Download presentation

Presentation is loading. Please wait.

Published byJamison Elmer Modified over 4 years ago

1
6/11/2014 Question: XYZ Furniture Ltd. is involved in producing Chairs and Tables. The firm makes a profit of Rs. 200 per chair and Rs. 300 per table. Each of these items is produced using three machines M1, M2 and M3. The labour hours required on each of these machines are as follows: Machine Hours Required Available Hour/Week ChairsTable M1M1 6672 M2M2 104100 M3M3 412120

2
6/11/2014 Formulate the above problem into Linear Programming Problem. Use graphical as well as Simplex Techniques to arrive at Optimum Solution

3
6/11/2014 Solution: a)Formulation Of Linear Programming Problem: Decision Variable: let there be two decision variables x1 and x2, such that x1 = number of chairs produced and x2 = number of tables produced. Objective Function: the objective of the producer is to determine the the number of chairs and tables produced, so as to maximize the total profit, ie., Z = 200x1 + 300x2

4
6/11/2014 (iii)Constraints: First Constraint: On machine M1, each chair requires 6 hours and each table requires 6 hours. The total of working hours is given by 6x1 + 6x2. Second Constraint: On machine M2, each chair requires 10 hours and each table requires 4 hours. The total of working hours is given by 10x1 + 4x2. Third constraint: On machine M1, each chair requires 4 hours and each table requires 12 hours. The total of working hours is given by 4x1 + 12x2.

5
6/11/2014 Further, since the manufacturer does not have more then 72 hours of working on machine M1, does not have more then 100 hours of working on machine M2 and does not have more then 120 hours of working on machine M3. So that the constraint are as follows: 06x1 + 06x2 72 10x1 + 04x2 100 04x1 + 12x2 120

6
6/11/2014 ivNon - Negativity Constraint: Since the production process can never be negative, we must have x1 0 and x2 0.

7
6/11/2014 Hence the manufacturers allocation problem can be put in the following mathematical form: Find two real numbers x1 and x2, such that to maximize the expression (Objective Function) Z = 200x1 + 300x2 Subject to the Constraints, 06x1 + 06x2 72 10x1 + 04x2 100 04x1 + 12x2 120 and x1; x2 0

8
6/11/2014 SIMPLEX ALGORITHIM: Introducing the three slack variables s 1, s 2, and s 3 to the left hand side of the three constraint inequalities to convert them into equality and assign a zero coefficient to these in the objective function: Z = 200x1 + 300x2 + 0s1 + 0s2 + 0s3

9
6/11/2014 Subject to the constraints, 06x 1 + 06x 2 + s 1 = 72 10x 1 + 04x 2 + s 2 = 100 04x 1 + 12x 2 + s 3 = 120 And x 1,x 2, s 1, s 2 ands 3 0.

10
6/11/2014 Design the initial Feasible Solution: An initial basic feasible solution is obtained by putting x1 = x2 = 0 Thus we get, s1 = 72, s2 = 100, s3 = 120.

11
6/11/2014 Setup the initial simplex table: For computational efficiency and simplicity, initial basic feasible solution, the constraints of the standard Linear programming problem as well as the objective function can be displayed in the tabular form, called Simplex Table.

12
6/11/2014 C j ( contribution/unit)200300000 Minimum ratio Contribut ion of basic variable/u nit Basic Varia ble Value of the Basic Variable Coefficient MatrixIdentify Matrix CBCB Bb (= X B )x 1 x 2 s1s1 s2s2 s3s3 b j / x j 0s1s1 b 1 = 726610072/6 = 12 0s2s2 b 2 =100104010100/4 = 25 0s3s3 b 3 = 120412 *001120/12 = 10 ** Contribution Loss per unit Z j = Σ C Bj a ij = 0 00000 Net Contribution per unit C j -Z j 200 300 *** 000

13
6/11/2014 *** Incoming Column ** Outgoing Row * Key Element

14
6/11/2014 Compute the new key row values by using the formula New Table Key row values = Old Table Key row values / Key Element New Table Key row values = 1/12 (120412001) =(120/124/1212/120/120/121/12) =(101/31001/12) Computing all other row values using the formula New Table Row values=Old row values-Corresponding Coefficient in Key ColumnxCorresponding new table key row valueNew First Row Values = Old First Row Values - 6 x Corresponding New Table Key row values. =(7266100) -(6026001/2) =(124010-1/2) New Second Row Values = Old Second Row Values - 4 x Corresponding New Table Key row values. =(100104010) -(404/34001/3) =(6026/3001-1/3)

15
6/11/2014 C j ( contribution/unit)200300000 Minimum ratio Contri bution of basic variabl e/unit Basic Variabl e Value of the Basic Variable Coefficient MatrixIdentify Matrix CBCB Bb (= X B )x 1 x 2 s1s1 s2s2 s3s3 b j / x j 0s1s1 b 1 = 124 *010-1/212/4 = 3 ** 0s2s2 b 2 =6026/3001-1/360/(26/3) = 90/13 300x2x2 b 3 = 101/31001/1210/(1/3) = 30 Contribution Loss per unit Z j = Σ C Bj a ij = 3000 1003000025 Net Contribution per unit C j -Z j 100 *** 00025

16
6/11/2014 Compute the new key row values by using the formula New Table Key row values = Old Table Key row values / Key Element New Table Key row values = 1/4(124010-1/2) =(12/44/40/41/40/4-1/2 x 1/4) =(3101/40-1/8) Computing all other row values using the formula New Table Row values=Old row values- Corresponding Coefficient in Key Column x Corresponding new table key row value

17
6/11/2014 New Second Row Values = Old Second Row Values - 26/3 x Corresponding New Table Key row values. =(6026/3001-1/3) -(2626/3013/60-13/12) =(3400-13/613/4) New Third Row Values =Old Third Row Values – 1/3 x Corresponding New Table Key row values =(101/31001/12) -(11/301/120-1/24) =(901-1/1201/8)

18
6/11/2014 C j ( contribution/unit) 200300000 Contributio n of basic variable/uni t Basic Variable Value of the Basic Variable Coefficient MatrixIdentify Matrix CBCB Bb (= X B )x 1 x 2 s1s1 s2s2 s3s3 200x1x1 b 1 = 3 10¼0-1/8 0s2s2 b 2 = 3400-13/613/4 300 x2x2 b 3 = 901-1/1201/8 Contribution Loss per unit Z j = Σ C Bj a ij = 3300 200300250 Net Contribution per unit Z j - C j 00250

19
6/11/2014 The above simplex table – 3 yields the optimum solution and it is x1 = 3 and x2 = 9 with maximum Z = c1 x1 + c2 x2 = 200 x 3 + 300 x 9 = 3300

Similar presentations

OK

Chapter 6 Linear Programming: The Simplex Method Section 3 The Dual Problem: Minimization with Problem Constraints of the Form ≥

Chapter 6 Linear Programming: The Simplex Method Section 3 The Dual Problem: Minimization with Problem Constraints of the Form ≥

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google