2. Linear Programming Problem

Transcription

1 . Linear Programming Problem. Introduction to Linear Programming Problem (LPP). When to apply LPP or Requirement for a LPP.3 General form of LPP. Assumptions in LPP. Applications of Linear Programming.6 Advantages of Linear Programming Techniques.7 Limitations of Linear Programming.8 Mathematical Formulation of Linear Programming Problem.9 Solution of LPP by Graphical method.0 Some Important Definitions. LPP: The Simplex Method. The Simplex Method - Maximization Case.3 The Simplex Method Examples. Problems Set. Introduction to Linear Programming Linear programming deals with the optimization of a function of variables known as objective function, subject to a set of linear equation and/or inequalities known as constraints. The objective function may be profit, cost, production capacity or any other measure of effectiveness, which is to be obtained in the best possible or optimal manner. The constraints may be different resources such as availability of raw material, storage capacity, man power, machine capacity, time etc. By linear, we mean a mathematical expression of the type a x + a x +. + a n x n, where a, a,, a n are constants and x, x x n are variables. By programming, we mean modelling and solving a problem mathematically that involve economic allocation of limited resources to achieve desired objective.

2 Definition The general LPP calls for optimizing (maximizing / minimizing) a linear function of variables called the Objective function subject to a set of linear equations and / or inequalities called the Constraints or Restrictions.. When to apply LPP or Requirement for a LPP All organization, big or small, have at their disposal, men, machines, money and materials, the supply of which may be limited. If the supply of these resources were unlimited, the need for the management tools like linear programming would not arise at all. Supply of resources being limited, the management must find the best allocation of its resources in order to maximize profit or to minimize the cost or use the production capacity to the maximum extent. LPP can be used for optimization problems if the following conditions are satisfied: Well defined objective function Alternative courses of action Decision variables should be interrelated and non- negative Constraints must be capable of being expressed as linear equations or inequalities in terms of variables Resources must be limited.3 General form of LPP The general LPP with n decision variables and m constraints can be stated in following form. Find the values of decision variables x, x x n so as to Optimize (max. or min.) Z = c x + c x +.+ c n x n subject to linear equations or inequalities, a x + a x +...+a n x n (,=, ) b and a x + a x +..+a n x n (,=, ) b... a m x + a m x +.+a mn x n (,=, ) b m x, x,, x n 0 Where Z = value of overall measure of performance x j = decision variable (for j =,,..., n)

3 c j = coefficients representing the per unit contribution of decision variable x j b i = total availability of i th resource (for i =,,, m) a ij = amount of resource i used by per unit of variable (activity) x j In Tabular form we can write as Resource usage per unit of variable Resource Decision variables.. n a a.a n a a.a n M a m a m.a mn Per unit Contribution of c c..c n decision variables Data needed for LP model Amount of resource available x, x x n are called decision variables.( The decision variables refer to any activity that is competing with other activities foe limited resources.) The values of the c j, b i, a ij (for i=, m and j=, n) are the input constants for the model. They are called as parameters of the model. The function being maximized or minimized Z = c x + c x +. +c n x n is called objective function. The limitations are normally called as constraints. The constraint a i x + a i x a in x n are sometimes called as functional constraint (L.H.S constraint). x j 0 restrictions are called non-negativity constraint.. Assumptions in LPP. Proportionality The contribution of each variable in the objective function or its usage of the resources is directly proportional to the value of the variable i.e. if resource availability increases by some percentage, then the output shall also increase by same percentage. b b... b m 3

4 . Additivity Sum of the resources used by different activities must be equal to the total quantity of resources used by each activity for all resources individually or collectively. 3. Divisibility The variables are not restricted to integer values. Deterministic Coefficients in the objective function and constraints are completely known and do not change during the period under study in all the problems considered.. Finiteness Variables and constraints are finite in number. 6. Optimality In LPP, we determine the decision variables so as to optimize the objective function of the LPP. 7. The problem involves only one objective, profit maximization or cost minimization.. Applications of Linear Programming Production Management: Linear programming can be applied in production management for determining product mix, production planning and assembly line-balancing. Product Mix: A company can produce several different products, each of which requires the use of limited production resources so as to maximize the total output subject to all constraints. Assembly-line balancing: This problem is likely to arise when an item can be made by assembling different components. The process of assembling requires some specific sequences. The objective is to minimize the total elapsed time. Agriculture Applications Linear programming can be applied in agricultural planning for allocating the limited resources such as labour, water supply and working capital etc, so as to maximize the net revenue.

5 Military Applications These applications involve the problem of selecting an air weapon system against gurillas so as to keep them pinned down and simultaneously minimize the amount of aviation gasoline used, a variation of transportation problem that maximizes the total tonnage of bomb dropped on a set of targets and the problem of community defence against disaster to find the number of defence units that should be used in the attack in order to provide the required level of protection at the lowest possible cost. Marketing Management Media Selection : LPP techniques helps in determining the advertising media mix so as to maximize the effective exposure, subject to limitation of budget, specified exposure rates to different market segments. Physical Distribution: LPP determines the most economic and efficient manner of locating manufacturing plants and distribution centre for physical distribution. Personnel Management Staffing problem: LPP is used to allocate optimum manpower to a particular job so as to minimize the total overtime cost or total manpower. Determination of Equitable salaries: LPP techniques has been used in determining equitable salaries and sales incentives. Job Evaluation and Selection: selection of suitable person for a specified job and evaluation of job in organizations has been done with the help of LPP. Other applications of linear programming lies in the area of administration, education, transportation, assignment of jobs to machines, hospital administration etc..6 Advantages of Linear Programming Techniques. It helps us in making the optimum utilization of productive resources.. The quality of decisions may also be improved by linear programming techniques. 3. Provides practically solutions.. In production processes, high lighting of bottlenecks is the most significant advantage of this technique.

6 .7 Limitations of Linear Programming Some limitations are associated with linear programming techniques. In some problems, objective functions and constraints are not linear. Generally, in real life situations concerning business and industrial problems constraints are not linearly treated to variables.. There is no guarantee of getting integer valued solutions. For example, in finding out how many men and machines would be required to perform a particular job, rounding off the solution to the nearest integer will not give an optimal solution. Integer programming deals with such problems. 3. Linear programming model does not take into consideration the effect of time and uncertainty. Thus the model should be defined in such a way that any change due to internal as well as external factors can be incorporated.. Sometimes large scale problems cannot be solved with linear programming techniques even when the computer facility is available. Such difficulty may be removed by decomposing the main problem into several small problems and then solving them separately.. Parameters appearing in the model are assumed to be constant. But, in real life situations they are neither constant nor deterministic. 6. Linear programming deals with only single objective, whereas in real life situation problems come across with multi objectives. Goal programming and multi-objective programming deals with such problems..8 Mathematical Formulation of Linear Programming Problem The steps for mathematical formulation of LPP are. Identify the decision variables of the problem. Non negativity constraints 3. Express constraints in terms of decision variables.. Write objective function to be optimized (maximize or minimize) as a linear function of decision variables. 6

7 Product Mix problem. A firm produces three products. These products are processed on three different machines. The time required manufacturing one unit of each of the three products and the daily capacity of the three machines are given in the table. Machine Time per unit (Mins) Product Product Product 3 Machine capacity Min /day M 3 0 M M3-30 It is required to determine the daily number of units to be manufactured for each product. The profit per unit for product, and 3 is Rs., Rs. 3 and Rs. 6 respectively. It is assumed that all the amounts produced are consumed in the market. Formulate the mathematical model for the model. Solution. Identify the Decision Variables Let x be the number of units of product x be the number of units of product x 3 be the number of units of product 3. Non negativity constraints x, x, x Express constraints in terms of decision variables. Here limitations are available machine capacity(mins/day). Time per unit (Mins) Machine capacity Machine Product Product Product 3 Min /day M 3 0 M M3-30 For Machine M As one unit of product requires mins Therefore, x unit of product requires x mins 7

8 As one unit of product requires 3 mins Therefore, x unit of product requires 3 x mins As one unit of product 3 requires mins Therefore, x 3 unit of product 3 requires x 3 mins minimum capacity of machine M is 0 mins per day therefore x + 3x + x 3 0 Similarly, For machine M x + 0x + 3x 3 70 For machine M3 x + x + 0x Write objective function Here objective is to maximize daily profit. As one unit of product gives profit of Rs. x unit of product gives profit of x Rs. As one unit of product gives profit of 3 Rs. x unit of product gives profit of 3x Rs. As one unit of product 3 gives profit of 6 Rs. x 3 unit of product 3 gives profit of 6x 3 Rs. Therefore objective function is Maximize Z = x + 3x + 6x 3 Hence find x, x, x 3 so as to Maximize Z = x + 3x + 6x 3 Subject to x + 3x + x 3 0 x + 0x + 3x 3 70 x + x + 0x 3 30 x, x, x 3 0 8

9 Diet problem. A person wants to decide the constituents of a diet which will fulfill his daily requirements of proteins, fats and carbohydrates at the minimum cost. The choice is to be made from four different types of foods. The yields per unit of these foods are given in the table. Food Type Yield/unit Proteins Fats Carbohydrates Cost/Unit Minimum Requirement Rs Formulate the LP for the problem. Solution. Identify the Decision Variables Let x be the number of units of food type l x be the number of units of food type x 3 be the number of units of food type 3 x be the number of units of food type. Non negativity constraints x, x, x 3, x 0 3. Express constraints in terms of decision variables. Here constraints are on the daily minimum requirement of the various constituents For protein: 3x + x + 8x 3 + 6x 800 For fats: x + x + 7x 3 + x 00 For carbohydrates: 6x + x + 7x 3 + x 700 9

10 . Write objective function Here objective is to minimize cost Minimize Z = x + 0x + 8x 3 + 6x Hence find x, x, x 3, x so as to Minimize Z = x + 0x + 8x 3 + 6x Subject to 3x + x + 8x 3 + 6x 800 x + x + 7x 3 + x 00 6x + x + 7x 3 + x 700 x, x, x 3, x 0 0

11 .9 Solution of LPP by Graphical method After formulating the Linear programming Model the next step is to solve the model. There are various mathematical method to solve the LPP. But when we have only two decision variables then we used Graphical method to solve LP model. The steps are. Express each constraint in terms of equation i.e. replaces inequality sign in each constraint by an equality sign. Assign arbitrary value to the decision variables x and x to get plotting points. x k 0 x 0 k (x, x ) = (k, k ) Plot these point on the graph and join them by straight line.. If the inequality constraint corresponding to that line is then the region below the line lying in the first quadrant is shaded. Similarly for the region above the line is shaded. The common region which satisfies all the constraints is called feasible region or solution space. 3. Determine coordinate of each corner point of the feasible region.. Compute and compare the value of the objective function at each corner point.. Identify corner point that gives optimal (max. or min.) value of the objective function..0 Some Important Definitions. Solution: A set of variables x, x x n is called a solution to a given LPP if it satisfies given constraints.. Feasible solution: A set of variables x, x x n is called a feasible solution to LPP if it satisfy constraints as well as non-negativity restrictions. 3. Infeasible solution: A set of variables x, x x n is called a infeasible solution to LPP if it does not satisfy either constraints or non-negativity restrictions.. Optimal solution: A feasible solution which optimize has the value of the objective function is called optimal solution to LPP.. Unbounded solution: A solution which can increase or decrease the value of objective function of given LPP indefinitely is called an unbounded solution.

12

13 3

14

15

16 6

17 7

18 8

19 9

20 0

21 . LPP:The Simplex Method For two variables, we used graphical method to solve LPP. But in real life problem when problems are formulated as a Linear programming model it has more than two variables. Thus we need a more efficient method to suggest an optimal solution for such problems. We shall discuss a procedure called the Simplex method for solving an LP model of such problems. It was developed by G. B. Dantzig in 97. Meaning of Simplex: In one dimension, a simplex is a line segment connecting two points. In two dimension, a simplex is a triangle formed by joining three points. In three dimension, it is a four sided pyramid having four corners. In general, simplex represents an object in n-dimensional space connecting n+ points. The concept of simplex is similar to the graphical method. The simplex method examines the extreme point in a systematic manner, repeating the same set of steps of the algorithm until an optimal solution is reached. It is for this reason that it is also called the iterative method. As numbers of extreme points or corner points are finite so by simplex method we get optimal solution in finite number of steps as we move from one corner point to another. Also it indicate an unbounded solution. Some Basic Terms: Slack variable: It is a variable that is added to the left-hand side of a ( less than or equal to) type constraint to convert the constraint into an equality. In economic terms, slack variables represent left-over or unused capacity. Specifically: a i x + a i x + a i3 x a in x n b i can be written as a i x + a i x + a i3 x a in x n + s i = b i Where i =,,..., m Surplus variable: It is a variable subtracted from the left-hand side of a ( greater than or equal to) type constraint to convert the constraint into an equality. It is also known as negative slack variable. In economic terms, surplus variables represent over fulfillment of the requirements.

22 Specifically: a i x + a i x + a i3 x a in x n b i can be written as a i x + a i x + a i3 x a in x n - s i = b i where i =,,..., m Artificial variable: It is a non negative variable introduced to facilitate the computation of an initial basic feasible solution. In other words, a variable added to the left-hand side of a greater than or equal to type constraint to convert the constraint into equality is called an artificial variable.. The Simplex Method - Maximization Case Consider the general linear programming problem involving n variables and m constraints: Maximize z = c x + c x + c 3 x c n x n subject to a x + a x + a 3 x a n x n b a x + a x + a 3 x a n x n b... a m x + a m x + a m3 x a mn x n b m x, x,..., x n 0 Where: c j (j =,,..., n) in the objective function are called the cost or profit coefficients. b i (i =,,..., m) are called resources. a ij (i =,,..., m; j =,,..., n) are called technological coefficients or input-output coefficients. In Standard form of LPP To solve LPP with the help of simplex method convert it into standard form. The standard form of LPP should have the following characteristics: Objective function is of function is of maximization type All constraints are expressed as equation by adding slack variables or subtracting surplus variables. Right hand side of each constraint is non- negative. All variables are non-negative.

23 Maximize z = c x + c x + c 3 x c n x n + 0s +0s +..+0s m subject to a x + a x + a 3 x a n x n + s = b a x + a x + a 3 x a n x n + s = b... a m x + a m x + a m3 x a mn x n + s m = b m x, x,..., x n 0 s, s,..., s m 0 An initial basic feasible solution is obtained by setting x = x =... = x n = 0 (non basic variables) We get basic variables s = b s = b... s m = b m The initial simplex table is formed by writing out the coefficients and constraints of a LPP in a systematic tabular form. The following table shows the structure of a simplex table. Structure of a simplex table c j c c c n Coefficient of basic variable(c B ) Basic Variables (B) x x x n s s s m Value of basic variables b (=X B ) s a a a n 0 0 b s a a a 0 0 b : : : : : : : : : s m a m a m a mn 0 0 b m z j c j z j c z c z c n z n Where: c j = coefficients of the variables (m + n) in the objective function. c B = coefficients of the current basic variables in the objective function. B = basic variables in the basis. X B = solution values of the basic variables. c j z j = index row. 3

24 The Simplex Algorithm Steps (Maximisation Case). Formulate the Problem Formulate the mathematical model of the given linear programming problem. If the objective function is given in minimization form then convert it into maximization form in the following way: Min z = - Max (-z) Any minimization problem can be converted into an equivalent maximization problem by multiplying the objective function with (-) Convert every inequality constraint in the Linear Programming Problem into an equality constraint by adding a slack variable to each constraint with zero cost coefficients to these in the objective function.. Find out the Initial Solution An initial basic feasible solution is obtained by setting decision variables x = x =... = x n = 0 (non basic variables), We get basic variables s = b s = b... s m = b m This solution is shown in the initial simplex table. Coefficient of basic variable(c B ) c j c c c n x x x n s s s m Value of basic variables b (=X B ) Basic Variables (B) s a a a n 0 0 b s a a a 0 0 b : : : : : : : : : s m a m a m a mn 0 0 b m z j c j z j c z c z c n z n

25 3. Test for Optimality Calculate the values of c j z j. There may arise three cases:- If all the values of c j z j 0, then the current basic feasible solution is the optimal solution. If at least one column a k of coefficient matrix for which c k z k > 0 and all elements of a k column are negative then there exist an unbounded solution to the given problem. If there are one or more positive values in c j z j row, then current initial basic feasible solution is not optimal.(i.e. it can be improved). Select the variable to enter the basis (incoming variables) and Test for Feasibility (variables to leave the basis(outgoing variables)) Incoming Variable:- select the most positive value in c j z j row and the element corresponding to these positive value is incoming variable, and column containing these incoming variables is called key column. Outgoing Variable:- Divide the values under b column by the corresponding positive element in the key column, and compare the ratios. The row that indicates the minimum ratio is called the key row (and corresponding variable is termed as outgoing variable). However, division by zero or negative coefficients in the key column is not allowed. In the case of a tie, break the tie arbitrarily. Important note: If all these ratio are negative then problem has an unbounded solution and we stop the procedure.. Identify the Pivot Element (Key Element) The number that lies at the intersection of the key column and key row of a given table is called the key element. It is always a non-zero positive number. 6. Finding the New Solution If key element is then row remains the same in the new simplex table.

26 If the key element is other than, then divide each element in the key row (including element in b- column) by the key element, to find the new values for that row. Convert all the element of the key column to zero (key element is ) by performing elementary row operation. 7. Repeat the Procedure Go to step 3 and repeat the procedure until all the values of are c j z j row either zero or negative..3 The Simplex Method Examples I. Use Simplex method to solve the following problem: Maximize Z = x + x Subject to x + x 3x + x x + x 9 and x, x 0 Solution Step Express the problem in Standard Form of LPP. Maximize Z = x + x + 0s + 0s + 0s 3 Subject to x + x + s = 3x + x + s = x + x + s 3 = 9 x, x, s, s, s 3 0 Step Find Initial Basic Feasible solution Set x = x = 0 (basic variables) We get s =, s =, s 3 = 9 (non- basic variables) 6

27 Coefficien t of basic variable(c B) We put these values in initial simplex table c j Basic Variable s(b) x x s s s 3 Value of basic variables b (=X B ) Minimum positive ratio θ = b/ KC 0 s 0 0 / = 6 (KR) 0 s / = 0 s / = 9 z j c j z j KC Since c j z j is positive under some column so current solution is not optimal. Step 3 Iterative towards an optimal solution Incoming Variable:- x (as it has most positive value of c j z j, and column containing these incoming variables is called key column) Outgoing Variable:- s (as it has minimum positive ratio θ = 6 and row containing these outgoing variable is called key row) Key Element:- (intersection of key column and key row) Step Find new solution Make key element = with the help of elementary row operation and other element of key column =0 with these key element. Row Operation for II nd Simplex table R (new) R (old) i.e. R (new) = b

28 Also to make key column value zero with the help of key element i.e. by row R (new), we have R (new) R (old) R (new) R (old) = R (new) = ( ) R (new) = b ( ) ( ) ( ) ( ) ( ) ( ) 0 0 R 3 (new) R 3 (old) R (new) R 3 (old) = R (new) = ( ) R 3 (new) = 3 b ( ) ( ) ( ) ( ) ( ) ( ) 0 We put these values in II nd simplex table Coefficien t of basic variable(c B) 0 3 c j Basic Variable s(b) x 0 s 0 s 3 3 x x s s s 3 Value of basic variables b (=X B ) 0 0 Minimum positive ratio θ = b/ KC /(/) = 0 / (/) = 60/ 0 3 3/(3/) = (KR) z j c j z j 3 KC

29 Since c j z j is positive under some column so current solution is not optimal. Step Iterative towards an optimal solution Incoming Variable:- x (as it has most positive value 3 of c j z j, and column containing these incoming variables is called key column) Outgoing Variable:- s 3 (as it has minimum positive ratio θ = and row containing these outgoing variable is called key row) Key Element:- 3 (intersection of key column and key row) Step 6 Find new solution Make key element = with the help of elementary row operation and other element of key column =0 with these key element. Row Operation for III rd Simplex table R 3 (new) 3 R 3(old) i.e. R 3 (new) = b Also to make key column value zero with the help of key element i.e. by row R 3 (new), we have R (new) R (old) R 3 (new) R (old) = R 3 (new) = ( ) 0 R (new) = 0 b ( ) ( ) ( ) ( ) ( ) ( )

30 R (new) R (old) R 3 (new) R (old) = R 3 (new) = ( ) 0 0 R (new) = 0 0 b ( ) ( ) ( ) ( ) ( ) ( ) 3 3 We put these values in III rd simplex table Coefficien t of basic variable(c B) c j Basic Variable s(b) x 0 0 s 0 0 x 0 x x s s s 3 Value of basic variables b (=X B ) Minimum positive ratio θ = b/ KC z j 0 c j z j Since all values of c j z j 0 so current solutions is optimal. x =, x = are optimal solution. and Maximize Z =

31 II. Use Simplex method to solve the following problem: Minimize Z = x 3x +x 3 Subject to 3x x +3x 3 7 x + x x + 3x +8x 3 0 and x, x, x 3 0 Solution Step Express the problem in Standard Form of LPP. Given Minimize Z = x 3x +x 3 So Maximize ( Z ) = x +3x x 3 Therefore, Maximize ( Z ) = x +3x x 3 Subject to 3x x +3x 3 + s = 7 x + x + s = x + 3x +8x 3 + s 3 = 0 x, x, x 3, s, s, s 3 0 Step Find Initial Basic Feasible solution Set x = x = x 3 = 0 (basic variables) We get s = 7, s =, s 3 = 0 (non- basic variables) 3

32 We put these values in initial simplex table Coefficie nt of basic variable(c B) c j Basic x x x 3 s s s 3 Value of Variable basic s(b) variables b (=X B ) Minimum positive ratio θ = b/ KC 0 s (divide by ive is not possible) 0 s 0 s / = 3 (KR) /3 = 3.33 z j c j z j KC Since c j z j is positive under some column so current solution is not optimal. Step 3 Iterative towards an optimal solution Incoming Variable:- x (as it has most positive value 3 of c j z j, and column containing these incoming variables is called key column) Outgoing Variable:- s (as it has minimum positive ratio θ = 3 and row containing these outgoing variable is called key row) Key Element:- (intersection of key column and key row) Step Find new solution Make key element = with the help of elementary row operation and other element of key column =0 with these key element. Row Operation for II nd Simplex table R (new) R (old) R (new) = 0 0 b 0 3 3

33 Also to make key column value zero with the help of key element i.e. by row R (new), we have R (new) R (old) + R (new) R (old) = R (new) = 0 0 b 0 3 (+) (+) (+) (+) (+) (+) (+) (+) R (new) = R 3 (new) R 3 (old) 3 R (new) b R 3 (old) = R (new) = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) R 3 (new) = We put these values in II nd simplex table Coefficie nt of basic variable(c B) c j Basic x x x 3 s s s 3 Value of Variable basic s(b) variables b (=X B ) 0 0 s 3 x 0 s Minimum positive ratio θ = b/ KC 0 0 0/(/) = (KR) 0 3 (divide by ive is not possible) (divide by ive is not possible) z j c j z j 3 KC

34 Since c j z j is positive under some column so current solution is not optimal. Step Iterative towards an optimal solution Incoming Variable:- x (as it has most positive value of c j z j, and column containing these incoming variables is called key column) Outgoing Variable:- s (as it has minimum positive ratio θ = and row containing these outgoing variable is called key row) Key Element:- (intersection of key column and key row) Step 6 Find new solution Make key element = with the help of elementary row operation and other element of key column =0 with these key element. Row Operation for III rd Simplex table R (new) R (old) b R (new) = Also to make key column value zero with the help of key element i.e. by row R (new), we have R (new) R (old) + R (new) R (old) = R (new) = b (+) (+) (+) (+) (+) (+) (+) (+) R (new) =

35 R 3 (new) R 3 (old) + R (new) R 3 (old) = R (new) = b 0 0 (+) (+) (+) (+) (+) (+) (+) (+) R 3 (new) = We put these values in III rd simplex table Coefficie nt of basic variable(c B) c j Basic x x x 3 s s s 3 Value of Variable basic s(b) variables b (=X B ) 0 x 0 3 x 0 0 s z j 3 c j z j Since all values of c j z j 0 so current solutions is optimal. x =, x =, x 3 = 0 are optimal solution. And Minimum value of objective function is given by Minimize Z =. 3

36 . Problems Set. What is Linear Programming Problem?. Write mathematical model of linear programming problem? 3. What are the steps involved in formulating a Linear programming problem?. Discuss applications of Linear Programming to managerial decision making?. A company makes two kinds of leather belts. Belt A is a high quality belt, and belt B is of lower quality. The respective profits are Rs..00 and Rs per belt. Each belt of type A requires twice as much time as a belt of type B, and if all belts were of type B, the company could make 000 per day. The supply of leather is sufficient for only 800 belts per day (Both A and B combined). Belt A requires a fancy buckle and only 00 per day are available. There are only 700 buckles a day available for belt B. Formulate the LP model to maximize product mix. 6. A dealer wishes to purchases a number of fans and sewing machines. He was only Rs. 760 to invest and has space for almost 0 items. A fan costs him Rs. 360 and a sewing machine is Rs. 0. His expectation is that he can sell a fan at profit of Rs. and a sewing machine at a profit of Rs. 8. Assuming that he can sell all the items that he can buy, how should be invest his money in order to maximize. 7. The standard weigh of a special purpose brick is kg and it contains two basic ingredients B and B. B costs Rs. per kg. and B costs Rs. 8 per kg. Strength considerations dictate that the brick should contain not more than kg of B and a minimum kg of B. Since the demand for the product is likely to be related to the price of the brick. Formulate the above as Linear Programming Problems to minimize cost of brick satisfying the above conditions. 8. Food X contains 6 units of vitamin A per gram and 7 units of vitamin B per gram and costs paise per gram. Food Y contains 8 units of vitamin A per gram and units of vitamin B per gram and costs 0 paise per gram. The daily minimum requirement of vitamin A and B is 00 units and 0 units respectively. Formulate the L.P.P. So that cost is minimized. 9. Explain graphical method to solve Linear Programming Problem? 0. Solve graphically Maximize Z = x + 80x Subject to x +0x 00 0x +x 0 36

37 and x, x 0 Ans. x =, x = and maximum z = 00. Solve graphically Minimize z =6x +x Subject to x + x 60 3x + 7x 8 x +x 8 and x,x 0. Ans. x = 8, x = 0 and minimum z = 08. Use the graphical method to solve the following Linear Programming Problem Minimize Z = 0x + 0x Subject to x + x 0 3x + x 30 x +3x 60 and x,x 0 Ans. Try yourself 3. Use the graphical method to solve the following L.P. problem Maximize Z = 3x + x Subject to x + x 00 x 80 x 60 x,x 0 Ans. Unbounded solution. Define feasible solution, infeasible solution, optimal solution in context with Linear programming problem?. Define Slack and surplus variable in context of LPP? When we use these variable. 6. Solve the following by Simplex method: Maximize Z = 3x +x Subject to x + x 0 x +x 600 x,x 0 Ans. x = 0, x = 0, and max z = Solve the following by Simplex method: Maximize Z = 3x +x Subject to x + x 37

38 x -x x,x 0 Ans. x = 3, x =, and max z = 8. Solve the following by Simplex method: Maximize Z = 00x +600x +00 x 3 Subject to x + x +6 x x +x +x 3 0 x,x, x 3 0 Ans. x = 0, x = 0, x 3 = 0 and max z = Solve the following by Simplex method: Maximize Z = 3x +x + x 3 Subject to x +3 x 8 x +x 3 0 3x +x +x 3 x,x, x 3 0 Ans. x = 89/, x =0/, x 3 = 6/ and max z = 76/ 38