HW6 Sol

# HW6 Sol - EE103 HW5 Solution Winter 2009 Prof SEJ EE103...

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EE103 HW5 Solution, Winter 2009, Prof SEJ EE103 Applied Numerical Computing Instructor: Prof. S. E. Jacobsen HW6 Solution Students: Distributed HW solutions are a component of the course and should be fully understood. SEJ 1) a) Note that this problem is a minimization problem. To “conform” with the “desk” problem, a maximization problem, we could multiply the objective by “ - 1” and convert to a “max” problem; however, we will not do so. We approach the minimization problem directly. 1 2 3 1 2 3 1 2 3 1 2 3 Minimize 12 10 30 subject to: 3 2 8 17 3 9 2 8 16 0 x x x x x x x x x x x x x After the introduction of slack variables, 4 5 6 , , x x x , we have the following: 1 2 3 4 1 2 3 5 1 2 3 6 1 2 3 4 5 6 Minimize z subject to: 3 2 8 17 3 9 2 8 16 12 10 30 0 0 0 0 0 x x x x x x x x x x x x x x x x x x z x And the initial tableau will be as follows: x 1 x 2 x 3 x 4 x 5 x 6 z b T0 = -3 2 8 1 0 0 0 17 -1 1 3 0 1 0 0 9 -2 1 8 0 0 1 0 16 12 -10 -30 0 0 0 1 0 An “obvious” first solution to the above system of equalities, in non-negative variables (since z is the value of the objective, there is no constraint that requires z to be non-negative), is to

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EE103 HW5 Solution, Winter 2009, Prof SEJ equate the “slack” variables and z (associated with the identity matrix), 4 5 6 , , , x x x z , with the right-hand- side, and to consider all other variables to be zero; such a solution is called a “basic solution”. That is, the current solution is ( , ) (0,0,0,17,9,16,0) x z Because we’re trying to decrease the value of the ob jective (its current value is 0 z ), we attempt to increase the value of one of the non-basic variables, in this case 1 2 3 x or x or x . However, if we increase the value of 1 x we’ll actually increase the value of the objective (note that, as a function of the increase of 1 x , we’ll have 1 1 1 ( ) 12 , 0 z x z x z for x (why?) ). However, if we increase the value of either 2 3 x or x , the value of the objective will, in fact, decrease (why?). Note that if we increase the value of 2 x , z will decrease at the rate of 10 per unit increase of 2 x , or if we increase the value of 3 x , z will decrease at the rate of 30 per unit increase of 3 x . While it is not necessary to choose the variable that provides the greatest “rate of decrease”, we’ll do so here. We choose to in crease the value of 3 x as much as possible. It is important to note that in this case the value of the current basic variables, 4 5 6 , , x x x will each decrease as 3 x increases. However, we require that all current basic variables remain non- negative; therefore, we need to determine the largest value of 3 x that maintains the non- negativity of the current basic variables. The largest such value is determined by requiring all of the below to remain non-negative as 3 x increases: 4 3 3 5 3 3 6 3 3 ( ) 17 8 0 ( ) 9 3 0 ( ) 16 8 0 x x x x x x x x x That is, the largest allowable increase is given by the so- called “min ratio test” min{17 / 8,9 / 3,16 / 8} 16 / 8 . That is, the current basic variable, 6 x , “blocks” the increase of 3 x . The Simplex Algorithm proceeds by replacing 6 x with 3 x in the basis. That is, we pivot on the element in the 3 rd
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