Lpp

LINEAR PROGRAMMING PROBLEMS

PROF. VINAY PANDIT

LINEAR PROGRAMMING - I : Formulation and Graphic Solution DEFINITION AND APPLICATIONS: The mathematical definition of linear programming (L.P.) can be stated as — “It is the analysis of problems in which a linear function of a number of variables is to be maximized (minimized), when those variables are subject to a number of restraints in the form of linear inequalities”. Linear programming models thus belong to a class of mathematical programming models concerned with efficient allocation of resources to known activities with the objective of meeting a desired goal. Organizations can have many goals. Hence, a wide variety of problems can be efficiently solved using L.P. technique. Here are a few examples: (1) A product mix problem: Decide the combination of various product quantities to maximise profit or to minimise production cost. (2) Allocation of bank funds: To achieve highest possible returns. This should be achieved within liquidity limits set by RBI and maintaining flexibility to meet the customers demand for loans. (3) Manufacturing problem: To manufacture goods (say furniture) so as to give maximum profits bearing in mind the time constrain and the market demand for the goods. (4) Advertising application: To achieve the best possible exposure to the client’s product at the lowest possible advertising cost. (5) Portfolio selection: Select specific investments among available alternatives so as to maximise return or to minimise risk.

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(6) Staffing problem: Develop a work schedule that allows say, a large restaurant or a hospital or a police station to meet their man power needs at all hours with minimum number of employees. (7) Trim loss problem: Find the combination of components to be produced from standard sheets in order to keep trim loss to a minimum. An extension of application of linear programming technique is found in transportation and assignment problems.

TERMINOLOGY OF LINEAR PROGRAMMING: A typical linear programme has the following components: (1) An objective function. (2) Constraints or restrictions. (3) Non-negativity restriction. And the following terms are commonly used to describe a typical L.P.P.

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Decision variables: Decision variables are the unknowns whose values are to be determined from the solution of the problem. E.g. decision variables in the furniture manufacturing problem are say the tables and chairs whose values or actual units of production are to be found from the solution of the problem. These variables should be inter-related in terms of consumption of resources. For example, both tables and chairs require carpenter’s time and also wood and other resources and any change in the quantity produced of table affects the production level of chairs. Secondly the relationship among the variables should be linear.

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Objective function: A firms objectives are expressed as a function of decision variables. It represents the mathematical equation of the goals of the firm in terms of unknown values of the decision variables. Thus if the objective is to maximise net profits in a furniture manufacturing problem, then profits are expressed as function of (dependent on) the net per unit profits of table and chair and the number of units produced (of tables and chairs). OR- MMS

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Constraints: A constraint represents the limitations imposed on the values of decision variables in the solution. These limitations exist due to limited availability of resources as well as the requirements of these resources in the production of each unit of the decision variable. For example manufacturer of a table requires certain amount of time in a certain department and the department works only for a given period, (say 8 hours in a day for 5 days in a week). The constraints may represent some other type of limitations also. As in the production of a commodity the market demand can put an upper limit on the value of the decision variable in the optimal solution. Thus, the constraints define the limits within which a solution to the problem must be found. These constraints must be capable of expression in mathematical form of an equality or inequality.

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Linear relationships: Linear programming deals with problems in which the objective function and the constraints can be expressed as linear functions. Hence, when the problem is solved graphically, in a two variable case the constraints the objective function, gives a straight line on a two dimensional graph.

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Equations and inequalities: Equations are represented by = (equality) sign. They are specific statements. But many business problems cannot be neatly expressed in equations (called strict equality). Instead of precise statements, we may have only minimum or maximum requirements or availability. For example we may state that available labour time is 40 hours per week, hence, labour time used in production should be less than or equal to 40 hours per week. We thus need inequalities. Less than or equal to relationship is written as () and () sign indicates greater than or equal to relationship. Most of the constraints in a LPP are expressed as inequalities. They indicate the upper or the lower limits of resource use or production level. They do not express exact levels. Thus, they allow for many possibilities of the optimal values of the decision variable i.e. more than one combination of the decision variables may give the same optimal value of the objective function.

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Non-negativity restrictions: Linear programming technique is used to obtain solution to real world problems. The solution to the problem implies finding values of the decision variables. These must be non-negative. As one cannot think of manufacture of -4 tables or -6 chairs i.e. negative production. Hence, decision variable should assume either zero or positive values. If we denote two decision variables as X1 and X2 then the non negativity restriction is expressed as X1 > = 0; X2 >= 0.

FORMULATION OF LINEAR PROGRAMMING PROBLEM (LPP): Formulation of a LPP involves constructing a mathematical model from the given data. This can be done only if the following requirements are met: (a) There should be a clearly identifiable objective and it should be measurable in quantitative terms. E.g. In a manufacturing problem the objective can be maximisation of profit or minimisation of cost. (b) The resources to be allocated in the problem should be identifiable and quantitatively measurable. E.g. The use of labour time, or raw material in the manufacturing process should be clearly stated. (c) The relationships representing the objective function and the constraints equations must be linear. (d) There should be a series of feasible alternative courses of action available to the decision maker. These are determined by the resource constraints. When all the above mentioned conditions are satisfied the problem can be expressed as L.P. problem. Then solve it for an optimal solution. Let us first illustrate the formulation of LPP. We have seen that a typical L.P.P. has three components: (1) Objective function (2) Constraints (3) Non-negativity restrictions Formulation of LPP therefore implies translating the given descriptive problem into these three sets of linear relationships between the decision variables. OR- MMS

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METHODS OF SOLUTION: Solving an LP problem involves: (i)

Selection of appropriate method of solution and

(ii)

Then obtain a solution to the problem with the help of selected method

(iii)

Test whether this solution is optimal.

The problem can be solved by using: (1) Graphical method: This method can be used if there are only t• decision variables in the LPP. (2) Simplex method: This method is useful in solving LP problems with two or more than two decision variables. We explain, in this chapter, the graphical method of solving LPP. The Graphical method of solution: This method can be used in case where LPP has only two decision variables. But there is no restriction on the number of constraints. The method uses the familiar graphical presentation with two axes. The method becomes unwieldy when there are three variables since we then need a three dimensional graph. The method cannot be used if the number of decision variables is more than three. In such a case we have to use a non graphical method to obtain a solution. The graphical method of solution to L.P. problem uses all the equations in a given problem, namely the equation expressing objective unction the constraints imposed in achieving the objective. These constraints can be of (i) greater than (ii) less than or (iii) strict equality type. There is also a non-negativity restriction on the values of the decision variables. It implies that the solution of the problem lies in the first quadrant of the graph. All these relations are linear.

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SPECIAL CASES IN LPP 1. Infeasibility 2. Unboundedness 3. Redundancy 3. Alternate optima (Alternate optimum solution)

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INFEASIBILITY

It is a case where there is no solution, which satisfies all the constraints at the same time. This may occur if the problem is not correctly formulated. Graphically, infeasibility is a case where there is no region, which satisfies all constraints simultaneously.

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UNBOUNDEDNESS

A LPP can fail to have an optimum solution if the objective can be made infinitely large without violating any of the constraints. If we come across unboundedness in solving real problems, then the problem is not correctly formulated. Since, no real situation permits any management to have infinite production of goods and infinite profits, unbounded solution results if in a maximization problem all constraints are of greater than or equal to type. In such a situation there is no upper limit on feasible region. Similarly, an unbounded solution occurs in a minimization problem if all constraints are of less than or equal to type.

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REDUNDANCY

A constraint, which does not affect the feasible region, is called a redundant constraint. Such a constraint is not necessary for the solution of the problem. It can therefore be omitted while formulating the problem. This will save the computation time. In many LP problems, redundant constraints are not recognized as being redundant until the problem is solved. However, when computers are used to solve LPP, redundant constraints do not cause any difficulty.

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ALTERNATIVE OPTIMA

[The slope of a line ax + by + c = 0 is defined as –b ] a The solution to a LPP shall always be unique if the slope of the objective function line is different from the slope of all of the constraint lines. Incase, the slope of objective function line is same as the slope of one of its constraint line, then multiple optimum solution might exist.

ISO-PROFIT/ISO-COST LINES: An iso-profit line can be obtained by putting the objective function equal to some numerical value and plotting that equation on the graph in the same way as the constraints are plotted. Each point on an iso-profit line yields the same profit. The lines parallel to the one obtained by moving away from the origin correspond to higher and higher profit levels. The line whose one point touches the extreme corner point is considered and the profit corresponding thereto is the highest profit attainable. And the variable values at that point are optimal values of the decision variables.

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LINEAR PROGRAMMING – II : Simplex Method THE SIMPLEX METHOD OF SOLUTION: The simplex method uses a simplex algorithm; which is an iterative, procedure for finding, in a systematic manner the optimal solution to a linear programming problem. The procedure is based on the observation that if a feasible solution to a linear programming exists; it is located at a corner point of the feasible region determined by the constraints of the problem. The simplex method, selects the optimal solution from among the set of feasible solution to the problem. The algorithm is very efficient as it considers only those feasible solutions, which are provided by the corner points. Thus, we need to consider a minimum number of feasible solutions to obtain an optimal one. The method is quite simple and the first step requires the determination of basic feasible solution. Then, with the help of a limited number of steps the optimum solution can be determined. Terminology of Simplex Method: 

Algorithm: A formalised systematic procedure for solving problem.

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Simplex Tableau: A table used to keep track of the calculations made b iteration of the simplex procedure and to provide basis for tableau revision.

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Basis: The set of basic variables which are not restricted to zero in the basic solution and are listed in solution column.

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The basic variables: The variables with non-zero positive values make up the basis are called basic variables and the remaining variables are called non-basic variables.

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Iteration: A sequence of steps taken in moving from one basic. The solution to another basic feasible solution.

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Cj row: A row in the simplex tableau which contains the co-efficients variables in the objective function.

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Zj row: A row in the simplex tableau whose elements represent the decrease (increase) of the value of the objective function if one unit of the jth variable is brought into the solution.

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Cj - Zj or

j row: A row whose elements represent net per unit contribution of the

jth variable in the objective function, if the variable is brought into the new basic solution. Positive value of

j therefore indicates gain and negative value indicates

loss in the total value Z obtained of the objective function.

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Key or pivot column: The column with the largest positive

j and it indicates which

variable will enter the next solution in a maximization case.

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Key or pivot row: The row with the smallest positive value of the, replacement ratio 0 of the constraint rows. The replacement ratio is obtained by dividing elements in the solution column by the corresponding elements in the key column. The key row indicates the variable that will leave the basis to make room for new entering variable.

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Key (pivot) element: The element at the intersection of key row and key column.

In addition to these terms in a simplex tableau we have the follow terms which are necessary to make a linear programming problem fit to be solved by simplex method. 

Slack variable: A variable used to convert a less than or equal constraint () into equality constraint. It is added to the left hand side of the constraint.

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Surplus variable: A variable used to convert a greater than or equal to (?) constraint into equality constraint. It is subtracted from the left hand side of the constraint.

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Artificial variable: It is a variable added to greater than or type () constraint. This is in addition to surplus variables used.

These variables are used to obtain an initial feasible solution in simplex method. These variables are reduced to zero at optimality.

SOME TECHNICAL ISSUES: In the earlier chapter, we considered some special problems encountered in solving LPP using graphical method. Here we discuss, how the presence of these problems - namely, infeasibility, unboundedness, multiple solutions, degeneracy, is indicated in a simplex tableau. 

Infeasibility: A solution is called feasible if it satisfies all the constraints and the nonnegativity conditions. Sometimes it is possible that the constraints may be inconsistent so that there is no feasible solution to the problem. Such a situation is called infeasibility. In a graphical solution, the infeasibility is evident when there is no feasible region in which all the constraints can be satisfied simultaneously. However, problem involving more than two variables cannot be easily graphed and it may not be immediately known that the problem is infeasible, when the model is constructed. The simplex method provides information as to where the infeasibility lies. If the simplex algorithm terminates with one or more artificial variables at a positive value, then there is no feasible solution to the original problem.

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Unboundedness: It occurs when there are no constraints on the solution. So that one or more of the decision variables can be increased indefinitely without violating any of the restrictions. Graphically the objective function line can be moved in the desired direction over the feasible region, without any limits. How do we recognize unboundedness in a simplex method? We know the replacement ratio determines the leaving variable in a simplex tableau. Now if there are no non-negative ratios (i.e. ratios are negative) or they are equal to

i.e. of

the type say 60/0, then we have unbounded solution.

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Alternative Optima: (Multiple optimum solution) A solution to a linear programming problem may or may not be unique. This is indicated in a graphical solution by the slope of the line of the objective function which may coincide with the slope of one of the constraints. In case of simplex method, Whenever a non basic variable (i.e. a variable which is not in the solution ) has a zero value in the

j (i.e. cj j – zj j) row of an optimal

tableau then bringing that variable into the solution will produce a solution which is also optimal.( Alternative Optimal solution )

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Degeneracy: It occurs when one or more of the basis variables assume zero value. In conditions of degeneracy, the solution would contain a smaller number of non – zero variables than the number of constraints i.e. if there are 3 constraints the number of non-zero variables in the solution is less than 3.

Some obvious examples of degeneracy occur if: a) One or more basic variable have a zero value in the optimal solution. b) There is a tie in the replacement ratios for determining the leaving variable. The next tableau gives the degenerate solution. c) When algorithm pivots in a degenerate row, the objective function value in the next tableau does not change i.e. there is the problem of cycling. - The system moves along the same route and the cycle would be repeated forever. There are sophisticated rules to handle the problem of cycling; however, they are outside the scope of this book. It is also observed that real life problems rarely cycle.

DUALITY IN LINEAR PROGRAMMING: Corresponding to every linear programming problem, there is another linear programming problem. The given problem is called the primal and the other its dual. Although the idea of duality is essentially mathematical, it has important interpretations.

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This can help managers in answering questions about alternative courses of action and their effect on values of the objective function. When the primal problem is of the maximisation type the dual is of the minimisation type and vice versa. It is an interesting feature of the simplex method that we can use it to solve either the original problem — the primal — or the dual. Whichever problem we start out to solve, it will also give us the solution to the other problem. Consider the following general linear programming problem. The primal Maximise Z = c1x1 + c2x2 Subject to a11x1 + a12x2 or = 0 We can write this problem in a short form as Maximise Z = cx Subject to

ax < or = b x> or = 0

The dual corresponding to this problem is Minimise Z* = b’y Subject to a’y > or = c’ y> or = 0 where b’ =transpose of matrix b a’ = transpose of matrix a c’ = column matrix of coefficients of objective function. Y’ = matrix of dual variables. Or Minimise Z* = b1y1 +b2y2 Subject to a11y1 +a21y2 < or = c1 a12y1 +a22y2 < or = c2 y1,y2> or = 0 OR- MMS

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The dual problem is constructed from the primal as follows: (1) Each constraint in primal problem will have a corresponding variable (dual variable y) in dual problem. (2) The elements of the right hand side of the constraints in the primal are equal to the respective co efficient of the variables in objective function in the dual [i.e. y1 will have a coefficient b1 in the objective function etc.]. (3)If the primal problem is that of maximisation the dual problem is of minimisation. (4) The maximisation problem has (< or =) type constraints and the minimisation problem has (> or =) type constraints. If the constraints in the primal are mixed type, they are converted into constraints of the same type before formulating the dual. (5) The coefficient matrix of the dual is the transpose of the coefficient matrix of the primal. (6) The variables in both the problems are non-negative. The primal has 5 constraints and 3 variable hence dual will have 5 variables and 3 constraints. The Dual is Minimise Z* = 4y1 + l2y2- 2y3 + 8y4 – 8y5 Subject to: y1 +2y2- y3+3y4 -3y5 > or = 8 4y2-y3+2y4-2y5 > or = 10 3y1 --y3--y4.+y5 > or = 5 y1, y2, y3, y4, y5> or = 0 Dual and the optimum simplex tableau: In the formulation of dual from a primal, we notice that dual and primal are structurally related. This helps us in obtaining information about the dual variables from the solution to the problem from the simplex algorithm. If we look at the optimal table of the simplex algorithm of the primal, the ∆j values corresponding to the slack variables give the optimum value of the dual variables. The optimal values of the structural variables in either problem are always equal to the values of the shadow prices of the corresponding constraints in the optimal solution of the other (dual) problem. OR- MMS

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The value of the objective function in both the cases remains the same if a feasible solution exists for both the primal and the dual. It is therefore not necessary to solve both the primal and its dual. Another relation between the primal and the dual is, if one problem has an unbounded solution its dual has no feasible solution.

ECONOMIC INTERPRETATION OF DUAL: We noted that the primal and the dual are related mathematically, we can now show that they are also related in economic sense. Consider the economic interpretation of the dual — first for a maximisation problem and then for a minimisation problem. Example 3.6: The maximisation problem: Consider the following problem. The optimal solution to this problem dives production of 18 units of Xi and 8 units of x2 per week. It yields the maximum prof of a Rs. 1000, Maximise Z = 40x1 + 35x2 Subject to 2x1 + 3X2 < or = 60

Raw materials constraint per week.

4x1 + 3X2 < or = 96

Capacity constraint per week.

x1,x2 > or = 0 The optimal solution to this problem gives production of 18 units of x 1 and 8 units of x2 per week. It yields the maximum profit of a Rs. 1000. Now, to rent the facilities of the firm for one week, the firm has 60 kg of raw material and 96 capacity hours. If we let yl represent the rent per kg of raw material and y2 the rent per capacity hour, the firm would receive a total rent equal to 6Oy1 + 96y2. We shall compute the minimum value of the rent so that the firm will know what minimum offer shall be economically acceptable to it. The lower limit can be set up after keeping in mind that the alternative to renting must be at least as favourable as using the capacity itself. The rent of the resources should be at least equal to the earnings from producing products x1 and x2. We know that production of one unit of x1 requires 2 kg of raw material and 4 capacity hours. Thus, the total rent for these amounts of resources should be greater than, or equal to, the profit obtainable from one unit of the product, i.e. Rs. 40. Hence 2y1+4y2 > or = 40 OR- MMS

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similarly, the resources consumed in producing one unit of product x2 are 3 kg of raw material and 3 capacity hours. The total rent of these resources should equal to atleast Rs. 35, the unit profit of product x2. i.e., 3y1 + 3Y2 > or = 35 Besides, the rent cannot be negative. Therefore, y1 and y2 should both be non negative. In complete form, the problem can be expressed as: Minimise: Z* = 60y1 + 96y2 Subject to: 2y1 + 4y2 > or = 40 3y1 + 3y2 > or = 35 y1, y2 > or = 0 This problem is absolutely the same as the dual to the given problem. These rates y 1 and y2 are obtainable from the solution of the dual as y 1 = 10/3 and y2 = 25/3. Also, these values can be obtained from ∆j row of simplex tableau showing optimal solution to the primal problem. As we have seen, values of the objective function of the primal and the dual are identical. Naturally, the minimum total rent acceptable to the firm is equal to the maximum profit that it can earn by producing the output itself using the given resources. The individual rents of y1 and y2, are called the shadow prices or imputed prices. They indicate the worth of the resources. These prices, of the two resources, materials and capacity hours, are imputed from the profit obtained front utilizing their services, and are not derived from the from the original cost of these resources. Now we know that each unit of product x1 contributes Rs. 40 to the profit. The imputed price of material and capacity is respectively, Rs. 10/3 per kg and 25/3 per hour, we can find the total imputed cost of the resources used in making a unit of the product as 2kg at Rs. 10/3 per kg i.e., 2 X 10/3 = 20/3 Rs. 4 hours at Rs. 25/3 per kg. = 4 X 25/3 = 100/3 Rs. The total cost is 120/3 = 40 Rs. Thus, the total imputed cost of producing one unit of product x1 equals the per unit profit obtainable from it.

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Similarly, from each unit of product x2, the total imputed cost of resources employed would be: 3kg at Rs.10/3 per kg = 3 x lO/3 = 10 Rs. 3 hours at Rs. 25/3 per hour 3 x 25/3 = 25 Rs. The total cost is 10 + 25 = 35 Rs. This obviously equals the profit per unit of the product. This proves that the valuation of the resources is such that their total value equals the total profit obtained at the optimum level of production. The shadow prices are also called the marginal value products or marginal profitability of the resources. Thus, if there were a market for renting resources, the firm would be willing to take some materials if the price of the material were less than Rs. 10/3 per kg, and capacity hours, if the price is less than Rs. 25/3 per hour. If we denote marginal profitability of resources as MPR and the marginal profitability of capacity as MPc, respectively, the shadow prices of the two resources, we can write the dual as follows: Minimise H = 50 MPR + 96 MPc Subject to: 2MPR + 4MPc > or = 40 3MPR + 3MPc > or = 35 MPR, MPc > or = 0 Now let us consider the economic significance of the surplus variables S1 and S2 in the dual. The numerical values of these variables can be obtained from ∆j row in the optimal solution of the primal. The value of S1 in the optimal solution represents the opportunity cost of the product x1 while the value of S2 represents the opportunity cost of product x2. Production of an additional unit of x1 will give the firm a profit of Rs. 40 and, at the same time, the firm would use up resources worth 2 x 10/3 + 4 x 25/3 = Rs. 40. Thus, the net effect of producing one unit of product would be 40 - 40 = 0. Similarly, for product x2 the opportunity cost equals zero. Further if x1, x2 is a feasible solution to the primal and y1, y2 is the feasible solution to its dual then c1x1 + c2x2 < or = b1y1 + b2y2 i.e. [profit obtained is less than or equal to the

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rents to be paid. This would induce the producer to rent the resources rather than produce the goods himself. The concept of dual and shadow prices help us in determining the upper and lower bounds for changes in requirement vectors and coefficients in the objective function. Such that the feasibility of the LPP is not disturbed.

SENSITIVITY ANALYSIS: REFER CLASS NOTES

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LALA LAJPATRAI INSTITUTE OF MANAGEMENT MMS –I SEMESTER – I FREQUENCY DISTRIBUTION Q1) Form a frequency distribution table for the following data. Marks scored by 50 students out of 10 are given below 4 7 3 0 1 6 5 4 5 5 8 6 9 4 3 1 10 8 8 7 5 3 5 8 5 4 2 2 6 2 2 7 6 5 5 6 1 4 9 7 6 0 1 2 5 4 3 6 5 6 Q2) Form a frequency distribution table for the following data by taking class intervals as 0-19, 20-39, 40-59.......... Marks scored by 50 students out of 100 are given below 37 28 65 59 54 12 29 60 63 58 59 64 38 34 58 82 79 52 8 49 23 21 38 43 54 61 18 91 89 67 71 53 57 55 25 39 44 48 59 63 36 65 71 68 56

70 61 73 45 54

Q3) Form a frequency distribution table for the following data by taking class intervals as 154-157, 157-160, 160-163............. Height (in cm.) Of 25 students from a class is given below. 158. 161. 159. 164. 159.5 0 160.

7 159.

8 158.

3 159.

166.0

3 162.

7 154.

4 166.

2 158.

163.7

6 162.

5 160.

7 151.

0 163.

161.6

3 157.

3 168.

9 159.

8 161.

162.6

8 4 9 2 Also prepare cumulative frequency of both the types, relative frequency, percentage frequency

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Q4) Write five class intervals of the weight (in kg) of width of 2.5 kg. each such that the lower boundary of the first class interval is 50kg. Also give the upper class boundary of the fourth class interval. Q5) Construct a frequency distribution table from the following data. Weight of 40 workers of a certain factory (in kg.) is given below. Given that log (40) = 1.6021 52.3 59.5 49.2 58.7 61.8 60.9 61.2 48.9 56.6 56.8 69.0 61.3 52.9 55.4 69.9 67.6 62.8 57.4 50.7 45.1 62.5 62.8 71.4 64.8 57.1 64.7 48.1 61.7 52.0 47.8 50.5 64.3 73.5 56.4 45.6 63.4 57.2 51.8 51.2 56.6 Q6) Construct a frequency distribution table from the following data. Marks scored by 50 students out of 100 are given below. Given that log(50) = 1.6990. Also prepare cumulative frequencies of both the types, relative frequency, percentage frequency 37 28 65 59 54 12 29 60 63 70 58 59 64 38 34 58 82 79 52 61 8 49 23 21 38 43 54 61 18 73 91 89 67 71 53 57 55 25 39 45 44 48 59 63 36 65 71 68 56 54 Q7) Construct a frequency distribution table from the following data by taking the variable as the number of letters in the word in the stanza below: “Heights by great men reached and kept Were not attained by sudden flight; But they while their companions slept Were toiling upwards in the night.” Q8) For the following data prepare cumulative frequencies of both the types frequencies and answer the following questions Monthly salary (in Rs.) No. Of employees 300-400 12 400-500 18 50-600 27 600-700 20 700-800 17 800-900 6 → How many employees have their salary less than 500 rupees? → How many employees have their salary greater than 600 rupees → Find the percentage of employees whose salary belongs to the group Rs. 400- Rs. 800 OR- MMS

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Q9) Form a frequency distribution table for the following AGE IN YEARS NO. OF PERSONS Less than 25 0 Less than 30 12 Less than 35 28 Less than 40 47 Less than 45 58 Less than 50 65 Less than 55 70 Q10) Form a frequency distribution table for the following Sales in Rs. No. Of shops Less than 10000 8 Less than 12000 22 Less than 14000 43 Less than 16000 60 Less than 18000 72 Less than 20000 80 Assume that no shop has registered a sale of less than 8000 rupees. Also find relative and percentage frequencies. Q11) The following paired data (x, y) represents the number of family members in a family (x) and monthly income in rupees (y) for 25 families staying in a building. (6, 1325) (2,840) (4, 1280) (4, 960) (3, 1380) (5, 1225) (6, 1125) (6,950) (2, 1220) (4, 1150) (6, 1375) (3, 1250) (6, 1300) (6, 1275) (3, 1050) (7, 1175) (2, 1100) (6, 1050) (5, 1025) (7, 980) (3, 1310) (6, 1075) (6, 900) (5, 840) (7, 1253). Taking the class intervals as 2-4, 5-7 for x and 800-1000, 1000-12000 and 1200- 1400 for y prepare bivariate frequency table for the following data. Also prepare 1) marginal frequency distribution for x and y 2) conditional frequency distribution table for x when income lies in the group of 1000 and above 3) and conditional frequency distribution table for y when numberof Family members is 2-4 Q12) the following data gives points scored in a tennis match by the two players as (x, y) at the end of 20 games. By taking class intervals as 5-9, 10-14, etc for both x and y construct: → Bivariate frequency table → Marginal frequency table for x and y → Conditional frequency table for y given x ≥ 15 x y x y x y x y x y 10 12 12 11 7 9 15 19 17 21 14 14 22 18 16 14 15 16 22 20 OR- MMS

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11 12

11 8

12 19

18 15

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10 10

10 10

5 16

13 10

11 7

7 18

Q13) Two dice are thrown at random obtain the frequency distribution of the sum of the numbers which appear on them. Q14) The following is the disribution of age of 100 employees of a company PQR prepare cumulative frq. Of less than type and answer the following questions AGE (in yrs) 20-25 25-30 30-35 35-40 40-45 45-50 50-55 55-60 NO. OF 5 9 13 28 20 12 10 3 EMPLOYES 1)How many employees are atleast 30 years old? 2) What percentage of employees are 20-45 years of age? 3) What proportion of employees arw younger than Mr. Forty who is forty years of age? 4)What is the age limit below which we have the youngest 27% of the employees? 5)If an additional increment is to be given to all those and only those who have not crossed 35 years mark, how many employees wil not be given an additional increment? 6) If 60 is the retirement age, how many of these employees wil retire during next 10 years provided they continue to be in the service till retirement? 7) what range of 15 years has the largest num,ber of employees?

OR- MMS

LINEAR PROGRAMMING PROBLEMS

PROF. VINAY PANDIT

OR- MMS

LINEAR PROGRAMMING PROBLEMS

PROF. VINAY PANDIT

OR- MMS