UNIT 6 MODELLING DECISION PROBLEMS (LP)

Transcription

1 UNIT 6 MODELLING DECISION This unit: PROBLEMS (LP) Introduces the linear programming (LP) technique to solve decision problems 1

2 INTRODUCTION TO LINEAR PROGRAMMING A Linear Programming model seeks to maximize or minimize a linear function, subject to a set of linear constraints. The linear model consists of the following components: A set of decision variables. These are the variables that affect the problem under investigation. An objective function, i.e. the equation that is to be maximised or minimised. A set of constraints. 2

3 INTRODUCTION TO LINEAR PROGRAMMING Linear programming can be used in construction management to solve problems such as: Optimizing use of resources. Determining most economic product mix. Determining optimum size of bid. Transportation and routing problems. Personnel assignment. 3

4 GRAPHING LINEAR INEQUALITIES A linear equation in two variables x and y ax + by + c = 0 has a solution set that may be exhibited graphically as points on a straight line in the xy-plane. There is also a simple graphical representation for linear inequalities of two variables ax + by + c 0 4

5 PROCEDURE FOR GRAPHING LINEAR INEQUALITIES 1. Draw the graph of the equation obtained for the given inequality by replacing the inequality sign with an equal sign. Use a dashed or dotted line if the problem involves a strict inequality, < or >. Otherwise, use a solid line to indicate that the line itself constitutes part of the solution. 5

6 PROCEDURE FOR GRAPHING LINEAR INEQUALITIES 2. Pick a test point lying in one of the half-planes determined by the line sketched in step 1 and substitute the values of x and y into the given inequality. Use the origin whenever possible. 3. If the inequality is satisfied, the graph of the inequality includes the half-plane containing the test point. Otherwise, the solution includes the half-plane not containing the test point 6

7 EXAMPLE 1 Determine the solution set for the inequality 2x + 3y 6. Solution Replacing the inequality with an equality =, we obtain the equation 2x + 3y = 6, whose graph is: 7

8 EXAMPLE 1 Picking the origin as a test point, we find 2(0) + 3(0) 6, or 0 6, which is false. Thus, the solution set is: 8

9 EXAMPLE 2 Graph x 3y > 0. Solution Replacing the inequality > with an equality =, we obtain the equation x 3y = 0, whose graph is:

10 EXAMPLE 2 Use a dashed line to indicate the line itself will not be part of the solution, since we are dealing with a strict inequality >.

11 EXAMPLE 2 Since the origin lies on the line, we cannot use the origin as a testing point. Picking instead (3, 0) as a test point, we find (3) 2(0) > 0, or 3 > 0, which is true. Thus, the solution set is:

12 GRAPHING SYSTEMS OF LINEAR INEQUALITIES The solution set of a system of linear inequalities (i.e. multiple linear inequalities) in two variables x and y is the set of all points (x, y) that satisfy each inequality of the system. The graphical solution of such a system may be obtained by graphing the solution set for each inequality independently and then determining the region in common with each solution set.

13 EXAMPLE 3 Determine the solution set for the system 4x + 3y 12 x - y 0 Solution The intersection of the solution regions of the two inequalities represents the solution to the system:

14 EXAMPLE 3 Solution The intersection of the solution regions of the two inequalities represents the solution to the system:

15 FEASIBLE SOLUTION SET & OPTIMAL SOLUTION Each point in the solution set S is a candidate for the solution of the linear programming problem and is referred to as a feasible solution. The set S itself is referred to as a feasible set. Among all the points in the set S, the point(s) that optimizes the objective function of the linear programming problem is called an optimal solution

16 THEOREMS OF LP If a linear programming problem has an optimal solution, then it must occur at a vertex, or corner point, of the feasible set S associated with the problem If the objective function P is optimized at two adjacent vertices of S, then it is optimized at every point on the line segment joining these vertices, in which case there are infinitely many solutions to the problem. If S is an empty set, then the linear programming problem has no solution: that is, P has neither a maximum nor a minimum value.

17 BOUNDED AND UNBOUNDED SET The solution set of a system of linear inequalities is bounded if it can be enclosed by a circle. Otherwise, it is unbounded. Unbounded Bounded

18 BOUNDED AND UNBOUNDED SET Given that the objection function is P = ax + by If the solution set is bounded, then the objective function has both a maximum and a minimum value in the solution set. If the solution set is unbounded and both a and b are nonnegative, then the objection function P has a minimum value on the solution set provided that the constraints defining the solution set include the inequalities x 0 and y 0.

19 FORMULATING DECISION PROBLEMS USING LP 1. Identify the decision problem clearly E.g. organizing the supply of ready-mix concrete to a number of construction sites 2. Identify Decision Variables E.g. total daily requirements, lorry loads, hourly rate, fuel consumption, transport costs, distance to be travelled etc. 3. Identify the relationship (Objective Function) between these variables E.g. Total cost = x 1 + x x n. For LP, the function should take the form of linear function. 19

20 FORMULATING DECISION PROBLEMS USING LP 4. Identify Constraints; other set of relationships that affect the behaviour of the decision variables These constraints can be represented in the form of linear equalities and/or inequalities functions (e.g. x 1 70, and x 2 100, etc.) 5. Obtain values for the decision variables that verify the objective function considering the given constraints (Optimisation), which works either to maximise or minimise the objective function. 20

21 PRODUCTION PROBLEM EXAMPLE Ace Novelty wishes to produce two types of souvenirs: type-a will result in a profit of $1.00, and type-b in a profit of $1.20. To manufacture a type-a souvenir requires 2 minutes on machine I and 1 minute on machine II. A type-b souvenir requires 1 minute on machine I and 3 minutes on machine II. There are 3 hours available on machine I and 5 hours available on machine II. How many souvenirs of each type should Ace make in order to maximize its profit?

22 PRODUCTION PROBLEM EXAMPLE Solution Let s first tabulate the given information: Type-A Type-B Time Available Profit/Unit $1.00 $1.20 Machine I 2 min 1 min 180 min Machine II 1 min 3 min 300 min Let x be the number of type-a souvenirs and y the number of type-b souvenirs to be made.

23 PRODUCTION PROBLEM EXAMPLE Type-A Type-B Time Available Profit/Unit $1.00 $1.20 Machine I 2 min 1 min 180 min Machine II 1 min 3 min 300 min Then, the total profit (in dollars) is given by P = x + 1.2y which is the objective function to be maximized.

24 PRODUCTION PROBLEM EXAMPLE Type-A Type-B Time Available Profit/Unit $1.00 $1.20 Machine I 2 min 1 min 180 min Machine II 1 min 3 min 300 min The total amount of time that machine I is used is 2x + y and must not exceed 180 minutes. Thus, we have the inequality 2x + y 180

25 PRODUCTION PROBLEM EXAMPLE Type-A Type-B Time Available Profit/Unit $1.00 $1.20 Machine I 2 min 1 min 180 min Machine II 1 min 3 min 300 min The total amount of time that machine II is used is x + 3y and must not exceed 300 minutes. Thus, we have the inequality x + 3y 300

26 PRODUCTION PROBLEM EXAMPLE Type-A Type-B Time Available Profit/Unit $1.00 $1.20 Machine I 2 min 1 min 180 min Machine II 1 min 3 min 300 min Finally, neither x nor y can be negative, so x 0 y 0

27 PRODUCTION PROBLEM EXAMPLE In short, we want to maximize the objective function P = x + 1.2y Subject to the system of inequalities 2x + y 180 x + 3y 300 x 0 y 0

28 PRODUCTION PROBLEM EXAMPLE Graphing the solution for the 4 inequality

29 PRODUCTION PROBLEM EXAMPLE There are 2 ways of solving: 1. By using the vertices of the solution set; or 2. By using the gradient of the objective function

30 PRODUCTION PROBLEM EXAMPLE (USING VERTICES) Next, find the vertices of the feasible set S. The vertices are found by solving the intersecting equations simultaneously The vertices are A(0, 0), B(90, 0), C(48, 84), and D(0, 100).

31 PRODUCTION PROBLEM EXAMPLE (USING VERTICES) Now find the values of P at the vertices and tabulate them:

32 PRODUCTION PROBLEM EXAMPLE (USING VERTICES) Finally, identify the vertex with the highest value for P: We can see that P is maximized at the vertex C(48, 84) and has a value of 148.8

33 PRODUCTION PROBLEM EXAMPLE (USING GRADIENT) You can also let P equal any value. Then you will have the equation of a straight line which will give you the same gradient. gradient constant

34 PRODUCTION PROBLEM EXAMPLE (USING GRADIENT) Plot the gradient -1/1.2 and parallel shift it until moving it any further would result in no feasible point. This would give the optimal solution at point C. C gradient

35 PRODUCTION PROBLEM EXAMPLE (USING GRADIENT) The values for point C is found by solving simultaneously the intersecting equations: x+3y=300 & 2x+y=180. Point C is (48,84). Substituting in the objective function P = x + 1.2y gives the maximum P = Important note: Your graph needs to be roughly to scale for this method to work!

36 PRACTICAL TIPS One practical issue you will face is to decide on the scale to use for both the x-axis and the y-axis. For example, what x & y axis values to use to graph the following inequalities: 150x + 100y 12,000 ; x+y 100 ; 1000x+600y 180,000 Make y the subject and simply the inequality 150x + 100y 12,000 simply to: y = x x+y 100 simply to: y = 100 x 1000x+600y 180,000 simply to: y = x When x = 0 for all 3 equations, the max y-axis is 300

37 VIDEOS Solving LP using vertices Solving LP using graphical method

38 CLASS EXERCISE 1 A nutritionist advises an individual who is suffering from iron and vitamin B deficiency to take at least 2400 milligrams (mg) of iron, 2100 mg of vitamin B1, and 1500 mg of vitamin B2 over a period of time. Two vitamin pills are suitable, brand-a and brand-b. Each brand-a pill costs 6 cents and contains 40 mg of iron, 10 mg of vitamin B1, and 5 mg of vitamin B2. Each brand-b pill costs 8 cents and contains 10 mg of iron and 15 mg each of vitamins B1 and B2. What combination of pills should the individual purchase in order to meet the minimum iron and vitamin requirements at the lowest cost?

39 CLASS EXERCISE 2 A building developer produces 2 types of houses: detached and semi-detached. The profit gained from a detached house and a semidetached house is $10,000 and $6,000 respectively. The developer has the capacity to build 100 houses per a year. The average area of a detached house is 150 sqm, while it is 100 sqm for a semidetached house. The developer got a land for housing development with area of 12,000 sqm. The supply of bricks is limited by 180,000 sqm. On average, a detached house requires 1000 sqm of bricks, while a semidetached house requires 600 sqm. Develop a LP model for this problem to determine how may detached and semidetached houses should be constructed in order to maximise profit.

40 CLASS EXERCISE 3 Sarah makes bracelets and necklaces to sell at a craft store. Each bracelet makes a profit of $7, takes 1 hour to assemble, and costs $2 for materials. Each necklace makes a profit of $12, takes 2 hour to assemble, and costs $3 for materials. Sarah has 48 hours available to assemble bracelets and necklaces. If she has $78 available to pay for materials, how many bracelets and necklaces should she make to maximize her profit?

41 CLASS EXERCISE 4 A contractor makes a decision about projects to bid for the coming financial year. Two types of projects are available for the contractor; short term (6-months max. in duration) and long term (over 1-year). Given the current market conditions and material costs, each short term project generates max. of 5000 profit on average, and each long term project generates max. of profit on average. A number of constraints are considered: Each project (short and long term) requires a project manager. The contractor has 10 posts of project manager. Each short term project requires 2 equipment and each long term project requires 5 equipment. The contractor has 41 different equipment in total. A short term project requires 8 days of design time on average, and a long term project requires 12 days of design time on average. The design team of the contractor works part-time of a total 144 days a year.

42 CLASS EXERCISE 5 A steel company produces beams in I and L shapes. Each I- beam can be sold for a profit of $100/m and each L-beam for a profit of $70/m. the production time is 40 hours per week and takes 6 hours to make a meter of I-shape beam and 3 hours to make a meter of L-shape beam. Market demand requires that the company makes at least 3 times as many L-shaped beams as I-shaped beams. I-shaped beams take up 4 times as much storage space as L-shaped beams and there is room for at most 4 I-shaped beams each week. Develop a LP model for this problem to determine how many meters for I-shaped beams and L-shaped beams should be produced each week in order to maximise profit.

43 THE END Any questions? 43