A11.4801i Chapter 1 Linear Programming

Chapter 2 Linear programming

  Introduction

• Many management decisions involve trying to make the most effective use of an organization’s resources.

• • Resources typically include machinery, labor, money, time,

  warehouse space, or raw materials.
• Resources may be used to produce products (such as machinery, furniture, food, or clothing) or services (such as schedules for shipping and production, advertising policies, or investment decisions).
• Linear programming (LP) is a widely used mathematical technique designed to help managers in planning and decision making relative to resource allocation .
• Despite the name, linear programming, and the more general category of techniques called “mathematical programming”, have very little to do with computer programming.

• • In the world of Operations Research, programming refers to

  modeling and solving a problem mathematically .

• • Computer programming has, however, played an important

  role in the advancement and use of LP to solve real-life LP

  Linear Programming Model

Most of the deterministic OR models can be formulated as

mathematical programs.

  "Program," in this context, has to do with a “plan,” not a computer program.

  General form of Linear programming model Maximize / Minimize z = f(x , x ,…, x ) _{1 2 n}

   g (x , x , …, x ) 

  Subject to _{i 1 2 n} b i =1,…,m _i 

  { } x ≥ 0, j = 1,…,n _j

• x _j

  are called decision variables . These are

things that you control and you want to

determine its values

• g

  {   

  } b i are called structural (or functional or technological) constraints

  Model Components

• x

  j ≥ 0 are nonnegativity constraints

• f(x

  1 , x

  2 ,…, x n

  ) is the objective function

  i (x

  1 , x

  2 ,…, x n

  )

  

Example: Giapetto woodcarving Inc.,

• Giapetto Woodcarving, Inc., manufactures two types of wooden toys: soldiers and trains. A soldier sells for $27 and uses $10 worth of raw materials. Each soldier that is manufactured increases Giapetto’s variable

  labor and overhead cost by $14. A train sells for $21

  and uses $9 worth of raw materials. Each train built

  increases Giapetto’s variable labor and overhead cost by $10. The manufacture of wooden soldiers and trains requires two types of skilled labor: carpentry and finishing. A soldier requires 2 hours of finishing labor and 1 hour of carpentry labor. A train requires 1

  hour of finishing and 1 hour of carpentry labor. Each

  week, Giapetto can obtain all the needed raw material

  but only 100 finishing hours and 80 carpentry hours.

  Demand for trains is unlimited, but at most 40 soldiers are bought each week. Giapetto wants to maximize weekly profit. Formulate a linear

  programming model of Giapetto’s situation that can

  Solution: Giapetto woodcarving Inc.,

• Step 1: Model formulation

1. Decision variables: we begin by finding the

  decision variables. In any LP, the decision

  variables should completely describe the

decisions to be made . Clearly, Giapetto

  must decide how many soldiers and trains should be manufactured each week. With this in mind, we define:

  X = number of soldiers produced each

  1

  week X = number of trains produced each

  2

  week

  

Solution: Giapetto woodcarving Inc.,

  2. Objective function: in any LP, the decision maker wants to maximize (usually revenue or profit) or minimize (usually costs) some function of the decision variables. The function to be maximized or minimized is called the objective function. For the Giapetto problem, we will maximize the net profit (weekly revenues – raw materials cost

• – labor and overhead costs). Weekly revenues and costs can be expressed in terms of the decision variables, X and X

  1

  2

  as following: Solution: Giapetto woodcarving Inc.,

• Weekly revenues = weekly revenues from soldiers + weekly revenues from trains = 27 X + 21 X

  1

  2 Also, Weekly raw materials costs = 10 X + 9 X

  1

  2 Other weekly variable costs = 14 X 1 + 10 X

  2 Therefore, the Giapetto wants to maximize: (27 X 1 + 21 X 2 ) – (10 X 1 + 9 X 2 ) – (14 X 1 + 10 X 2 ) = 3 X + 2 X

  1

2 Hence, the objective function is:

  

Solution: Giapetto woodcarving Inc.,

3. Constraints: as X and X increase,

  1

  2 Giapetto’s objective function grows larger.

  This means that if Giapetto were free to choose any values of X and X , the

  1

  2

  company could make an arbitrarily large profit by choosing X and X to be very

  1

  2

  large. Unfortunately, the values of X and X

  1

  2

  are limited by the following three restrictions (often called constraints):

  Constraint 1: each week, no more than 100 hours of finishing time may be used.

  Constraint 2: each week, no more than 80 hours of carpentry time may be used.

  Constraint 3: because of limited demand, at

  Solution: Giapetto woodcarving Inc.,

• The three constraints can be expressed in terms of the decision variables X and X as

  1

  2

  follows:

  Constraint 1:

2 X + X

  1 2  100 Constraint 2:

  X + X

  1 2  80 Constraint 3:

  X

  1  40 Note:

  The coefficients of the decision variables in the constraints are called technological

  

coefficients . This is because its often reflect

  the technology used to produce different products. The number on the right-hand side of each constraint is called Right-Hand Side

  

Solution: Giapetto woodcarving Inc.,

• • Sign restrictions: to complete the formulation

  of the LP problem, the following question must be answered for each decision variable: can the decision variable only assume nonnegative values, or it is allowed to assume both negative and positive values?

  If a decision variable X can only assume a i nonnegative values, we add the sign restriction (called nonnegativity constraints)

  X i  0.

  If a variable X can assume both positive and i negative values (or zero), we say that X is i unrestricted in sign (urs).

  

In our example the two variables are restricted

  

Solution: Giapetto woodcarving Inc.,

• Combining the nonnegativity constraints with the objective function and the structural constraints yield the following optimization model (usually called LP model):

  Max Z = 3 X + 2 X (objective function)

  1

  2 subject to (st)

2 X + X

  _{1 2  100 (finishing constraint)} X + X _{1 2  80 (carpentry constraint)}

  X _{1  40 (soldier demand constraint)}

  X _{1 2}  0 and X  0 (nonnegativity constraint) The optimal solution to this problem is : What is Linear programming

problem (LP)?

• LP is an optimization problem for which we do the

  following:

  1. We attempt to maximize (or minimize) a linear function of the decision variables. The function that is to be maximized or minimized is called objective function.

  2. The values of decision variables must satisfy a set of constraints. Each constraint must be a linear equation or linear inequality.

3. A sign restriction is associated with each variable.

  for any variable X , the sign restriction specifies _i either that X i must be nonnegative (X i > 0) or that X i

  Linear Programming Assumptions

  (i) proportionality linearity (ii) additivity (iii) divisibility (iv) certainty

  

(i) activity j’s contribution to objective function is

c j x j and usage in constraint i is a ij x j both are proportional to the level of activity j

(volume discounts, set-up charges, and nonlinear

efficiencies are potential sources of violation) (ii) “cross terms” such as x

  1 x

  5 may not appear in the objective or constraints.

  Explanation of LP Assumptions

  Explanation of LP Assumptions (iii) Fractional values for decision variables are permitted (iv) Data elements a , c , b , u are known with certainty _{ij j i j}

• Nonlinear or integer programming models should be used when some subset of assumptions (i), (ii) and (iii) are not satisfied.
• Stochastic models should be used when a problem

  has significant uncertainties in the data that must be explicitly taken into account [a relaxation of assumption (iv)].

  Applications Of LP

  1. Product mix problem

  2. Diet problem

  3. Blending problem

  4. Media selection problem

  5. Assignment problem

  6. Transportation problem

  7. Portfolio selection problem

  8. Work-scheduling problem

  9. Production scheduling problem

  10. Inventory Problem

  11. Multi period financial problem

1. Product Mix Problem

  Example Formulate a linear programming model for this

problem, to determine how many containers of

each product to produce tomorrow in order to

maximize the profits. The company makes four

types of juice using orange, grapefruit, and

pineapple. The following table shows the price

and cost per quart of juice (one container of juice) as well as the number of kilograms of fruits required to produce one quart of juice.

  Product Price/quart Cost/quart Fruit needed Orange juice

  3

  1 1 Kg. Grapefruit juice

  2

  0.5 2 Kg. Pineapple juice

  2.5

  1.5 1.25 Kg. Product Mix Problem Example (cont.) On hand there are 400 Kg of orange, 300 Kg.

  of grapefruit, and 200 Kg. of pineapples. The manager wants grapefruit juice to be used for no more than 30 percent of the number of containers produced. He wants the ratio of the number of containers of orange juice to the number of containers of pineapples juice to be at least 7 to 5. pineapples juice should not exceed one-third of the total product.

  

Product Mix Problem

Solution Decision variables X1 = # of containers of orange juice X2 = # of containers of grapefruit juice X3 = # of containers of pineapple juice X4 = # of containers of All-in-one juice Objective function Max Z = 2 X1 + 1.5 X2 + 1 X3 + 2 X3 Constraints

  Orange constraints

  X 1 .

  25 X 4 400   Grapefruit constraint

2 X 2 .

  25 X 4 300   Pineapple constraints 1 .

25 X 3 .

  25 X 4 200   Max. of grapefruit

  X 2 .

  3 (

  X

  1 X

  2 X

  3 X 4 )    

  X

  1

  7 

  X

  3

  5 Ratio of orange to pineapple

  1 Max. of pineapple

  X 2 (

  X

  1 X

  2 X

  3 X 4 )    

  3

2. Diet problem

  Example

My diet requires that all the food I eat come from one

of the four “basic food groups” (chocolate cake, ice cream, soda, and cheesecake). At present, the following four foods are available for consumption: brownies, chocolate ice cream, cola, and pineapple cheesecake. Each brownie costs 50 cents, each

scoop of chocolate ice cream costs 20 cents, each

bottle of cola costs 30 cents, and each piece of

pineapple cheesecake costs 80 cents. Each day, I

must ingest at least 500 calories, 6 oz of chocolate, 10 oz of sugar, and 8 oz of fat. The nutritional content per unit of each food is shown in the following table. Formulate a linear programming model that can be used to satisfy my daily

  

Diet problem

Calories Chocolat Sugar Fat e

  Brownie 400 3 ounce 2 ounce 2 ounce Chocolate ice cream 200

  2

  2

  4 (1 scoop) Cola (1 bottle) 150

  4

  1 Pineapple cheesecake 500

  4

  5 (1piece)

  

Diet problem

Solution

• Decision variables: as always, we begin by

  determining the decisions that must be made by the decision maker: how much of each food type should be eaten daily. Thus, we define the decision variables:

  X

  1 = number of brownies eaten daily

  X

  2

= number of scoops of chocolate ice cream

eaten daily

  X

  3 = number of bottles of cola drunk daily

  X

  4 = number of pieces of pineapple cheesecake

  

Diet problem

• Objective function: my objective

  

function is to minimize the cost of my

diet. The total cost of my diet may be

determined from the following relation: Total cost of diet = (cost of brownies) +

(cost of ice cream) + (cost of cola) +

(cost of cheesecake)

  Thus, the objective function is: Min Z = 50 X + 20 X + 30 X + 80 X

  1

  2

  3

  4

  

Diet problem

• Constraints: the decision variables must satisfy the

  following four constraints: Constraint 1: daily calorie intake must be at least 500 calories.

  Constraint 2: daily chocolate intake must be at least 6 oz.

  

Constraint 3: daily sugar intake must be at least 10 oz.

Constraint 4: daily fat intake must be at least 8 oz.

  To express constraint 1 in terms of the decision variables, note that (daily calorie intake) = (calorie in brownies) + (calories in chocolate ice cream) + (calories in cola) + (calories in pineapple cheesecake)

  Therefore, the daily calorie intake = 400 X ₁ + 200 X ₂ + 150 X ₃ + 500 X ₄ must be greater than 500 ounces By the same way the other three constraints can be

  Diet problem The four constraints are: 400 X + 200 X + 150 X + 500 X

  

  1

  2

  3

  4 500

  3 X + 2 X

  1 2  6

  2 X + 2 X + 4 X + 4 X

  1

  2

  3 4  10

  2 X + 4 X + X + 5 X

  1

  2

  3 4  8

Nonnegativity constraints: it is clear that

all decision variables are restricted in sign, i.e., X i  0, for all i = 1, 2, 3, and Diet problem

• Combining the objective function, constraints, and nonnegativity constraints, the LP model is as follows:

  Min Z = 50 X + 20 X + 30 X + 80 X

  1

  2

  3

  4 st.

  400 X + 200 X + 150 X + 500 X

  1

  2

  3 4  500

  3 X + 2 X

  1 2  6

  2 X + 2 X + 4 X + 4 X

  1

  2

  3 4  10

  2 X + 4 X + X3 + 5 X

  1

  2 4  8

  X i  0, for all i = 1, 2, 3, and 4

  

The optimal solution to this LP is X = X = 0,

  1

  4 X = 3, X = 1 and Z = 90 cents

  2

  3

3. Blending problem

  Example The Low Knock Oil company produces two grades of cut rate gasoline for industrial distribution. The grades, regular and economy, are produced by refining a blend of two types of crude oil, type X100 and

type X220. each crude oil differs not only in cost per barrel, but in

composition as well. The accompanying table indicates the percentage of crucial ingredients found in each of the crude oils and the cost per barrel for each. Weekly demand for regular grade of Low Knock gasoline is at least 25000 barrels, while demand for the economy is at least 32000 barrels per week. At least 45% of each barrel of regular must be ingredient A. At most 50% of each barrel of

economy should contain ingredient B. the Low Knock management

must decide how many barrels of each type of crude oil to buy each

week for blending to satisfy demand at minimum cost .

  Crude oil Ingredient Ingredient Cost/barrel type A % B % ($)

  X100

  35

  55

  30

  

Blending problem

Solution Let

  X ₁

= # of barrels of crude X100 blended to produce the refined regular

  X

^{2 = # of barrels of crude X100 blended to produce the refined economy}

  X ₃

= # of barrels of crude X220 blended to produce the refined regular

  X ₄ = # of barrels of crude X220 blended to produce the refined economy Min Z = 30 X ₁ + 30 X ₂ + 34.8 X ₃

+ 34.8 X

₄ St.

  X ₁ + X _{3  25000}

  X ^{2 + X 4}  32000

• 0.10 X
₁ + 0.15 X _{3  0}

0.05 X

  ₂ – 0.25 X _{4  0} X i  0, i = 1, 2, 3, 4

  The optimal solution is: X ₁ = 15000, X ₂ = 26666.6, X ₃ = 10000,

  X ₄ = 5333.3, and Z = 1,783,600

4. Media selection problem

  Example

A company has budgeted up to $8000 per week

for local advertisement. The money is to be

allocated among four promotional media: TV

spots, newspaper ads, and two types of radio advertisements. The company goal is to reach

the largest possible high-potential audience

through the various media. The following table presents the number of potential customers reached by making use of advertisement in each of the four media. It also provides the cost per advertisement placed and the maximum number of ads that

  

Media Selection

Medium Audience Cost per Maximum ad ads per

  Reached per week ad TV spot (1 minute) 5000 800

  12 Daily newspaper 8500 925

  5 (full-page ad) Radio spot

  2400 290

  25 (30 second, prime time) Radio spot

  2800 380

  20 (1 minute, afternoon)

The company arrangements require that at least five radio spots be placed each week. To ensure a board-scoped promotional

campaign, management also insists that no more than $1800 be Media selection Solution Let X = number of 1-miute TV spots taken Each week ₁ X ^{2 = number of full-page daily newspaper ads taken each week.} X = number of 30-second prime-time radio spots taken each week. ₃ X ^{4 = number of 1-minute afternoon radio spots taken each week.}

  Max Z = 5000 X + 8500 X + 2400 X + 2800 X _{1 2 3 4} st X1  12 (maximum TV spots/week) X2  5 (maximum newspaper ads/week) X3  25 (maximum 30-second radio spots/week) X4  20 (maximum 1-minute radio spots/week) 800 X1 + 925 X2 + 290 X3 + 380 X4  8000 (weekly budget) X3 + X4  5 (minimum radio spots contracted) 290 X3 + 380 X4  1800 (maximum dollars spent on radio) X1, X2, X3, X4  0

5. Assignment problem

  Example A law firm maintains a large staff of young attorneys who hold the title of junior partner. The firm concerned with the effective utilization of this personnel resources, seeks some objective means of

making lawyer-to-client assignments. On march 1,

four new clients seeking legal assistance came to the firm. While the current staff is overloads and

identifies four junior partners who, although busy,

could possibly be assigned to the cases. Each young lawyer can handle at most one new client.

  Furthermore each lawyer differs in skills and specialty interests.

  Seeking to maximize the overall effectiveness of the new client assignment, the firm draws up the following table, in which he rates the estimated effectiveness (of a scale of 1 to 9) of each lawyer on

  

Assignment problem

Client case

  

Lawyer Divorce Corporate embezzlement exhibitionism

merger

  Adam

  6

  2

  8

  5 Brook

  9

  3

  5

  8 Carter

  4

  8

  3

  4 Darwin

  6

  7

  6

  4

  

Assignment problem

Solution

  Decision variables: 1 if attorney i is assigned to case j Let X =

  ij

  0 otherwise Where : i = 1, 2, 3, 4 stands for Adam, Brook,

  Carter, and Darwin respectively j = 1, 2, 3, 4 stands for divorce, merger, embezzlement, and exhibitionism respectively.

  The LP formulation will be as follows:

  

Assignment problem

_{11 12 13 14 21 22} + Max Z = 6 X + 2 X + 8 X + 5 X + 9 X + 3 X
• 5 X + 8 X + 4 X + 8 X + 3 X + 4 X
_{23 24 31 32 33 34}

  6 X +7 X + 6 X + 4 X _{41 42 43 44} St.

  X + X + X + X = 1 (divorce case) _{11 21 31 41} X + X + X + X = 1 (merger) _{12 22 32 42} X13 + X23 + X33 + X43 = 1 (embezzlement) X14 + X24 + X34 + X44 = 1 (exhibitionism) X11 + X12 + X13 + X14 = 1 (Adam) X21 + X22 + X23 + X24 = 1 (Brook) X31 + X32 + X33 + X34 = 1 (Carter) X41+ X42 + X43 + X44 = 1 (Darwin) The optimal solution is: X ^{13 = X 24 = X 32 = X41 = 1. All other variables are equal to zero .}

6. Transportation problem

  Example

The Top Speed Bicycle Co. manufactures and markets a

line of 10-speed bicycles nationwide. The firm has final assembly plants in two cities in which labor costs are low, New Orleans and Omaha. Its three major warehouses are located near the larger market areas of New York, Chicago, and Los Angeles.

  The sales requirements for next year at the New York warehouse are 10000 bicycles, at the Chicago warehouse 8000 bicycles, and at the Los Angeles warehouse 15000 bicycles. The factory capacity at

each location is limited. New Orleans can assemble

and ship 20000 bicycles; the Omaha plant can produce 15000 bicycles per year. The cost of shipping one bicycle from each factory to each warehouse

  Transportation problem New Chicago Los York Angele s New Orleans $2

  3

  5 Omaha

  3

  1

  4 The company wishes to develop a shipping schedule that will minimize its total annual transportation cost Transportation problem Solution

To formulate this problem using LP, we again employ the

concept of double subscribed variables. We let the

first subscript represent the origin (factory) and the

second subscript the destination (warehouse). Thus,

in general, X refers to the number of bicycles shipped _ij

from origin i to destination j. Therefore, we have six

decision variables as follows:

  X = # of bicycles shipped from New Orleans to New York ₁₁ X = # of bicycles shipped from New Orleans to Chicago ₁₂ X = # of bicycles shipped from New Orleans to Los Angeles ₁₃ X = # of bicycles shipped from Omaha to New York ₂₁ X = # of bicycles shipped from Omaha to Chicago ₂₂

  

Transportation problem

  Min Z = 2 X

• 3 X
• 5 X
• 3 X
• X
• 4

  11

  12

  13

  21

  22

  X

23 St

  X ^{11 + X 21 = 10000 (New York demand)}

  X ₁₂ + X ₂₂ = 8000 (Chicago demand)

  X ₁₃ + X ₂₃ = 15000 (Los Angeles demand)

  X ₁₁ + X ₁₂ + X ₁₃  20000 (New Orleans Supply

  X ₂₁ + X ₂₂ + X ₂₃  15000 (Omaha Supply) X ij  0 for i = 1, 2 and j = 1, 2, 3

7. Portfolio selection

  Example

The International City Trust (ICT) invests in short-

term trade credits, corporate bonds, gold stocks, and construction loans. To encourage a diversified portfolio, the board of directors has placed limits on the amount that can be committed to any one type of investment. The

  ICT has $5 million available for immediate investment and wishes to do two things: (1) maximize the interest earned on the

investments made over the next six months,

and (2) satisfy the diversification requirements

as set by the board of directors. The specifics

of the investment possibilities are:

  

Portfolio selection

Investment Interest earned

  % Maximum investment ($ Million)

  Trade credit

  7

  1 Corporate bonds

  11

  2.5 Gold stocks

  19

  1.5 Construction loans

  15

  1.8 In addition, the board specifies that at least 55% of the funds invested must be in gold stocks and construction loans, and that no less

  

Portfolio selection

Solution To formulate ICT’s investment problem as a linear programming model, we assume the following decision variables: X = dollars invested in trade credit

  1 X = dollars invested in corporate bonds

  2 X = dollars invested in gold stocks

  3 X = dollars invested in construction

  4

  

Portfolio selection

  Max Z = 0.07 X + 0.11 X + 0.19 X + 0.15 X

  1

  2

  3

  4 St.

  X  1

  1 X _{2  2.5}

  X _{3  1.5}

  X ₄  1.8 X _{3 + X4  0.55(X}

₁

+ X + X + X ) _{2 3 4} X _{1  0.15(X 1} + X + X + X ) _{2 3 4} X ^{1 + X 2 + X 3 + X 4}  5

  X _{i  0 , i = 1, 2, 3, 4} The optimal is: X = 75,000, X = 950,000, X = _{1 2 3} 1,500,000, and X = 1,800,000, and total interest ₄ Z = 712,000

8. Work Scheduling Problem

  Example Microsoft has a 24-hour-a-day, 7-days-a-week toll free hotline that is being set up to answer questions regarding a new product. The following table summarizes the number of full-time equivalent employees (FTEs) that must be on duty in each time block.

  Shift Time FTEs 1 0-4

  15 2 4-8 10 3 8-12

  40 4 12-16 70 5 16-20

  40 Work Scheduling problem

• Microsoft may hire both full-time and part-time

  employees. The former work 8-hour shifts and the

  latter work 4-hour shifts; their respective hourly wages are $15.20 and $12.95. Employees may start work only at the beginning of one of 6 shifts.

  Part-time employees can only answer 5 calls in the

time a full-time employee can answer 6 calls. (i.e.,

a part-time employee is only 5/6 of a full-time employee.)

  Decision Variables

x = # of full-time employees that begin work in shift t

_t y = # of part-time employees that work shift t _t

  (4  12.95) (8  15.20)

  Min

  121.6 (x + x ) +51.8 (y + _{1 6 1} + y ) ₆

  5 s.t. y 

  15 _{6 1} + x + x ₁

  6

  5 y 

  10 _{1 2} + x + x ₂

  6 All shifts

  5 y 

  40 x + x _{2 3} ³

• 6 must be

  5 covered y 

  70 x + x _{3 4} ⁴

• 6

  5 y 

  40 _{4 5} + x + x ₅

  6

  5 y 

  35 _{5 6} + x + x ₆

  6 x , y _{t t}

   PT employee is 5/6 FT employee Terminology for solution of LP

• A

  feasible solution is a solution for which all the constraints are satisfied.

• A

  

corner point feasible solution (CPF) is a

feasible solution that lies at a corner point.
• An

  infeasible solution is a solution for which at least one constraint is violated.

• The

  feasible region is the collection of all feasible solution.

• An Optimal solution is a feasible solution that has the most favorable value of the objective function. (it is always one of the CPF solution
• The

  

most favorable value is the largest

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

  Graphical solution A Graphical Solution Procedure (LPs with 2 decision variables can be solved/viewed this way.)

  1. Plot each constraint as an equation and then decide which side of the line is feasible (if it’s an inequality).

  2. Find the feasible region.

3. find the coordinates of the corner (extreme) points of the feasible

region.

  4. Substitute the corner point coordinates in the objective function

5. Choose the optimal solution

  

Example 1: A Minimization Problem

• LP Formulation

  Min z = 5x + 2x

  1

  

2

s.t. 2x + 5x > 1

  1

  2 4x - x > 12

  1

  2 x + x > 4

  1

  2 x , x > 0

  1

  2

  

Example 1: Graphical Solution

• Graph the Constraints

  = 0, then x = 2; when x

• Constraint 1: When x

¹

^{2 2} = 0, then x = 5. Connect (5,0) and (0,2). The ₁ ">" side is above this line.

  = 0, then x = 3. But

• Constraint 2: When x

²

¹ setting x to 0 will yield x = -12, which is not on _{1 2}

the graph. Thus, to get a second point on this

line, set x to any number larger than 3 and ₁

solve for x : when x = 5, then x = 8. Connect

_{2 1 2} (3,0) and (5,8). The ">" side is to the r
• Constraint 3: When x = 0, then x = 4; when x

₁

_{2 2} = 0, then x = 4. Connect (4,0) and (0,4). The ₁ ">" side is above this line.

  Example 1: Graphical Solution

• Constraints Graphed

  x x x x _{2 2 2 2} Feasible Region

  Feasible Region

• 4 x x >

  12

• 4 - 4 x x x x > >

  12

  12 4 x x > _{1 1 1 1 2 2 2 2} 12 -

  5

  5

  5

  5

  4 x x >

  4

  x x + + x > >

• x

  4 _{1 1 1 1} + x x > _{2 2 2 2}

  4

  4

  3

  3

  3

  3 2 x + 5 x >

  10 2 x x + 5 + 5 x x > >

  2

  10 2 x + 5 x > _{1 1 1 1 2 2 2 2}

  10

  10

  2

  2

  2

  2 (16/5,4/5) (10/3, 2/3)

  1

  1

  1

  1 x x x x _{1 1 1 1}

  1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 Example 1: Graphical Solution

• Solve for the Extreme Point at the Intersection of the second and third Constraints

  4x - x = 12 _{1 2} x + x = 4 _{1 2} Adding these two equations gives: 5x ^{1 = 16 or x 1 = 16/5.}

  Substituting this into x ^{1 + x 2 = 4 gives: x 2 = 4/5}

• • Solve for the extreme point at the intersection of the first and third constraints

  2x + 5x =10 _{1 2} x + x = 4 ₁

₂

Multiply the second equation by -2 and add to the first equation, gives

  3x = 2 or x = 2/3 _{2 2} Substituting this in the second equation gives x1 = 10/3 Point Z (16/5, 4/5) 88/5

  (10/3, 2/3)

  18

  

Example 2: A Maximization Problem

Max z = 5x + 7x

  1

  2 s.t. x < 6

  1 2x + 3x < 19

  1

  2 x + x < 8

  1

  2 x , x > 0

  1

  2

  

Example 2: A Maximization

Problem
• Constraint #1 Graphed

  8 ⁸ _{7 7}

  6 ⁶ _{5 5}

  4 ⁴ _{3 3}

  2 ² _{1 1} 1 2 3 4 5 6 7 8 9 10

^{1 2 3 4 5 6 7 8 9 10}

x x _{2 2} x x _{1 1} x x _{1 1}

  < <

  6

  6 (6, 0)

  (6, 0)

  

Example 2: A Maximization

Problem
• Constraint #2 Graphed

  x x _{2 2 8}

  8 1/3

  1/3 (0, 6 ) _{7 7} (0, 6 ) ₆

  6 2 x + 3 x <

  19 _{5 5} 2 x + 3 x < _{1 1 2 2}

  19 _{3 3} ⁴

  4 ²

  2 1/2 1/2

  (9 , 0) _{1 1} (9 , 0) x _{1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10} x _{1 1} Example 2: A Maximization

Problem

• Constraint #3 Graphed

  x x _{2 2}

  (0, 8) _{8 8} (0, 8) ⁷

  7 x x < 8 + _{6 6 5 5} x + x < _{1 1 2 2}

  8 _{3 3 4}

  4 ₁ ²

  2 ₁ (8, 0)

  (8, 0) x _{1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10} x _{1 1}

  

Example 2: A Maximization Problem

• Combined-Constraint Graph

  x x _{2 2} x x < 8 + _{7 8 7 8} x x < _{1 1 2 2} 8 + ⁶

  6 x <

  6 _{5 5} x < _{1 1}

  6

  4 _{3 3 4} 2 x + 3 x <

  19 _{2 1 1 2} 2 x + 3 x < _{1 1 2 2}

  19 x _{1 2 3 4 5 6 7 8 9 10}

_{1 2 3 4 5 6 7 8 9 10}

x _{1 1}

  

Example 2: A Maximization Problem

• Feasible Solution Region

  8 ⁸ _{7 7}

  6 ⁶ _{5 5}

  4 ⁴ _{3 3}

  2 ² _{1 1}

1 2 3 4 5 6 7 8 9 10

^{1 2 3 4 5 6 7 8 9 10} x x _{1 1} Feasible

  Feasible Region

  Region x x _{2 2}

• The Five Extreme Points

  2

  5 (0, 19/3)

(5, 3)

(6, 2)

  5

  5

  5

  4

  4

  4

  4

  3

  3

  3

  3

  2

  Example 2: A Maximization Problem

  2

  2

  1

  1

  1

  1

  Region

  Feasible Region

  2 ² _{1 1} 1 2 3 4 5 6 7 8 9 10 ^{1 2 3 4 5 6 7 8 9 10} x x _{1 1} Feasible

  4 ⁴ _{3 3}

  6 ⁶ _{5 5}

  8 ⁸ _{7 7}

  (0, 0)

  Example 2: A Maximization Problem

• Having identified the feasible region for the problem, we now search for the optimal solution, which will be the point in the feasible region with the largest (in case of maximization or the smallest (in case of minimization) of the objective function.
• To find this optimal solution, we need to evaluate the objective function at each one of the corner points of the feasible region.

  

Example 2: A Maximization Problem

• Optimal Solution

  8 ⁸ _{7 7}

  6 ⁶ _{5 5}

  4 ⁴ _{3 3}

  2 ² _{1 1} 1 2 3 4 5 6 7 8 9 10

^{1 2 3 4 5 6 7 8 9 10}

x x _{1 1} x x _{2 2} Optimal Solution

  Optimal Solution Optimal Solution Optimal Solution

  Point Z (0,0) 0 (6,0) 30 (6,2) 44 (5,3) 46 (0,19/3) 44.33

  

Extreme Points and the Optimal Solution

• The corners or vertices of the feasible region are referred to as the extreme points.
• An optimal solution to an LP problem can be found at an extreme point of the feasible region.
• When looking for the optimal solution, you do not have to evaluate all feasible solution points.
• You have to consider only the extreme points of the feasible region.

  

Feasible Region

• The feasible region for a two-variable linear programming problem can be nonexistent, a single point, a line, a polygon, or an unbounded area.
• Any linear program falls in one of three categories:
• – is infeasible

• – has a unique optimal solution or alternate optimal

  solutions
• – has an objective function that can be increased without bound

• • A feasible region may be unbounded and yet there may

  be optimal solutions. This is common in minimization

  problems and is possible in maximization problems.

  Special Cases

• Alternative Optimal Solutions