Resource Allocation Models)
Problem 2.12. This problem is of mathematical interest
Maximise Z = 3a + 4b S.T. Converting and writing in the form of equations, 1a – 1b ≤ –1. Maximise Z = 3a + 4b S.T
– 1a + 1b ≤ 0 1a – 1b = 0
And both a and b are ≥ 0 –1a + 1b = 0
And both a and b are ≥ 0
Referring to figure 2.11, the straight line for equation 1 starts in second quadrant and extended in to first quadrant. The line for equation 2 passes through the origin. We see that there is no point, which satisfies both the constraints simultaneously. Hence there is no feasible solution. Given l.p.p. has no feasible solution.
Figure 2.11. Graph for the problem 2.11
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Figure 2.12. Graph for the problem 2.12
Problem 2.13. Solve the l.p.p. by graphical method.
Maximise Z = 3a + 2b S.T.
1a + 1b ≤ 4
1a – 1b ≤ 2 and both a and b are ≥0.
Solution: The figure 2.13 shows the graph of the equations.
Equations are: Maximise Z = 3a + 2b S.T.
1a + 1b = 4
1a – 1b = 2 and both a and b are ≥ 0.
In the figure the Isoprofit line passes through the point N (3,1). Hence optimal Profit Z = 3 × 3 + 2 × 1 = Rs.11 /-
Figure 2.13. Graph for the problem 2.13
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Problem 2.14: Formulate the l.p.p. and solve the below given problem graphically.
Old hens can be bought for Rs.2.00 each but young ones costs Rs. 5.00 each. The old hens lay 3 eggs per week and the young ones lay 5 eggs per week. Each egg costs Rs. 0.30. A hen costs Rs.1.00 per week to feed. If the financial constraint is to spend Rs.80.00 per week for hens and the capacity constraint is that total number of hens cannot exceed 20 hens and the objective is to earn a profit more than Rs.6.00 per week, find the optimal combination of hens.
Solution: Let x be the number of old hens and y be the number of young hens to be bought. Now the old hens lay 3 eggs and the young one lays 5 eggs per week. Hence total number of eggs one get is 3x + 5y.
Total revenues from the sale of eggs per week is Rs. 0.30 (3x + 5y) i.e., 0.90 x + 1.5 y Now the total expenses per week for feeding hens is Re.1 (1x + 1y) i.e., 1x + 1y.
Hence the net income = Revenue – Cost = (0.90 x + 1.5 y) – (1 x + 1 y) = – 0.1 x + 0.5 y or 0.5y – 0.1 x. Hence the desired l.p.p. is
Maximise Z = 0.5 y – 0.1 × S.T.
2 x + 5 y ≤ 80
1x + 1y ≤ 20 and both x and y are ≥ 0 The equations are:
Maximise Z = 0.5 y – 0.1 × S.T.
2 x + 5 y = 80
1x + 1y = 20 and both x and y are ≥ 0
In the figure 2.13, which shows the graph for the problem, the isoprofit line passes through the point C. Hence Zc = Z (0,16) = Rs.8.00. Hence, one has to buy 16 young hens and his weekly profit will be Rs.8.00
Figure 2.14. Graph for the problem 2.14
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Point to Note: In case in a graphical solution, after getting the optimal solution, one more constraint is added, we may come across following situations.
(i) The feasible area may reduce or increase and the optimal solution point may be shifted depending the shape of the polygon leading to decrease or increase in optimal value of the objective function.
(ii) Some times the new line for the new constraint may remain as redundant and imposes no extra restrictions on the feasible area and hence the optimal value will not change.
(iii) Depending on the position of line for the new constraint, there may not be any point in the feasible area and hence there may not be a solution. OR the isoprofit line may coincide with a line and the problem may have innumerable number of solutions.
SUMMARY
1. The graphical method for solution is used when the problem deals with 2 variables.
2. The inequalities are assumed to be equations. As the problem deals with 2 variables, it is easy to write straight lines as the relationship between the variables and constraints are linear. In case the problem deals with three variables, then instead of lines we have to draw planes and it will become very difficult to visualize the feasible area.
3. If at all there is a feasible solution (feasible area of polygon) exists then, the feasible area region has an important property known as convexity Property in geometry. (Convexity means: Convex polygon consists of a set points having the property that the segment joining any two points in the set is entirely in the convex set. There is a mathematical theorem, which states, “The points which are simulations solutions of a system of inequalities of the
≤ type form a polygonal convex set”.
The regions will not have any holes in them, i.e., they are solids and the boundary will not have any breaks. This can be clearly stated that joining any two points in the region also lies in the region.
4. The boundaries of the regions are lines or planes.
5. There will be corners or extreme points on the boundary and there will be edges joining the various corners. The closed figure is known as polygon.
6. The objective function of a maximisation is represented by assuming suitable outcome (revenue) and is known as Isoprofit line. In case of minimisation problem, assuming suitable cost, a line, known as Isocost line, represents the objective function.
7. If isoprofit or isocost line coincides with one corner, then the problem has unique solution.
In case it coincides with more than one point, then the problem has alternate solutions. If
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the isoprofit or isocost line coincides with a line, then the problem will have innumerable number of solutions.
8. The different situation was found when the objective function could be made arbitrarily large. Of course, no corner was optimal in that case.
QUESTIONS
1. An aviation fuel manufacturer sells two types of fuel A and B. Type A fuel is 25 % grade 1 gasoline, 25 % of grade 2 gasoline and 50 % of grade 3 gasoline. Type B fuel is 50 % of grade 2 gasoline and 50 % of grade 3 gasoline. Available for production are 500 liters per hour grade 1 and 200 liters per hour of grade 2 and grade 3 each. Costs are 60 paise per liter for grade 1, 120 paise for grade 2 and100 paise for grade 3. Type A can be sold at Rs. 7.50 per liter and B can be sold at Rs. 9.00 per liter. How much of each fuel should be made and sold to maximise the profit.
2. A company manufactures two products X1 and X2 on three machines A, B, and C. X1 require 1 hour on machine A and 1 hour on machine B and yields a revenue of Rs.3/-. Product X2 requires 2 hours on machine A and 1 hour on machine B and 1 hour on machine C and yields revenue of Rs. 5/-. In the coming planning period the available time of three machines A, B, and C are 2000 hours, 1500 hours and 600 hours respectively. Find the optimal product mix.
3. Maximize Z = 1 x + 1 y S.T.
1 x + 2 y ≤ 2000 1 x + 1 y ≤ 1500
0 x + 1 y ≤ 600 and both x and y are ≥ 0.
4. Maximise Z = 8000a + 7000b S.T.
3 a + 1 b ≤ 66 1 a + 1 b ≤ 45 1 a + 0 b ≤ 20
0 a + 1 b ≤ 40 and both a and b are ≥ 0.
5. Minimise Z = 1.5 x + 2.5 y S.T.
1 x + 3 y ≥ 3
1 x + 6 y ≥ 2 and both x and y ≥ 0 6. Maximise Z = 3 a + 2 b S.T.
1a – 1 b ≤ 1
1 a + 1b ≥ 3 and both x and y are ≥ 0 7. Maximise Z = –3 x + 2 y S.T.
1 x + 0 y ≤ 3
1 x – 1 y ≤ 0 and both x and y are ≥ 0
Linear Programming Models (Resource Allocation Models) 4343434343 8. Maximize Z = – 1 a + 2b S.T.
– 1 a + 1 b ≤ 1
– 1 a + 2 b ≤ 4 and both a and b are ≥ 0.
9. Maximise Z = 3 x – 2 y S.T.
1x + 1 y ≤ 1
2 x + 2 y ≥ 4 and both x and y are ≥ 0 10. Maximize Z = 1x + 1 y S.T.
1 x – 1y ≥ 0
–3x + 1 y ≥ 3 and both x and y ≥ 0
3.1. INTRODUCTION
As discussed earlier, there are many methods to solve the Linear Programming Problem, such as Graphical Method, Trial and Error method, Vector method and Simplex Method. Though we use graphical method for solution when we have two problem variables, the other method can be used when there are more than two decision variables in the problem. Among all the methods, SIMPLEX METHOD is most powerful method. It deals with iterative process, which consists of first designing a Basic Feasible Solution or a Programme and proceed towards the OPTIMAL SOLUTION and testing each feasible solution for Optimality to know whether the solution on hand is optimal or not.
If not an optimal solution, redesign the programme, and test for optimality until the test confirms OPTIMALITY. Hence we can say that the Simplex Method depends on two concepts known as Feasibility and optimality.
The simplex method is based on the property that the optimal solution to a linear programming problem, if it exists, can always be found in one of the basic feasible solution. The simplex method is quite simple and mechanical in nature. The iterative steps of the simplex method are repeated until a finite optimal solution, if exists, is found. If no optimal solution, the method indicates that no finite solution exists.
3.2. COMPARISION BETWEEN GRAPHICAL AND SIMPLEX METHODS
1. The graphical method is used when we have two decision variables in the problem. Whereas in Simplex method, the problem may have any number of decision variables.
2. In graphical method, the inequalities are assumed to be equations, so as to enable to draw straight lines. But in Simplex method, the inequalities are converted into equations by:
(i) Adding a SLACK VARIABLE in maximisation problem and subtracting a SURPLUS VARIABLE in case of minimisation problem.
3. In graphical solution the Isoprofit line moves away from the origin to towards the far off point in maximisation problem and in minimisation problem, the Isocost line moves from far off distance towards origin to reach the nearest point to origin.
4. In graphical method, the areas outside the feasible area (area covered by all the lines of constraints in the problem) indicates idle capacity of resource where as in Simplex method, the presence of slack variable indicates the idle capacity of the resources.