Assignment #3 - Solutions MATH 3300A (01) Optimization Fall 2015

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1 Assignment #3 - Solutions MATH 33A (1) Optimization Fall 15 Section Typical isoprofit line is 3x 1 +c x =z. This has slope -3/c. If slope of isoprofit line is <-, then Point C is optimal. Thus if -3/c <- or c <1.5 the current basis is no longer optimal. Also if the slope of the isoprofit line is >-1 Point A will be optimal. Thus if -3/c >-1 or c >3 the current basis is no longer optimal. Thus for 1.5 c 3 the current basis remains optimal. For c =.5 x 1 =, x = 6, but z = 3() +.5(6) = $1.. Currently Number of Available Carpentry Hours = b =. If we reduce the number of available carpentry hours we see that when the carpentry constraint moves past the point (4,) the carpentry and finishing hours constraints will be binding at a point where x 1 >4. In this situation b <4 + = 6. Thus for b <6 the current basis is no longer optimal. If we increase the number of available carpentry hours we see that when the carpentry constraint moves past (,1) the carpentry and finishing hours constraints will both be binding at a point where x 1 <. In this situation b >1.Thus if b >1 the current basis is no longer optimal. Thus the current basis remains optimal for 6 b 1. If 6 b 1, the number of soldiers and trains produced will change. 3. If b 3, the demand for soldiers, is increased then the current basis remains feasible and therefore optimal. If, however, b 3 < then the point where the finishing and carpentry constraints are binding is no longer feasible (it has s 3 <). Thus for b 3 > the current basis remains optimal. For b 3 > the point where the carpentry and finishing constraints are still binding remains at (, 6) so producing soldiers and 6 trains remains optimal. 4a. If isocost line is flatter than HIM constraint, point C is optimal. If isocost line is steeper than HIW, point B is optimal. Thus current basis remains optimal if -7/ -c 1 /1-1/6 or 5/3 c b. Current basis remains optimal if -7/ -5/c -1/6 or 1/7 c 3. 4c. If we decrease HIW requirement optimal solution moves towards D = (, ). We lose feasibility if HIW requirement is () = 4. Increase HIW requirement and optimal solution moves towards C = (1, ). We lose feasibility when HIW requirement = 7(1) = 4. Thus current basis remains optimal for 4,, HIW Requirement 4,,. If HIW = +, the optimal solution is where 7x 1 + x = + and x 1 + 1x = 4. This yields x 1 = and x = d. If we decrease HIM requirement optimal solution moves towards A = (4, ). We lose feasibility if HIM requirement is (4) =. Increase HIM requirement and optimal solution moves towards B = (, 14). We lose feasibility when HIM requirement = 14(1) = 16. Thus current basis remains optimal for,, HIW Requirement 16,,. If HIM = +, the optimal solution is where 7x 1 + x = and x 1 + 1x = 4 +. This yields x 1 = /4 and x = /.

2 4e. If HIW requirement is +, z = 5x 1 + 1x = so HIW shadow price is -5 (thousand). If HIM requirement is 4 +, z = 5x 1 + 1x = , so HIM shadow price is f. New z-value = 3 - (-) (-5) = $31, Section BV = {x,s 1 }, B =, B -1 =, c BV = [1 ] 1 c BV B -1 = [1 ] 1 = [ 1] Coefficient of x 1 in row = c BV B -1 a 1 -c 1 = [ 1] +1= Coefficient of s in row = c BV B -1 = 1 RHS of row = c BV B -1 b = [ 1] 4 = Column for x 1 = B -1 a 1 = 1 = Column for s = B -1 = 1 RHS of Optimal Tableau = B -1 b = 1 4 = Thus optimal tableau is z + x 1 + s =

3 x 1 + x + s = x 1 + s 1 - s = Section Change c 3 to + Δ. Then c BV B -1 = [ +Δ 6] = [ 1 + Δ 1-4Δ]. Coefficient of x in row =c BV B -1 a -c = [ 1 + Δ 1-4Δ ] 6-3 = 5 - Δ. 1.5 Thus New row is z + (5-Δ)x + (1 + Δ)s + (1-4Δ)s 3. Current basis remains optimal if satisfies the following 3 inequalities: 5-Δ iff Δ Δ iff Δ Δ iff Δ.5 Thus current basis remains optimal if -5 Δ.5 or -5 c If c 3 = 1 the current solution is still optimal but z increases by to z =. If c 3 = 5, Δ = 5. Then the tableau for the current optimal basis becomes z -5x + s - 1s 3 = 3 -x + s 1 + s - s 3 = 4 -x + x 3 + s - 4s 3 = x x -.5s + 1.5s 3 = Eventually we obtain the new optimal solution z = 1/3, x 3 = 4/3, x 1 = x =.. If c 1 = 55 then Δ = -5. Then c BV B -1 = [ 55].5 4 = [ 1.5.5] 1.5 c BV B -1 b = [ 1.5.5] 4 = 7

4 6 Coefficient of x in row = [ 1.5.5] - 3 = Tableau for current optimal basis is now z x 1 x x 3 s 1 s s 3 RHS After pivoting in x we obtain the new optimal solution z = 7, s 1 = 7., x 3 = 11., x = 1.6. Thus reduction in price of desks has resulted in no desks being manufactured. 3. Denote the RHS of first constraint by 4 + Δ. As long as B Δ the current basis remains optimal. B Δ = 4 + Δ Thus for Δ -4 or b = 4 the current basis remains optimal. If b 1 = 3 the current basis is still optimal. Thus s x x = B -1 = 6 and new optimal solution is s 1 = 6, x 3 =, x 1 =. The new optimal z-value may be obtained from

5 3 z = c BV B -1 b = [ 1 1] = 4. Coefficient of Tables in row = [ 1 1] 3-5 = -5. Thus 1.5 current basis is no longer optimal. To compute the new optimal solution we need to determine the column for tables in the current optimal tableau. This is given by 5 B -1 a = 4 3 = We can now find the new optimal solution from the following tableau: z x 1 x x 3 s 1 s s 3 RHS Entering x in the basis in row 3 yields the new optimal solution z = 93 1/3, s 1 = 37 1/3, x 3 =, x = /3. 5. The coefficient of home computer tables in row will be [ 1 1] made = 4. Thus current solution remains optimal and no home computer tables will be Section 6.5. max z = 4x 1 + x + 3x 3 s.t. x 1 + x + x 3 1 x 1 + x + x 3-1 x 1, x, x 3

6 3. min w = 5y 1 + 7y + 6y 3 + 4y 4 s.t. y 1 + y + y 4 4 y 1 + y + y 3 = -1 y 3 + y 4 = y 1, y, y 3 y 4 u.r.s. Section The dual objective function is w =.5y 1 +.5y. The optimal dual solution is y 1 =.4 y = 1.4 w =.5(.4)+.5(1.4) =.9. Thus rhs of optimal primal tableau's row must be From row of the optimal primal tableau we find that the optimal dual solution is y 1 = y = 1. But then Optimal Primal Obj. Function Value = Optimal Dual Obj. Func. Value = 6() + 1(1) = 1 /= /3 so something is wrong.