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Operations Research 1 – 06/07 – Chapter 3 1 Operations Research 1 Chapter 3 Introduction to Operations Research Hillier & Liebermann Eighth Edition McGrawHill

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AACIMP 2009 Summer School lecture by Boris Goldengorin. "Logistics" course. 3rd - 4th hour. Part 2.

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Page 1: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 1

Operations Research 1

Chapter 3Introduction to Operations Research

Hillier & LiebermannEighth EditionMcGrawHill

Page 2: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 2

Contents• Linear Programming

– Example– Formulations– Graphical solution– The linear programming model– Additional examples

Page 3: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 3

The Wyndor Glass Co. Example• Three plants for glass products (windows and doors):

– Aluminium frames and hardware are made in Plant 1.– Wood frames are made in Plant 2.– Plant 3 produces the glass and assembles the products.

• Capacity is released voor new products:– Two products are an option: product 1 and product 2.

• Product 1 requires capacity in Plants 1 and 3 only.• Product 2 requires capacity in Plants 2 and 3 only.

• Both products need capacity in plant 3. What is the most profitable mix to produce?

Page 4: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 4

Formulation as a Linear Programming Problem

Plant

Production Time per batch, Hours

Production Time Available per Week,

Hours

Product 1 2

1 1 0 42 0 2 123 3 2 18

Profit per batch $ 3,000 $ 5,000

X1 = number of batches of product 1 produced per week

X2 = number of batches of product 2 produced per week

Z = total profit per week (in thousands of dollars) from producing these two products

X1 and X2 are the decision variables for the model. Using the bottom row of table 3.1 we obtain Z = 3X1 + 5X2

Table 3.1

Page 5: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 5

Mathematical formulation of the LP problemMaximize Z = 3X1 + 5X2

Subject to restrictions X1 ≤ 4

2X2 ≤ 123X1 + 2X2 ≤ 18

AndX1 ≥ 0, X2 ≥ 0.

Page 6: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 6

Graphical Solution figure 3.1

Figure 3.1

Shaded area shows the values of (X1,X2) allowed by:

X1 ≥ 0,

X2 ≥ 0,

X1 ≤ 4

X1 ≤ 4

X2 ≥ 0

X1 ≥ 0

Page 7: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 7

Graphical Solution figure 3.2

Figure 3.2

Shaded area shows the set of permissible values of (X1,X2), called the feasible region.2X2 = 12

X1 =4

3X1 + 2X2 = 18

Page 8: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 8

Graphical Solution figure 3.3

Z = 10 = 3X1 + 5X2

Z = 20 = 3X1 + 5 X2

Z = 36 = 3X1 + 5X2 (2,6)

Page 9: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 9

Terminology for Linear Programming

Prototype Example General ProblemProduction capacities of plants3 plants

Production of products2 productsProduction rate of product j, xj

Profit Z

Resourcesm resources

Activitiesn activitiesLevel of activity j, Xj

Objective function (Overall measure of performance Z)

TABLE 3.2

Page 10: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 10

Data needed for a LP Model

Resource

Resource Usage per Unit of Activity

Amount of Resource available

Activity 1 2 … n

12..

m

a11 a12 … a1n

a21 a22 … a2n

… … … … am1 am2 … amn

b1

b2

.

.bm

Contribution to Z per unit of activity

c1 c2 … cn

TABLE 3.3 Data needed for a linear programming model involving the allocation of resources to activities

Page 11: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 11

Symbols used as notation of the various components commonly used for a LP Model

• Below certain Symbols are listed, along with their interpretation of the general problem.– Z = Value of overall measure of performance.– xj = level of activity j (for j = 1, 2, …, n).– cj = increase in Z that would result from each unit increase in level of

activity j.– bi = amount of resource i that is available for allocation to activities (for i

= 1, 2, …, m).– aij = amount of resource i consumed by each other unit of acivity j.

• The model poses the problem in terms of making decisions about the levels of the activitys, so x1, x2, …, xn are called the decision variables. In Table 3.3 the values of ci, bi, and aij (for i = 1, 2, …, m and j = 1, 2, … , n) are the input constants also referred to as the parameters of the model.

Page 12: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 12

A standard form of the model• We can now formulate the mathematical model for this

general problem of allocation resources to activities. The model is to select the values of x1, x2, …, xn so as to

Maximize Z = c1x1 + c2x2 + … + cnxn, subject to restrictions

a11x1 + a12x2 + … + a1nxn ≤ b1

a21x1 + a22x2 + … + a2nxn ≤ b2

… … … … … … … … … …am1x1 + am2x2

+ … + amnxn ≤ bm,and

x1 ≥ 0, x2 ≥ 0 , …, xn ≥ 0

Page 13: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 13

Summary of terminology in LP 1objective function = the function being maximized: c1x1 +

c2x2 + … + cnxn.constraints = restrictions for the objective function.functional constraints = the first m constraints with a

function of all the variables (ai1x1 + ...+ ainxn) on the left-hand side.

nonnegativity constraints = the xj ≥ 0 restrictionsFeasible solution = a solution for which all the constraints

are satisfied.Infeasible solution = a solution for which at least one

constraint is violated.Feasible region = the collection of all feasible solutions.

Page 14: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 14

Summary of terminology in LP 2Optimal solution = a feasible solution that has the most

favorable value of the objective function.Most favorable value = the largest value if the objective

function is to be maximized or the smallest value if the objective function is to be minimized.

It is possible for a problem to have no feasible solutions, multiple optimal solutions. If there are no optimal solutions we speak of an unbounded Z or an unbounded objective

Page 15: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 15

Figure 3.4

2x2 ≤ 12

3x1 + 2x2 ≤ 18

x1 ≤ 0

X1 ≤ 4

x2 ≥ 0

3x1 + 5x2 ≥ 50

Maximize Z = 3x1 + 5x2,

Subject to X1 ≤ 4

2x2 ≤ 12

3x1 + 2x2 ≤ 18

3x1 + 5x2 ≥ 50

And x1 ≥ 0, x2 ≥ 0

Figure 3.4

The Wyndor Glass Co. Problem would have no solutions if the constraint 3x1 + 5x2 ≥ 50 were added to the problem.

Page 16: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 16

Figure 3.5

Every point on this darker line segment is optimal, each with Z = 18

Maximize Z = 3x1 + 2x2,Subject to X1 ≤ 4 2x2 ≤ 12 3x1 + 2x2 ≤ 18And x1 ≥ 0, x2 ≥ 0

Figure 3.5

The Wyndor Glass Co. Problem would have multiple solutions if the objective function were changed to Z = 3x1 + 2x2

Feasible region

Page 17: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 17

Figure 3.6

Feasible region

(4,10), Z = 62

(4,8), Z = 52

(4,6), Z = 42

(4,4), Z = 32

(4,2), Z = 22

Maximize

Z = 3x1 + 5x2

Subject to x1 ≤ 4

And x1 ≥ 0, x2 ≥ 0

Figure 3.6

The Wyndor Glass Co. Problem would have no optimal solutions if the only functional constraint were x1 ≤ 4, because x2 then could be increased indefinitely in the feasible region without ever reaching the maximum value of Z = 3x1 + 5x2

Page 18: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 18

Figure 3.7

(0,6) (2,6)

(4,3)

(0,0)

(4,0)

Figure 3.7

The five dots are the five CPF solutions for the Wyndor Glass Co. Problem.

Page 19: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 19

Assumptions of LPProportionality assumption: The contribution of each activity to the value of the objective function Z is proportional to the level of the activity xj, as represented by the cjxj term in the objective function. Similarly, the contribution of each activity to the left-hand side of each functional constraint is proportional to the level of activity xj, as represented by the aijxj term in the constraint.

Consequently, this assumption rules out any exponent other than 1 for any variable in any term of any function (whether the objective function or the function on the left-hand side of a functional constraint) in a linear programming model.

Page 20: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 20

Graphs of Table 3.4

0

2

5

8

11

0

3

7

12

18

0

3

56 6

0

2

4

6

8

10

12

14

16

18

20

0 1 2 3 4

ProportionalitySatisfiedCase 1

Case 2

Case 3

In case 1 through 3 proportionality is violated.

x1

x2

Page 21: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 21

Additivity assumption & Table 3.5Additivity Assumption: Every function in a linear programming model (whether the objective function or the function on the left-hand side of a functional constraint) is the sum of the individual contributions of the respective activities.

(x1,x2)

Value of Z

Additivity SatisfiedAdditivity Violated

Case 1 Case 2

(1,0)(0,1)

35

3 35 5

(1,1) 8 9 7

• TABLE 3.5 Examples of satisfying or violating additivity for the objective function

Page 22: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 22

Table 3.6

(x1,x2)

Amount of Resource Used

Additivity SatisfiedAdditivity Violated

Case 3 Case 4

(2,0)(0,3)

66

6 66 6

(2,3) 12 15 10.8

• TABLE 3.6 Examples of satisfying or violating additivity for a functional constraint

Page 23: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 23

Divisibility assumptionDivisibility Assumption: Decision variables in a linear programming model are allowed to have any values, including noninteger values, that satisfy the functional and nonnegativity constraints. Thus, these variables are not restricted to just integer values. Since each decision variable presents the level of some activity, it is being assumed that the activities can be run at fractional levels.

Page 24: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 24

Certainty assumptionCertainty assumption: The value assigned to each parameter of a linear programming model is assumed to be a known constant.

Page 25: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 25

Regional Planning: The southern confederation of kibbutzim

Kibbutz Usable Land (Acres) Water Allocation (Acre Feet)123

400600300

600800375

Crop Maximum Quota (Acres)

Water Consumption (Acre Feet/ Acre)

Net Return ($ / Acre)

Sugar beetsCottonSorghum

600500325

321

1,000750250

• TABLE 3.8 Resource data for the Southern Confederation of Kibbutzim

• TABLE 3.9 Crop data for the Southern Confederation of Kibbutzim

Page 26: Logistics. Example

Operations Research 1 – 06/07 – Chapter 3 26

Regional Planning (continued): Table 3.10

Crop

Allocation (Acres)Kibbutz

1 2 3

Sugar beetsCottonSorghum

X1

X4

X7

X2

X5

X8

X3

X6

X9

• TABLE 3.10 Decision variables for the Southern Confederation of Kibbutzim problem