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Session 4

LINEAR PROGRAMMING

Pre University-Mathematics for Business 1

Learning Objectives • represent geometrically a linear inequality in two variables. • state a linear programming problem and solve it geometrically. • consider situations in which a linear programming problem exists.

2

Introduction • Linear programming is constrained optimization, where the constraints and the objective function are all linear. • It is called "programming" because the goal of the calculations help you choose a "program" of action.

3

Examples of LP Problems (1) 1. A Product Mix Problem • A manufacturer has fixed amounts of different resources such as raw material, labor, and equipment. • These resources can be combined to produce any one of several different products. • The quantity of the ith resource required to produce one unit of the jth product is known. • The decision maker wishes to produce the combination of products that will maximize total income. 4

Examples of LP Problems (2) 2. A Blending Problem • Blending problems refer to situations in which a number of components (or commodities) are mixed together to yield one or more products. • Typically, different commodities are to be purchased. Each commodity has known characteristics and costs. • The problem is to determine how much of each commodity should be purchased and blended with the rest so that the characteristics of the mixture lie within specified bounds and the total cost is minimized. 5

Examples of LP Problems (3) 3. A Production Scheduling Problem • A manufacturer knows that he must supply a given number of items of a certain product each month for the next n months. • They can be produced either in regular time, subject to a maximum each month, or in overtime. The cost of producing an item during overtime is greater than during regular time. A storage cost is associated with each item not sold at the end of the month. • The problem is to determine the production schedule that minimizes the sum of production and storage costs. 6

Examples of LP Problems (4) 4. A Transportation Problem • A product is to be shipped in the amounts al, a2, ..., am from m shipping origins and received in amounts bl, b2, ..., bn at each of n shipping destinations. • The cost of shipping a unit from the ith origin to the jth destination is known for all combinations of origins and destinations. • The problem is to determine the amount to be shipped from each origin to each destination such that the total cost of transportation is a minimum. 7

Examples of LP Problems (5) 5. A Flow Capacity Problem • One or more commodities (e.g., traffic, water, information, cash, etc.) are flowing from one point to another through a network whose branches have various constraints and flow capacities. • The direction of flow in each branch and the capacity of each branch are known. • The problem is to determine the maximum flow, or capacity of the network.

8

Introduction • A linear function in x and y has the form

Z ax by • The function to be maximized or minimized is called the objective function.

9

Start with……… • A manufacturer makes wooden desks (X) and tables (Y). Each desk requires 2.5 hours to assemble, 3 hours for buffing, and 1 hour to crate. Each table requires 1 hour to assemble, 3 hours to buff, and 2 hours to crate. The firm can do only up to 20 hours of assembling, 30 hours of buffing, and 16 hours of crating per week. Profit is $3 per desk and $4 per table. Maximize the profit. 10

The linear programming model, for a manufacturing problem, involves: • Processes or activities that can be done in different amounts

11

• Constraints – resource limits – The constraints describe the production process – how much output you get for any given amounts of the inputs. – The constraints say that you cannot use more of each resource than you have of that resource. – Linear constraints means no diminishing or increasing returns. Adding more input gives the same effect on output regardless of how much you are already making. – Non-negativity constraints – the process levels cannot be less than 0. This means that you cannot turn your products back into resources. 12

• Objective function -- to be maximized or minimized – A linear objective function means that you can sell all you want of your outputs without affecting the price.

13

Solution The objective function is Profit = 3x + 4y x is the number of desks y is the number of tables

14

Solution Constraints: assembling

2.5x + y ≤ 20

buffing

3x + 3y ≤ 30

crating

x + 2y ≤ 16

non-negativity

x ,y ≥ 0 15

Solution The optimum is at x=4, y=6, profit=36

16

Example 2 Maximize the objective function Z = 3x + y subject to the constraints

2x y 8 2 x 3 y 12 x0 y 0 17

Solution: The feasible region is nonempty and bounded. Evaluating Z at these points, we obtain Z A 3 0 0 0

Z B 3 4 0 12 Z C 3 3 2 11 Z D 3 0 4 4

The maximum value of Z occurs when x = 4 and y = 0. 18

Example 3 Example: Product Mix Problem The N. Dustrious Company produces two products: I and II. The raw material requirements, space needed for storage, production rates, and selling prices for these products are given in Table 1.

The total amount of raw material available per day for both products is 15751b. The total storage space for all products is 1500 ft2, and a 19 maximum of 7 hours per day can be used for production.

Example 3 All products manufactured are shipped out of the storage area at the end of the day. Therefore, the two products must share the total raw material, storage space, and production time. The company wants to determine how many units of each product to produce per day to maximize its total income. Solution • The company has decided that it wants to maximize its sale income, which depends on the number of units of product I and II that it produces. • Therefore, the decision variables, x1 and x2 can be the number of units of products I and II, respectively, produced per day. 20

Example 3 • The object is to maximize the equation: Z = 13x1 + 11x2 subject to the constraints on storage space, raw materials, and production time. • Each unit of product I requires 4 ft2 of storage space and each unit of product II requires 5 ft2. Thus a total of 4x1 + 5x2 ft2 of storage space is needed each day. This space must be less than or equal to the available storage space, which is 1500 ft2. Therefore, 4X1 + 5X2 1500 • Similarly, each unit of product I and II produced requires 5 and 3 1bs, respectively, of raw material. Hence a total of 5x l 21 + 3x2 Ibs of raw material is used.

• This must be less than or equal to the total amount of raw material available, which is 1575 Ib. Therefore, 5x1 + 3x2 1575 • Product I can be produced at the rate of 60 units per hour. Therefore, it must take I minute or 1/60 of an hour to produce I unit. Similarly, it requires 1/30 of an hour to produce 1 unit of product II. Hence a total of x1/60 + x2/30 hours is required for the daily production. This quantity must be less than or equal to the total production time available each day. Therefore, x1 / 60 + x2 / 30 7 or x1 + 2x2 420 • Finally, the company cannot produce a negative quantity of any product, therefore x1 and x2 must each be greater than 22 or equal to zero.

• The linear programming model for this example can be summarized as:

23

Graphical Solution to LP Problems

24

Graphical Solution to LP Problems • An equation of the form 4x1 + 5x2 = 1500 defines a straight line in the x1-x2 plane. An inequality defines an area bounded by a straight line. Therefore, the region below and including the line 4x1 + 5x2 = 1500 in the Figure represents the region defined by 4x1 + 5x2 1500. • Same thing applies to other equations as well. • The shaded area of the figure comprises the area common to all the regions defined by the constraints and contains all pairs of xI and x2 that are feasible solutions to the problem.

25

Graphical Solution to LP Problems • This area is known as the feasible region or feasible solution space. The optimal solution must lie within this region. • There are various pairs of x1 and x2 that satisfy the constraints such as:

26

Graphical Solution to LP Problems • Trying different solutions, the optimal solution will be: X1 = 270 X2 = 75 • This indicates that maximum income of $4335 is obtained by producing 270 units of product I and 75 units of product II. • In this solution, all the raw material and available time are used, because the optimal point lies on the two constraint lines for these resources. • However, 1500- [4(270) + 5(75)], or 45 ft2 of storage space, is not used. Thus the storage space is not a constraint on the optimal solution; that is, more products could be produced before the company ran out of storage space. Thus this constraint is said to be slack. 27

Graphical Solution to LP Problems • If the objective function happens to be parallel to one of the edges of the feasible region, any point along this edge between the two extreme points may be an optimal solution that maximizes the objective function. When this occurs, there is no unique solution, but there is an infinite number of optimal solutions. • The graphical method of solution may be extended to a case in which there are three variables. In this case, each constraint is represented by a plane in three dimensions, and the feasible region bounded by these planes is a polyhedron. 28

LINEAR PROGRAMMING

Pre University-Mathematics for Business 1

Learning Objectives • represent geometrically a linear inequality in two variables. • state a linear programming problem and solve it geometrically. • consider situations in which a linear programming problem exists.

2

Introduction • Linear programming is constrained optimization, where the constraints and the objective function are all linear. • It is called "programming" because the goal of the calculations help you choose a "program" of action.

3

Examples of LP Problems (1) 1. A Product Mix Problem • A manufacturer has fixed amounts of different resources such as raw material, labor, and equipment. • These resources can be combined to produce any one of several different products. • The quantity of the ith resource required to produce one unit of the jth product is known. • The decision maker wishes to produce the combination of products that will maximize total income. 4

Examples of LP Problems (2) 2. A Blending Problem • Blending problems refer to situations in which a number of components (or commodities) are mixed together to yield one or more products. • Typically, different commodities are to be purchased. Each commodity has known characteristics and costs. • The problem is to determine how much of each commodity should be purchased and blended with the rest so that the characteristics of the mixture lie within specified bounds and the total cost is minimized. 5

Examples of LP Problems (3) 3. A Production Scheduling Problem • A manufacturer knows that he must supply a given number of items of a certain product each month for the next n months. • They can be produced either in regular time, subject to a maximum each month, or in overtime. The cost of producing an item during overtime is greater than during regular time. A storage cost is associated with each item not sold at the end of the month. • The problem is to determine the production schedule that minimizes the sum of production and storage costs. 6

Examples of LP Problems (4) 4. A Transportation Problem • A product is to be shipped in the amounts al, a2, ..., am from m shipping origins and received in amounts bl, b2, ..., bn at each of n shipping destinations. • The cost of shipping a unit from the ith origin to the jth destination is known for all combinations of origins and destinations. • The problem is to determine the amount to be shipped from each origin to each destination such that the total cost of transportation is a minimum. 7

Examples of LP Problems (5) 5. A Flow Capacity Problem • One or more commodities (e.g., traffic, water, information, cash, etc.) are flowing from one point to another through a network whose branches have various constraints and flow capacities. • The direction of flow in each branch and the capacity of each branch are known. • The problem is to determine the maximum flow, or capacity of the network.

8

Introduction • A linear function in x and y has the form

Z ax by • The function to be maximized or minimized is called the objective function.

9

Start with……… • A manufacturer makes wooden desks (X) and tables (Y). Each desk requires 2.5 hours to assemble, 3 hours for buffing, and 1 hour to crate. Each table requires 1 hour to assemble, 3 hours to buff, and 2 hours to crate. The firm can do only up to 20 hours of assembling, 30 hours of buffing, and 16 hours of crating per week. Profit is $3 per desk and $4 per table. Maximize the profit. 10

The linear programming model, for a manufacturing problem, involves: • Processes or activities that can be done in different amounts

11

• Constraints – resource limits – The constraints describe the production process – how much output you get for any given amounts of the inputs. – The constraints say that you cannot use more of each resource than you have of that resource. – Linear constraints means no diminishing or increasing returns. Adding more input gives the same effect on output regardless of how much you are already making. – Non-negativity constraints – the process levels cannot be less than 0. This means that you cannot turn your products back into resources. 12

• Objective function -- to be maximized or minimized – A linear objective function means that you can sell all you want of your outputs without affecting the price.

13

Solution The objective function is Profit = 3x + 4y x is the number of desks y is the number of tables

14

Solution Constraints: assembling

2.5x + y ≤ 20

buffing

3x + 3y ≤ 30

crating

x + 2y ≤ 16

non-negativity

x ,y ≥ 0 15

Solution The optimum is at x=4, y=6, profit=36

16

Example 2 Maximize the objective function Z = 3x + y subject to the constraints

2x y 8 2 x 3 y 12 x0 y 0 17

Solution: The feasible region is nonempty and bounded. Evaluating Z at these points, we obtain Z A 3 0 0 0

Z B 3 4 0 12 Z C 3 3 2 11 Z D 3 0 4 4

The maximum value of Z occurs when x = 4 and y = 0. 18

Example 3 Example: Product Mix Problem The N. Dustrious Company produces two products: I and II. The raw material requirements, space needed for storage, production rates, and selling prices for these products are given in Table 1.

The total amount of raw material available per day for both products is 15751b. The total storage space for all products is 1500 ft2, and a 19 maximum of 7 hours per day can be used for production.

Example 3 All products manufactured are shipped out of the storage area at the end of the day. Therefore, the two products must share the total raw material, storage space, and production time. The company wants to determine how many units of each product to produce per day to maximize its total income. Solution • The company has decided that it wants to maximize its sale income, which depends on the number of units of product I and II that it produces. • Therefore, the decision variables, x1 and x2 can be the number of units of products I and II, respectively, produced per day. 20

Example 3 • The object is to maximize the equation: Z = 13x1 + 11x2 subject to the constraints on storage space, raw materials, and production time. • Each unit of product I requires 4 ft2 of storage space and each unit of product II requires 5 ft2. Thus a total of 4x1 + 5x2 ft2 of storage space is needed each day. This space must be less than or equal to the available storage space, which is 1500 ft2. Therefore, 4X1 + 5X2 1500 • Similarly, each unit of product I and II produced requires 5 and 3 1bs, respectively, of raw material. Hence a total of 5x l 21 + 3x2 Ibs of raw material is used.

• This must be less than or equal to the total amount of raw material available, which is 1575 Ib. Therefore, 5x1 + 3x2 1575 • Product I can be produced at the rate of 60 units per hour. Therefore, it must take I minute or 1/60 of an hour to produce I unit. Similarly, it requires 1/30 of an hour to produce 1 unit of product II. Hence a total of x1/60 + x2/30 hours is required for the daily production. This quantity must be less than or equal to the total production time available each day. Therefore, x1 / 60 + x2 / 30 7 or x1 + 2x2 420 • Finally, the company cannot produce a negative quantity of any product, therefore x1 and x2 must each be greater than 22 or equal to zero.

• The linear programming model for this example can be summarized as:

23

Graphical Solution to LP Problems

24

Graphical Solution to LP Problems • An equation of the form 4x1 + 5x2 = 1500 defines a straight line in the x1-x2 plane. An inequality defines an area bounded by a straight line. Therefore, the region below and including the line 4x1 + 5x2 = 1500 in the Figure represents the region defined by 4x1 + 5x2 1500. • Same thing applies to other equations as well. • The shaded area of the figure comprises the area common to all the regions defined by the constraints and contains all pairs of xI and x2 that are feasible solutions to the problem.

25

Graphical Solution to LP Problems • This area is known as the feasible region or feasible solution space. The optimal solution must lie within this region. • There are various pairs of x1 and x2 that satisfy the constraints such as:

26

Graphical Solution to LP Problems • Trying different solutions, the optimal solution will be: X1 = 270 X2 = 75 • This indicates that maximum income of $4335 is obtained by producing 270 units of product I and 75 units of product II. • In this solution, all the raw material and available time are used, because the optimal point lies on the two constraint lines for these resources. • However, 1500- [4(270) + 5(75)], or 45 ft2 of storage space, is not used. Thus the storage space is not a constraint on the optimal solution; that is, more products could be produced before the company ran out of storage space. Thus this constraint is said to be slack. 27

Graphical Solution to LP Problems • If the objective function happens to be parallel to one of the edges of the feasible region, any point along this edge between the two extreme points may be an optimal solution that maximizes the objective function. When this occurs, there is no unique solution, but there is an infinite number of optimal solutions. • The graphical method of solution may be extended to a case in which there are three variables. In this case, each constraint is represented by a plane in three dimensions, and the feasible region bounded by these planes is a polyhedron. 28