Unit 2 lpp tp
- 1. QUANTITATIVE TECHNIQUES FOR MANAGERS UNIT-II RMB 207 DR. ZIAUL HASSAN BAKHSHI
- 2. Linear Programming Models A model, which is used for optimum allocation of scarce or limited resources to competing products or activities under such assumptions as certainty, linearity, fixed technology, and constant profit per unit, is linear programming.
- 3. PROPERTIES OF LINEAR PROGRAMMING MODEL (a)The relationship between variables and constraints must be linear. (b) The model must have an objective function. (c)The model must have structural constraints. (d) The model must have non-negativity constraint.
- 4. Transportation Problem The Transportation problem is to transport various amounts of a single homogeneous commodity that are initially stored at various origins, to different destinations in such a way that the total transportation cost is a minimum. It can also be defined as to ship goods from various origins to various destinations in such a manner that the transportation cost is a minimum 12/25/2016
- 5. Basic Definitions • Feasible Solution A set of non-negative individual allocations (xij ≥ 0) which simultaneously removes deficiencies is called as feasible solution. • Basic Feasible Solution A feasible solution to ‘m’ origin, ‘n’ destination problem is said to be basic if the number of positive allocations are m+n-1. If the number of allocations is less than m+n-1 then it is called as Degenerate Basic Feasible Solution. Otherwise it is called as Non-Degenerate Basic Feasible Solution. • Optimum Solution A feasible solution is said to be optimal if it minimizes the total transportation cost. 12/25/2016
- 6. Simple Network Representation 1 2 m 1 2 n Sources Destinations … … Supply s1 Supply s2 Supply sm Demand d1 Demand d2 Demand dn xij Costs cij 12/25/2016
- 7. The Transportation Simplex Tableau Destination Supply ui Source 1 2 … n 1 c11 c12 … c1n s1 2 c21 c22 … c2n s2 … … … … … … m cm1 cm2 … cmn sm Demand d1 d2 … dn Z = vj 12/25/2016
- 8. Feasible Solutions A transportation problem will have feasible solutions if and only if n j j m i i ds 11
- 9. North-West Corner Rule Step 1 The first assignment is made in the cell occupying the upper left-hand (north-west) corner of the table. The maximum possible amount is allocated here i.e. x11 = min (a1, b1). This value of x11 is then entered in the cell (1,1) of the transportation table. Step 2 If b1> a1, move vertically downwards to the second row and make the second allocation of amount x21 = min (a2, b1- x11) in the cell (2, 1). If b1< a1, move horizontally right side to the second column and make the second allocation of amount x12 = min (a1 - x11, b2) in the cell (1, 2). If b1 = a1, there is tie for the second allocation. One can make a second allocation of magnitude x12 = min (a1 - a1, b2) in the cell (1, 2) or x21 = min (a2, b1 - b1) in the cell (2, 1) Step 3 Start from the new north-west corner of the transportation table and repeat steps 1 and 2 until all the requirements are satisfied.
- 10. Lowest Cost Entry Method (Matrix Minima Method) Step 1 Determine the smallest cost in the cost matrix of the transportation table. Allocate xij = min (ai, bj) in the cell (i, j) Step 2 If xij = ai, cross out the ith row of the table and decrease bj by ai. Go to step 3. If xij = bj, cross out the jth column of the table and decrease ai by bj. Go to step 3. If xij = ai= bj, cross out the ith row or jth column but not both. Step 3 Repeat steps 1 and 2 for the resulting reduced transportation table until all the requirements are satisfied. Whenever the minimum cost is not unique, make an arbitrary choice among the minima. 12/25/2016
- 11. Vogel’s Approximation Method (Unit Cost Penalty Method) Step1 For each row of the table, identify the smallest and the next to smallest cost. Determine the difference between them for each row. These are called penalties. Put them aside by enclosing them in the parenthesis against the respective rows. Similarly compute penalties for each column. Step 2 Identify the row or column with the largest penalty. If a tie occurs then use an arbitrary choice. Let the largest penalty corresponding to the ith row have the cost cij. Allocate the largest possible amount xij = min (ai, bj) in the cell (i, j) and cross out either ith row or jth column in the usual manner. Step 3 Again compute the row and column penalties for the reduced table and then go to step 2. Repeat the procedure until all the requirements are satisfied. 12/25/2016
- 12. MOVING TOWARDS OPTIMALITY (1) DETERMINE THE NET-EVALUATIONS ( U-V METHOD) Since the net evaluation is zero for all basic cells, it follows that zij - cij = ui +vj - cij , for all basic cells (i, j). So we can make use of this relation to find the values of ui and vj . Using the relation ui +vj = cij , for all i and j which (i,j) is a basic cell, we can obtain the values of ui `s and vj `s. After getting the values of ui `s and vj `s, we can compute the net-evaluation for each non- basic cell and display them in parenthesis in the respective cells. 12/25/2016
- 13. MOVING TOWARDS OPTIMALITY (2) SELECTION OF THE ELECTING VARIABLES Choose the variable xrs to enter the basis for which the net evaluation zrs - crs = max { zij - cij > 0} . After identifying the entering variable xrs , form a loop which starts at the non-basic cell (r,s) connecting only basic cells . Such a closed path exists and is unique for any non- degenerate solution. Allocate a quantity alternately to the cells of the loop starting + to the entering cell. The value of is the minimum value of allocations in the cells having -. Now compute the net-evaluation for new transportation table and continue the above process till all the net-evaluations are positive for non-basic cells. 12/25/2016
- 14. DEGENERACY IN TRANSPORTATION PROBLEM Transportation with m-origins and n-destinations can have m+n-1 positive basic variables, otherwise the basic solution degenerates. So whenever the number of basic cells is less than m + n-1, the transportation problem is degenerate. To resolve the degeneracy, the positive variables are augmented by as many zero-valued variables as is necessary to complete m +n –1 basic variables. 12/25/2016
- 15. W→ F ↓ W1 W2 W3 W4 Factory Capacity F1 19 30 50 10 7 F2 70 30 40 60 9 F3 40 8 70 20 18 Warehouse Requirement 5 8 7 14 34 Find the initial basic feasible solution by using North-West Corner Rule , Matrix Minima Method and Vogel’s Approximation Method 12/25/2016