Example for Solution of Transportation Problem

EXAMPLE FOR SOLUTION OF TRANSPORTATION PROBLEMAn organization has four destinations and three sources for supply of goods. The transportation cost per unit is given below. The entire availability is 700 units which exceeds the cumulative demand of 600 units. Decide the optimal transportation scheme for this case. Solution Step 1: Check for balance of supply and demand S Supply = 250 + 200 + 250 = 700 units S Demand = 100 + 150 + 250 + 100 = 600 units Decision Rul (i) If S Supply = S Demand then go to next step. (ii) Else; if S Supply > S Demand then, add a ―dummy destination‖ with zero transportation cost. (iii) Or else; if S supply < S Demand then, add a ―dummy source‖ with zero transportation cost. Since, in this problem S supply > S Demand Hence; add a ―dummy destination‖ (say D5) with zero transportation cost and balance demand which is difference in supply and demand (= 100 units). The initial transportation matrix is now formulated with transportation cost in the small box of each route. Note that each cell of the transportation matrix represents a potential route. Introducing dummy column for balancing the supply and demand Step 2: (i) Decide the nature of problem : Minimization of transportation-cost (ii) Make initial assignment Initial assignment may be done by using any of the following approaches : (i) Least-cost method (ii) North-West corner method (iii) Vogel's approximation method We would demonstrate all the three methods. (i) Initial Solution by Least Cost method Select the lowest transportation (or shipping) cost cell (or route) in the initial matrix. For example: it is route S1D5, S2D5 and S3D5 in our problem with zero shipping cost. Allocate the minimum of remaining balance of supply (in last column) and demand (in last row). Let us select S1D5 route. One can also select other route (S2D5 or S3D5) in case of tie. For S1D5, available supply is 250 and available demand is 100 units. The lower is 100 units. Hence, allocate 100 units-through this route (i.e, S1D5). With this allocation, entire demand of route S1D5 is consumed but supply of corresponding source, S1, is still (250-100) or 150 units left. This is marked in last column of supply. The entire demand of destination, D5, is consumed. We get the following matrix (Fig. 12.6) by crossing out the consumed destination (D5): Now, we leave the consumed routes (i.e., column D5) and work for allocation of other routes. Next, least cost route is S1D1, with 13 per unit of shipping cost. For this route, the demand is 100 units and remaining supply is 150 units. We allocate minimum of the two, i.e., 100 units in this route. With this destination, D1 is consumed but source S1 is still left with (150-100) = 50 units of supply. So, now leave the destination D1 and we get the following matrix. With 100 units allocation in route S1D5 Assignment for destination D1 and D5 consumed Now, we work on remaining matrix, which excludes first column (D1) and last column (D5). Next assignment is due in the least cost route, which is route S2D4. For this route, we can allocate 100 units which is lesser of the corresponding demand (100 units) and (200 units). By this allocation in route S2D4, the demand of destination D4 is consumed. So, this column is now crossed out. Assignment with destination D1, D4 and D5 consumed Now, we work on the remaining matrix which excludes, column, D1, D4 and D5. Next assignment is due in the least cost route of the remaining routes. Note that we have two potential routes: S1D2 and S2D3. Both have 16 units of transportation cost. In case of any tie (such as this), we select any of the routes. Let us select route, S1D2, and allocate 50 units (minimum of demand of 150 and supply of remaining 50 units). With this, all supply of source S1 is consumed. Therefore, cross out row of S1. We get the following matrix: Destination D1, D4 and D5 source S1 are consumed Now, remaining allocation is done in route S2D3 (as 100 units). With this source, S2 is consumed. Next allocation of 100 units is done in route S3D2 and 150 units in route S3D3. Final initial assignment is as follows: Total cost in this assignment is (13 × 100 + 16 × 50 + 100 × 0 + 16 × 100 + 15 × 100 + 17 × 150) or Rs. 9450. Initial assignment by least cost method if any For our problem (m + n. m = number of destinations. Hence. remove degeneracy and go to step 4. Hence. step 5). problem is not degenerate.Step 3: Count the number of filled (or allocated) routes. Decision rule (i) If filled route = (m + n – 1) then go for optimality check (i. row. proceed to step 5. as we have done in the least cost method. The number of filled route is equal to 7. In the North-West Corner (NWC) method. including dummy. we start with the top-left (corner- .e. including dummy column. this approach should not be applied if initial assignment has already been made by any other method.l) = 5 + 3-1 = 7. (ii) If filled route < (m + n – 1) then the solution is degenerate. Here. if any n = number of source. Initial Assignment by North-West Corner Method (an alternative to least cost method) This approach is also for making initial assignment. Therefore. Therefore. D1 and D2. allocation in the top-left cell is due in route S2D3. With this. it is cell S1D2. Remaining allocations are done in S2D3. S3D4 and S3D5 in sequential order.most) route. Allocate in the same way. Note that. demand for this route is 100 and supply is 250. We get the initial solution by north-west corner method as follows (Fig. Here. allocate 100 units in this route. Therefore. the total cost is (13 × 100 + 16 × 150 + 16 × 100 + 0 × 100) or Rs. . Remaining matrix excludes S1. allocation is made in this route for the minimum of supply or demand. 12. 150 units are allocated in this route. which excludes column Dr Again. Hence. are consumed. Thus. go to step 4 for removing degeneracy. with this. column corresponding to D1 is consumed. both D2 and S. work on the remaining matrix. select the top-left route. 200 units may be allocated and S2 is now consumed. In our case. Now.11): Initial assignment by North-West corner method For this assignment. Hence. Now. which is one less than (m + n – 1). Step 3: Check for degeneracy (m + n – 1) = 5 + 3 – 1 = 7 Number of filled cells = 6. which is S:DrIrrespective of cost. 9350. We get the following allocation after removing degeneracy. Initial assignment by North-West corner method after removing degeneracy Initial Assignment by Vogel’s Approximation Method (VAM) This is the third alternative method for doing initial assignment of a transportation problem. the least cost un-filled cells are S1D5 and S2D5. This route S3D5. Hence. allocation is to made in row of source S3. (which is zero for all calculation purposes). which one must select. allocate a very-very small quality. Therefore. . The difference is called as penalty cost for not using the least-cost route. should be the lowest cost of this row. we calculate the difference between the two least-cost routes for each row and column. first allocation is as follows. Let us select S1D5 and allocate in this.Step 4: In case of degeneracy. in the least cost of un-filled cells. First calculation of Penalty cost in VAM Highest of all calculated penalty costs is for S3 and (S2). The route (or cell). In the above figures of North-West corner method allocation. In this method. ) .First calculation in Vogel’s method Now. since destination D1 is consumed. Since there is tie between all routes. For this row. the least cost route is S1D1. Hence. Now for this. we break the tie by arbitrarily selecting any route (S1D2 in this case. with the first allocation. destination D5 is consumed. next assignment is due in this route: Second calculation of Penalty cost in VAM Second allocation in Vogel’s method After second allocation. source S1 has highest penalty cost. We exclude this column and work on the remaining matrix for calculating the penalty cost. We get the following matrix. we leave this column and proceed for calculation of next penalty cost. Allocation is done in route S1D2. . column D4 is consumed. we get the following allocations in the Vogel’s approximation method.Third calculation of Penalty cost Third allocation in Vogel’s method Fourth calculation of Penalty cost in VAM Fourth allocation in Vogel’s method With the fourth allocation. the allocations of 100 units and 150 units are done in route S2D3 and S4D3 respectively. In the only left column D3. Thus. 9350: Step 3: Check for degeneracy (m + n – 1) = 7 Number of filled cell = 6. Final allocation after removing degeneracy in Vogel’s method Optimization of Initial Assignment . This cell is route S1D5 or S2D5. which is one less than (m + n + 1). go to step 4 for removing the degeneracy. Thus. we get following matrix after removing degeneracy. Hence.Final allocation through Vogel’s method The initial cost for this allocation is (13 × 100 + 16 × 150 + 16 × 100 + 15 × 100 + 17 × 150 + 0 × 100) or equal to Rs. Step 4: We allocate in the least-cost un-filled cell. Let us select route S1D5. u1 = 0. using following equation: u1 + v1 = Cij (For any filled route) where u1 = row value vj = column value Cij = unit cost of assigned route Once first set of column values (vj is known.degenerate initial solution. Now consider all filled routes of this row. for all rows and columns. First. Hence.The initial feasible assignment is done by using least-cost method or North-West corner method or Vogel's approximation method. let us define row value. However. next step is to check the optimality of the initial solution. Let us select row 1. we have to calculate the opportunity cost of un-occupied routes. i. For these routes.e. calculate column values v. we start with any row (or column). locate other routes of filled cells in these columns. ui and vj values are determined for a non. Calculate next of ui (or vj values using above equation. For this row.. In this way. source S1. none of these methods guarantees optimal solution. Step 6: Check the optimality . Step 5: Check the optimality of the initial solution For this. Note that all the corner points of the loop are either filled cells or positive opportunity cost un-assigned cells. Now. We show the algorithm with our previous problem. which has a positive opportunity cost. transfer the minimal of all allocations at the filled cells to the positive opportunity cost cell. then go to next step. the row and column addition of demand and supply is maintained. .Calculate the opportunity of non-allocated orunfilled routes. successive corner points from unfilled cell are subtracted with this value. In this way. For this. the initial solution is optimal. use the following equation: Opportunity unassigned route = ui + vj – Cij where ui = row value vj = column value Cij = unit cost of unassigned route If the opportunity cost is negative for all unassigned routes. Step 7: Make a loop of horizontal and vertical lines which joins some filled routes with the unfilled route. ¥or this. Corresponding addition is done at alternate cells. If in case any of the opportunity costs is positive. 10) : For this. S1and take u1 = 0. v1 = 13 – 0 = 13 v2 = 16 – 0 = 16 v5 = 0 – 0 = 0 Calculation for ui and vj in least cost initial assignment Now. For this cell v4 = 15 – 0 = 0 .Let us consider the initial allocation of least-cost method (Fig. 12. Now S1DpS1D2.and S1D5are filled cells. as it has a ui value. cell S2D4 is selected. For this cell u3 = 17 – 16 = 1 Now. for filled cells. we start with row. as this has a vj value. as this has a ui value. cell S3D2 is taken. as it has a vj value. cell S3D3 is selected. For this cell u2 = 16 – 16 = 0 Now. Hence. cell S2D3 is selected. (vj = Cij – ui). For this cell v3 = 17 – 1 = 16 Now. we go to next step and make a loop as follows. Closed loop for cell S3D5 .Thus. all ui and vj are known. Step 6: Calculate opportunity cost of un-assigned routes. the solution is non-optimal. Unassigned route S1D3 S1D4 S2D1 S2D2 S2D5 S3D1 S3D4 S3D5 Opportunity cost (ui + vj – Cij) 0 + 16 – 19 = –3 0 + 15 – 17 = –2 0 + 13 – 17 = –4 0 + 16 – 19 = –3 0+0–0=0 1 + 13 – 15 = –2 1 + 15 – 16 = 0 1 + 0 – 0 = +1 Since route S3D5 has positive opportunity cost. hence. the opportunity cost of unassigned cells is as follows.The revised allocation involves 100 units transfer from cells S1D5 and S3D2 to cells S3D5 and S1D2. Fresh calculation of ui and vj is also done in the similar way as explained in Step 5. 9450). we allocate to the least-cost un-filled cell S1D5. Thus. For this assignment. since un-allocated routes have negative (or zero) opportunity cost. optimal allocation of route is given in Figure. Now. the present assignment is the optimal one. Note that total cost is less than the initial assignment cost of least-cost method (= Rs. Thus. revised allocation is as follows: Revised allocation in least-cost assignment Since above solution is degenerate now. . Unassigned route S1D3 S1D4 S2D1 S2D2 S2D5 S3D1 S3D2 S3D4 Opportunity cost (ui + vj – Cij) 0 + 17 – 19 = –2 0 + 16 – 17 = –1 –1 + 13 – 17 = –5 –1 + 16 – 19 = –4 –1 + 0 – 0 = –1 0 + 13 – 15 = –0 0 + 16 – 17 = –1 0 + 16 – 16 = 0 Opportunity cost Route S1D1 S1D2 S2D3 S2D4 S3D3 S3D5 Unit 100 150 100 100 150 100 Cost in this route 13 × 100 = 1300 16 × 150 = 2400 16 × 100 = 1600 15 × 100 = 1500 17 × 150 = 2550 0 × 100 = 0 . optimality of North-West corner method solution is done.Similarly. 100 units in S3D4. the optimality of Vogel’s method’s initial solution is done. 9350. the initial assignment is optimal one with total cost of Rs. Similarly. The optimal assignment of routes is 100 units is S1D1. 200 units in S2D3. 50 units in S3D3. 150 units in S1D2.Total cost = Rs.9350 Optimal allocation in different routes Calculation of ui and vj for N-W corner method’s initial solutions Opportunity cost of above assignment is as follows: Since all opportunity costs are negative or zero. Opportunity cost of above N-W corner assignment is as follows Unassigned route S1D3 S1D4 S2D1 Opportunity cost (ui + vj – Cij) 0 + 17 – 19 = –2 0 + 16 – 17 = –1 –1 + 13 – 17 = –5 . S2D2 S2D4 S2D5 S3D1 S3D2 –1 + 16 – 19 = –4 –1 + 16 – 15 = 0 –1 + 0 – 0 = –1 0 + 13 – 15 = –2 0 + 16 – 17 = –1 Calculation of ui and vj for Vogel method’s initial solutions Opportunity cost of above assignment is as follows: Unassigned route S1D3 S1D4 S2D1 S2D2 S2D5 S3D1 Opportunity cost (ui + vj – Cij) 0 + 17 – 19 = –2 0 + 16 – 17 = –1 –1 + 13 – 17 = –5 –1 + 16 – 19 = –4 –1 + 0 – 0 = –1 0 + 13 – 15 = –2 . college or university level homework or assignment to us and we will make sure that you get the answers you need which are timely and also cost effective.S3D2 S3D4 0 + 16 – 17 = –1 0 + 16 – 16 = 0 Since all opportunity costs are negative or zero. vogel and northwest corner methods with examples. 9350. 100 units in S2D3. and 150 units in S3D3. Note that this solution is different from North-West corner solution but total cost is same and minimum. You can submit your school. The transportation problem solution approach Email Based Homework Assignment Help in Solved Example for Transportation Problem Transtutors is the best place to get answers to all your doubts regarding the standard representation of transport problem. 100 units in S2D4. The optimal assignment of routes is 100 units in S1D2. Our tutors are available round the clock to help you out in any way with industrial management. methods to solve TP like least cost. the initial assignment of Vogel’s solution is optimal with total cost of Rs. . You can also interact directly with our industrial management tutors for a one to one session and get answers to all your problems in your school. GCSE. The costs associated with the dummy origin are equal to zero. MathStat: Nikhil M. the sum of the supplies at the origins must equal the sum of the demands at the destinations (Balanced transportation problem). We will make sure that you get the best help possible for exams such as the AP. a dummy destination is added with demand equal to the excess supply. Our tutors will make sure that you achieve the highest grades for your industrial management assignments.Live Online Tutor Help for Solved Example for Transportation Problem Transtutors has a vast panel of experienced industrial management tutors who specialize in the methods to solve transportation problems and can explain the different concepts to you effectively. Round Square etc. Each cell represents a shipping route (which is an arc on the network and a decision variable in the LP formulation). college or university level industrial management. To solve the transportation problem by its special purpose algorithm. and the unit shipping costs are given in an upper right hand box in the cell. and shipping costs from all origins are zero. if total supply is less than total demand. IB. Joshi Transportation Problem Transportation Method A transportation tableau is given below. IGCSE. The supply at the dummy origin is equal to the difference of the total supply and the total demand. • If the total supply is greater than the total demand. • Similarly. a dummy origin is added. AS. . A level. 1) from Origin 1 to Destination 1. Develop an Initial Solution Two methods to get the initial solution: • Northwest Corner Rule • Minimum Cell-Cost Method Northwest Corner Rule 1.2) for next allocation. move on to cell (1. within the supply constraint of source 1 and the demand constraint of destination 1. ƒ If the demand requirement for destination 1 is satisfied but the supply capacity for Source 1 is not exhausted.2). move to cell (2. 2.1) ƒ If the demand requirement for Destination 1 is satisfied and the supply capacity for Source 1 is also exhausted. Starting from the northwest corner of the transportation tableau. unacceptable shipping routes are given a cost of +M (a large number). The first allocation will satisfy either the supply capacity of Source 1 or the destination requirement of Destination 1. 3.When solving a transportation problem by its special purpose algorithm. Initial tableau developed using Northwest Corner Method . Continue the allocation in the same manner toward the southeast corner of the transportation tableau until the supply capacities of all sources are exhausted and the demands of all destinations are satisfied. allocate as much quantity as possible to cell (1. move on to cell (2. ƒ If the demand requirement for destination 1 is not satisfied but the supply capacity for Source 1 is exhausted. the tie can be broken by selecting the cell that can accommodate the greater quantity.Total Cost = 12(400)+13(100)+4(700)+9(100)+12(200)+4(500)= 142. • Step 2: Calculate the remaining ui’s and vj’s by solving the relationship cij = ui + vj for occupied cells. Select the cell with the minimum cell cost in the tableau and allocate as much to this cell as possible. • Step 1: Set u1 = 0.000 MODI Method (for obtaining reduced costs) Associate a number. • Step 3: For unoccupied cells (i. the reduced cost = cij – ui – vj. it is not the most attractive because our objective is not included in the process. Initial tableau developed using Minimum Cell-Cost Method Total Cost = 12(300)+4(200)+4(700)+10(100)+9(200)+4(500)= 120. ui. Continue the procedure until all of the supply and demand requirements are satisfied. with each row and vj with each column. 3. but within the supply and demand constraints.000 Minimum Cell-Cost Method Although the North-west Corner Rule is the easiest. Select the cell with the next minimum cell-cost and allocate as much to this cell as possible within the demand and supply constraints. 2. Steps of Minimum Cell-Cost Method 1.j). . In a case of tied minimum cell-costs between two or more cells. ) If none. generate a stepping stone path by forming a closed loop with this cell and occupied cells by drawing connecting alternating horizontal and vertical lines between them.) GO TO STEP 1. STOP. Delivery cost per ton from each plant to each suburban location is shown below. and Eastwood — 10 tons. . Example: Acme Block Co. (For maximization problems select the unoccupied cell with the largest reduced cost. Step 3: Add this allocation to all cells where additions are to be made. Westwood — 45 tons. Acme has two plants. make the others occupied with 0′s. make only one unoccupied. (Note: An occupied cell on the stepping stone path now becomes 0 (unoccupied).Step 1: For each unoccupied cell. If more than one cell becomes 0. a dummy destination is created with demand of 20 and 0 unit costs. Select the unoccupied cell with the most negative reduced cost. (ABC) Acme Block Company has orders for 80 tons of concrete blocks at three suburban locations as follows: Northwood — 25 tons. calculate the reduced cost by the MODI method. and subtract this allocation to all cells where subtractions are to be made along the stepping stone path. Step 2: For this unoccupied cell. each of which can produce 50 tons per week. Determine the minimum allocation where a subtraction is to be made along this path. How should end of week shipments be made to fill the above orders? Since total supply = 100 and total demand = 80. v2 = 30. Allocate 20. Allocate 5.3) 40 – 0 – 32 = 8 (2. Iteration 4: Since there is only one row with two cells left. Allocate 25. v4 = 0. Set u1 = 0 2. Iteration 3: Of the remaining cells the least cost is 30 for x12. Since u2 + vj = c2j for occupied cells in row 2. then 10 + v3 = 42. then v1 = 24. Reduce s1 by 25 to 5 and eliminate the Northwood column. 3.4) 0 – 10 – 0 = -10 . hence v3 = 32. hence u2 = 10. then u2 + 30 = 40. Calculate the reduced costs (circled numbers on the previous slide) by cij – ui – vj. Unoccupied Cell Reduced Cost (1. Since ui + v2 = ci2 for occupied cells in column 2. make the final allocations of 40 and 10 to x22 and x23. 4. Reduce the Westwood column to 40 and eliminate the Plant 1 row. 1. Since u1 + vj = c1j for occupied cells in row 1. Iteration 2: Of the remaining cells the least cost is 24 for x11.Iteration 1: Tie for least cost (0).1) 30 – 24 -10 = -4 (2. Reduce s1 by 20 to 30 and delete the Dummy column. respectively. arbitrarily select x14. 1. or u2 = 10. then 10 + v3 = 42 or v3 = 32. Thus for the next tableau: x24 = 0 + 20 = 20 (0 is its current allocation) x14 = 20 – 20 = 0 (blank for the next tableau) x12 = 5 + 20 = 25 x22 = 40 – 20 = 20 The other occupied cells remain the same. and. v2 = 30. (1. 2. respectively.4) is (2. Set u1 = 0. The allocations in the subtraction cells are 20 and 40. and hence reallocate 20 along this path. Since u1 + vj = cij for occupied cells in row 1. (1. Since u2 + vj = c2j for occupied cells in row 2.2). 3. Since ui + v2 = ci2 for occupied cells in column 2. The minimum is 20.Iteration 1: The stepping stone path for cell (2. Iteration 2 Calculate the reduced costs (circled numbers on the previous slide) by cij – ui – vj. then v1 = 24. (2.4).4).2). . 10 + v4 = 0 or v4 = -10. then u2 + 30 = 40. 4. 2). then 6 + v3 = 42 or v3 = 36.Unoccupied Cell Reduced Cost (1.(1.1) 30 – 10 – 24 = -4 The most negative reduced cost is = -4 determined by x21. Unoccupied Cell Reduced Cost (1. Since u2 + vj = c2j for occupied cells in row 2. Thus the new solution is obtained by reallocating 20 on the stepping stone path. 4. and 6 + v4 = 0 or v4 = -6.4) 40 – 0 – 36 = 0 – 0 – (-6) = 4 6 .3) 40 – 0 – 32 = 8 (1. 1.(2.1).(1. Since ui + v1 = ci1 for occupied cells in column 2.3) (1. then u2 + 24 = 30 or u2 =6. Set u1 = 0 2.1). The stepping stone path for this cell is (2. Thus for the next tableau: x21 = 0 + 20 = 20 (0 is its current allocation) x11 = 25 – 20 = 5 x12 = 25 + 20 = 45 x22 = 20 – 20 = 0 (blank for the next tableau) The other occupied cells remain the same. 3. Iteration 3 Calculate the reduced costs (circled numbers on the previous slide) by cij – ui – vj. then v1 = 24 and v2 = 30.4) 0 – 0 – (-10) = 10 (2. The allocations in the subtraction cells are 25 and 20 respectively.2). Since u1 + vj = c1j for occupied cells in row 1. any set of values of x1. These non-negativity constraints are sometimes known as reality constraints. x2. x2. .….1) but not (1. The problem of finding (x1. that satisfies the constraints (1.1) where in each line either . the non-negativity restrictions xj 0 (1. xn satisfying the constraints (1. Definition : Suppose that one is given a linear function of n real variables z = f (x1. We shall assume that every Linear Programming Problem has included in its constraints. xn satisfying the constraints (1. x2. xn) = c1x1 + c2x2 +…+ cnxn and a set of linear inequalities and/or equations.1) and inequalities (1.2) 40 – 6 – 30 = 4 Since all the reduced costs are non-negative. x2.(2. Somewhat perversely.2) is called a feasible solution. In examples they typically represent quantities that for physical reasons are non-negative.…. called constraints (1. xn).…. this is the optimal tableau. The set of feasible solutions is the feasible region.2) is called a non-feasible solution.2) and these will be written separately from the other constraints. Any set of values of x1.…. = or occurs.1) and makes z a maximum (or minimum) is called a Linear Programming Problem . A diet is to contain at least 4000 units of carbohydrates. 500 units of fat and 300 units of protein. (3) Identify the objective function and express it as a linear function of decision variables. Food A costs 2 dollars per unit and food B costs 4 dollars per unit. Two foods A and B are available.The function f is called the objective function and z the objective variable. A unit of food B contains 25 units of carbohydrates. There are many real life situations where an LPP may be formulated. x2. We will discuss formulation of those problems which involve only two variables.…. The following examples will help to explain the mathematical formulation of an LPP. 10 units of fat and 20 units of protein. (4) Add the non-negativity restrictions on the decision variables. (1) Identify the decision variables and assign symbols x and y to them. Mathematical Formulation of Linear Programming Problems There are mainly four steps in the mathematical formulation of linear programming problem as a mathematical model. negative values of decision variables have no valid interpretation. xn is a feasible solution that makes f (x1. xn) a maximum (or minimum) then x is an optimal solution and the corresponding value of z is the optimal value. (2) Identify the set of constraints and express them as linear equations/inequations in terms of the decision variables. A unit of food A contains 10 units of carbohydrates. These constraints are the given conditions. Formulate the problem as an LPP so as to find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimum requirements. If set of values of x1. 20 units of fat and 15 units of protein.…. It might take the form of maximizing profit or production or minimizing cost. 01. These decision variables are those quantities whose values we wish to determine. Suggested answer: The above information can be represented as . as in the physical problems. B and C respectively. The maximum available time (in hours) for the machines A. Find the number of toys of each type that should be produced to get maximum profit. C are 380. Let x = number of toys of type-I to be produced y = number of toys of the type – II to be produced . B. Total cost = 2x + 4y The LPP formulated for the given diet problem is Minimize Z = 2x + 4y subject to the constraints 02. 4 hours and 4 hours in machines A. The time required to produce the first type of toy is 6 hours.Let the diet contain x units of A and y units of B. 8 hours and 12 hours in machines A. In the production of 2 types of toys. The time required to make the second type of toy is 8 hours. a factory uses 3 machines A. B and C respectively. B and C. Suggested answer: Mathematical Formulation The data given in the problem can be represented in a table as follows. 300 and 404 respectively. The profit on the first type of toy is 5 dollars while that on the second type of toy is 3 dollars. Total profit = 5x + 3y The LPP formulated for the given problem is: Maximise Z = 5x + 3y subject to the constraints Graphical Solution of Linear Programming Problem : Let us take the following example : minimise Z = 180x + 160y subject to 6x + y >= 12 3x + y >= 8 4x + 6y >= 24 x <= 5 y <= 5 x.y >= 0 Since there are only two variables in this LP problem we have the graphical representation of the LP given below with the feasible region (region of feasible solutions to the constraints associated with the LP) outlined. . Once a line has been drawn then it is a simple matter to work out which side of the line corresponds to all feasible solutions to the original inequality constraint (e.To draw the diagram above we turn all inequality constraints into equalities and draw the corresponding lines on the graph (e. We determine the optimal solution to the LP by plotting (180x + 160y) = K (K constant) for varying K values (iso-profit lines). Note here that it is inaccurate to attempt to read the values of x and y off the graph and instead we solve the simultaneous equations   3x + y = 8 4x + 6y = 24 to get x = 12/7 = 1.86 and hence the value of the objective function is given by 180x + 160y = 180(12/7) + 160(20/7) = 765. all feasible solutions to 6x + y >= 12 lie to the right of the line 6x + y = 12). Hence we can see that the optimal solution to the LP occurs at the vertex of the feasible region formed by the intersection of 3x + y = 8 and 4x + 6y = 24.g.71 and y = 20/7 = 2. the constraint 6x + y >= 12 becomes the line 6x + y = 12 on the graph). The smallest value of K (remember we are considering a minimisation problem) such that 180x + 160y = K goes through a point in the feasible region is the value of the optimal solution to the LP (and the corresponding point gives the optimal values of the variables). One such line (180x + 160y = 180) is shown dotted on the diagram.g.71 . 4. Importance of study of Micro Economics in modern society: It is very important to read micro economics in modern life. Proper use of wealth: For moving the scarcity of human being how to improve the wealth and how to improve the use of wealth know from the micro economics. Micro economics provides those forecasting for produce goods. 5. So. leaders of laborers should have the knowledge about micro economics. Because there is relation between labor work right and micro economics. So. social problems sufficient knowledge about micro economics needed. To the leaders of laborers: To consist trade union and to be aware about the right of laborers. To the social workers: To give solution about the problem of unemployment. 6. businessmen and industrialists always have to produce goods. society alliances people faces problem economic problem. for that it has importance a lot. In daily life of human being: Various economic problems have an influence in daily life of human being.71 It is clear that the above graphical approach to solving LP’s can be used for LP’s with two variables but most LP’s have more than two variables. To the businessmen and industrialists: By getting the forecasting about future. 1. . poverty etc. Below discussing the importance of study of micro economics in modern society. it is important for social worker to study micro economics. From the knowledge of economics as well as micro economics men find the solution of these problems. 3. Because today’s main problem is found in economics. This brings us to the simplex algorithm for solving LP’s. It means in daily life men face limited wealth and unlimited scarcity.Hence the optimal solution has cost 765. 2. Expand the scope of intellectuality: Reading micro economics the thinking and intellectuality level of human being can be expanded. for implementing good and effective plan study of micro economics is needed. it teaches to be a good saver of money as well as wealth. money market. plan is implemented. real market etc.7. For implementing plan: For nation organizational and individual life. The knowledge about demand aid supply of goods and raw materials. To the politicians: To do politics politician must have the knowledge about the micro economics. how it will be produced. . keeping in mind. This knowledge can be gained only by micro economics. Without the knowledge of micro economics politicians can not get prosperity in politics. money market. So. Because they should have the knowledge about general market. 10. Government administration: In government administration the employees should have the knowledge about micro economics. 9. How much a product will produce. So. plan is very important. 8. how it will be distributed etc. from above discussion we are now able to understand that how much micro economics is needed for modern society. principles of revenue etc. is needed for being an employee of government administration. In case of saving tendency: Micro economics does not teach for being frugal.