Ch04 - 1 MP CHAPTER 4 SOLUTIONS SECTION 4.1 1. max z = 3x1...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
MP CHAPTER 4 SOLUTIONS SECTION 4.1 1. max z = 3x 1 + 2x 2 s.t. 2x 1 + x 2 + s 1 = 100 x 1 + x 2 + s 2 = 80 x 1 + s 3 = 40 2. min z = 50x 1 + 100x 2 s.t. 7x 1 + 2x 2 - e 1 = 28 2x 1 + 12x 2 - e 2 = 243. 3. min z = 3x 1 + x 2 s.t. x 1 - e 1 = 3 x 1 + x 2 + s 2 = 4 2x 1 - x 2 = 3 SECTION 4.4 1. From Figure 2 of Chapter 3 we see that the extreme points of the feasible region are Basic Feasible Solution H = (0, 0) s 1 = 100, s 2 = 80, s 3 = 40 x 1 = x 2 = x 3 = 0 E = (40, 0) x 1 = 40, s 1 = 20, s 2 = 40 x 2 = x 3 = s 3 = 0 F = (40, 20) x 1 = 40, x 2 = 20, s 2 = 20 x 3 = s 1 = s 3 = 0 G = (20, 60) x 1 = 20, x 2 = 60, s 3 = 20 x 3 = s 1 = s 2 = 0 D = (0, 80) s 1 = 20, x 2 = 80, s 3 = 40 s 2 = x 1 = x 3 = 0 2. From Figure 4 of Chapter 3 we see that the correspondence is as follows: Extreme Point Basic Feasible Solution E = (3.6, 1.4) x 1 = 3.6, x 2 = 1.4, e 1 = e 2 = 0 B = (0, 14) x 2 = 14, e 2 = 144, e 1 = x 1 = 0 C = (12, 0) x 1 = 12, e 1 = 56, x 2 = e 2 = 0 3. Basic Variables Basic Feasible Solution Corner Point x 1 , x 2 x 1 =150 x 2 =100 s 1 =s 2 =0(150, 100) x 1 , s 1 x 1 =200, s 1 =150, x 2 =s 2 =0(200, 0) x 1 , s 2 x 1 =350, s 2 =-300, x 2 =s 1 =0 Infeasible x 2 , s 1 x 2 =400, s 1 =-450, x 1 =s 2 =0 Infeasible x 2 , s 2 x 2 =175, s 2 = 225, x 1 =s 1 =0(0, 175) s 1 , s 2 s 1 =350, s 2 =400, x 1 =x 2 =0 (0, 0) 4. Let σ A = 1/6, σ C = 1/3 and σ B = ½. Then 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
). 20 , 10 ( ) 40 , 0 ( 2 1 ) 0 , 30 ( 3 1 ) 0 , 0 ( 6 1 = + + 5. + 2 . 33 2 . 12 6 . 2 1 10 144 0 14 0 will do the trick. 6. Suppose we are given that A d = 0 and d>=0 and that x is feasible . Then for all c>=0, x+ c d>=0. Also A( x +c d)=Ax=b. Therefore for all c>=0 x +c d is feasible. We now show that if Ad =0 and d>=0 are not both true, then d is not a direction of unboundedness. This is equivalent to showing that if d is a direction of unboundedness then both Ad =0 and d>=0 must be true. Clearly if Ad =0 is not true then x +c d is not feasible so d is not a direction of unboundedness. If d>=0 is not true then d must have an I’th component that is negative. Then for large enough c>=0 the I’th component of x +c d will be negative, so x +c d will not be feasible and d will not be a direction of unboundedness. 7. 3 0 1 1 is a direction of unboundedness. As we move in this direction z increases and we stay feasible. SECTION 4.5 1. z x 1 x 2 s 1 s 2 s 3 RHS Ratio ------------------------------------------ 1 -3 -2 0 0 0 0 ------------------------------------------ 0 2 1 1 0 0 100 50 ----------------------------------------- 0 1 1 0 1 0 80 80 x 1 enters basis in ------------------------------------------ 0 1 0 0 0 1 40 40* row 3 ------------------------------------------ 2
Background image of page 2
z x 1 x 2 s 1 s 2 s 3 RHS Ratio ------------------------------------------ 1 0 -2 0 0 3 120 ------------------------------------------ 0 0 1 1 0 -2 20 20* Enter x 2 in row 1
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 56

Ch04 - 1 MP CHAPTER 4 SOLUTIONS SECTION 4.1 1. max z = 3x1...

This preview shows document pages 1 - 4. Sign up to view the full document.

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