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UNIT 2

    Major steps in the solution of L.P.P by Graphical Major steps in the solution of L.P.P by Graphical Method. Step:1 draw x1 and x2 axes (which are mutually perpendicular) on a graph sheet. Step:2 draw a line and identity the region connected with its corresponding to each  constraints. Step:3 identity the solutions space which is the region that is common to all constraints  including the non negative restrictions. Step:4 find the value of Z at each vertex of the solution space. Step:5 identity the optimum solution. Example:1 Solve by graphical method. Max Z = 3x1 + 4x2 Subject to constraints 4x1 + 2x2 ≤ 80 2x1 + 5x2 ≤ 180 and x1, x2 ≥ 0 Solution : Consider the first constraint 4x1 + 2x2 ≤ 80  Treating as a strict equation, we have 4x1 + 2x2 = 80 Put x1 = 0  0 + 2x2 = 80 x2 =  80 2 = 40 put x2 = 0  4x1 + 0 = 80  x1 =  80 4 = 20 x1 0 20 x2 40 0 Consider the second constraint 2x1 + 5x2 ≤ 180 Treating as a strict equation, we have 2x1 + 5x2 = 18...
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I.BCOM

                              UNIT–1                                              DE-MORGANS LAW De Morgan’s Laws: For any two sets A and B,  (A∪B) |=A|∩B| and  (A∩B) |=A |∪B| For any three sets A, B, and C. A-(B∪C)=(A-B)∩(A-C) A-(B∩C)=(A-B)∪(A-C). 1. Verify the distributive law  A∪(B∩C)=(A∪B)∩(A∪C)   Example 1 If A={1,2,3,4}, B={2,4,5,6} and C={1,3,5}, verify  that A∩(B∪C)=(A∩B)∪(A∩C). Proof: Given that A={1,2,3,4}, B={2,4,5,6} and C={1,3,5} Consider, LHS A∩(B∪C) (B∪C)={2,4,5,6}∪{1,3,5}  ={1,2,3,4,5,6} A∩(B∪C)={1,2,3,4}…………(1) (A∩B)= {1,2,3,4}∩{2,4,5,6}={2,4} (A∩C)= {1,3} (A∩B)∪(A∩C)= {1,2,3,4}………(2) A∩(B∪C)=(A∩B)∪(A∩C) Hence Proved. (ii)   A-(B∩C)=(A-B)∪(A-C). Consider, A-(B∩C) (B∩C)={2,3,4,5,6}∩{4,5,6,7,8,9}  ={4,5,6} A-(B∩C)={0,1,3,4,6,7,9,1...
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