viterbi school of engineering daniel j. epstein department...

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Viterbi School of Engineering Daniel J. Epstein Department of Industrial and Systems Engineering ISE 330: Introduction to Operations Research Fall 2006 (October 25): Midterm

SOLUTIONS

1 hour 15 minutes

( 9 pages total )

Question Total Score

(a) 15

(b) 20

(c) 20

(d) 20

(e) 15

(f) 25

(g) 15

(h) 10

(i) 20

160

Hermione: " "I hope you're pleased with yourselves. We could have been all killed

-- or worse, expelled."

Your situation is not as dire. Have a good test! ~ E.C.

Name: _____________________________________________________

ISE 330: Midterm (October 25, 2006) Initial: _______________

E.C. 2

Bertie Bott’s every flavor jellybeans is a favorite amongst Harry Potter and his friends, who frequent Honeydukes in Hogsmead to purchase these tasty treats. "When they say every flavor, they mean every flavor..." ~ Ron Weasley "Bertie Bott's Every Flavor Beans – A Risk With Every Mouthful!" ~ Quidditch World Cup ad

Bertie Bott’s would like to make some decisions on the quantity of Brussel Sprouts, Marmalade, Earwax, and Sardine flavor jellybeans to produce, each of which requires special potions and labor to create. Each batch of Brussel Sprouts, Marmalade, Earwax, and Sardine jellybeans sell for 4, 2, 3, and 5 sickles respectively. Bertie Bott’s has no problems selling every jelly bean it makes. Each batch of Brussel Sprouts jellybeans requires 2 units of potions and 8 units of labor. Each batch of Marmalade jellybeans requires 3 units of potions and 1 unit of labor. Each batch of Earwax jellybeans requires 4 unit of potions and 1 unit of labor. Each batch of Sardine jellybeans requires 2 units of potions and 5 units of labor to produce. Assume that exactly 300 units of potions and exactly 300 units of labor have to be consumed in the production. (a) Formulate a linear programming model that will help Bertie Bott’s determine how it

can maximize profits subject to some constraints. [ 15 points ] Max Z = 4x1 + 2x2 + 3x3 + 5x4

s.t. 2x1 + 3x2 + 4x3 + 2x4 = 300

8x1 + x2 + x3 + 5x4 = 300

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,

Picture excerpted from harrypotter.warnerbros.com/bertiebotts


E.C. 3

(b) Using the Big M method, construct the complete first simplex tableau for the simplex

method and identify the corresponding initial basic feasible solution. Also identify the initial entering basic variable and the leaving basic variable.

[20 points] Max Z = 4x1 + 2x2 + 3x3 + 5x4 – M

!

x 5 – M

!

x 6

s.t. 2x1 + 3x2 + 4x3 + 2x4 +

!

x 5 = 300

8x1 + x2 + x3 + 5x4 +

!

x 6= 300

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,

!

x 5 ≥ 0,

!

x 6 ≥ 0

Z x1 x2 x3 x4

!

x 5

!

x 6 RHS

Z 1 -4 -2 -3 -5 M M 0

!

x 5 0 2 3 4 2 1 0 300

!

x 6 0 8 1 1 5 0 1 300

Initial tableau in standard form:

Z x1 x2 x3 x4

!

x 5

!

x 6 RHS

Z 1 -4-10M -2-4M -3-5M -5-7M 0 0 -600M

!

x 5 0 2 3 4 2 1 0 300

!

x 6 0 8 1 1 5 0 1 300

The initial basic feasible solution is x = [ 0, 0, 0, 0, 300, 300 ].


E.C. 4

(c) Using the two-phase method, construct the complete first simplex tableau for phase I

and identify the corresponding initial basic feasible solution. Also, identify the initial entering basic variable and the leaving basic variable.

[20 points] Min Z =

!

x 5 +

!

x 6

s.t. 2x1 + 3x2 + 4x3 + 2x4 +

!

x 5 = 300

8x1 + x2 + x3 + 5x4 +

!

x 6= 300

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0,

!

x 5 ≥ 0,

!

x 6 ≥ 0

Z x1 x2 x3 x4

!

x 5

!

x 6 RHS

Z -1 0 0 0 0 1 1 0

!

x 5 0 2 3 4 2 1 0 300

!

x 6 0 8 1 1 5 0 1 300

Initial tableau in standard form:

Z x1 x2 x3 x4

!

x 5

!

x 6 RHS

Z -1 -10 -4 -5 -7 0 0 -600

!

x 5 0 2 3 4 2 1 0 300

!

x 6 0 8 1 1 5 0 1 300

The initial basic feasible solution is x = [ 0, 0, 0, 0, 300, 300 ].


E.C. 5

(d) You are given the final tableau of Phase I below. Construct the initial tableau for

Phase II, and identify the corresponding initial basic feasible solution. Also identify the initial entering basic variable and the leaving basic variable.

[20 points]

Z x1 x2 x3 x4

!

x 5

!

x 6 RHS

Z -1 0 0 0 0 1 1 0

x3 0 0 11/15 1 1/5 4/15 -1/15 60

x1 0 1 1/30 0 3/5 -1/30 2/15 30

Remove the artificial variable columns and add the original objective function:

Z x1 x2 x3 x4 RHS

Z 1 -4 -2 -3 -5 0

x3 0 0 11/15 1 1/5 60

x1 0 1 1/30 0 3/5 30

Put into proper form:

Z x1 x2 x3 x4 RHS

Z 1 0 -28/15 -3 -13/15 120

x3 0 0 11/15 1 1/5 60

x1 0 1 1/30 0 3/5 30

Z x1 x2 x3 x4 RHS

Z 1 0 1/3 0 -2 300

x3 0 0 11/15 1 1/5 60

x1 0 1 1/30 0 3/5 30

The initial basic feasible solution is x = [ 30, 0, 60, 0, 0, 0 ]. ( x = [ 30, 0, 60, 0 ] is also OK. )


E.C. 6

(e) Construct the dual of the problem in part (a). [ 15 points ] Min W = 300y1 + 300y2

s.t. 2y1 + 8y2 ≥ 4

3y1 + y2 ≥ 2

4y1 + y2 ≥ 3

2y1 + 5y2 ≥ 5

y1, y2 unconstrainted.


E.C. 7

(f) Use the graphical method to find the solution to the dual. Identify the solution and the

binding constraints and label them with on the graph. [ 25 points ]

If you need to re-do your graph, you can find extra graph paper on page 11.


E.C. 8

(g) Now, augment the dual problem with only surplus variables, and use your findings in

part (f) to propose an optimal solution, with values for all the surplus variables, for the problem.

[ 15 points ] Min W = 300y1 + 300y2

s.t. 2y1 + 8y2 – y3 = 4

3y1 + y2 – y4 = 2

4y1 + y2 – y5 = 3

2y1 + 5y2 – y6 = 5

y1, y2 unconstrainted.

The defining equations are: 4y1 + y2 = 3

2y1 + 5y2 = 5

Solving the system of equations, we have that y = [ 5/9, 7/9, 4/9, 5/3, 0, 0 ], and W* = 400. (h) Given the solution to the dual you have derived in part (f), use the complementary

slackness property to identify the basic and non-basic variables in the optimal solution to the original problem in part (a).

[ 10 points ] The complementary slackness property, tells us that only x3 and x4 are basic in the optimal solution; all other variables are non-basic.


E.C. 9

(i) Now, return to the initial tableau for Phase II given in part (d). You can either

continue from your results from part (d) or use your results in part (h) to find the optimal solution for the primal problem.

[ 20 points ]

Z x1 x2 x3 x4 RHS

Z 1 0 1/3 0 -2 300

x3 0 0 11/15 1 1/5 60

x1 0 1 1/30 0 3/5 30

Z x1 x2 x3 x4 RHS

Z 1 2 2/5 0 0 400

x3 0 -1/3 13/18 1 0 50

x4 0 5/3 1/18 0 1 50

The optimal solution is x = [ 0, 0, 50, 50, 0, 0 ], with Z* = 400.

viterbi school of engineering daniel j. epstein department...

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