Final Exam Solved Questions - Linear Programming and Network Flows | MATH 444, Exams of Mathematics

Download Final Exam Solved Questions - Linear Programming and Network Flows | MATH 444 and more Exams Mathematics in PDF only on Docsity! 1. Dual: 1 2 1 2 3 1 2 3 1 3 1 2 3 8 2 1 2 4 2 min 12 6 8 when 2 3 , , 0 λ λ λ λ λ λ λ λ λ λ λ λ λ + ≥ + − ≥ −+ +  + ≥ ≥ Initial tableau: 8 2 1 1 0 0 12 2 1 0 0 1 0 6 0 4 2 0 0 1 8 1 2 3 0 0 0 0 − − . Second tableau: 8 4 0 1 0 1/2 8 2 1 0 0 1 0 6 0 2 1 0 0 1/2 4 1 4 0 0 0 3/2 12 − − − − . Final tableau: 2 1 0 1/4 0 1/ 8 2 0 0 0 1/ 4 1 1/ 8 4 4 0 1 1/2 0 1/ 4 8 7 0 0 1 0 1 20 − − − − − − . The primal solution is given by 1 2 3ˆ ˆ ˆ0, 2, 8x x x= = = , the maximum value is 20 . The dual solution is given by 1 2 3ˆ ˆ ˆ1, 0, 1λ λ λ= = = , the minimum value is 20 . 2. Using two bottom rows the initial tableau associated with the first phase is given by: 1 0 1 1 0 1 6 0 1 1 0 1 0 12 0 0 0 0 0 1 0 1 0 1 1 0 0 6 − − − − − Using three bottom rows the final tableaus of both the first and the second phase are given by: 1 0 1 1 0 1 6 1 1 0 1 1 1 6 0 0 0 0 0 1 0 1 2 0 0 0 0 1 0 0 2 0 12 − − − − − − − . The feasible point 1 2 30, 6, 6x x x= = = given by the first phase is also a global maximum. 3. The slack 2̂ 1 0s = > , so 2̂ 0λ = . From 1 3ˆ ˆ0, 0x x> > conclude that 1 3ˆ ˆ0, 0µ µ= = . It follows that 1 3 3 ˆ ˆ 8 ˆ 7 λ λ λ  + = = and hence 1̂ 1λ = . 4. ˆ 1 1 0Tλ  =    , 1 2 ˆ 1 1 0 1 1 2 3 0 0 3 2 T T T Tc A c cλ           − = − = − =                 so 2 3Tc  =    . The basic variables are given by 2 1 3, ,x x y so 3 2 0 T Bc  =    . Note that 1T̂ T Bc Bλ −= , 2 1 0 1 1 0 2 3 1 B      =         and 2 1 0 1 1 0 1 1 0 2 3 1 T Bc       =           . 5. (a) The only part of the tableau that changes is the ( ) 1 1 12 10 0 0 0 T T T T T B Bc c c B A c c B A c c − −      + ∆ − = − + ∆ = − + ∆ ≤           so the increase allowed is at most 12 . (b) The basic variables are given by 2 2,x y so 2 0 T Bc  =    . This time ( ) ( ) 1 1 1 1 21 1 1 2 2 2 2 2 27 2 1 0 0 0 0 0 0 0 T T T T T T T T B B B Bc c c c B A c c B A c c B A c c c − − −+ ∆ − + ∆ = − + ∆ − ∆ =             − + ∆ − ∆ = − − ∆ ≤                      It is also necessary to check ( ) 1 61 1 1 1 2 2 2 6 21 6 0 0 0 0 0 0 1 T B Bc c B c c −           − + ∆ = − − ∆ = − − ∆ ≤               −   . The decrease allowed is at most 1. 6. (a) ( )1 1 1 1 90/7 75/7 0 95/7 B b b B b B b B b− − − −      + ∆ = + ∆ = + ∆ ≥        leads to 90 37 7 1 0b+ ∆ ≥ , 75 3 7 7 1 0b− ∆ ≥ , 95 8 7 21 1 0b− ∆ ≥ , so 130 25b− ≤ ∆ ≤ (b) ( ) 4575 65 15 62007 7 7 7 25 ˆ ˆ ˆ 0 0 0 T T Tb b b bλ λ λ       + ∆ = + ∆ = + =         