hwk7_6&7Solutions - P 1 , P 2 , B excess > 0...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework 7 Solutions Fall 2010 6. Winston, page 115, Problem 15. Objective function in terms of products: Decision Variables: Let P 1 = Units of process 1 run P 2 = Units of process 2 run A = Amount of chemical A produced B sold = Amount of chemical B sold B excess = Amount of chemical B to dispose of Objective function: max z = 16*A + 14*B sold – 2* B excess Constraints: 2P 1 + 3P 2 < 60 (labor constraint) P 1 + 2P 2 < 40 (materials constraint) A = 2P 1 + 3P 2 (production equation for A) B sold + B excess = P 1 + 2P 2 (production equation for B) B sold < 20 (demand constraint) P 1 , P 2 , A, B sold , B excess > 0 (non-negativity constraint) Alternatively in terms of processes: Decision Variables: Let P 1 = Units of process 1 run P 2 = Units of process 2 run B excess = Amount of chemical B to dispose of Objective function: max z = 16*(2P 1 + 3P 2 ) + 14*(P 1 + 2P 2 – B excess ) – 2*B excess Constraints: 2P 1 + 3P 2 < 60 (labor constraint) P 1 + 2P 2 < 40 (materials constraint) P 1 + 2P 2 – B excess < 20 (demand constraint)
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: P 1 , P 2 , B excess > 0 (non-negativity constraint) Homework 7 Solutions Fall 2010 7. Winston, page 116, Problem 20 Decision Variables: Let X 1 = Acres invested in tract 1 for Spruce X 2 = Acres invested in tract 1 for Hunting X 3 = Acres invested in tract 1 for Both X 4 = Acres invested in tract 2 for Spruce X 5 = Acres invested in tract 2 for Camping X 6 = Acres invested in tract 2 for Both Objective Function: max z = .2*X 1 + .4*X 2 + .5*X 3 + .06*X 4 + .09*X 5 + 1.1*X 6 (in thousands of dollars) Constraints: X 1 + X 2 + X 3 < 300 (allocation constraint for tract 1) X 4 + X 5 + X 6 < 100 (allocation constraint for tract 2) 3*X 1 + 3*X 2 + 4*X 3 + X 4 + 30*X 5 + 10*X 6 < 1,500 (capital constraint in hundreds of dollars) .1*X 1 + .2*X 2 + .2*X 3 + .05*X 4 + 5*X 5 + 1.01*X 6 < 200 (labor constraint in worker-days) X 1 , X 2 , X 3 , X 4 , X 5 , X 6 > 0 (non-negativity constraint)...
View Full Document

This note was uploaded on 09/07/2011 for the course OR 1101 taught by Professor Trotter during the Fall '09 term at Cornell University (Engineering School).

Page1 / 2

hwk7_6&amp;7Solutions - P 1 , P 2 , B excess > 0...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online