This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Engineering Optimization: ISYE3133C Solution to Practice Midterm #2
1. (4 pts each for a total of 20 pts) For each of these statements, mark True or False (no justification needed): a) F  Every dual feasible basis for a standard form LP is optimal.. A dual feasible basis might be primal infeasible. b) T  There is a one to one correspondence between the extreme points of a standard form linear program and its basic feasible solutions. c) F  Every transportation problem is balanced. d) F  In a basic feasible solution only the nonbasic variables can have value zero. Non basic variables always have zero value. Basic variables are the solution to a linear system of equations, which could set some of the variables to zero. e) F  The dual of a standard form LP always has nonnegative variables. 2. (50 pts Total) Consider the following linear program and its graph: Extreme point for part b) Maximize Subject to z = 3x1 + x2 x1 + x2 2 (1) x1 + x2 4 (2) x1 3 (3) x1, x2 0 a) Convert the problem into standard form. Maximize Subject to z = 3x1 + x2 x1 + x2 + x3 =2 x1 + x2 + x4 = 4 x1 + x5 = 3 x1, x2, x3, x4,x5 0 b) Write a basis for the standard form problem. Indicate what are the associated primal and dual solutions. Indicate if the basis is primal/dual feasible/infeasible. Write the reduced costs of the nonbasic variables. Identify the associated extreme point in the graph. Since we can choose lets chose the simples basis. B={3,4,5}. Non basic variables are x1=0, x2=0 and x3, x4,x5 solve x3 = 2 x4 = 4 x5 = 3 So the primal solution is x1=0, x2=0, x3=2, x4=4,x5=3. The primal solution has all values >=0 so the basis is Primal Feasible. The dual solution y1, y2, y3 solves y1 = 0 y2 = 0 y3 = 0 so the dual solution is y1=y2=y3=0. The reduced costs of non basic variables are: c1 = 3(1)y1(1)y2(1)y3=3>0 c2 = 1(1)y1(1)y2(0)y3=1>0 So the basis is dual infeasible. The corresponding extreme point is indicated in the figure. c) Find the optimal basis for the standard form problem. Indicate what are the associated primal and dual solutions. Why is it optimal? Optimal basis is B={1,2,3}. Non basic variables are x4=0, x5=0 and x1, x2,x3 solve x1 + x2 + x3 x1 + x2 x1 =2 =4 =3 So the primal solution is x1=3, x2=1, x3=4, x4=0,x5=0. The primal solution has all values >=0 so the basis is Primal Feasible. The dual solution y1, y2, y3 solves y1 + y2 + y3 = 3 y1 + y2 =1 y1 =0 so the dual solution is y1=0, y2=1,y3=2. The reduced costs of non basic variables are: c4 = 0(0)y1(1)y2(0)y3=1<=0 c5 = 0(0)y1(0)y2(1)y3=1<=0 So the basis is dual feasible. The basis is primal and dual feasible, hence it is optimal. d) For the standard form problem, suppose we include a new variable into the formulation with objective coefficient 2 and whose coefficients for constraints (1), (2) and (3) are 1, 1 and 2 respectively. Is the basis from c) still optimal? Justify your answer. The reduced cost for this new variable is c6 = 2(1)y1(1)y2(2)y3 for the dual optimal solution y1=0, y2=1,y3=2, then c6 = 2(1)0(1)1(2)2=3<=0. The reduced cost of the new variable is nonpositive so the basis is still optimal. 3. (30 pts) Touche Young has three auditors. Each can work as many as 160 hours during the next month. During this time three projects must be completed. Project 1 will take 130 hours; project 2, 140 hours; and project 3, 160 hours. The amount per hour that can be billed for assigning each auditor to each project is given bellow. Auditor 1 2 3 Project 1 120 140 160 Project 2 150 130 140 Project 3 190 120 150 Remember to describe (in words) the variables, constraints and objective function. a. Formulate a transportation problem to maximize total billings during next month. Clearly indicate the supply and demand constraints. Maximize z = 120x11 + 150x12+190x13 +140x21 + 130x22+120x23 +160x31 + 140x32+150x33 x11 + x12 + x13 160 x21 + x22 + x23 160 x31 + x32 + x33 160 x11 + x21 + x31 130 x12 + x22 + x32 140 x13 + x23 + x33 160 x11, x21, x31, x12, x22, x32, x13, x23,x33 0 b. Draw the Graph or diagram of your transportation problem. Clearly label the nodes indicating if they are supply or demand nodes and their respective capacities or demands. Write the profit/cost per unit of flow besides each one of the arcs. Subject to S1=160 Auditor 1 120
15 0 D1=130 Project 1 19 0 S2=160 D2=140 14
Auditor 2 0
130
12 0 Project 2 16 0 S2=160 Auditor 3 0 14
150 D3=160 Project 3 Supply Nodes Demmand Nodes c. Is the transportation problem from a)b) balanced? The problem is not balanced because: 160+160+160=480430=130+140+160 redit Question
You can use the following in any of the problems. normal max problem of Standard Form problem Credit Question The Dual
normal max problem max
n z=
j=1 n cj xj
cj xj
j=1 n max s.t. z = s.t.
n j=1 j=1 aij xj = bi
aijx j 0 i x =b
j i = 1, . . . , m
i = 1, . .. .. ., , n m j = 1, j = 1, . . . , n (P)
(P) is m xj 0 min w =
min w = s.t. s.t. m i=1 bi y i
bi y i i=1 m m i=1 aij yi cj
aij yi cj j = 1, . . . , n
j = 1, . . . , n (D)
(D) (1) (1) i=1 hat that asible solution to (D) Also, the dual of (D) is (P). easible solution to (D) of (P) which can take the following values: optimal objective value he optimal objective value of = + if (P) is unbounded. (P) which can take the following values: = = + if (P) is unbounded.  if (P) is infeasible. s =  if (P) (P) has an optimal solution (i.e. (P) is feasible and is not unbounded). a number if is infeasible. is a number if (P) has an optimal solution (i.e. (P) is feasible and is not unbounded). + for any number d. d + for any number d.
m i=1 d consider the following three possibilities for (P): uld consider the following three possibilities for (P): nbounded. By lemma 3 if the primal is unbounded then the dual is infeasible. Then this unbounded. By lemma if the primal y unbounded solution to (D). not happen because we3assumed that is is a feasible then the dual is infeasible. Then this nnot happen because we assumed that y is a feasible solution to (D). feasible. Then z = . (2) holds because  is less than or equal to any number and nfeasible. Then z = . (2) holds because  is less than or equal to any number and i is a number. i y i is a number. an optimal solution. Lemma 1 states that s an optimal solution. Lemma 1 states that z m bi y i . z i=1 bi y i . (2) (2) m m ...
View
Full
Document
This test prep was uploaded on 04/18/2008 for the course ISYE 3133 taught by Professor Juanpablovielma during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 JuanPabloVielma
 Optimization

Click to edit the document details