Lesson Topics - Pepperdine Universityseaver-faculty.pepperdine.edu/jburke2/ba452/PowerP2/ReviewQB_3.pdf · Lesson Topics Rounding Off (5) ... ( 2) are Transportation Problems of

B.3 Integer Programming Review Questions

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Lesson Topics

Rounding Off (5) solutions in continuous variables

to the nearest integer (like 2.67 rounded off to 3) is an unreliable way to solve a linear programming problem when decision variables should be integers. Sensitivity Analysis with Integer Variables is more important than with continuous variables because a small change in a constraint coefficient can cause a relatively large change in the optimal solution. Assignment Problems with Valuable Time

(2) minimize the total time of assigning workers to jobs. Minimizing total time is appropriate when each worker has the same value of time. Assignment Problems with Supply and

Demand (2) are Transportation Problems of

suppliers to demanders except that each demand is assigned to exactly one supplier.


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Rounding Off

Question. Consider the following all-integer linear program:

Max 1x1 + 1x2

s.t. 4x1 + 6x2 ≤ 22 1x1 + 5x2 ≤ 15 2x1 + 1x2 ≤ 9 x1, x2 ≥ 0 and integer

a. Graph the constraints for this problem. Use dots to indicate all

feasible integer solutions.

b. Find the optimal integer solution.

c. Solve the LP Relaxation of this problem (that is, the problem without requiring decision variables to be integers).


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Answer to Question:

a.

b. The optimal solution to the LP Relaxation is shown on the above graph to be x1 = 4, x2 = 1. Its value is 5.

c. The optimal integer solution is the same as the optimal solution to the

LP Relaxation. This is always the case whenever all the variables take on integer values in the optimal solution to the LP Relaxation.


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Rounding Off

Question. Muir Manufacturing produces two

popular grades of commercial carpeting among its

many other products. In the coming production

period, Muir needs to decide how many rolls of each

grade should be produced in order to maximize profit. Each roll of Grade X

carpet uses 3 units of synthetic fiber, requires 1 hour of production time,

and needs 1 unit of foam backing. Each roll of Grade Y carpet uses 1 unit

of synthetic fiber, requires 3 hours of production time, and needs 1 units of

foam backing.

The profit per roll of Grade X carpet is $3 and the profit per roll of Grade Y

carpet is $2. In the coming production period, Muir has 9 units of synthetic

fiber available for use. Workers have been scheduled to provide up to 7

hours of production time. The company has 10 units of foam backing

available for use.

a. Develop a linear programming model for this problem to determine

how many rolls of each carpet should be produced.

b. Graphically solve the linear-programming problem from Part a if you

require that the number of rolls of each carpet be integers.

c. Graphically solve the linear-programming problem from Part a if you

do not require that the number of rolls of each carpet be integers

(instead, the number of rolls of each carpet are continuous variables).

d. Compare your solutions in Parts b and c.

Tip: Your written answer should define the decision variables, formulate the

objective and constraints, and solve for the optimum. --- You will not earn

full credit if you just solve for the optimum; you must also define the

decision variables, and formulate the objective and constraints.


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Answer to Question:

Part a:

Let X = the number of rolls of Grade X carpet to make

Let Y = the number of rolls of Grade Y carpet to make

Max 3X + 2Y

s.t. 3X + Y < 9

X + 3Y < 7

X + Y < 10

X , Y > 0

Part c:

A graph of the feasible set reveals that the third costraint is redundant --- it

does not affect the feasible set. Adding a graph of isovalue lines reveals

the optimum occurs where the first and second constraints bind (the third

constraint is redundant). Solving the binding form of those two constraints

yields the optimal solution: X = 2.5, Y = 1.5

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8 9 10


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Part b:

A graph of the feasible set reveals that the third costraint is redundant --- it

does not affect the feasible set. Use dots to indicate all feasible integer

solutions. Adding a graph of isovalue lines reveals the optimum occurs at

X = 3, Y = 0

Part d: Comparing solutions, the optimal integer solution in Part b is not the

result of rounding off the optimal continuous solution in Part c.

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8 9 10


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Rounding Off

Question. The Iron Works, Inc. seeks to maximize

profit by making two products from steel and labor.

It just received this day's allocation of 9 pounds of

steel, and there is 8 hours of labor available.

It takes 3 pounds of steel to make a unit of Product 1, and 1 pound of steel

to make a unit of Product 2. It also takes 2 hours of labor to make a unit of

Product 1, and 5 hours of labor to make a unit of Product 2. The physical

plant has the capacity to make up to 5 units of total product (Product 1 plus

Product 2). Product 1 has unit profit 3 dollars, and Product 2 has 9 dollars.


how much should be produced.

b. Graphically solve the linear-programming problem from Part a you

require that production units be integers.


do not require that production units be integers (instead, production

units are continuous variables).







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Answer to Question:

Part a:

Let X = units of Product 1 produced.

Let Y = units of Product 1 produced.

Max 3X + 9Y

s.t. 3X + Y < 9 (steel)

2X + 5Y < 8 (labor)

X + Y < 5 (capacity)

X, Y 0

Part c:

A graph of the feasible set and isovalue lines reveals the optimum occurs

where the non-negativity of X and the second constraint binds. Solving the

binding form of those two constraints yields the optimal solution: X = 0, Y =

8/5 = 1.6

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8 9 10

X on horizontal axis

Y on vertical axis

First constraint, through (3,0) and (0,9)

Second constraint, through (4,0) and (0,1.6)

Third constraint, (5,0) and (0,5)

Feasible set of solutions is the region bounded

by first, second, and non-negativity constraints


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Part b:

A graph of the feasible set with dots to indicate all feasible integer solutions

and isovalue lines reveals the optimum occurs at X = 1, Y = 1

Part d: Comparing solutions, the optimal integer solution in Part b is not the

result of rounding off the optimal continuous solution in Part c.

0

1

2

3

4

5

6

7

8

9

10

0 1 2 3 4 5 6 7 8 9 10


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Rounding Off

Question. Exxon Mobil Corporation seeks to

maximize profit by making two grades of gasoline

from crude oil and additives. It just received this

day's allocation of 4 thousand gallons of crude oil,

and 2 thousand gallons of additives.

It takes 0.8 gallons of crude oil to make a gallon of Premium gasoline, and

0.4 gallons of crude oil to make a gallon of Regular gasoline. It also takes

0.2 gallons of additives to make a gallon of Premium gasoline, and 0.6

gallons of additives to make a gallon of Regular gasoline. Premium

gasoline has unit profit 3 of dollars, and Regular gasoline has 1 dollar.














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Answer to Question:

Part a:

Let X = thousands of units of Premium gasoline produced.

Let Y = thousands of units of Regular gasoline produced.

Max 3X + 1Y

s.t. 0.8X + 0.4Y < 4 (crude oil)

0.2X + 0.6Y < 2 (additives)

X, Y 0

Part c:

A graph of the feasible set and isovalue lines (dashed lines above) reveals

the optimum occurs where the non-negativity of Y and the first constraint

binds. Solving the binding form of those two constraints yields the optimal

solution: X = 5, Y = 0

Part b: Since the continuous solution is discrete, it is also the solution to

the discrete case. (Since (X,Y) are in thousands of units, (X,Y) would still

be an integer solution as long as 1000X and 1000Y were integers.

0

1

2

3

4

5

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7

8

9

10

0 1 2 3 4 5 6 7 8 9 10


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Rounding Off

Question. Exxon Mobil Corporation seeks to

maximize profit by making two grades of gasoline

from crude oil and additives. It just received this

day's allocation of 1.5 thousand gallons of crude oil,

and 0.8 thousand gallons of additives.

It takes 0.6 gallons of crude oil to make a gallon of Premium gasoline, and

0.3 gallons of crude oil to make a gallon of Regular gasoline. It also takes

0.2 gallons of additives to make a gallon of Premium gasoline, and 0.3

gallons of additives to make a gallon of Regular gasoline. Premium

gasoline has unit profit 3 of dollars, and Regular gasoline has 4 dollars.














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Answer to Question:

Part a:

Let X = thousands of units of Premium gasoline produced.

Let Y = thousands of units of Regular gasoline produced.

Max 3X + 4Y

s.t. 0.6X + 0.3Y < 1.5 (crude oil)

0.2X + 0.3Y < 0.8 (additives)

X, Y 0

Part c:


the isovalue lines are steeper than the second constraint, and the optimum

occurs where the first and the second constraint bind. Solving the binding

form of those two constraints yields the optimal solution: X = 1.75, Y = 1.50

0

1

2

3

4

5

0 1 2 3 4 5


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Part b: You would earn full credit if you noted that since (X,Y) are in

thousands of units, X = 1.75 thousand and Y = 1.50 thousand are integer

solutions. You would also earn full credit by solving the problem if you

require that X and Y are integers.

Max 3X + 4Y

s.t. 0.6X + 0.3Y < 1.5 (crude oil)

0.2X + 0.3Y < 0.8 (additives)

X, Y 0

A graph of the feasible set with dots to indicate all feasible integer solutions

and isovalue lines reveals the optimum occurs at X = 1, Y = 2

Part d: The integer solution in Part b is not the result of rounding off the

continuous solution in Part c.

0

1

2

3

4

5

0 1 2 3 4 5


15

Rounding Off

Question. Jacuzzi produces two types of hot tubs:

Fuzion and Torino.

There are 5 pumps, 36 hours of labor, and 60 feet of

tubing available to make the tubs per week. Here are

the input requirements, and unit profits:

Fuzion Torino

Pumps 1 1

Labor 9 hours 4 hours

Tubing 8 feet 15 feet

Unit Profit $6 $9














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Answer to Question:

Part a:

Let F = Fuzion units produced per week.

Let Y = Torino units produced per week.

Max 6 F + 9 T

s.t. F + T < 5 (pumps)

9 F + 4 T < 36 (labor hours)

8 F + 15 T < 60 (tubing)

F, T 0

Part c:


the optimum occurs where the first and the third constraint bind. Solving

the binding form of those two constraints yields the optimal solution: F =

2.143, T = 2.857

0

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2

3

4

5

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7

8

9

10

0 1 2 3 4 5 6 7 8 9 10


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Part b:

Max 6 F + 9 T

s.t. F + T < 5 (pumps)

9 F + 4 T < 36 (labor hours)

8 F + 15 T < 60 (tubing)

F, T 0

A graph of the feasible set dots and isovalue lines (dashed lines above)

reveals the optimum occurs at either (0,4) or (1,3) or (3,2). It turns out

there are two optima: (0,4) and (3,2).

Part d: The integer solutions in Part b are not the result of rounding off the

continuous solution in Part c.

0

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3

4

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8

9

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0 1 2 3 4 5 6 7 8 9 10


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Assignment with Valuable Time

Question. Sibeau Custom Tailoring has five idle

tailors and two custom garments to make. The

estimated time (in hours) it would take each tailor to

complete a garment is shown in the next slide.

Tailor

Garment 1 2 3 4 5

Polyester Suit 9 11 6 11 8

Silk Suit 11 10 9 12 9

Formulate an appropriate binary-integer program for determining the tailor-

garment assignments that minimize the total estimated time spent making

the two garments. No tailor is to be assigned more than one garment, and

each garment is to be worked on by exactly one tailor. Formulate the

problem, but you need not solve for the optimum.

Tip: Your written answer should define the decision variables, and

formulate the objective and constraints.


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Answer to Question: Define the decision variables xij = 1 if garment i is

assigned to tailor j, and = 0 otherwise.

Minimize total time spent making garments:

Min 9x11 + 11x12 + 6x13 + 11x14 + 8x15

+ 11x21 + 10x22 + 9x23 + 12x24 + 9x25

Define the constraints of exactly one tailor per garment:

1) x11 + x12 + x13 + x14 + x15 = 1

2) x21 + x22 + x23 + x24 + x25 = 1

Define the constraints of no more than one garment per tailor:

3) x11 + x21 < 1 4) x12 + x22 < 1

5) x13 + x23 < 1 6) x14 + x24 < 1

7) x15 + x25 < 1


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Assignment with Valuable Time

Question. Sibeau Custom Tailoring has five idle

tailors and two custom garments to make. The

estimated time (in hours) it would take each tailor to

complete a garment is shown in the next slide. (An 'X'

in the table indicates an unacceptable tailor-garment assignment.)

Tailor

Garment 1 2 3 4 5

Polyester Suit 9 11 6 11 8

Silk Suit 11 10 X 12 9

Formulate and solve a binary-integer program for determining the tailor-

garment assignments that minimize the total estimated time spent making

the two garments. No tailor is to be assigned more than one garment, and

each garment is to be worked on by exactly one tailor.






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Answer to Question: Define the decision variables

xij = 1 if garment i is assigned to tailor j, and = 0 otherwise.

Minimize total time spent making garments:

Min 9x11 + 11x12 + 6x13 + 11x14 + 8x15

+ 11x21 + 10x22 + 12x24 + 9x25

Define the constraints of exactly one tailor per garment:

1) x11 + x12 + x13 + x14 + x15 = 1

2) x21 + x22 + x24 + x25 = 1

Define the constraints of no more than one garment per tailor:

3) x11 + x21 < 1 4) x12 + x22 < 1

5) x13 < 1 6) x14 + x24 < 1

7) x15 + x25 < 1


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Interpretation: Assign tailor 1 to garment 3 and tailor 2 to garment 5. Time

spent is 15 hours.


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Assignment with Supply and Demand

Question. Dow Chemical uses the chemical Rbase

in production operations at two divisions. Only three

suppliers of Rbase meet Dow’s quality standards.

The quantity of Rbase needed by each Dow division

and the price per gallon charged by each supplier are as follows:

The cost per gallon ($) for shipping from each supplier to each division are

as follows:

Cij Div 1 Div 2

Sup 1 1 0.80

Sup 2 2.50 0.20

Sup 3 3.15 5.40

Dow wants to diversify by spreading its business so that each division’s

demand is assigned to exactly one supplier.

Formulate the optimal assignment of suppliers to divisions as a linear-

programming problem. Formulate the problem, but you need not solve the

problem.

Demand (1000s of gallons)

Price per Gallon ($)

Div 1 40 Sup 1 12.60

Div 2 45 Sup 2 14.00

Sup 3 10.20


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Answer to Question:

Linear programming formulation (supply inequality, demand equality).

Variables: Xij = 1 if Supplier i is assigned to Division j, else 0

Assignment Costs:

The total cost is the sum of the purchase cost and the transportation

cost.

Supplier 1 assigned to Division 1 (cost in $1000s):

o Purchase cost: (40 x $12.60) = $504

o Transportation Cost: (40 x $1) = $40

o Total Cost: $544

Assignment Costs: Cij = Cost of assigning Supplier i to Division j

Cij Div 1 Div 2

Sup 1 544 603

Sup 2 660 639

Sup 3 534 702


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Objective (minimize cost):

Min 544X11 + 603X12 + 660X21 + 639X22 + 534X31 + 702X32

Demand Constraints (since each division’s demand is assigned to exactly

one supplier):

X11 + X21 + X31 = 1

X12 + X22 + X32 = 1

Optional: There is no mention of supply constraints, which are common in

assignment problems. Here is what those common constraints would be in

this problem.

Supply Constraints (Each supplier can supply at most 1 Division):

X11 + X12 < 1

X21 + X22 < 1

X31 + X32 < 1


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Assignment with Supply and Demand

Question. The Goodyear Tire and Rubber Company

uses rubber in its American, Asian, and European tire

manufacturing plants. Goodyear can buy rubber from

either India, or Indonesia, or Malaysia, or Thailand.

The number of tons of rubber needed daily by each tire plant and the price

per ton charged by each supplier are as follows:

The cost (dollars per ton) for shipping from each supplier to each

manufacturing plant are as follows:

American Asian European

India 3 6 4

Indonesia 5 9 8

Malaysia 2 6 4

Thailand 7 1 2

To reduce fixed costs, Goodyear wants each manufacturing plant’s

demand to be assigned to exactly one rubber supplier. And because of

capacity constraints, each rubber supplier can supply at most one tire plant.

Formulate the optimal assignment of rubber suppliers to manufacturing

plant s as a linear-programming problem. Formulate the problem, but you

need not solve the problem.

Demand (tons) Price (dollars per ton)

American 5 India 3

Asian 4 Indonesia 4

European 3 Malaysia 2

Thailand 5


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Answer to Question:


Variables: Xij = 1 if Supplier i is assigned to Plant j, else 0

Assignment Costs:

The total cost is the sum of the purchase cost and the transportation

cost.

Supplier 1 (India) assigned to Plant 1 (American) (cost in dollars):

o Purchase cost: (5 x $3) = $15

o Transportation Cost: (5 x $3) = $15

o Total Cost: $30

Assignment Costs: Cij = Cost of assigning Supplier i to Plant j

Cij Plant 1 Plant 2 Plant 3

Sup 1 5(3+3)=30 4(3+6)=36 3(3+4)=21

Sup 2 5(4+5)=45 4(4+9)=52 3(4+8)=36

Sup 3 5(2+2)=20 4(2+6)=32 3(2+4)=18

Sup 4 5(5+7)=60 4(5+1)=24 3(5+2)=21


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Objective (minimize cost):

Min 30X11 + 36X12 + 21X13

+ 45X21 + 52X22 + 36X23

+ 20X31 + 32X32 + 18X33

+ 60X41 + 24X42 + 21X43

Subject to:

Demand Constraints (each demand is assigned to exactly one supplier):

X11 + X21 + X31 + X41 = 1

X12 + X22 + X32 + X42 = 1

X13 + X23 + X33 + X43 = 1

Supply Constraints (each supplier can supply at most one demander):

X11 + X12 + X13 < 1

X21 + X22 + X23 < 1

X31 + X32 + X33 < 1

X41 + X42 + X43 < 1

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Lesson Topics - Pepperdine Universityseaver-faculty.pepperdine.edu/jburke2/ba452/PowerP2/ReviewQB_3.pdf · Lesson Topics Rounding Off (5) ... ( 2) are Transportation Problems of