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Linear Prog COP4355_Part_55

# Linear Prog COP4355_Part_55 - Chapter 18 Markov Analysis...

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224 Chapter 18 Markov Analysis 18.13 18.14 Replacing bulbs when they fail policy () From: Gd Fn Mg Fa Gd .6 0 0 1 Fn .2 .75 0 0 To: Mg .15 .15 .2 0 Fa .05 .10 .8 0    The equilibrium equations will be .6Gd + 0Fn + 0Mg + 1Fa = Gd .2Gd + .75Fn + 0Mg + 0Fa = Fa .15Gd + .15Fn + .2Mg + 0Fa = Mg .05Gd + .10Fn + .8Mg + 0Fa = Fa or 18.14 (continued) -.4Gd + 0Fn + 0Mg + 1Fa = 0 .2Gd - .25Fn + 0Mg + 0Fa = 0 .15Gd + .15Fn - .8Mg + 0Fa = 0 .05Gd + .1Fn + .8Mg – 1Fa = 0 Gd + Fn + Mg + Fa = 1 0 .3 .1 0 .2 -.3 1 1 1 -.11 %current @ equilibrium = = = 68.75% -.1 .3 .1 -.16 0 .2 -.3 1 1 1 -.1 0 .1 0 0 -.3 1 1 1 -.03 %overdue @ equilibrium = = = 18.75% -.16 -.16 %Bad [email protected] equilibrium -.1 .3 0 0 .2 0 1 1 1 -.02 = = = 12.5% -.16 -.16

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Solutions Manual and Workbook 225 Deleting the fourth equation yields: .4 0 0 1 Gd 0 .2 -.25 0 0 Fn 0 = .15 .15 -.8 0 Mg 0 1 1 1 1 Fa 1    Solving using Cramer’s rule yields: () ( ) ) 44 14 -.4 0 0 0 .2 -.25 0 0 .15 .15 .8 0 1 1 1 1 Fa = -.4 0 0 1 .2 -.25 0 0 .15 .15-.8 0 1 1 1 1 .4 0 0 -1 1 .2 -.25 0 .15 .15-.8 = .2 -.25 0 -1 1 .15 .15-.8 1 + + ) .4 0 0 + -1 1 .2 -.25 0 1 1 .15 .15-.8 -.08 = = 0.158 -.4275 - .08 + Expected replacement cost = .158 x \$20
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Linear Prog COP4355_Part_55 - Chapter 18 Markov Analysis...

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