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### graphical method

1. 1. A PRESENTATION ON SOLVING LPP BYGRAPHICAL METHOD Submitted By: Kratika Dhoot MBA- 2nd sem
2. 2. What is LPP ???• Optimization technique• To find optimal value of objective function, i.e. maximum or minimum• “LINEAR” means all mathematical functions are required to be linear…• “PROGRAMMING” refers to Planning, not computer programming… KRATIKA DHOOT
3. 3. What is graphical method ???• One of the LPP method• Used to solve 2 variable problems of LPP… KRATIKA DHOOT
4. 4. Steps for graphical method… FORMULATE THE OUTLINE THE PROBLEM SOLUTION AREA ( for objective & ( area which satisfiesconstraints functions) the constraints) CIRCLE POTENTIAL FRAME THE GRAPH PLOT THE GRAPH SOLUTION POINTS ( one variable on ( one variable on ( the intersection horizontal & other at horizontal & other points of all vertical axes) at vertical axes) constraints) PLOT THE CONSTRAINTS (inequality to be as equality; SUBSTITUTE & FIND give arbitrary value to variables OPTIMIZED SOLUTION & plot the point on graph ) KRATIKA DHOOT
5. 5. LET US TAKE AN EXAMPLE!!! SMALL SCALE ACCOMPLISHED BY BUT NUMBER OF ELECTRICAL SKILLED MEN & WORKERS CAN’T REGULATORS WOMEN WORKERS EXCEED 11 INDUSTRYSALARY BILL NOT MALE WORKERS ARE DATA COLLECTEDMORE THAN Rs. PAID Rs.6,000pm & FOR THE 60,000 pm FEMALE WORKERS ARE PERFORMANCE PAID Rs.5,000pmDATA INDICATED MALE MEMBERS DETERMINE No. OF MALES & CONTRIBUTES Rs.10,000pm & FEMALES TO BE EMPLOYED INFEMALE MEMBERS CONTRIBUTES ORDER TO MAXIMIZE TOTAL Rs.8,500pm RETURN KRATIKA DHOOT
6. 6. STEP 1-FORMULATE THE PROBLEMObjective Function :-Let no. of males be x & no. of females be yMaximize Z = Contribution of Male members + contribution of Female members Max Z = 10,000x + 8,500ySubjected To Constraints :- x + y ≤ 11 ………..(1) 6,000x + 5,000y ≤ 60,000 ………..(2) KRATIKA DHOOT
7. 7. STEP 2- FRAME THE GRAPH• Let no. of Male Workers(x) be on horizontal axis & no. of Female Workers (y) be vertical axis.. No. of females No. of males KRATIKA DHOOT
8. 8. STEP 3- PLOT THE CONSTRAINTS• To plot the constraints, we will opt an arbitrary value to the variables as:-x + y ≤ 11:- converting as x + y = 11 x 0 11 y 11 06,000x+5,000y≤60,000:- converting as 6x + 5y= 60 x 0 10 y 12 0 KRATIKA DHOOT
9. 9. STEP 4- PLOT THE GRAPHNo. of No. offemales females 14 14 12 ● ( 0 , 11 ) 12 ● ( 0 , 12 ) 10 10 8 8 6 6 4 4 2 ( 11 , 0 ) 2 ( 10 , 0 ) 0 ● 0 ● 2 4 6 8 10 12 No. of 2 4 6 8 10 12 No. of males males x + y ≤ 11 6x + 5y ≤60 KRATIKA DHOOT
10. 10. STEP 5- FIND THE OPTIMAL SOLUTION No. of females 12 ● OPTIMAL ● SOLUTION POINT 10 8 ( 5, 6 ) 6 ● 4 FEASIBLE REGION 2 0 ● ● 2 4 6 8 10 12 No. of males KRATIKA DHOOT
11. 11. STEP 6- CIRCLE POTENTIAL OPTIMAL POINTSNo. of females 12 ( 0 , 11 ) 10 8 ( 5, 6 ) 6 4 2 ( 10 , 0 ) 0 ( 0,0) 2 4 6 8 10 12 No. of males KRATIKA DHOOT
12. 12. STEP 7- SUBSTITUE & OPTIMIZE Max Z = 10,000x + 8,500y POTENTIAL Z = 10,000x + 8,500y MAXIMUM ZOPTIMAL PTS. (0,0) 10,000(0) + 8,500(0) 0 (0,11) 10,000(0)+8,500(11) 93,500 (5,6) 10,000(5)+8,500(6) 1,01,000 1,01,000 (10,0) 10,000(10)+8,500(0) 1,00,000 KRATIKA DHOOT
13. 13. CONCLUSION• Thus, maximum total return is about Rs.1,01,000 by adopting 5 male workers & 6 female workers.• Hence, optimal solution for LPP is :- No. of male workers = 5 No. of female workers = 6 Max. Z = Rs. 1,01,000 KRATIKA DHOOT
14. 14. Let us take other example!!!• Find the maximum value of objective function Z= 4x + 2ys.t. x + 2y ≥ 4 3x + y ≥ 7 -x + 2y ≤ 7 &x≥0&y≥0 KRATIKA DHOOT
15. 15. PLOT THE CONSTRAINTS x 0 4x + 2y = 4 y 2 0 x 0 7/33x + y = 7 y 7 0 x 0 -7-x + 2y = 7 y 7/2 0 KRATIKA DHOOT
16. 16. PLOTTINGCONSTRAINTS TO GRAPH KRATIKA DHOOT
17. 17. x 0 4x + 2y ≥ 4 y 2 0 Y 7 6 5 4 3 2 1 • (0,2) (4,0) 0 1 2 3 •4 5 6 7 X KRATIKA DHOOT
18. 18. 3x + y ≥ 7 x 0 7/3 y 7 0 (0,7) Y 7 6 • 5 4 3 2 1 ( 7/3 , 0 ) 0 1 2 • 3 4 5 6 7 X KRATIKA DHOOT
19. 19. x 0 -7-x + 2y ≤ 7 y 7/2 0 7 Y 6 5 ( 0 , 7/2 ) 4 • 3 2 ( -7 , 0 ) 1 • 0 -7 -6 -5 -4 -3 -2 -1 X KRATIKA DHOOT
20. 20. There is a common portion or common points which intersects by all 3 regions of lines 3x + y ≥ 7 -x + 2y ≤ 7 Y 7 6 5 x + 2y ≥ 4 4 3 2 1 0 1 2 3 4 5 6 7 -7 -6 -5 -4 -3 -2 -1 X KRATIKA DHOOT
21. 21. CIRCLE THE POTENTIAL POINTS!!! Y 7 6 (1,4) 5 4 3 2 (2,1) 1 ( 4 ,0 ) 0 1 2 3 4 5 6 7 -7 -6 -5 -4 -3 -2 -1 X KRATIKA DHOOT
22. 22. STEP 7- SUBSTITUE & OPTIMIZE Max Z = 4x + 2y POTENTIAL Z = 4x + 2y MAXIMUM ZOPTIMAL PTS. (1,4) 4(1) + 2(4) 12 (2,1) 4(2)+2(1) 10 (4,0) 4 (4)+ 2(0) 16 16 KRATIKA DHOOT
23. 23. CONCLUSIONHence, the optimal solution is: X=4 Y=0 Max z = 16 KRATIKA DHOOT
24. 24. PRACTICE QUESTIONS …(1) Maximize f(x) = x1 + 2x2 subject to: x1 + 2x2 ≤ 3 x1 + x2 ≤ 2 x1 ≤ 1 & x1 , x2 ≥ 0(2) Maximize z = 4x+2y subject to: 4x+6y≥12 2x+4y≤4 & x≥0 ; y≥0 KRATIKA DHOOT
25. 25. THANK YOU !!! KRATIKA DHOOT