain the graph of the line as given in the figure 59.4 below. y B(0,4) x + 2y = 8 x A(8,0) x y Fig. 59.4
4 20 :: Mathematics We have already seen that x and y represents the first quadrant. The graph given by x + 2y 8 lies towards that side of the line x + 2y = 8 in which the origin is situated because any point in this region will satisfy the inequality. Hence the shaded region in the figure (59.5) represents x, y and x + 2y 8 simultaneously. y (0,4) x + 2y = 8 (8,0) x x y Similarly, if we have to consider the regions bounded by x, y and x + 2y 8, then it will lie in the first quadrant and on that side of the line x + 2y = 8 in which the origin is not located. The graph is shown by the shaded region, in figure (59.6) y Fig (0,4) x + 2y = 8 x (8,0) x y Fig The shaded region in which all the given constraints are satisfied is called the feasible region Feasible Solution A set of values of the variables of a linear programming problem
5 Linear Programming Problem: A Geometric Approach:: 21 which satisfies the set of constraints and the non negative restrictions is called a feasible solution of the problem ptimal Solution A feasible solution of a linear programming problem which optimizes its objective function is called the optimal solution of the problem. Note : If none of the feasible solutions maximize (or minimize) the objective function, or if there are no feasible solutions, then the linear programming problem has no solution. In order to find a graphical solution of the linear programming problem following steps be employe: Step 1: Step 2: Step 3: Steps 4: Step 5: Step 6: Formulate the linear programming problem. Graph the constraints inequalities (by the method discussed above) Identify the feasible region which satisfies all the constraints simultaneously. For less than or equal to constraints the region is generally below the lines and for greater than or equal to constraints, the region is above the lines. Locate the solution points on the feasible region. These points always occur at the vertex of the feasible region. Evaluate the objective function at each of the vertex (corner point) Identify the optimum value of the objective function. Example A : Minimize the quantity z = Solution : The objective function to be minimized is
6 22 :: Mathematics z = First of all we draw the graphs of these inequalities, which is as follows: B (0,1) x1 =1 As we have discussed earlier that the region satisfied by and is the first quadrant and the region satisfied by the line + along with will be on that side of the line + =1 in which the origin is not located. Hence, the shaded region is our feasible solution because every point in this region satisfies all the constraints. Now, we have to find optimal solution. The vertex of the feasible region are A(1,0) and B(0,1). The value of z at A=1 The value of z at B = 2 Take any other point in the feasible region say (1,1), (2,0), (0,2) etc. We see that the value of z is minimum at A(1,0). Example B : Fig Minimize the quantity z = A(1,0)
7 Linear Programming Problem: A Geometric Approach:: 23 Solution : The objective function to be minimized is z = First of all we draw the graphs of these inequalities (as discussed earlier) which is as follows: D E(0,1) ( 03, 4 ) C (½,½) 3, 0 2 x1 =1 A(1,0) 2 +4 =3 B ( 3, 0 2 ) Fig The shaded region is the feasible region. Every point in the region satisfies all the mathematical inequalities and hence the feasible solution. Now, we have to find the optimal solution. The value of z at B is 3 2 The value of z at C (½, ½) is The value of z at E (0,1) is 2 If we take any point on the line 2 +4 =3 between B and C 3 we will get and elsewhere in the feasible region greater than 2. f course, the reason any feasible point (between B and C) on 2 +4 =3 minimizes the objective function (equation) z = + 2 is that the two lines are parallel (both have slope
8 24 :: Mathematics 1 ). Thus this linear programming problem has infinitely many 2 solutions and two of them occur at the vertices. Example C : Maximize z = Solution : The objective function is to maximize z = First of all we draw the graphs of these inequalities, which is as follows D(0,240) C (0,150) B(90,105) A(160,0) 3 +2 =480 E (300,0) +2 =300 Fig The shaded region ABC is the feasible region. Every point in the region satisfies all the mathematical inequations and hence the feasible solutions.
9 Now, we have to find the optimal solution. The value of z at A(160, 0) is The value of z at B (90, 105) is The value of z at C (0, 150) is The value of z at (0, 0) is 0. Linear Programming Problem: A Geometric Approach:: 25 If we take any other value from the feasible region say (60, 120), (80,80) etc. we see that still the maximum value is obtained at the vertex B (90, 105) of the feasible region. Note: For any linear programming problem that has a solution, the following general rule is true. If a linear programming problem has a solution it is located at a vertex of the set of feasible solutions; if a linear programming problem has multiple solutions, at least one of them is located at a vertex of the set of feasible solutions. In either case, the value of the objective function is unique. Check point 1 : Example D : Maximize z = In a small scale industry a manufacturer produces two types of book cases. The first type of book case requires 3 hours on machine A and 2 hours on machines B for completion, whereas the second type of book case requires 3 hours on machine A and 3 hours on machine B. The machine A can run at the most for 18 hours while the machine B for at the most 14 hours per day. He earns a profit of Rs.30/ on each book case of the first type and R.40/ on each book case of the second type. How many book cases of each type should he make each day so as to have a maximum profit?
10 26 :: Mathematics Solution : Let be the number of first type book cases and be the number of second type book cases that the manufacturer will produce each day. Since and are the number of book cases so (1) Since the 1 st type of book case requires 3 hours on machine A, therefore, book cases of first type will require 3 hours on machine A. 2 nd type of book case also requires 3 hours on machine A, therefore, book cases of 2 nd type will require 3 hours on machine A. But the working capacity of machine A is at most 18 hours per day, so we have or + 6 (2) Similarly, on the machine B, 1 st type of book case takes 2 hours and second type of book case takes 3 hours for completion and the machine has the working capacity of 14 hours per day, so we have (3) Profit per day is given by z = (4) Now, we have to determine and such that Maximize z = (objective function) subject to the conditions } constraints We use the graphical method to find the solution of the problem. First of all we draw the graphs of these inequalities, which is as follows:
11 Linear Programming Problem: A Geometric Approach:: 27 (0,6) C B(4,2) (0,0) A(6,0) Fig =14 =6 The shaded region ABC is the feasible region. Every point in the region satisfies all the mathematical inequations and hence known as feasible solution. We know that the optimal solution will be obtained at the vertices (0,0), A(6,0), B(4,2). Since the co ordinates of C are not integers so we don t consider this point. Co ordinates of B are calculated as the intersection of the two lines. Now the profit at is zero. Profit at A = = 180 Profit at B = = = 200 Thus the small scale manufacturer gains the maximum profit of Rs.200/ if he prepares 4 first type book cases and 2 second type book cases. Example E : Solve Example B of lesson 58 by graphical method. Solution : From Example B of lesson 58, we have (minimize) z = (objective function) subject to the conditions
12 28 :: Mathematics } 16 (Constraints) Fist we observe that any point (, ) lying in the first quadrant clearly satisfies the constraints, and, Now, we plot the bounding lines or 3 + = 24 x1 x2 + = B (0,24) Any point on or above the line 3 =24 satisfies the constraint 3 24 (Fig ) A(8,0) Fig Similarly any point on or above the line =16 satisfies the constraint 16 (Fig ) B (0,16) A(16,0) Fig Again any point on or above the line +3 =24 satisfies the constraints (Fig ) B(0,8) Fig A(24,0)
13 Thus combining all the above figures we get Linear Programming Problem: A Geometric Approach:: 29 D (0,24) (0,16) C (4,12) (0,8) B(12,4) (8,0) A(24,0) (16,0) 3 =24 =16 +3 =24 The region which is the common shaded area unbounded above. Now the minimum value of Fig z = is at one of the points A(24,0), B(12,4) C(4,12) and D(0,24) At A : z = = 144,000 At B : z = = 88,000 At C : z = = 72,000 At D : z = = 96,000 Thus, we see that z is minimum at C (4,12) where =4 and = 12. Hence, for minimum cost the firm should run plant P for 4 days and plant Q for 12 days. The minimum cost will be Rs.72, Check point 2 : Solve Example C of lesson 58 by graphical method. Example F : Maximize the quantity
14 30 :: Mathematics z = +2 1, Solution : First we graph the constraints 1, B (0,1) The shaded portion is the set of feasible solution. Now, we have to maximize the objective function. The value of z at A(1,0) is 1. The value of z at B(0,1) is 2. If we take the value of z at any other point from the feasible region, say (1,1) or (2,3) or (5,4) etc., then we notice that every time we can find another point which gives the larger value than the previous one. Hence, there is no feasible point that will make z largest. Since there is no feasible point that makes z largest, we conclude that this linear programming problem has no solution. Example G : A (1,0) Fig Solve the following problem graphically. =1 Minimize z =
15 Linear Programming Problem: A Geometric Approach:: 31 Solution : First we graph the constraints, 5 5, or 0, 5 5 (0,1) A +5 =5 feasible region Fig The shaded region is the feasible region. Here, we see that the feasible region is unbounded from one side. But it is clear from the figure (59.16) that the objective ( 5, 5 function attains its minimum value at the point A which is the 4 4 ) 4 point of intersection of the two lines =0 and +5 =5. Solving these we get = = Hence, z is minimum when =, =, and its minimum value is 2 10 = 10. Note: If we want to find max. z with these constraints then it is not possible in this case because the feasible region is unbounded from one side. INTEXT QUESTINS Minimize z = subject to the conditions
16 32 :: Mathematics 2. Maximize z = subject to the conditions , , 3. Solve 1(a) of the Intext Questions Solve 1(b) of the Intext Questions 58.1 WHAT YU HAVE LEARNT A set of values of the variables of a linear programming problem which satisfies the set of constraints and the non negative restrictions is called a feasible solution. A feasible solution of a linear programming problem which optimizes its objective function is called the optimal solution of the problem. If none of the feasible solutions maximizes (or minimizes) the objective function, or if there are no feasible solutions, then the linear programming problem has no solutions. If a linear programming problem has a solution, it is located at a vertex of the set of feasible solution. If a linear programming problem has multiple solutions, at least one of them is located at a vertex of the set of feasible solutions. But in all the cases the value of the objective function remains the same. TERMINAL QUESTINS 1. A dealer has Rs.1500/ only for the purchase of rice and wheat. A bag of rice costs Rs.180/ and a bag of wheat costs Rs.120/. He has a storage capacity of ten bags only and the dealer gets a profit of Rs.11/ and Rs.8/ per bag of rice and wheat respectively. How he should spend his money in order to get maximum profit? 2. Solve the following problem : Max. z =
17 Linear Programming Problem: A Geometric Approach:: Solve the following problem : Maximize z = A machine producing either product A and B can produce A by using 2 units of chemicals and 1 unit of a compound and can produce B by using 1 unit of chemicals and 2 units of the compound. nly 800 units of chemicals and 1000 units of the compound are available. The profits available per unit of A and B are respectively Rs.30 and Rs.20. Find the optimum allocation of units between A and B to maximize the total profit. Find the maximum profit. 5. Maximize z = , ANSWERS T CHECK PINTS Check point 1: (0,40) C (0,30) B (20,20) feasible region Fig Max z = 140 at B(20,20) A (40,0) (60,0) + 2 =60 + =40
18 feasible region 34 :: Mathematics Check point 2 : (0,600) feasible region C (0,400) B (200,200) ANSWERS T INTEXT QUESTINS (400,0) A(300,0) =400 Fig =600 Max z = 1200 at C (0,400) D(0,24) (0,16) C(4,12) (0,8) B(12,4) (8,0) (16,0) A(24,0) 3 =24 =16 +3 =24 Fig Min. z =720 at C(4,12), =4, =12
19 Linear Programming Problem: A Geometric Approach:: (0,25) Feasible region D (0,10) C (3,9) B A(10,0) 5 +2 =50 Fig =12 Max. z = 330 at C(3,9), = 3, = 9 +3 = (, ) C(0,60) B(10,50) Feasible region A(20,0) 5 =100 =60 Fig Max. z = 1250 at B(10,50), = 10, = 50
20 36 :: Mathematics 4. B (0, ) Feasible region 2 +3 =8 Fig A(10,0) 4 +3 =40 Min. z = 40,000 at A(10,0), =10, =0 ANSWERS T TERMINAL QUESTINS 1. ( 0, 25 2) C (0,10) Feasible region B(5,5) ( ) (10,0) A25, 0 x 3 1 = =50 Fig Max z = 95 at B(5,5), = = 5
21 Linear Programming Problem: A Geometric Approach:: ( 0, 25 3) C(0,5) Feasible region B(4,3) (10,0) A 25, 0 4 Fig =50 +2 =10 Max z = 160 at B(4,3), = 4 = 3 3. (0,1100) (0,700) D (0,600) C B Feasible region =9600 A (900,0) =8400 Fig =9900 A (900,0) D(0,600) B (626, 335) & (0, 0) C(480, 420) Max z=8984 at B(626, 335) = 626, = 335
22 38 :: Mathematics 4. Max. z = (0,800) C(0,500) B(200,400) Feasible region ABC A(400,0) (1000,0) 2 = =1000 Fig Max z = at B (200,400). =200, = (0,24) C(0,20) B(8,12) Feasible region ABC A(16,0) (20,0) =20 Fig =5760 Max z = 392 at B(8,12). = 8, = 12
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