UploadLanguage (EN)Support
BusinessMobileSocial MediaMarketingTechnologyArt & PhotosCareerDesignEducationPresentations & Public SpeakingGovernment & NonprofitHealthcareInternetLawLeadership & ManagementAutomotiveEngineeringSoftwareRecruiting & HRRetailSalesServicesScienceSmall Business & EntrepreneurshipFoodEnvironmentEconomy & FinanceData & AnalyticsInvestor RelationsSportsSpiritualNews & PoliticsTravelSelf ImprovementReal EstateEntertainment & HumorHealth & MedicineDevices & HardwareLifestyle
Upload
Sign in
Uploaded byRushikeshKadam71
PPTX, PDF61 views

MFCS2-Module1.pptx

The document provides an example to formulate a linear programming problem (LPP) and solve it graphically. It first defines the steps to formulate an LPP which includes identifying decision variables, writing the objective function, mentioning constraints, and specifying non-negativity restrictions. It then gives an example problem on maximizing profit from production of two products with machine hours and input requirements. This example problem is formulated as an LPP and represented graphically to arrive at the optimal solution.

Education◦
Related topics:
Operations Research•
Download to read offline
1 / 88
2 / 88
3 / 88
4 / 88
5 / 88
6 / 88
7 / 88
8 / 88
9 / 88
10 / 88
11 / 88
12 / 88
13 / 88
14 / 88
15 / 88
16 / 88
17 / 88
18 / 88
19 / 88
20 / 88
Mathematical Foundation for Computer Science 2
Linear Programming Problem Module-1
Syllabus Introduction, Formulation of linear programming problem and basic feasible solution: graphical method, Simplex method, artificial variables, Big M method, Two Phase method. Self Learning Topics:special cases of LPP
Introduction
Introduction (contd…)
History of OR
Linear Programming
Linear Programming
Linear Programming Let us look at the steps of defining a Linear Programming problem generically: 1. Identify the decision variables 2. Write the objective function 3. Mention the constraints 4. Explicitly state the non-negativity restriction For a problem to be a linear programming problem, the decision variables, objective function and constraints all have to be linear functions. If all the three conditions are satisfied, it is called a Linear Programming Problem.
Example Suppose an industry is producing two types of products P1 and P2. The profit per kg of Rs. 30 and Rs. 40 for the two products respectively. These two products require processing in three types of machines. Following table shows the available machine hours per day and time required on each machine to produce 1 kg of P1 and P2. Formulate the problem in LP model
Example
Problem Consider a chocolate manufacturing company that produces only two types of chocolate – A and B. Both the chocolates require Milk and Choco only. To manufacture each unit of A and B, the following quantities are required: ● Each unit of A requires 1 unit of Milk and 3 units of Choco ● Each unit of B requires 1 unit of Milk and 2 units of Choco The company kitchen has a total of 5 units of Milk and 12 units of Choco. On each sale, the company makes a profit of ● Rs 6 per unit A sold ● Rs 5 per unit B sold. Now, the company wishes to maximize its profit. Formulate LPP? Profit: Max Z = 6X+5Y X+Y ≤ 5 3X+2Y ≤ 12 X ≥ 0 & Y ≥ 0
Problem A farmer has recently acquired a 110 hectares piece of land. He has decided to grow Wheat and barley on that land. Due to the quality of the sun and the region’s excellent climate, the entire production of Wheat and Barley can be sold. He wants to know how to plant each variety in the 110 hectares, given the costs, net profits and labor requirements according to the data shown below: The farmer has a budget of Rs. 10,000 and availability of 1,200 man-days during the planning horizon. Formulate the LPP. Max Z = 50X + 120Y 100X + 200Y ≤ 10,000 10X + 30Y ≤ 1200 X + Y ≤ 110 X ≥ 0, Y ≥ 0
Graphical Method Graphical solution is limited to linear programming models containing only two decision variables. Procedure Step I: Convert each inequality as equation Step II: Plot each equation on the graph Step III: Shade the ‘Feasible Region’. Highlight the common Feasible region. ❑ Feasible Region: Set of all possible solutions. Step IV: Compute the coordinates of the corner points (of the feasible region). These corner points will represent the ‘Feasible Solution’. ❑ Feasible Solution: If it satisfies all the constraints and non negativity restrictions.
Procedure (Cont...) Step V: Substitute the coordinates of the corner points into the objective function to see which gives the Optimal Value. That will be the ‘Optimal Solution’. ❑ Optimal Solution: If it optimizes (maximizes or minimizes) the objective function. ❑ Unbounded Solution: If the value of the objective function can be increased or decreased indefinitely, Such solutions are called Unbounded solution. ❑Inconsistent Solution: It means the solution of problem does not exist. This is possible when there is no common feasible region. Graphical Method
Convert into Standard form
Outgoing variable Incoming variable
Final Table Optimal Solution: X1 = 2 X2 =2 Z = 26
Problem
Problem
Example
Cj -1 3 -2 0 0 0 Min. Ratio CB B.V X1 X2 X3 S1 S2 S3 b 0 S1 3 -1 3 1 0 0 7 - 0 S2 -2 4 0 0 1 0 12 12/4=3 0 S3 -4 3 8 0 0 1 10 10/3=3.33 Zj 0 0 0 0 0 0 Cj - Zj -1 3 -2 0 0 0
R1 R2 R3 Cj -1 3 -2 0 0 0 Min. Ratio CB B.V X1 X2 X3 S1 S2 S3 b 0 S1 5/2 0 3 1 1/4 0 10 4 3 X2 -1/2 1 0 0 1/4 0 3 - 0 S3 -5/2 0 8 0 -3/4 1 1 - Zj -3/2 3 0 0 3/4 0 Cj - Zj 1/2 0 -2 0 -3/4 0 NR2= R2/4 NR1= R1 + NR2 NR3= R3- 3 NR2
R1 R3 R2 Cj -1 3 -2 0 0 0 CB B.V X1 X2 X3 S1 S2 S3 b -1 X1 1 0 6/5 2/5 1/10 0 4 3 X2 0 1 3/5 1/5 3/10 0 5 0 S3 0 0 11 1 -1/2 1 11 Zj -1 3 3/5 1/5 4/5 0 Cj - Zj 0 0 -13/5 -1/5 -4/5 0 NR1=2R1/5 NR2= R2 + NR1/2 NR3= R3+ 5 NR2/2 All Δj are -ve or 0 so stop.
Solution X1 =4 , X2 = 5 , X3=0 Z’ = 11 Z = -11 It is the case of degenerate solution as one of the decision variable is equal to 0.
Example
Δ
Big M Method
Pivot row= row/key element
Example
Table -1 Pivot row= row/key element
Table -2
Table-3
MIXED CONSTRAINTS
Simplex Table II Cj 2 3 4 0 0 -M -M Basic variable solution Value X1 X2 X3 S1 S2 A1 A2 Min Ratio 0 S1 480 5/2 0 7/2 1 1/4 -1/4 0 137.1 3 X2 120 1/2 1 1/2 0 -1/4 1/4 0 240 -M A2 180 1/2 0 3/2 0 3/4 -3/4 1 120 Zj 360- 180M - M/2 +3/2 3 - 3M/2 + 3/2 0 -3M/4 -3/4 3M/4 + 3/4 -M Cj-Zj M/2 +1/2 0 3M/2 + 5/2 0 3M/4+ 3/4 - 7M/4- 3/4 0 NR2=R2/4 NR1=R1 - NR2 NR3= R3 - 3xNR2
Simplex Table III Cj 2 3 4 0 0 -M -M Basic variable solutio n Value X1 X2 X3 S1 S2 A1 A2 0 S1 20 4/3 0 0 1 -3/2 3/2 -7/3 3 X2 60 1/3 1 0 0 -1/2 1/2 -1/3 4 X3 120 1/3 0 1 0 1/2 -1/2 2/3 Zj 660 7/3 3 4 0 1/2 -1/2 5/3 Cj-Zj -1/3 0 0 0 -1/2 -M+ 1/2 -M -5/3 NR3=2xR3/3 NR1=R1 - 7xNR3/2 NR2= R2 - NR3/2 Solution: X1=0 X2=60 X3= 120 Z= 660
Two Phase Method
Two Phase Method contd….
Example
Standard form Min Z = x1 + x2 +0.s1 +0.s2 + A1 + A2 2x1 + x2 - s1 + A1 = 4 x1 + 7x2 - s2 + A2 =7 x1,x2,s1,s2,A1,A2 >=0 Phase 1 Min Z = A1 + A2 Table 1 Cj 0 0 0 0 1 1 Basic variable solution Value X1 X2 S1 S2 A1 A2 Min. Ratio 1 A1 4 2 1 -1 0 1 0 4 1 A2 7 1 7 0 -1 0 1 1 Zj 11 3 8 -1 -1 1 1 Cj-Zj -3 -8 1 1 -1 -1
Cj 0 0 0 0 1 1 Basic variable solution Value X1 X2 S1 S2 A1 A2 Min. Ratio 1 A1 3 13/7 0 -1 1/7 1 -1/7 21/13 0 X2 1 1/7 1 0 -1/7 0 1/7 7 Zj 3 13/7 0 -1 1/7 1 -1/7 Cj-Zj -13/7 0 1 -1/7 0 6/7 Table 2
Cj 0 0 0 0 1 1 Basic variable solution Value X1 X2 S1 S2 A1 A2 0 X1 21/13 1 0 -7/13 1/13 7/13 -1/13 0 X2 10/13 0 1 1/13 -2/13 -1/13 2/13 Zj 0 0 0 0 0 0 0 Cj-Zj 0 0 0 0 1 1 Table 3
Cj 1 1 0 0 Basic variable solution Value X1 X2 S1 S2 1 X1 21/13 1 0 -7/13 1/13 1 X2 10/13 0 1 1/13 -2/13 Zj 31/13 1 1 -6/13 -1/13 Cj-Zj 0 0 6/13 1/13 Table 1 Phase 2 Standard form Min Z = x1 + x2 +0.s1 +0.s2 2x1 + x2 - s1 = 4 x1 + 7x2 - s2 =7 x1,x2,s1,s2,A1,A2 >=0 Solution X1 = 21/13 X2 = 10/13 Z= 31/13
Example
Phase I Min
Table 1
Table 2
Table 3
Phase II
Problem