{"id":12877,"date":"2019-12-18T15:21:45","date_gmt":"2019-12-18T09:51:45","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=12877"},"modified":"2021-07-02T15:35:57","modified_gmt":"2021-07-02T10:05:57","slug":"2nd-puc-statistics-question-bank-chapter-8","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-statistics-question-bank-chapter-8\/","title":{"rendered":"2nd PUC Statistics Question Bank Chapter 8 Operation Research"},"content":{"rendered":"

You can Download Chapter 8 Operation Research Questions and Answers, Notes, 2nd PUC Statistics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Statistics Question Bank Chapter 8 Operation Research<\/h2>\n

2nd PUC Operation Research One and Two Marks Questions and Answers<\/h3>\n

Question 1.
\nDefine Operations research
\nAnswer:
\n“It is the method of application of scientific methods, techniques and tools to problems involving the operations of a system, so as to provide those in control of the system with Optimum solutions to the problem.”<\/p>\n

Question 2.
\nMention a method of solving Linear Programmings Problems
\nAnswer:
\nGraphical method.<\/p>\n

Question 3.
\nDefine L.P.P
\nAnswer:
\nLinear programming is a mathematical technique, which deals with the optimization (Maximization or Minimization) of activities subject to the available resources.<\/p>\n

Question 4.
\nMention two areas where the economic operations research can be applied\/Uses of Oprations of Research
\nAnswer:
\nLinear programming problems are used in Industries for effective use of men, machine, money and Material. Game theory is applied in conflicting situations, to suggest the best strategy. Replacement Problems helps in suggesting the best time period to replace a specific part in manufacturing process, and Inventory problems help in organizing stocks and maintaining costs, and so on.<\/p>\n

Question 5.
\nDefine a solution for a LPP?
\nAnswer:
\nA set of real values X = (x1<\/sub> x2<\/sub>, x3<\/sub>, …xn<\/sub>), which satisfies the constraints AX(\u2264 = \u2265)b, is called a Solution.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWhat is a feasible solution in a L.P.P?
\nAnswer:
\nA set of real values X = (x1<\/sub>,\u00a0x2<\/sub>, x3<\/sub>, ……. xn<\/sub>), which satisfies-<\/p>\n

    \n
  • The constraints AX(\u2264 = \u2265)b and<\/li>\n
  • The non-negativity restriction x > 0<\/li>\n<\/ul>\n

    Question 7.
    \nDefine optimal solution in an L.P.P.
    \nAnswer:
    \nA set of real values X = (x1<\/sub>, x2<\/sub>, x3<\/sub>, …xn<\/sub>), which satisfies,<\/p>\n

      \n
    • The constraints AX(\u2264 = \u2265)b<\/li>\n
    • The non-negativity restriction x \u2265 0 and<\/li>\n
    • Optimizes the objective function Z = CX is called optimal Solution
      \nOR If a feasible which optimizes the objective solution Z = CX is called optimal solution<\/li>\n<\/ul>\n

      Question 8.
      \nDefine Multiple Optimal solution in LPP.
      \nAnswer:
      \nAn LPP is said to have Multiple Optimal Solution if has more one Optimal Solutions.<\/p>\n

      Question 9.
      \nWhen unbounded solution exist for a given L.P.P?
      \nAnswer:
      \nFor some LPP the optimum value of Z may be infinity, then LPP is said to have unbounded solution.<\/p>\n

      Question 10.
      \nWhen an L.P.P is said to have no solution?
      \nAnswer:
      \nIn a LPP if all the constraints cannot be satisfied simultaneously then there exists no solution<\/p>\n

      Question 11.
      \nWhy the feasible solution can exists only on xy-plane \/ Why the solution to the LPP lies in the first Quadrant?
      \nAnswer:
      \nFor the given LPP, the solution can exists only on xy-plane, because of the non-negativity restrictions, x and y both are positive on xy-plane<\/p>\n

      Question 12.
      \nIn an L.P.P objective functtin is Max Z = 20x + 70y and if solution are A(0,8) and B(12,5). Find optimal solution.
      \nAnswer:
      \nZA<\/sub> = 20(0) + 70(8) = 560; and ZB<\/sub> = 20(12) + 70(5) = 590 (max)
      \n\u2234 Optimal soplution is at B(12,5); is x = 12 and y = 5<\/p>\n

      Question 13.
      \nIs an optimal solution always a feasible solution? (In L.P.P)
      \nAnswer:
      \nYes<\/p>\n

      Question 14.
      \nDefine\/ what is a game?
      \nAnswer:
      \n\u2019\u2019Whenever there is a situation of conflict and competition between two or more opposing teams, We refer to the situation as a game”<\/p>\n

      Question 15.
      \nWrite any two properties of a Game\/Rectangular Game
      \nAnswer:<\/p>\n

        \n
      1. there are finite numbers of players<\/li>\n
      2. Each player have finite number of courses of action OR The game is said to be played, one of the player should adopt one of course action while playing the game.<\/li>\n<\/ol>\n

        Question 16.
        \nDefine\/Explain \/What is meant by a rectangular game\/2-person zero-sum game?.
        \nAnswer:
        \nIn a game of two players, the gain of one player is the loss of the other.<\/p>\n

        Question 17.
        \nDefine \u2019strategy\u2019 in a game.
        \nAnswer:
        \nThe strategy of player is the predetermined rule by which he chooses his course of action while playing the game<\/p>\n

        Question 18.
        \nDefine the term pure strategy
        \nAnswer:
        \nWhile playing a game, the pure strategy of a player is the predetermined decision to adopt a specified course of action irrespective of the course of action of the opponent<\/p>\n

        Question 19.
        \nWhat do you mean by a mixed strategy in 2-person-zero-sum game?
        \nAnswer:
        \nWhile playing a game, mixed strategy of a player is his pre-decision to choose his course of action according to certain pre-assigned probabilities<\/p>\n

        Question 20.
        \nExplain the maximin and Minimax of the game
        \nAnswer:<\/p>\n

          \n
        • Maximin is the maximum of row minimums of the payoff matrix, (\u03b1)<\/li>\n
        • Minimax is the minimum of the column maximums of the payoff matrix (\u03b2)<\/li>\n<\/ul>\n

          \"KSEEB<\/p>\n

          Question 21.
          \nWhat is a saddle point?
          \nAnswer:
          \nSaddle point is the position where Maximin and Minimax coincide<\/p>\n

          Question 22.
          \nWhat is the Principle of Dominance?
          \nAnswer:
          \nIf the strategy of a player dominates over the other strategy in all conditions, then the latter Strategy can be ignored, because it cannot affect the solution in any way.<\/p>\n

          Question 23.
          \nIn a two person zero-sum game, the value \u03bd = -1, is the game fair?
          \nAnswer:
          \nThe game is not fair, (the game is said to be fair if \u03bd = 0)<\/p>\n

          Question 24.
          \nWhat is meant by pay-off in a rectangular game?
          \nAnswer:
          \nIt is the gain or loss of the players while playing a game.<\/p>\n

          Question 25.
          \nIn a game if \u03bd = 2 then what are the values of \u03b1 & \u03b2?
          \nAnswer:
          \nHere \u03b1 – Maximin and \u03b2 – Minimax are equal and are equal to \u03bd = 2.<\/p>\n

          Question 26.
          \nIn a rectangular game, the gain of player – A is 5. Then what is the Gain\/Loss of the other player – B?
          \nAnswer:
          \nLoss = – 5<\/p>\n

          Question 27.
          \nCalculate E.O.Q Given D\/R = 5000 units\/month, C1<\/sub> = Rs.lO\/month, C3<\/sub> = Rs.200\/ month.
          \nAnswer:
          \n\"2nd<\/p>\n

          Question 28.
          \nIn a rectangular game pay-off matrix of player A is:<\/p>\n

          \"2nd
          \nWrite down the pay-off matrix of player B
          \nAnswer:
          \nThe pay-off matrix of player B is:
          \n\"2nd<\/p>\n

          Question 29.
          \nFor the following pay-off matrix of A. Write down the pay-off matrix of B
          \n\"2nd
          \nAnswer:
          \nThe pay-off matrix of player B is:
          \n\"2nd<\/p>\n

          Question 30.
          \nWhat do you mean by a replacement problem?
          \nAnswer:
          \nReplacement problem is the problem of deciding the age at which old equipments (machine and their spare parts, trucks etc.) are replaced.<\/p>\n

          Question 31.
          \nGive any two needs\/uses of replacement of equipment which deteriorate with age
          \nAnswer:
          \nNew equipments are more efficient. Maintenance\/Operating cost and production cost is less than old ones.<\/p>\n

          Question 32.
          \nWhen the equipment, which deteriorates with age, is replaced? \/What is the principle adopted for the replacement of equipment which deteriorates with age?
          \nAnswer:
          \nWhen its Annual average cost: A(n) is minimum, at that age the equipment which deteriorates with age should be replaced.<\/p>\n

          Question 33.
          \nWhen do you suggest the optimal age for replacement of equipment?
          \nAnswer:
          \nWhen its Annual average cost A(n) ceases to decrease arid starts to increase, that is point of time is to replace. OR Where Annual average cost A(n) is least we suggest to replace the old machine by new one.<\/p>\n

          Question 34.
          \nSpecify the needs for replacement of capital equipment
          \nAnswer:<\/p>\n

            \n
          • As the equipment grow older, its maintenance cost would be costlier.<\/li>\n
          • Production cost would be lesser in new equipments.<\/li>\n<\/ul>\n

            Question 35.
            \nIf the depreciation cost and the cumulative maintenance cost for equipment for the second year are Rs.10000 and Rs.10200 respectively. What is the annual average cost?
            \nAnswer:
            \n\"2nd<\/p>\n

            \"KSEEB<\/p>\n

            Question 36.
            \nWhat do you mean by Inventory?
            \nAnswer:
            \nAn Inventory is a physical stock of goods, which is held for the purpose of future production or sales.<\/p>\n

            Question 37.
            \nWhat is an Inventory Model?
            \nAnswer:
            \nA set of mathematical equations required to solve the inventory problems is called an Inventory model.<\/p>\n

            Question 38.
            \nWhat is EOQ\/E.L.S?
            \nAnswer:
            \nIt is the size\/lot of the order in which the ‘set up cost\u2019 and ‘holding cost’ of the Inventory is minimum.<\/p>\n

            Question 39.
            \nMention any two essentials\/needs\/advantages of an Inventory.
            \nAnswer:<\/p>\n

              \n
            • It helps in smooth and efficient running of the business.<\/li>\n
            • It provides adequate and satisfactory service to the customers.<\/li>\n
            • It facilitates bulk purchase of raw materials at discounted rates.<\/li>\n<\/ul>\n

              Question 40.
              \nGive any two disadvantages of the Inventory?
              \nAnswer:
              \nThe Disadvantages of Inventory are\u00a0 warehouse rent, labour of maintenance, loss due to fall in price, deterioration of goods should be incurred<\/p>\n

              Question 41.
              \nExplain
              \n1. Lead time-t
              \n2. Holding cost-C1<\/sub>
              \n3. Set up cost-C3<\/sub>
              \n4. shortage cost-C2<\/sub>
              \nAnswer:
              \nLead time – t: The time gap between placing of order and arrival of goods at the inventory is the Lead time<\/p>\n

              Holding cost – C1<\/sub>: It is the cost associated with carrying or holding the goods in stock until the goods are sold or used. It includes rent for space, interest on capital, maintenance of records, taxes, insurance etc.<\/p>\n

              Setup cost – C3<\/sub>: It is the cost of setting up the machines for production or the cost of 4 placing the order for the goods. It includes labour cost, transportation cost etc. It is also called as Ordering cost.<\/p>\n

              Shortage cost – C2<\/sub>: The cost associated with delay or inability to meet the demand, because ^ of shortage of Stock is called shortage cost Or penalty cost.<\/p>\n

              Question 42.
              \nWhen an L.P.P. is said to have no solution?
              \nAnswer:
              \nIn an L.P.P. if all the constraints cannot be satisfied simultaneously then there exists no solution.<\/p>\n

              Question 43.
              \nComment: The graphical solution to the L.P.P. lies in the I Quadrant.
              \nAnswer:
              \nDue to the non-negative restriction X \u2265 0 which satisfy that feasible solution can lie only in the I Quadrant.<\/p>\n

              Question 44.
              \nIn a Two-person-zero-sum game, the value of the game V = -1 is it a fair game?
              \nAnswer:
              \nThe game is said to be fair if V = 0. Here V = -1.
              \n\u2234 It is not a fair game.<\/p>\n

              Question 45.
              \nWhat is meant by value of a game?
              \nAnswer:
              \nThe payoff at the saddle point is known as value of the game denoted by ‘V’.<\/p>\n

              \"KSEEB<\/p>\n

              Question 46.
              \nWhen is a game said to be fair?
              \nAnswer:
              \nWhen value of the game V = 0.<\/p>\n

              Question 47.
              \nWhat is principle of dominance?
              \nAnswer:
              \nIf the strategy of a player dominates over the other strategy in all conditions, then the latter strategy can be ignored.<\/p>\n

              Question 48.
              \nWhat is meant by pay-off in a rectangular game?
              \nAnswer:
              \nThe gain or loss between the players while playing a game is called pay off.<\/p>\n

              Question 49.
              \nWrite down the formula for calculating Annual average cost of an item\/equipment.
              \nAnswer:
              \nThe annual average cost : A(n) = \\(\\frac { T }{ n }\\)
              \nTotal cost T = (P – Sn<\/sub>) + \u03a3 Ci.
              \nWhere P – Purchase cost
              \nSn<\/sub> – Resale\/ {Scrap Value}
              \nCi<\/sub> – Maintenance cost.<\/p>\n

              Question 50.
              \nWhat are the uses of replacement of equipment which deteriorate with age?
              \nAnswer:
              \nNew equipments are more efficient. With adoption of new technology, maintenance cost (Operating cost) and production cost would be less than the old ones.<\/p>\n

              Question 51.
              \nIf in a game the -saddle point\u2019 is 5, what is the value of \u2018mini-max\u2019?
              \nAnswer:
              \nValue of \u2018mini-max\u2019 is \u20185\u2019 (because saddle point is the position where maximin and minimax coincide i.e. \u03b1 = \u03b2 = \u03bd.<\/p>\n

              Question 52.
              \nMention two areas where the economic operations research can be applied.
              \nAnswer:
              \nOperations Research can be applied in L.P.P as production management, marketing management, replacement problems helps in suggesting the best time to replace a machine etc.<\/p>\n

              Question 53.
              \nWrite down the L.P.P. model.
              \nAnswer:
              \nOptimize Z = CX
              \nSubject to constraints AX (\u2264 = \u2265) b.
              \nand non-negativity restriction X \u2265 0.<\/p>\n

              Question 54.
              \nSolve the game.
              \n\"2nd
              \nAnswer:
              \nBy maximin – minimax principle.
              \n(i) Row minimums are circled. (ii) Column maximums are boxed.
              \nHere the payoff 5 is circled as well as boxed.
              \nThe saddle point occurs at (1, 1)
              \nThe suggested solution is pure strategy
              \ni.e., For player A – A1<\/sub>
              \nFor player B – B1<\/sub>,
              \nvalue of the game V = 5.<\/p>\n

              Question 55.
              \nSolve the following game Player A<\/p>\n

              \"2nd
              \nAnswer:
              \nUsing principle of dominance.
              \nAll elements (payoffs) of A1<\/sub> row are more than A2<\/sub>
              \ni.e. A1<\/sub> dominates A2<\/sub> delete A2<\/sub>
              \n\"2nd
              \n6 is less than 8, i.e., B1<\/sub> dominates B2<\/sub> delete B2<\/sub>.
              \nValue of the game V = 6.
              \nThe suggested strategy is pure strategy.
              \nFor player A – A1<\/sub> and for B – B1<\/sub> OR
              \nSolve using maximin – minimax priniple.<\/p>\n

              \"KSEEB<\/p>\n

              Question 56.
              \nIn a two-person-zero-sum game the pay off matrix of player A is given below.
              \n\"2nd
              \nWrite the payoff matrix of player A.
              \nAnswer:
              \nHere Row elements are written as columns elements by changing signs, with respective strategies and players.
              \n\u2234 The pay off matrix of player A is:
              \n\"2nd<\/p>\n

              Question 57.
              \nIn a rectangular game the pay off matrix of player A is –
              \n\"2nd
              \nWrite down the pat off matrix of player B is –
              \nAnswer:
              \nHere row elements are written as column elements by changing signs, with response strategies and players. The pay off matrix of player B is –
              \nPlayer A
              \n\"2nd<\/p>\n

              Question 58.
              \nWhat is meant by equipment which deteriorates with age? Give an example.
              \nAnswer:
              \nAs equipment become older, they deteriorate i.e., they become less efficient, and their maintenance cost increases. Ex : Motor Bike.<\/p>\n

              Question 59.
              \nThe cumulative maintenance cost of a machine during fourth year is Rs.13,400. Its purchase cost is Rs. 8,500. Find the average annual cost assuming that the machine has no resale value.
              \nAnswer:
              \nAverage annual cost : A(n) = \\(\\frac { T }{ n }\\)
              \nHere T = (P – Sn<\/sub> + \u03a3Ci
              \n\\(A(4)=\\frac{8500+13400}{4}=R s .5475\\)<\/p>\n

              Question 60.
              \nWhat is the need of Inventory in a business concern?
              \nAnswer:
              \nAn inventory is essential for smooth and efficient running of the business. By bulk purchasing of goods required at discounted rates, adequate and satisfactory service can be given to the customers and can earn more profit.<\/p>\n

              Question 61.
              \nWhat are the disadvantages of the inventory?
              \nAnswer:
              \nThe inventory has disadvantages such as ware house rent, labour of maintenance, loss due to fall in price, deterioration of goods, etc., should be incurred.<\/p>\n

              Question 62.
              \nWrite down the formulae for the following in Inventory model I (without shortages) (a) EOQ (b) Minimum Average Cost.
              \nAnswer:
              \n(a) EOQ \/ ELS = Q\u00b0 = \\(\\sqrt{\\frac{2 \\mathrm{C}_{3} \\mathrm{R}}{\\mathrm{C}_{1}}}\\)
              \n(b) Minimum Average cost = \\(\\sqrt{2 \\mathrm{C}_{1} \\mathrm{C}_{3} \\mathrm{R}}\\)<\/p>\n

              Question 63.
              \nWrite down the formulae for the following inventory model II (with shortage cost)
              \n(a) EOQ
              \n(b) Minimum average cost
              \nAnswer:
              \n\"2nd<\/p>\n

              Question 64.
              \nWhat are optimal lot size and maximum inventory level in the Inventory model II.
              \nAnswer:
              \nELS\/EOQ : Q\u00b0= \\(\\sqrt{\\frac{2 \\mathrm{C}_{3} \\mathrm{R}}{\\mathrm{C}_{1}}} \\sqrt{\\frac{\\mathrm{C}_{1}+\\mathrm{C}_{2}}{\\mathrm{C}_{2}}}\\) and Maximum inventory level : S\u00b0 = \\(\\frac{\\mathrm{c}_{2} \\mathrm{Q}^{0}}{\\mathrm{c}_{1}+\\mathrm{C}_{2}}\\)<\/p>\n

              Question 65.
              \nIf R = 7000\/year, C1<\/sub> = Rs. 2 per unit\/year and C3<\/sub> = Rs. 75. Find the optimum lot size and optimum number of orders.
              \nAnswer:
              \nSince shortages are not given use Model II .
              \nOptimum lot size are;
              \n\\(Q^{0}=\\sqrt{\\frac{2 C_{3} R}{C_{1}}}=\\sqrt{\\frac{2 \\times 75 \\times 7000}{2}}\\)
              \nQ\u00b0 = 724.57 \u2248 725 units.
              \nOptimum no of orders : \\(\\mathrm{n}^{0}=\\frac{\\mathrm{R}}{\\mathrm{Q}^{0}}=\\frac{7000}{725}\\) \u2234 n\u00b0 = 9.67 \u2248 10 orders<\/p>\n

              \"KSEEB<\/p>\n

              Question 66.
              \nFind the EOQ and minimum average cost for the following data.
              \nAnnual demand = 1000 units.
              \nCost of an unit = Rs. 20
              \nOrdering cost = Rs. 2 per order
              \nHolding cost = 15% per year
              \nAnswer:
              \nGiven: R= 1000, P = 20, C3<\/sub> = 2, I = 15%
              \n\u2234 C1<\/sub> = PI = 20 \u00d7 15% = 3.
              \nEOQ : \\(Q^{0}=\\sqrt{\\frac{2 C_{3} R}{C_{1}}}=\\sqrt{\\frac{2 \\times 2 \\times 1000}{3}}=36.51\\)
              \n\u2234 Q\u00b0 = 37 units
              \nMinimum average cost C(Q\u00b0) = \\(\\sqrt{2 \\mathrm{C}_{1} \\mathrm{C}_{3} \\mathrm{R}}=\\sqrt{2 \\times 3 \\times 2 \\times 1000}=109.54 \\simeq 110\\)<\/p>\n

              Question 67.
              \nFind EOQ for the data given below.
              \nR = 100 units per day
              \nC3<\/sub> = Rs. 50 per order
              \nA1<\/sub> = Re. 1 per unit\/day
              \nC2<\/sub> = Rs. 10 per unit\/day.
              \nAnswer:
              \nHere shortage cost (C2<\/sub>) is given, use EOQ model II and all information is given per day.
              \n\"2nd
              \n\u2234 Q\u00b0= 104.9 = 105 units.<\/p>\n

              Question 68.
              \nFind minimum average inventory cost for the following data :
              \nR = Rs. 150 per month
              \nC1<\/sub> = Rs. 1.50 per unit per month
              \nC2<\/sub> = Rs. 5 per unit per month
              \nC3<\/sub> = Rs. 25 per order.
              \nAnswer:
              \nMinimum average inventory cost is
              \n\"2nd<\/p>\n

              Question 69.
              \nGiven :
              \nSet up cost = Rs. 50 \/ cycle
              \nStorage cost = Rs. 3 \/ item\/ year
              \nDemand = 5000 items\/year
              \nCalculate \u2018minimum average inventory cost\u2019.
              \nAnswer:
              \nGiven R = 5000 items \/ year
              \nC1<\/sub> = Rs. 3 \/ item \/ year
              \nC3<\/sub> = Rs. 50\/ cycle
              \nMinimum average inventory cost is :<\/p>\n

              \"2nd
              \nQuestion 70.
              \nThe objective function and two solutions of an L.P.P. are : maxz = 200x + 100y and A (0, 5), B(10, 7). Find the optimal value of z.
              \nAnswer:
              \nFor A (0,5): Max z =200(0)+ 100(5)
              \nz =500
              \nFor B(10, 7): Max z =200(10)+100(7)
              \nz =2700
              \nThe optimal value of z =2700.<\/p>\n

              Question 71.
              \nIf the depreciation cost and the cumulative maintenance cost for an equipment for the third year are Rs. 6,000 and Rs. 6,200 respectively, what is \u2018Annual Average Cost\u2019?
              \nAnswer:
              \nAnnual Average Cost
              \n\"2nd<\/p>\n

              2nd PUC Operation Research Five Marks Questions With Answers<\/h3>\n

              Linear Programming Problems<\/strong><\/p>\n

              Question 1.
              \nA manufacturer produces 2 products A and B which needs two machines P and Q. Product A requires 6 hours on machine P and 2 hours on machine Q. Product B requires 4 hours on machine P and 4 hours on machine Q. There are 100 hours of time available on machine P and 80 hours on machine Q. Profit earned by the manufacturer on selling one unit of A is Rs.10 and on selling one unit of B is Rs.15. Formulate L.P.P. ,
              \nAnswer:
              \nLet x be the number of units of product A Sold
              \nand y be the number of units of product B sold
              \n\"2nd
              \nThen, the L.P.P is –
              \nMaximize Z = 10x + 15y
              \nSubject to 6x + 4y \u2264100
              \n2x + 4y \u2264 80
              \nand x \u2265 0, y \u2265 0.<\/p>\n

              Question 2.
              \nA firm produces two types of cloth, printed and plain. Three kinds of yarns are required to produce the cloth – red, green and blue. One unit length of printed cloth needs 2 metres of red yarn, 6 metres of blue yarn and 4 metres of green yarn. One unit length of plain cloth needs 5 metres red yarn. The firm has a stock of 8 metres of red yarn, 10 metres of green yarn and 15 metres of blue yarn. It is assumed that income obtained from one length of printed cloth is Rs. 10 and one length of plain cloth is Rs. 6. Formulate the L.P.P. so as to maximise the income obtained from the sale of finished product.
              \nAnswer:
              \nThe L.P.P. is :
              \nmaximize Z = 10x + 6y
              \nSubject to constraints : 2x + 5y \u2264 8
              \n6x \u2264 10
              \n4x \u2264 15
              \nand x \u2265 0, y \u2265 0.<\/p>\n

              \"KSEEB<\/p>\n

              Question 3.
              \nA company manufactures two products X & Y. Due to deficit of raw materials not more than TOO units of X and 200 units of Y can be produced in a month. Total labour available is 600 hrs per month. Each unit of X requires 2 hrs and each unit of Y requires 4 hrs to be produced. If the profit obtained by selling one unit of X is Rs. 50 and one unit ofY is Rs. 100, formulate an LPP.
              \nAnswer:
              \nThe linear programming problem is :
              \nMax. Z = 50x + 100y
              \nSubject to x \u2264 100
              \ny \u2264 200
              \n2x + 4y \u2264 600 and
              \nx \u2265 0, y \u2265 0.<\/p>\n

              Question 4.
              \nSolve the following L.P.P graphically.
              \nMaximize Z = lOx + 5y
              \nSubject to 2x + 4y \u2265 20.
              \nx + 2y \u2264 30
              \nand x, y \u2264 0.
              \nAnswer:
              \nThe constraints should be represented graphically.
              \nWrite each constraints as equalities and are represented by straight lines by finding two points on each of these lines.
              \nConsider,
              \n\"2nd
              \nThe equations are graphically represented as below :
              \n\"2nd
              \nThe common shaded area is the feasible region. The corner points of the feasible region A(0,5), B(0,15) C(30,0) and D(10,0)
              \nThe Z values at these corners points are tabulated as below:
              \n\"2nd
              \nSince ‘Z’ is to maximize and which maximizes at these corner points C(30,0)
              \n\u2234 The optimum solution is x = 30, y = 0
              \nThe optimum (maximum) value of Z is 300.<\/p>\n

              Question 5.
              \nFor the following. L.P.P obtain the solution graphically. .
              \nMaximize Z = 50x1<\/sub> + 100 x2<\/sub>
              \nSubject to the constraints
              \n10x1<\/sub> + 5x2<\/sub> \u2264 2500
              \n4x1<\/sub> + 10x2<\/sub> \u2264 2000
              \nand x1<\/sub> \u2265 0, X2<\/sub> \u2265 0.
              \nAnswer:
              \nThe constraints should be represented graphically.
              \nWrite each constraints (inequalities) as equalities (equation) and represent these equations as straight lines by finding two points on each of these lines.
              \nConsider,
              \n10x1<\/sub> + 5x2<\/sub> = 2500 ;
              \nput x1<\/sub> = 0 ;
              \nwe get x2<\/sub> = 500,
              \nSo, the points (x, y)\/(x1<\/sub> x2<\/sub>) on the line is (0, 500)
              \n\"2nd
              \nThe equations are graphically represented as below:
              \n\"2nd
              \nThe Common shaded area is the feasible region. The corner points of the feasible region are 0(0, 0), A (0,20), B( 188, 125), C(250,0)
              \nThe Z-values at these corner points are tabulated as below:
              \n\"2nd
              \nSince Z is to maximize and which maximizes at the comer point B (188, 125)
              \n\u2234 The optimum solutions is x1<\/sub> = 188, x2<\/sub> = 125.
              \nThe optimum (maximum) value of Z is 21900.<\/p>\n

              \"KSEEB<\/p>\n

              Question 6.
              \nGraphically solve the L.P.P.
              \nMaximize Z = 2000x + 4000y
              \nSubject to x – 2y \u2265 10.
              \nx + 2y \u2264 30 ;
              \nx, y \u2265 0.
              \nAnswer:
              \nThe given constraints should be represented graphically.
              \nWrite each constraints as equalities (equations) and represent as straight lines after finding two points on each of the lines.
              \nConsider,
              \nx – 2y = 10 ; Put x = 0 ; we get y = -5, So, the points (x, y) on the line is (0, – 5)<\/p>\n

              \"2nd\"2nd
              \nEquations are graphically represented as below:
              \n(Here one of the point of y is -ve. So, consider the -ve y-axis and feasible region can occur only at x, y axis (as x and y are +ve) due to the non-negativity restriction)
              \n\"2nd
              \nThe common shaded area is the feasible region. The corner points of the feasible region as A(10,0) B(20, 6.7) and C(30, 0)
              \nThe Z – values at these 7 corner points are tabulated as below:
              \n\"2nd
              \nsince Z is to maximize and is maximizes at the corner points B(20,6.7)
              \n\u2234 The optimum solution is x = 20 y = 67
              \nThe optimum value of Z is 66800.<\/p>\n

              Question 7.
              \nSolve the following linear programming problem, graphically
              \nMax z = 10x + 3y
              \nSubject to constraints 2x + 3y \u2265 3
              \nx + 2y \u2264 4
              \nand x \u2265 0, y \u2265 0
              \nAnswer:
              \nConsider the constraints as equalities and represent by graphically as straight line.<\/p>\n

              \"2nd
              \n\"2nd
              \nFrom the graph the corner points of the feasible region are A(0,1), B(0,2), C(4,0) and D(1.5,0)
              \nThe values of z at these corner points are :
              \n\"2nd
              \nThe maximum value of z occurs at C(4,0)
              \nThe optimum solution to the L.P.P is x = 4, y = 0 ; The optimum cost = 40<\/p>\n

              Transportation Problems<\/strong><\/p>\n

              Question 1.
              \nFor a transportation problem with cost matrix as given below:
              \n\"2nd
              \nFind a solution by north west corner rule and find transportation cost.
              \nAnswer:
              \nHere the availability; ai = requirement; bj = 112, thus the T.P. is balanced. Therefore there exists a feasible solution.
              \nConsider the north west corner cell at (1, 1) allocate as; minimum (a,, b,) = min (100, 30) x11<\/sub> = 30.
              \n\"2nd
              \nHere a1<\/sub> > b1<\/sub> and replace b1<\/sub> = 100 – 30 = 70.
              \n\"2nd
              \nNext, Allocate at (1,2) as: Min (70, 50) = 50; X12<\/sub> = 50.
              \nHere again a1<\/sub> > b1<\/sub> and replace, b1<\/sub> = 70 – 50 = 20.
              \n\"2nd
              \nNext allocations meat (1,3) as; min (20, 32) = 20 and replace (32 – 20) – 12 A; So, X13<\/sub> = 20.
              \n\"2nd
              \nNext allocation is mode at (2, 3) as:
              \nHere b3<\/sub> > a1<\/sub> So, Min (12, 12) = 12 ; X23<\/sub> = 12.
              \nAll allocations are made. Thus, initial feasible solutions is:-
              \nX11<\/sub> = 30, X12<\/sub> = 50, X13<\/sub> = 20, X23<\/sub> = 12
              \nThe transportation cost is: Z = \u03a3\u03a3Cij.Xij
              \n= 3 \u00d7 30 + 7 \u00d7 50 + 4 \u00d7 20 + 2 \u00d7 12 = 90 + 350 + 80 + 24 = Rs. 544.
              \nNote: In the problem m(rows) = 2, n(Columns) = 3
              \n\u2234 m + n – 1 = 2 + 3 – 1 =4
              \nThe number of allocations = 4 is same as (m + n – 1). The initial feasible solutions a basic feasible solution.<\/p>\n

              \"KSEEB<\/p>\n

              Question 2.
              \nFind the Initial basic feasible solution by matrix minimum method for the following T.P.
              \n\"2nd
              \nAnswer:
              \nBy matrix minimum method \/ least cost entry method let allocation is made at the least cost at (2, 1) as: X21<\/sub> = Min (90, 75) = 75
              \n\"2nd<\/p>\n

              Write down the reduced matrix after deleting \u2018D1<\/sub>\u2019 – destination.
              \nReplace. b2<\/sub> = 90 by (90 – 75) = 15
              \n\"2nd
              \nThe lowest cost in the reduced cost matrix is 4 at (1, 2) \/ (D1<\/sub> D3<\/sub>).
              \nAllocate as X13<\/sub> = Min (70, 30) = 30
              \nDelete D3<\/sub> – destination and replace 70 by (70 – 30) = 40
              \n\"2nd
              \nThe lowest cost in the reduced cost matrix is 5 at (2, 2) OR (O2<\/sub>, D2<\/sub>) Allocate as X22<\/sub> – Min (15, 55) = 15 and replace 55 by (55 – 15) = 40.
              \nSo, the Final allocation is at the cell (1, 2) \/ (D1<\/sub>, D2<\/sub>):
              \nAllocate as X12<\/sub> = min (40, 40) = 40. Thus all allocations are made as below:
              \nFrom the original table.
              \n\"2nd
              \nObserve O, availabilities are distributed to D2<\/sub> and D3<\/sub> and O2<\/sub> availabilities are distributed to D1<\/sub> and D2<\/sub>, total availabilities are allocated to required destinations as : X12<\/sub> = 40, X13<\/sub> = 30, X21<\/sub> = 75, X22<\/sub> = 15 and the total transportation cost is:
              \n\"2nd
              \n\u2234 Total transportation cost is Rs. 660\/- is the basic feasible solution for the given T.P.
              \nNote: In the above problem: number of rows: m = 2; number of columus ;
              \nn = 3(m + n- l) = (2 + 3- l) = 4.
              \nNumber of allocations = 4. Both are some, the solution is degenerate.<\/p>\n

              Question 3.
              \nFor the following transportation problem, obtain the Initial basic feasible solution by:-
              \n1. N.W.C.R and
              \n2. Matrix minima method (M.M.M)
              \n\"2nd
              \nAnswer:
              \n1. By N.W.CR:- Start allocating at north corner cell at the cell (1, 1):
              \nX11<\/sub> = Min (a1<\/sub> b1<\/sub>)
              \n= min (5, 7) = 5<\/p>\n

              Here O1<\/sub> is satisfied, next allocate at (2, 1) as:
              \nX21<\/sub> = Min (a2<\/sub>, b1<\/sub> – X11<\/sub>) = Min (8, 7-5)
              \n= Min (8, 2) = 2<\/p>\n

              D1<\/sub> satisfied and next allocation is made at (2, 2):
              \nX22<\/sub> = Min (a2<\/sub> – X21<\/sub>, b2<\/sub>)
              \n= Min (8-2, 9)
              \n= Min (6, 9) = 6<\/p>\n

              O2<\/sub> availabilities are allocated, next allocate at (3, 2):
              \nX32<\/sub> = Min (a3<\/sub>, b32<\/sub> – X22<\/sub>)
              \n= Min (7, 9 – 6) = Min (7, 3) = 3<\/p>\n

              D2<\/sub> Satisfied, next allocate at (3, 3) as:
              \nX33<\/sub> = Min (a3<\/sub> – X32<\/sub>, b3<\/sub>)
              \n= Min (7-3, 18) = min (4, 18) = 4<\/p>\n

              O3<\/sub> Satisfied, Allocate at (4, 3) as:
              \nX43<\/sub> = Min (a4<\/sub>, b3<\/sub> – X33<\/sub>)
              \n= Min (14, 18, -4) = Min (14,14) = 14
              \n\"2nd
              \nThe Initial Basic feasible solution (I. B. F.S) is:
              \n\"2nd
              \nAnswer:
              \n2. Solution by Matrix -minima method: From the given T.P:
              \n\"2nd
              \nBy M.M.M \/ least OR lowest cost entry Method: The lowest cost in the cost matrix is 1 at (4, 1): The allocation is X41<\/sub> = Min (a4<\/sub>, b1<\/sub>)X41 <\/sub> = Min (14,7) = 7
              \nD1 <\/sub>is satisfied so, delete D1<\/sub> and replace (O14<\/sub>– X41<\/sub>)= 14-7 = 7.
              \n\"2nd
              \nThe lowest cost is 2 at (4,3) in the first table. Allocate X43<\/sub> = Min (7,18) = 7 and replace 18 by (18 – 7) = 11. O4<\/sub> is satisfied and delete and the reduced table is:
              \n\"2nd
              \nIn the above reduced cost matrix the least cost is 3 at (2,2) of first table. So, allocate as X22<\/sub> = Min (8, 9) = 8 and replace. (9 – 8) = 1 and delete O2<\/sub>.
              \nThere ducad cost matrix is:
              \n\"2nd
              \nThe next allocation is made at the cell (3,3) which has least cost 7.
              \nAllocate X33<\/sub> = Min (7,11) = 7. Replace 11 by (11 – 7) = 4 O3<\/sub> is satisfied, since remaing only one origin, allocate all allocation in the same cost matrix.
              \nAllocate, at next least cost 174, as X12<\/sub> = Min (5,1)= 1 and replace 5 by (5- 1) = 4 as D2<\/sub> is satisfied next allocate as X13<\/sub> = Min (4,4) = 4.
              \nAll availabilities and requirements are satisfied.
              \nThe initial basic feasible solution is:
              \n\"2nd
              \nThe total transportation cost = Rs. 236.<\/p>\n

              \"KSEEB<\/p>\n

              Game Theory<\/strong><\/p>\n

              Question 1.
              \nFind the value for the following game using maximin minimax method. Also write the best strategies for A and B. Player B
              \n\"2nd
              \nAnswer:
              \nThe minimum payoff in each row of the pay off matrix is circled.
              \nThe maximum payoff in each column is boxed.
              \n\"2nd
              \nThe pay off 7 is circled as well as boxed.
              \n\u2234 Saddle point occurs at (2, 1)
              \nThe suggested solution is (Pure strategy)
              \nThe best strategy for A is A2<\/sub> and for B is B1<\/sub> value of the game V = 7.<\/p>\n

              Question 2.
              \nSolve the following game. In the game fair?
              \n\"2nd
              \nAnswer:
              \nBy maximin – minimax principle
              \nThe minimum pay off in each row is Circled.
              \nThe maximum pay off in each column is boxed.
              \n\"2nd
              \nHere, minimum payoffs and maximum payoffs are repeated and hence both are circled and boxed.
              \nThe payoffs \u20180\u2019 is circled as well as boxed.
              \n\u2234 Saddle point occurs at (1, 3)
              \nThe suggested solution is (The pure strategy)
              \nThe best strategies are A1<\/sub>, B3<\/sub>.
              \nValue of the game V = 0
              \nThe game is fair.<\/p>\n

              Question 3.
              \nTwo persons A and B without showing each other, place a coin each on the table. If the coin happens to be of the same denomination player A will take both of them. If they happen to be of different denominations, Player B will take both of them. Suppose Player A has a few one rupee coins and two rupee coins and suppose player B has few one – rupee, two-rupee and five-rupee coins.
              \nWrite down the payoff matrix of A. Does the game have saddle point ? If so, write down the solution.
              \nAnswer:
              \nThe pay off matrix of player A is
              \n\"2nd
              \nThe minimum pay off in each row is Circled.
              \nThe maximum pay off in each column is boxed.
              \nHere row minimums are repeated. .
              \nThe pay off \u2018-1\u2019 at (1, 3) is circled as well as boxed, which is the saddle point.
              \nThe solution to the games is:
              \nThe best strategies for player A is A1<\/sub>, and for B is B3<\/sub> i.e., player A always use to play with 1 Re coin and player B with 5 Re. coin.
              \nValue of the game V = -1.<\/p>\n

              Question 4.
              \nPlayer A writes one of the numbers 1, 2 and 3 on a piece of paper and keeps the figures hidden in his paper. And without knowing which number he has written, player B writes one of the numbers 2 and 5 on the same paper. Then if the two numbers written by the two players are added. If this sum is odd, A pays Rs. 5 to for B otherwise B pays Rs. 10 to A. Write down the pay off matrix of A. Does the game have a saddle point.
              \nAnswer:
              \nThe payoffs matrix of player A is:
              \n\"2nd
              \nBy maximin-minimax principle.
              \nThe minimum payoff in each row is Circled.
              \nThe maximum payoff in each column is boxed.
              \nHere columns maximums are repeated.
              \nHere none of the payoff is circled as well as boxed.
              \n\u2234 The game has no saddle point.
              \nThe suggested strategies is mixed strategy.<\/p>\n

              Question 5.
              \nSolve the following game using dominance principle. Also find the best strategies of A and B.
              \n\"2nd
              \nAnswer:
              \nBy principle of dominance, In the pay off matrix, each pay off strategy A3<\/sub> are greater than, each pay off strategy of A2<\/sub> Therefore A3<\/sub> dominates A1<\/sub> and A1<\/sub> is deleted.
              \n\"2nd
              \nHere, each pay off (in 2nd column) of B2<\/sub> strategy are less than, each pay off of strategy B1<\/sub>.
              \n\"2nd
              \nA3<\/sub> dominate A2<\/sub>, so delete A2<\/sub>
              \n\"2nd
              \nB2<\/sub> dominates B3<\/sub> so, delete B3<\/sub>
              \n\"2nd
              \nThus, the saddle point occurs at (3, 2). The suggested solution to the game is:
              \nStrategies for players A and B are (A3<\/sub>, B2<\/sub>)
              \nValue of the game V = 16<\/p>\n

              \"KSEEB<\/p>\n

              Question 6.
              \nFind the solution of the game by the principle of dominance for the following payoffs matrix of A.
              \n\"2nd
              \nAnswer:
              \nBy principle of dominance, A4<\/sub> dominates A3<\/sub>, and A1<\/sub>. So, delete A1<\/sub>, A3<\/sub>.
              \n\"2nd
              \nB2<\/sub> dominates B1<\/sub> and B3<\/sub>. So, delete B1<\/sub>, B3<\/sub>
              \n\"2nd
              \nA4<\/sub> dominates A2<\/sub>, delete A2<\/sub>,
              \n\"2nd
              \nB2<\/sub>, dominates B4<\/sub> delete B4<\/sub>.
              \n\"2nd
              \nSaddle Point occurs ar (4, 2).
              \n\u2234 Suggested pure strategies for player A and B are (A4<\/sub>, B2<\/sub>) value of the game V = 2.<\/p>\n

              Question 7.
              \nSolve the game using dominance principle :
              \n\"2nd
              \nAnswer:
              \nBy principle of dominance, A2<\/sub> dominates A1<\/sub>, A3<\/sub> & A4<\/sub>. So delete A1<\/sub>, A3<\/sub> & A4<\/sub>
              \n\"2nd<\/p>\n

              B2<\/sub> dominates B1<\/sub> B3<\/sub> & B4<\/sub>, so, delete B1<\/sub> B3<\/sub> & B4<\/sub>.
              \n\"2nd
              \nSaddle point occurs at (2, 2)
              \n\u2234 The suggested pure strategy for the players are (A2<\/sub>, B2<\/sub>).
              \nValue of the game V = 1.<\/p>\n

              Question 8.
              \nFollowing is the pay-off matrix of player A.
              \n\"2nd
              \nFind the solution of the game using the principle of Dominance.
              \nAnswer:
              \nBy Principle of dominance, 2nd strategy of player A dominates, 1st strategy. So, delete 1.
              \n\"2nd
              \n2nd column dominates 1st column, So delete column 1.
              \n\"2nd
              \n2nd row dominates 3rd row, delete 3rd row.
              \n\"2nd
              \n2nd column dominates 3rd column, so delete 3rd column.
              \n\"2nd
              \nSaddle point occurs at (2, 2)
              \n\u2234 The suggested pure strategies for player A and B are, for player A -2, for player B- 2.
              \nValue of the game V = 6.<\/p>\n

              Replacement Problems<\/strong><\/p>\n

              Question 1.
              \nThe cost of a machine is Rs. 12000. The maintenance cost and resale value per year of the machine is given below.
              \n\"2nd
              \nWhen should the machine be replace
              \nAnswer:
              \nHere, purchase price ; P = Rs. 12000
              \nLet Ci<\/sub> be the maintenance cost during the year and Sn<\/sub> be the resale value.
              \nThen, Average annual cost is A(n) = \\(\\frac { T }{ n }\\) , Where T = (P – Sn<\/sub>) + \u03a3C ; and n = no. of years.
              \n\"2nd
              \nHere, Average annual cost A(n) is minimum for n = 3.
              \n\u2234 The optimal replacement age is 3 years,
              \ni. e., The machine should be replaced for every 3 years.
              \nMinimum annual average cost is Rs. 3600<\/p>\n

              Question 2.
              \nA machine costs Rs. 36,000. Its resale value and maintenance cost at different years are given below:
              \n\"2nd
              \nResale Value (Rs.) 28,000 22,000 20,000 18,000 17,000 15,000
              \nFind out when the machine should be replaced.
              \nAnswer:
              \nGiven : cost\/purchase price \u2018P\u2019 = Rs. 36,000
              \nLet Ci<\/sub> be the maintenance cost, Sn<\/sub> be the resale value and n be the number of years.
              \nAverage annual cost : A(n) = \\(\\frac { T }{ n }\\) ; where T = (p – Sn<\/sub> ) + \u03a3Ci<\/sub>
              \n\"2nd
              \nHere, average annual cost A (n) is minimum for n = 5.
              \n\u2234 The optimal replacement age is 5 years i.e. the machine should be replaced for every 5 years.
              \nMinimum annual average cost is Rs. 5,960.<\/p>\n

              \"KSEEB<\/p>\n

              Question 3.
              \nA machine costs Rs. 80,000 and the operating cost is estimated to be Rs. 5000 for the first year and increases by Rs. 7000 every year for next five years. Determine the optimum period for replacement of the machine, assuming that the machine has no scrap (Resale) value.
              \nAnswer:
              \nHere the machine cost = P = Rs. 80,000 and since resale value Sn<\/sub> = 0 ; the depreciation (P – Sn<\/sub>) = 80000 for all the years and let Ci<\/sub> be the operating cost.
              \nThe Average annual cost A(n) = \\(\\frac { T }{ n }\\); Where T = Total cost = (P – Sn<\/sub>) + \u03a3Ci and n = no. of years.
              \n\"2nd
              \nHere, Average annual cost A(n) is minimum for n = 5.
              \n\u2234 The optimum replacement age is 5 years,
              \ni.e., the machines should be replaced at the end of 5 years.
              \nMinimum annual average cost = Rs. 35000<\/p>\n

              Question 4.
              \nA tractor owner finds from the past records that maintenance cost and depreciation cost per year of the tractor whose purchase price is Rs. 1,50,000 are as follows:
              \n\"2nd
              \nDetermine at which time is it possible to replace the tractor.
              \nAnswer:
              \nGiven: P = Rs. 1,50,000, let maintenance cost be Ci<\/sub> during the year and depreciation be (P – Sn<\/sub>), where. Sn<\/sub> be the resale\/ scrap value.
              \n\"2nd
              \nHere, Average annual cost A(n) is minimum for n = 6.
              \n\u2234 The optimal replacement age is n = 6 year, i.e., the tractor should be replaced after using it for 6 years. The optimum Average annual cost is Rs. 18583.33.<\/p>\n

              Inventory Problems<\/strong><\/p>\n

              Question 1.
              \nWhat are the advantages and disadvantages of maintenance of inventory?
              \nAnswer:
              \nAdvantages:<\/p>\n

                \n
              • Bulk production decreases costs of production, transportation, maintenance, etc.<\/li>\n
              • Instant\/uninterrupted supply of goods to customers.<\/li>\n
              • Inventory gives profit due to increase in market price.<\/li>\n
              • Inventory decreases the setup cost.<\/li>\n<\/ul>\n

                Disadvantages:<\/p>\n

                  \n
                • Ware house rent.<\/li>\n
                • Interest on invested capital.<\/li>\n
                • Labour on maintenance of inventory.<\/li>\n
                • Maintenance of records.<\/li>\n
                • Loss due to fall in price, deterioration and depreciation, etc.<\/li>\n<\/ul>\n

                  Question 2.
                  \nA firm has an annual demand of 3000 units of raw materials. The setup cost is Rs. 25 per order and holding cost is Rs. 2 per unit\/year. Determine
                  \n1. optimal order quantity
                  \n2. number of order per year
                  \n3. Annual average inventory cost.
                  \nAnswer:
                  \nGiven: R = 3000, C3<\/sub> = Rs. 25\/order ; C1<\/sub> = Rs. 2 per unit\/year.
                  \nHere shortage cost C2<\/sub> is not given use EOQ model I
                  \n1. (\/) EOQ : Optimal order Quantity = Q\u00b0 = \\(\\sqrt{\\frac{2 \\mathrm{C}_{3} \\mathrm{R}}{\\mathrm{C}_{1}}}=\\sqrt{\\frac{2 \\times 25 \\times 3000}{2}}\\) = 273.86 = 274 units<\/p>\n

                  2. Number of orders n\u00b0 = \\(\\frac{1}{t^{0}}=\\frac{R}{Q^{0}}=\\frac{3000}{273.86}\\) , Here t\u00b0 = \\(=\\frac{Q^{0}}{R}\\)
                  \nn\u00b0 = 10.95 = 11 orders per year<\/p>\n

                  3. Annual average inventory cost.
                  \nC (Q\u00b0) = \\(\\sqrt{2 \\mathrm{C}_{1} \\mathrm{C}_{3} \\mathrm{R}}=\\sqrt{2 \\times 2 \\times 25 \\times 3000}\\) = Rs. 547.72\/year<\/p>\n

                  Question 3.
                  \nA dealer has to supply 15 computer to a retail customer every week. He maintain an inventory in which the maintenance cost is Rs. 100 per unit per year. The ordering cost is Rs. 350 per order. Find EOQ, the optimum re-scheduling time and the optimum average cost.
                  \nAnswer:
                  \nGiven: R = 15\/week ; and
                  \nR = 15 \u00d7 52 per year.
                  \nC1<\/sub> = Rs. 100 per unit\/year.
                  \nC3<\/sub> = Rs. 350 per order.
                  \n\"2nd
                  \nQ\u00b0 = 73.89 units
                  \nRe-scheduling time t\u00b0 = \\(\\frac{Q^{0}}{R}=\\frac{73.89}{15 \\times 52}\\) = 0.0947 years.
                  \ni.e., 66 units should be ordered approximately every 1 month.
                  \nThe optimum average cost\/minimum average cost :
                  \nC(Q\u00b0) = \\(\\sqrt{2 \\mathrm{C}_{1} \\mathrm{C}_{3} \\mathrm{R}}=\\sqrt{2 \\times 100 \\times 350 \\times 15 \\times 52}\\) = Rs. 7389.18 per year.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 4.
                  \nA manufacture has to supply 80 units of his product every month to a customer. Capital cost is Rs. 800 per unit. Holding cost is 12% per year of capital cost. Shortages are not allowed. Cost of setting up machines for production is Rs. 300. How frequently should production be made so that inventory is most economical?
                  \nAnswer:
                  \nGiven: R = 80 units per month and R = 80 \u00d7 12 = 960 units per year.
                  \nP = Capital cost = Rs. 800 per unit.
                  \nI = Holding cost per unit per year = 12%.
                  \n\u2234 C1<\/sub> = carrying cost per unit = PI = 800 \u00d7 12% = Rs. 96 per year.
                  \nC3<\/sub> = Set up cost = Rs. 300 per order.
                  \nEconomic order quantity : Q\u00b0 =\\(\\sqrt{\\frac{2 C_{3} R}{C_{1}}}=\\sqrt{\\frac{2 \\times 300 \\times 960}{96}}\\)
                  \n\u2234 Q\u00b0 = 77.46 = 77 order.
                  \nReorder time is t\u00b0 = \\(\\frac{Q^{0}}{R}=\\frac{77.46}{960}\\) = 0.0807 years.
                  \ni.e., t\u00b0 =0.0897 \u00d7 365 = 29 days (approx)
                  \nFrequency of order is n\u00b0 = \\(\\frac{1}{t^{0}}\\) = 12.39 per year
                  \nMinimum average cost is
                  \nC(Q\u00b0) = \\(\\sqrt{2 \\mathrm{C}_{1} \\mathrm{C}_{3} \\mathrm{R}}=\\sqrt{2 \\times 96 \\times 300 \\times 960}\\) =7436.13
                  \nC(Q\u00b0) = Rs. 7436.13 per year<\/p>\n

                  Question 5.
                  \nIf the demand is 300 units per year, inventory holding cost is Rs. 8 per unit per year, shortage cost is Rs. 10 per unit per year and setup cost is Rs. 500 per replenishment. Determine EOQ and the maximum shortage level.
                  \nAnswer:
                  \nGiven : R = 300 unit per year,
                  \nC1<\/sub> = Rs. 8 per unit per year.
                  \nC2<\/sub> = Shortage cost = Rs. 10 per unit per year,
                  \nC3<\/sub> – Setup cost = Rs. 500 per replenishment. .
                  \nThe economic order quantity is
                  \n\"2nd
                  \nQ\u00b0 = 259.80 = 260 units.
                  \nThe maximum shortage level is –
                  \n(Q\u00b0 – S\u00b0) Here S\u00b0: Maximum inventory level.
                  \n\"2nd
                  \n\u2234 (Q\u00b0-S\u00b0) = 2.59.80- 144.33 =115.46
                  \nMaximum shortage level is 115 units.<\/p>\n

                  Question 6.
                  \nA dealer supplies the following information with regard to a product dealt by him. Annual demand = 4500 units. Ordering cost = Rs. 20. Holding cost = Rs. 2.00. Shortage Cost = Rs. 12. Determine EOQ and Total variable cost.
                  \nAnswer:
                  \nGiven: R = 4500 units ;
                  \nC3<\/sub> = Rs. 20
                  \nC1<\/sub>=Rs. 2,
                  \nC2<\/sub> = Rs 12
                  \nThe economic order quantity is –
                  \nQ\u00b0 = 1052 units (Approx)
                  \n\"2nd
                  \ni.e., Q\u00b0= 324 Units.
                  \nTotal variable cost\/minimum average cost is
                  \n\"2nd
                  \nC(Q\u00b0, S\u00b0) =Rs. 555.49<\/p>\n

                  Question 7.
                  \nA Factory requires 2000 units for an item per year. Purchase cost is Rs. 30, the setup cost is Rs. 200. The carrying cost is Rs. 8 per item\/year. If the shortage cost is Re. 1 per item\/month. Determine
                  \n1. EOQ
                  \n2. Minimum average cost.
                  \nAnswer:
                  \nGiven : R = 2000 ; P = Rs. 30 C3<\/sub> = Rs. 200
                  \nC2<\/sub> = Re. 1 per item\/month.
                  \ni.e\u201e C2<\/sub> =1><12= 12\/year. C1<\/sub> = Rs. 8 per item per year.
                  \n(Here shortage cost is given, so use EOQ model II)
                  \n1. Economic order quantity is –
                  \n\"2nd
                  \nQ\u00b0 = 408.25 units<\/p>\n

                  2. Minimum average cost is –
                  \n\"2nd<\/p>\n

                  Question 8.
                  \nThe annual demand for an item is 3000 units. Capital cost is Rs. 7 per unit. Inven\u00actory carrying cost is 20% of capital cost per annum. If set up cost is Rs. 150, determine
                  \n1. EOQ
                  \n2. Number of orders per year
                  \n3. Optimal cost
                  \nAnswer:
                  \nGiven R = 3000, p = 7,1 = 20%
                  \nc3<\/sub> = 150. (Shortages are not given use EOQ model I)
                  \n1. EOQ : \\(Q^{0}=\\sqrt{\\frac{2 C_{3} R}{C_{1}}}\\)
                  \nC1<\/sub> = p1 = 7 \u00d7 \\(\\frac { 20 }{ 100 }\\)
                  \n\\(Q^{0}=\\sqrt{\\frac{2 \\times 150 \\times 3000}{1.4}}\\) = 801.78 units<\/p>\n

                  2. n\u00b0 = \\(\\frac{R}{Q^{0}}=\\frac{3000}{801.78}\\) = 3.4 orders<\/p>\n

                  3. C(Q\u00b0) = \\(\\sqrt{2 \\times 150 \\times 3000 \\times 1.4}\\) = Rs. 1122.50<\/p>\n

                  \"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

                  You can Download Chapter 8 Operation Research Questions and Answers, Notes, 2nd PUC Statistics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Statistics Question Bank Chapter 8 Operation Research 2nd PUC Operation Research One and Two Marks Questions and …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12877"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=12877"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/12877\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=12877"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=12877"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=12877"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}