## Unit II Linear Programming For Quantitative Techniques For Business Decisions Mcom Sem 2 Delhi University Complete Notes

Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University : Here Cakart.in website team members provide direct download links for Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University Complete notes in pdf format. Download these Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University Complete notes in pdf format and read well.

### Unit II Linear Programming For Quantitative Techniques For Business Decisions Mcom Sem 2 Delhi University Complete Notes

Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University : A linear programming model can be used to solve the assignment problem. Consider the example shown in the previous table, to develop a linear programming model. Let,

x11 represent the assignment of operator A to job 1 x12 represent the assignment of operator A to job 2 x13 represent the assignment of operator A to job 3 x21 represent the assignment of operator B to job 1 and so on.

Formulating the equations for the time taken by each operator, 10 x11 + 16 x12 + 7 x13 = time taken by operator A. 9 x21 + 17 x22 + 6 x23 = time taken by operator B. 6 x31 + 13 x32 + 5 x33 = time taken by operator C.

The constraint in this assignment problem is that each operator must be assigned to only one job and similarly, each job must be performed by only one operator. Taking this constraint into account, the constraint equations are as follows:

x11 + x12 + x13< 1 operator A x21 + x22 + x23< 1 operator B x31 + x32 + x33< 1 operator C x11 + x21 + x31 = 1 Job 1 x12 + x22 + x32 = 1 Job 2 x13 + x23 + x33 = 1 Job 3

**Objective function: **The objective function minimizes the time taken to complete all the jobs. Using the cost data table, the following equation can be arrived at:

The objective function is, Minimize Z = 10 x11 + 16 x12 + 7 x13 +9 x21 + 17 x22 + 6 x23 +6 x31 + 13 x32 + 5 x33

The linear programming model for the problem will be,

Minimize Z = 10 x11 + 16 x12 + 7 x13 +9 x21 + 17 x22 + 6 x23 +6 x31 + 13 x32 + 5 x33

subject to constraints

x11 + x12 + x13< 1 ………………..(i) x21 + x22 + x23< 1 ………………..(ii) x31 + x32 + x33< 1 ………………..(iii) x11 + x12 + x13 = 1 ………………..(iv) x12 + x22 + x32 = 1 ………………..(v) x13 + x23 + x33 = 1 ………………..(vi) where, xij> 0 for i = 1,2,3 and j = 1,2,3.

### Unit II Linear Programming For Quantitative Techniques For Business Decisions Mcom Sem 2 Delhi University Complete Notes

Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University : Linear programming is a powerful quantitative technique (or operational research technique) designs to solve allocation problem. The term **‘linear programming’** consists of the two words ‘Linear’ and ‘Programming’.

The word ‘Linear’ is used to describe the relationship between decision variables which are directly proportional. For example, if doubling (or tripling) the production of a product will exactly double (or triple) the profit and required resources, then it is linear relationship.

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Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University : Definitions Linear Programming is one of the important Techniques of OR It is useful in solving decision making problems which involves optimising a linear objective function subject to a set of linear constraints Varsha Varde2 Examples Selection of product mix which maximises profits subject to production, material, marketing, personnel and financial constraints Determination of capital budget which maximises NPV of the firm subject to financial, managerial, environmental and other constraints Choice of mixing short term financing which minimises cost subject to certain funding constraints Varsha Varde3

Formulation of LP problem DECISION VARIABLES Are the variables whose optimum values are to be found out by applying LP technique OBJECTIVE FUNCTION The part of a linear programming model that expresses what needs to be either maximised or minimised depending on the objective for the problem

Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University : Formulation of LP problem CONSTRAINTS It is an inequality or equation that expresses some restriction on the values that can be assigned to decision variables The constraints which represent non negativity conditions are called non negativity constraints The other constraints which represent restrictions on availability of resources etc are called structural constraints

Formulation of LP problem FEASIBLE SOLUTION A solution represents specific combination of values of decision variables A feasible solution is one that satisfies all constraints whereas an infeasible solution violates at least one constraint The optimal solution is the best feasible solution according to the objective function

New Office Furniture Ltd The new office furniture produces Desks, Chairs and Moulded Steel with the profit and raw material usage per unit as given below. The total availability of raw material for production is 2000kg. To satisfy contract commitments at least 100 desks must be produced. Due to the availability of seat cushions no more than 500 chairs must be produced Find out the optimal product mix. Products Profit Raw Steel Used Desks Rs500 7 kg per Desk Chairs Rs300 3 kg Per Chair Moulded Steel Rs60/ Kg 1.5 kg per Kg of moulded Steel Varsha Varde7

Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University : OBJECTIVE FUNCTION D: amount of desks (number) C: amount of chairs (number) M: amount of moulded steel (Kgs) Maximise Total Profit = 500 D + 300 C + 60 M CCNSTRAINTS New Office has only 2000 Kgs of raw steel available for production. 7 D + 3 C + 1.5 M ≤ 2000 To satisfy contract commitments; at least 100 desks must be produced. D ≥ 100 Due to the availability of seat cushions, no more than 500 chairs must be produced C ≤ 500 No production can be negative; D ≥ 0, C ≥ 0, M ≥0 Varsha Varde 9

Example Mathematical Model MAXIMIZE Z = 50 D + 30 C + 6 M (Total Profit) SUBJECT TO: 7 D + 3 C + 1.5 M ≤ 2000 (Raw Steel) D ≥100 (Contract) C ≤500 (Cushions) D, C, M ≥0 (Nonnegativity) D, C are integers Best or Optimal Solution of New Office Example 100 Desks, 433 Chairs, 0 Molded Steel Total Profit:Rs180000 Varsha Varde10

Example Graphical Solution Method

Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University : It is applicable when there are two decision variables The decision variables are represented by horizontal & vertical axis Straight lines are used to demarcate the feasible region The feasible region shows the solutions that satisfy all constraints Optimal solution lies at one of the corner points

Example A firm produces Two types of frames,Type 1 and Type 2.Each type 1 frame contributes a profit of Rs.225, whereas each type 2 frame contributes a profit of Rs.260.There are 4000 labor hours available. Each type 1 frame required 2 labor hours, and each type 2 frame requires 1 labor hour. There are 5000Kgs of metal available. Each type 1 frame requires 1Kg of metal and each type 2 frame requires 2 Kg of metal. Formulate the linear programming problem assuming that the demand exists for both the products. How many frames of each type should be produced to realize the optimal profit? (Use graphical method).What is the optimal profit?

Background Information Each type 1 frame contributes a profit of Rs.225, whereas each type 2 frame contributes a profit of Rs.260. The first constraint is a labor hour constraint. There are 4000 hours available. Each type 1 frame required 2 labor hours, and each type 2 frame requires 1 labor hour. Similarly, the second constraint is a metal constraint. There are 5000Kgs of metal available. Each type 1 frame requires 1Kg of metal and each type 2 frame requires 2 Kg of metal.

Model Formulation Let X 1 be number of frames of type I to be produced Let X 2 be number of frames of type II to be produced The algebraic model is given below: max 225x 1 + 260x 2 (profit objective) subject to 2x 1 + x 2 4000 (labor constraint) x 1 + 2x 2 5000 (metal constraint) x 1, x 2 0 (non negativity constraint) Solution The idea is to graph the constraints on a two-dimensional graph to see which points (x 1, x 2 ) satisfy all of the constraints. This set of points is labeled the feasible region. Then we see which point in the feasible region provides the largest profit. The graphical solution appears on the next slide.

Graphical Solution

Solution — continued To produce the graph, we first locate the lines where the constraints hold as equalities. For example, the line for labor is 2x 1 + x 2 = 4000. The easiest way to graph this is to find the two points where it crosses the axes. Joining the points (0,4000) and (2000,0), we get the line where the labor constraint is satisfied exactly, that is, as an equality. All points below and to the left of this line are also feasible; there are these are the points where less than the maximum number of 4000 labor hours are used.

Solution — continued We indicate the feasible side of the line by the short arrows pointing down to the left from the labor constraint line. Similarly the metal constraint line crosses the axes at the points (0,2500) and (5000,0), so we join these two points to find the line where all 5000 ounces of metal are used. Finally the points on or below both of these lines constitute the feasible region. These are the point below the heavy lines.

Solution — continued The crucial point, however, is that only three points can be optimal: (2000,0), (0, 2500), or (1000, 2000), the three “corner” points (other than (0,0)) in the feasible region. To find out the best of these three optimal points calculate profit at each point and select that point which gives maximum profit It is found that profit is maximum at x 1 = 1000 and x 2 = 2000, with a corresponding profit of P = Rs.7450. Thus the optimal solution is to produce 1000 of type I frames and 2000 of type II frames

Solution — continued You can think of the feasible region as all points on or inside the figure formed by four points: (0,0), (0,2500), (2000,0), and the point where labor hour and metal constraint lines intersect. The next step is to bring profit into the picture. We do this by constructing “isoprofit” lines – that is lines where total profit is a constant. Any such line can be written as 2.25x 1 + 2.60x 2 = P where P is a constant profit level. Solving for x2, we can put this equation in slope-intercept form: x 2 = P/2.60 – (2.25/2.60)x 1

Solution — continued This shows that any iso profit line has slope –2.25/2.60, and it crosses the vertical axis at the value P/2.60. Three of these isoprofit lines appear in the chart as dotted lines. Therefore, to maximize profit, we want to move the dotted line up and to the right until it just barely touches the feasible region. Graphically, we can see that the last feasible point it will touch is the point indicated in the figure, where the labor hour and metal constraint lines cross.

Solution — continued We can then solve two equations in two unknowns to find the coordinates of this point. They are x 1 = 1000 and x 2 = 2000, with a corresponding profit of P = Rs.7450. Note that if the slope of the isoprofit lines were much steeper, the the optimal point would be (2000,0). On the other hand,m if the slope were mush less steep, the optimal point would be (0,2500).These statements make intuitive sense. If the isoprofit lines are steep, this is because the unit profit from frame type 1 is large relative to the unit profit from frame type 2.

Solution — continued The crucial point, however, is that only three points can be optimal: (2000,0), (0, 2500), or (1000, 2000), the three “corner” points (other than (0,0)) in the feasible region. The best of these depends on the relative slopes of the constraint lines and isoprofit lines in the graph.

Unit II Linear Programming For Quantitative Techniques For Business Decisions MCOM Sem 2 Delhi University : Transportation, Assignment and Transshipment Problems Applications Physical analog of nodes Physical analog of arcs Flow Communication systems phone exchanges, computers, transmission facilities, satellites Cables, fiber optic links, microwave relay links Voice messages, Data, Video transmissions Hydraulic systems Pumping stations Reservoirs, Lakes Pipelines Water, Gas, Oil, Hydraulic fluids Integrated computer circuits Gates, registers, processors WiresElectrical current Mechanical systemsJoints Rods, Beams, Springs Heat, Energy Transportation systems Intersections, Airports, Rail yards Highways, Airline routes Railbeds Passengers, freight, vehicles, operators Applications of Network Optimization

Description A transportation problem basically deals with the problem, which aims to find the best way to fulfill the demand of n demand points using the capacities of m supply points. While trying to find the best way, generally a variable cost of shipping the product from one supply point to a demand point or a similar constraint should be taken into consideration.

Formulating Transportation Problems Example 1: Powerco has three electric power plants that supply the electric needs of four cities. The associated supply of each plant and demand of each city is given in the table 1. The cost of sending 1 million kwh of electricity from a plant to a city depends on the distance the electricity must travel. Transportation tableau A transportation problem is specified by the supply, the demand, and the shipping costs. So the relevant data can be summarized in a transportation tableau. The transportation tableau implicitly expresses the supply and demand constraints and the shipping cost between each demand and supply point.

Shipping costs, Supply, and Demand for Powerco Example FromTo City 1City 2City 3City 4Supply (Million kwh) Plant 1$8$6$10$935 Plant 2$9$12$13$750 Plant 3$14$9$16$540 Demand (Million kwh) 452030 Transportation Tableau Solution 1.Decision Variable: Since we have to determine how much electricity is sent from each plant to each city; X ij = Amount of electricity produced at plant i and sent to city j X 14 = Amount of electricity produced at plant 1 and sent to city 4

Objective function Since we want to minimize the total cost of shipping from plants to cities; Minimize Z = 8X 11 +6X 12 +10X 13 +9X 14 +9X 21 +12X 22 +13X 23 +7X 24 +14X 31 +9X 32 +16X 33 +5X 34

Supply Constraints Since each supply point has a limited production capacity; X 11 +X 12 +X 13 +X 14 <= 35 X 21 +X 22 +X 23 +X 24 <= 50 X 31 +X 32 +X 33 +X 34 <= 40

Demand Constraints Since each demand point requires minimum supply; X 11 +X 21 +X 31 >= 45 X 12 +X 22 +X 32 >= 20 X 13 +X 23 +X 33 >= 30 X 14 +X 24 +X 34 >= 30

Sign Constraints Since a negative amount of electricity can not be shipped all Xij’s must be non negative; Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

36 LP Formulation of Powerco’s Problem Min Z = 8X 11 +6X 12 +10X 13 +9X 14 +9X 21 +12X 22 +13X 23 +7X 24 +14X 31 +9X 32 +16X 33 +5X 34 S.T.:X 11 +X 12 +X 13 +X 14 <= 35 (Supply Constraints) X 21 +X 22 +X 23 +X 24 <= 50 X 31 +X 32 +X 33 +X 34 <= 40 X 11 +X 21 +X 31 >= 45 (Demand Constraints) X 12 +X 22 +X 32 >= 20 X 13 +X 23 +X 33 >= 30 X 14 +X 24 +X 34 >= 30 Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

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