Combined
Upcoming SlideShare
Loading in...5
×
 

Combined

on

  • 3,875 views

 

Statistics

Views

Total Views
3,875
Slideshare-icon Views on SlideShare
3,875
Embed Views
0

Actions

Likes
1
Downloads
72
Comments
0

0 Embeds 0

No embeds

Accessibility

Upload Details

Uploaded via as Microsoft Word

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Combined Combined Document Transcript

    • Operations Research (This is material is not full-fledged material. This covers only few topics) Procedure of Simplex Method The steps for the computation of an optimum solution are as follows: Step-1: Check whether the objective function of the given L.P.P is to be maximized or minimized. If it is to be minimized then we convert it into a problem of maximizing it by using the result Minimum Z = - Maximum(-z) Step-2: Check whether all right hand side values of the constrains are non-negative. If any one of values is negative then multiply the corresponding inequation of the constraints by -1, so as to get all values are non-negative. Step-3: Convert all the inequations of the constraints into equations by introducing slack/surplus variables in the constraints. Put the costs of these variables equal to zero. Step-4: Obtain an initial basic feasible solution to the problem and put it in the first column of the simplex table. Step-5: Compute the net evolutions Δj= Zj – Cj (j=1,2,…..n) by using the relation Zj – Cj = CB Xj – Cj . Examine the sign (i) If all net evolutions are non negative, then the initial basic feasible solution is an optimum solution. (ii) If at least one net evolution is negative, proceed on to the next step. Step-6: If there are more than one negative net evolutions, then choose the most negative of them. The corresponding column is called entering column. (i) If all values in this column are ≤ 0, then there is an unbounded solution to the given problem. (ii) If at least one value is > 0, then the corresponding variable enters the basis. Step-7:Compute the ratio {XB / Entering column} and choose the minimum of these ratios. The row which is corresponding to this minimum ratio is called leaving row. The common element which is in both entering column and leaving row is known as the leading element or key element or pivotal element of the table. Step-8:Convert the key element to unity by dividing its row by the leading element itself and all other elements in its column to zeros by using elementary row transformations. Step-9: Go to step-5 and repeat the computational procedure until either an optimum solution is obtained or there is an indication of an unbounded solution.
    • Artificial Variable Technique – Big M-method If in a starting simplex table, we don’t have an identity sub matrix (i.e. an obvious starting BFS), then we introduce artificial variables to have a starting BFS. This is known as artificial variable technique. There is one method to find the starting BFS and solve the problem i.e., Big M method. Suppose a constraint equation i does not have a slack variable. i.e. there is no ith unit vector column in the LHS of the constraint equations. (This happens for example when the ith constraint in the original LPP is either ≥ or = .) Then we augment the equation with an artificial variable Ai to form the ith unit vector column. However as the artificial variable is extraneous to the given LPP, we use a feedback mechanism in which the optimization process automatically attempts to force these variables to zero level. This is achieved by giving a large penalty to the coefficient of the artificial variable in the objective function as follows: Artificial variable objective coefficient = - M in a maximization problem, = M in a minimization problem where M is a very large positive number. Procedure of Big M-method The following steps are involved in solving an LPP using the Big M method. Step-1: Express the problem in the standard form. Step-2:Add non-negative artificial variables to the left side of each of the equations corresponding to constraints of the type ≥ or =. However, addition of these artificial variable causes violation of the corresponding constraints. Therefore, we would like to get rid of these variables and would not allow them to appear in the final solution. This is achieved by assigning a very large penalty (-M for maximization and M for minimization) in the objective function. Step-3:Solve the modified LPP by simplex method, until any one of the three cases may arise. 1. If no artificial variable appears in the basis and the optimality conditions are satisfied, then the current solution is an optimal basic feasible solution.
    • 2. If at least one artificial variable in the basis at zero level and the optimality condition is satisfied then the current solution is an optimal basic feasible solution. 3. If at least one artificial variable appears in the basis at positive level and the optimality condition is satisfied, then the original problem has no feasible solution. The solution satisfies the contains but does not optimize the objective function, since it contains a very large penalty M and is called pseudo optimal solution. Artificial Variable Technique – Big M-method Consider the LPP: Minimize Z = 2 x1 + x2 Subject to the constraints 3 x 1 + x2 ≥ 9 x1 + x2 ≥ 6 x1, x2 ≥ 0 Putting this in the standard form, the LPP is: Minimize Z = 2 x1 + x2 Subject to the constraints 3 x 1 + x 2 – s1 =9 x1 + x2 – s2 = 6 x1, x2 ,s1 , s2 ≥ 0 Here s1 , s2 are surplus variables. Note that we do not have a 2x2 identity sub matrix in the LHS. Introducing the artificial variables A1, A2 in the above LPP The modified LPP is as follows: Minimize Z = 2 x1 + x2 + 0. s1 + 0. s2 + M.A1 + M.A2 Subject to the constraints 3 x 1 + x 2 – s1 + A1 = 9
    • x1 + x2 – s2 + A2 = 6 x1, x2 , s1 , s2 , A1 , A2 ≥ 0 Note that we now have a 2x2 identity sub matrix in the coefficient matrix of the constraint equations. Now we can solve the above LPP by the Simplex method. But the above objective function is not in maximization form. Convert it into maximization form. Max Z = -2 x1 – x2 + 0. s1 + 0. s2 – M A1 – M A2 Cj: -2 -2 0 0 -M -M B.V CB XB X1 X2 S1 S2 A1 A2 MR XB/X1 A1 -M 9 1 -1 0 1 0 3 -M -1 A2 6 1 1 0 0 1 6 Zj -15M -4M -2M M M -M -M Δj -4M+2 -2M+1 M M 0 0 B.V CB XB X1 X2 S1 S2 A1 A2 MR XB/X1 A1 -M 9 1 -1 0 1 0 3 -M -1 A2 6 1 1 0 0 1 6 Zj -15M -4M -2M M M -M -M Δj -4M+2 -2M+1 M M 0 0
    • TRANSPORTATION PROBLEM Transportation problem is one of the subclasses of LPP’s in which the objective is to transport various quantities 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 minimum. To achieve this objective we must know the amount and location of available supplies and the quantities demanded. In addition, we must know the costs that result from transporting one unit of commodity from various origins to various destinations. Mathematical Formulation of the Transportation Problem A transportation problem can be stated mathematically as a Linear Programming Problem as below: Minimize Z = subject to the constraints = ai , i = 1, 2,…..,m = bj , j = 1, 2,…..,m xij ≥ 0 for all i and j Where, ai = quantity of commodity available at origin i bj= quantity of commodity demanded at destination j cij= cost of transporting one unit of commodity from ith origin to jth destination xij = quantity transported from ith origin to jth destination Tabular form of the Transportation Problem To D1 D2 ……. Dn Supply From O1 c11 c12 ……. c1n a1 O2 c21 c22 ……. c2n a2 . . . ……. . . . . . . . . . . . . Om cm1 cm2 ……. cmn am Demand b1 b2 ……. bn
    • NORTH - WEST CORNER RULE Step1:Identify the cell at North-West corner of the transportation matrix. Step2:Allocate as many units as possible to that cell without exceeding supply or demand; then cross out the row or column that is exhausted by this assignment Step3:Reduce the amount of corresponding supply or demand which is more by allocated amount. Step4:Again identify the North-West corner cell of reduced transportation matrix. Step5:Repeat Step2 and Step3 until all the rim requirements are satisfied. Vogel’s Approximation Method (VAM) Step-I: Compute the penalty values for each row and each column. The penalty will be equal to the difference between the two smallest shipping costs in the row or column. Step-II: Identify the row or column with the largest penalty. Find the first basic variable which has the smallest shipping cost in that row or column. Then assign the highest possible value to that variable, and cross-out the row or column which is exhausted. Step-III: Compute new penalties and repeat the same procedure until all the rim requirements are satisfied. An example for Vogel’s Method Find the IBFS of the following transportation problem by using Penalty Method. D1 D2 D3 Supply 10 6 7 8 15 15 80 78 15 5 5 Step 1: Compute the penalties in each row and each column . Supply Row Penalty 10 7-6=1 6 7 8 15 78-15=63 15 80 78 Demand 15 5 5 Column Penalty 15-6=9 80-7=73 78-8=70
    • Step 2: Identify the largest penalty and choose least cost cell to corresponding this penalty Supply Row Penalty 10 7-6=1 6 7 8 15 78-15=63 15 80 78 Demand 15 5 5 Column Penalty 15-6=9 80-7=73 78-8=70 Step-3: Allocate the amount 5 which is minimum of corresponding row supply and column demand and then cross out column2 Supply Row Penalty 5 10 7-6=1 6 7 8 15 78-15=63 15 80 78 Demand 15 5 5 Column Penalty 15-6=9 80-7=73 78-8=70 Step-4: Recalculate the penalties Supply Row Penalty 5 5 8-6=2 6 7 8 15 78-15=63 15 80 78 Demand 15 X 5 Column Penalty 15-6=9 78-8=70
    • Step-5: Identify the largest penalty and choose least cost cell to corresponding this penalty Supply Row Penalty 5 5 8-6=2 6 7 8 15 78-15=63 15 80 78 Demand 15 X 5 Column Penalty 15-6=9 78-8=70 Step-6: Allocate the amount 5 which is minimum of corresponding row supply and column demand, then cross out column3 Supply Row Penalty 5 5 5 8-6=2 6 7 8 15 78-15=63 15 80 78 Demand 15 X X Column Penalty 15-6=9 Step-7: Finally allocate the values 0 and 15 to corresponding cells and cross out column 1 D1 D2 D3 Supply 0 5 5 X O1 6 7 8 15 X O2 15 80 78 Demand X X X Solution of the problem Now the Initial Basic Feasible Solution of the transportation problem is X11=0, X12=5, X13=5, and X21=15 and Total transportation cost = (0x6)+(5x7)+(5x8)+(15x15) = 0+35+40+225 = 300.