Operations Research First Year Computer Text Book Operations

Operations Research First Year Computer Text Book: Operations Research 7 th Edition Hamdy A. Taha

Course Outline Part 1: Deterministic Models Chapter 2: Introduction to Linear Programming 2. 1 Introduction. 2. 2 Construction of the LP Model. 2. 3 Graphical LP Solution. 2. 3. 1 Solution of a Maximization Model. 2. 3. 2 Solution of a Minimization Model. 2. 3. 3 Slack, Surplus, and Unrestricted Model. 2. 4 Graphical Sensitivity Analysis. 2. 4. 1 Changes in the Objective Function Coefficient. 2. 4. 2 Unit Worth of a Resource. 2. 5 Computer Solution of Linear Programming Problems. 2. 6 Analysis of Selected Linear Programming Model. Dr. Farouk Sha'ban 2

Chapter 3: The Simplex Method 3. 1 Introduction. 3. 2 Standard Linear Programming Form and Its Solutions. 3. 2. 1 Standard Linear Programming Form. 3. 2. 2 Determination of Basic Solutions. 3. 2. 3 Unrestricted Variables and Basic Solution. 3. 3 The Simplex Algorithm. 3. 4 Simplex Method Application. 3. 5 Special Cases in Simplex Method Application. 3. 5. 1 Degeneracy. 3. 5. 2 Alternative Optima. 3. 5. 3 Unbounded Solution. 3. 5. 4 Infeasible Solution. Dr. Farouk Sha'ban 3

Chapter 4: Duality. 4. 1 Introduction. 4. 2 Definition of the Dual Problem. 4. 3 Relationship between the Optimal Primal and Dual. Chapter 5: Transportation Model. 5. 1 Definition of the Transportation Model. 5. 3 The Transportation Algorithm. 5. 3. 1 Determination of the Starting Solution. 5. 3. 2 Iterative Computation of the Algorithm. 5. 4 The Assignment Model. Dr. Farouk Sha'ban 4

Chapter 6: Network Models 6. 1 Scope of Network Application. 6. 2 Network Definitions. 6. 3 Minimal Spanning Tree Algorithm. 6. 4 Shortest Route Problem. 6. 5 Maximal Flow Model. 6. 7 CPM and PERT. 6. 7. 1 Network Representation. 6. 7. 2 Critical Path Computation. 6. 7. 3 Construction of the Time Schedule. Dr. Farouk Sha'ban 5

Introduction Operations Research (OR) It is a scientific approach to determine the optimum (best) solution to a decision problem under the restriction of limited resources. Using the mathematical techniques to model, analyze, and solve the problem. Phases of OR 1. Definition of the problem includes: • The description of the decision variables (alternatives) • The determination of the objective of the study • The specification of the limitations under which 6 Dr. Farouk Sha'banthe modeled system operates.

2. Model Construction Translating the real world problem into mathematical relationships (the most suitable model to represent the system, LP, dynamic programming, integer programming, ………. . ) 3. Solution of the model Using well-defined optimization techniques. An important aspect of model solution is sensitivity analysis. Dr. Farouk Sha'ban 7

4. Model validity Testing and evaluation of the model. A common method for testing a validity of a model is to compare its performance with some past data available for the actual system. 5. Implementation of the solution of validated model involves the translation of the model's results into instructions issued in understandable form to the individual Dr. Farouk Sha'ban 8

Basic components of the model 1. Decision Variables It is the unknown to determined from the solution of a model (what does the model seek to determine). It is one of the specific decisions made by a decision maker (DM). 2. Objective Function It is the end result (goal) desired to be achieved by the system. A common objective is to maximize profit or minimize cost. It is expressed as a mathematical function of the system decision variables. Dr. Farouk Sha'ban 9

3. Constraints These are the limitations imposed on the variables to satisfy the restriction of the modeled system. They must be expressed as mathematical functions of the system decision variables (D. V. ). Dr. Farouk Sha'ban 10

Example 1: A company manufactures two products A&B. with 4 & 3 units. A&B take 3&2 minutes respectively to be machined. The total time available at machining department is 800 hours (100 days or 20 weeks). A market research showed that at least 10000 units of A and not more than 6000 units of B are needed. It is required to determine the number of units of A&B to be produced to maximize profit. Dr. Farouk Sha'ban 11

Problem Formulation Decision variables X 1= number of units produced of A. X 2= number of units produced of B. Objective Function Maximize Z= 4 X 1 + 3 X 2 Constraints Dr. Farouk Sha'ban 3 X 1 + 2 X 1 X 2 X 1, X 2 <= 800 x 60 >=10000 <=6000 >=0 12

Example 2: Feed mix problem A farmer is interested in feeding his cattle at minimum cost. Two feeds are used A&B. Each cow must get at least 400 grams/day of protein, at least 800 grams/day of carbohydrates, and not more than 100 grams/day of fat. Given that A contains 10% protein, 80% carbohydrates and 10% fat while B contains 40% protein, 60% carbohydrates and no fat. A costs 2 L. E/kg, and B costs 5 L. E/kg. Formulate the problem to determine the optimum amount of each feed to minimize cost. Dr. Farouk Sha'ban 13

Problem Formulation Decision variables X 1= weight of feed A kg/day/animal X 2= weight of feed B kg/day/animal Objective Function Minimize Z= 2 X 1 + 5 X 2 Constraints Protein Carbohydrates Fats 0. 1 X 1 + 0. 4 X 2 0. 8 X 1 + 0. 6 X 2 0. 1 X 1, X 2 >=0. 4 >=0. 8 <= 0. 1 >=0 Dr. Farouk Sha'ban 14

Example 3: Blending Problem An iron ore from 4 mines will be blended. The analysis has shown that, in order to obtain suitable tensile properties, minimum requirements must be met for 3 basic elements A, B, and C. Each of the 4 mines contains different amounts of the 3 elements (see the table). Formulate to find the least cost blend for one ton of iron ore. Dr. Farouk Sha'ban 15

Problem Formulation Decision variables X 1= Fraction of ton to be selected from mine number 1 X 2= Fraction of ton to be selected from mine number 2 X 3= Fraction of ton to be selected from mine number 3 X 4= Fraction of ton to be selected from mine number 4 Objective Function Minimize Z= 800 X 1 + 400 X 2 + 600 X 3 + 500 X 4 Constraints 10 X 1 + 3 X 2 + 8 X 3 + 2 X 4 90 X 1 + 150 X 2 + 75 X 3 + 175 X 4 45 X 1 + 25 X 2 + 20 X 3 + 37 X 4 X 1 + X 2 + X 3 + X 4 X 1, X 2, X 3, X 4 Dr. Farouk Sha'ban >= 5 >= 10 >= 30 =1 >= 0 16

Example 4: Inspection Problem A company has 2 grades of inspectors 1&2. It is required that at least 1800 pieces be inspected per 8 hour day. Grade 1 inspectors can check pieces at the rate of 25 per hour with an accuracy of 98%. Grade 2 inspectors can check at the rate of 15 pieces per hour with an accuracy of 95%. Grade 1 costs 4 L. E/hour, grade 2 costs 3 L. E/hour. Each time an error is made by an inspector costs the company 2 L. E. There are 8 grade 1 and 10 grade 2 inspectors available. The company wants to determine the optimal assignment of inspectors which will minimize the total cost of inspection/day. Dr. Farouk Sha'ban 17

Problem Formulation Decision variables X 1= Number of grade 1 inspectors/day. X 2= Number of grade 2 inspectors/day. Objective Function Cost of inspection = Cost of error + Inspector salary/day Cost of grade 1/hour = 4 + (2 X 25 X 0. 02) = 5 L. E Cost of grade 2/hour = 3 + (2 X 15 X 0. 05) = 4. 5 L. E Minimize Z= 8 (5 X 1 + 4. 5 X 2) Constraints = 40 X 1 + 36 X 2 X 1 X 2 8(25) X 1+ 8(15) X 2 200 X 1 + 120 X 2 X 1 , X 2 Dr. Farouk Sha'ban <= 8 <= 10 >= 1800 >=0 18

Example 5: Trim-loss Problem. A company produces paper rolls with a standard width of 20 feet. Each special customer orders with different widths are produced by slitting the standard rolls. Typical orders are summarized in the following tables. Possible knife settings Dr. Farouk Sha'ban 19

Formulate to minimize the trim loss and the number of rolls needed to satisfy the order. Dr. Farouk Sha'ban 20

Problem Formulation Decision variables Xj = Number of standard rolls to be cut according to setting j j = 1, 2, 3, 4, 5, 6 Number of 5 feet rolls produced = 2 X 2 + 2 X 3 + 4 X 4 + X 5 Number of 7 feet rolls produced = X 1 + X 2+ 2 X 5 Number of 9 feet rolls produced = X 1 + X 3+ 2 X 6 Let Y 1, Y 2, Y 3 be the number of surplus rolls of the 5, 7, 9 feet rolls thus Y 1= 2 X 2 + 2 X 3 + 4 X 4 + X 5 - 150 Y 2= X 1 + X 2+ 2 X 5 - 200 Y 3= X 1 + X 3+ 2 X 6 - 300 The total trim losses = L (4 X 1 +3 X 2+ X 3 + X 5 + 2 X 6 + 5 Y 1+ 7 Y 2+ 9 Y 3) Where L is the length of the standard roll. Objective Function Minimize Z= L(4 X 1 +3 X 2+ X 3 + X 5 + 2 X 6 + 5 Y 1+ 7 Y 2+ 9 Y 3) Constraints 2 X 2+ 2 X 3+ 4 X 4+ X 5 - Y 1 = 150 X 1 + X 2 +2 X 5 - Y 2 = 200 X 1 + X 3 + 2 X 6 - Y 3 = 300 X 1, X 2, X 3, X 4, X 5, X 6 >= 0 Y 1, Y 2, Y 3 >= 0 Dr. Farouk Sha'ban 21

General form of a LP problem with m constraints and n decision variables is: Maximize Z = C 1 X 1+ C 2 X 2+ …………………. + Cn. Xn Constraints A 11 X 1 + A 12 X 2+…………+ A 1 n. Xn <= B 1 A 21 X 1 + A 22 X 2+…………+ A 2 n. Xn <= B 2. . Am 1 X 1 + Am 2 X 2+…………+ Amn. Xn <= Bm X 1, X 2, …………………, Xn >= 0 Dr. Farouk Sha'ban 22

OR Maximize Z= Σnj=1 Cj. Xj Constraints Σnj=1 aij. Xi <= bi Xi >=0 i = 1, …. , m j = 1, …. , n Where n = Number of activities. Xj = Level of activity j Cj =Contribution of the objective function/unit of activity j m = Number of resources Bi = Amount of resource i available. Aij=Amount of resource i consumed by one unit of activity j Other forms 1. 2. 3. 4. Minimize Z= Σnj=1 Cj. Xj Σnj=1 Aij. Xi >= Bi for some values of i Σnj=1 Aij. Xi = Bi for some values of i Xi unrestricted in sign for some values of i Dr. Farouk Sha'ban 23

Terminology of solutions for a LP model: A Solution Any specifications of values of X 1, X 2, ………, Xn is called a solution. A Feasible Solution Is a solution for which all the constraints are satisfied. An Optimal Solution Is a feasible solution that has the most favorable value of the objective function (largest for maximize or smallest for minimize). Dr. Farouk Sha'ban 24

Note • If there is exactly one optimal solution it must be a corner point feasible solution. • If there are multiple optimal solutions, then at least two of them must be adjacent corner- point feasible solutions. Two corner-point feasible solutions are said to be adjacent if the line segment connecting them lies on the boundary of the feasible region (one of the constraints). Dr. Farouk Sha'ban 25

Graphical Solution Construction of the LP model Example 1: The Reddy Mikks Company Reddy Mikks produces both interior and exterior paints from two raw materials, M 1&M 2. The following table provides the basic data of the problem. Dr. Farouk Sha'ban 26

A market survey indicates that the daily demand for interior paint cannot exceed that of exterior paint by more than 1 ton. Also, the maximum daily demand of interior paint is 2 ton. Reddy Mikks wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit. Dr. Farouk Sha'ban 27

Problem Formulation Decision variables X 1= Tons produced daily of exterior paint. X 2= Tons produced daily of interior paint. Objective Function Maximize Z= 5 X 1 + 4 X 2 Constraints 6 X 1 + 4 X 2 X 1 + 2 X 2 - X 1 + X 2 X 1, X 2 <= 24 <= 6 <= 1 <= 2 >=0 Any solution that satisfies all the constraints of the model is a feasible solution. For example, X 1=3 tons and X 2=1 ton is a feasible solution. We have an infinite number of feasible solutions, but we are interested in the optimum feasible solution that yields the maximum total profit. Dr. Farouk Sha'ban 28

Graphical Solution The graphical solution is valid only for two-variable problem which is rarely occurred. The graphical solution includes two basic steps: 1. The determination of the solution space that defines the feasible solutions that satisfy all the constraints. 2. The determination of the optimum solution from among all the points in the feasible solution space. Dr. Farouk Sha'ban 29

. ABCDEF consists of an infinite number of points; we need a systematic procedure that identifies the optimum solutions. The optimum solution is associated with a corner point of the solution space. Dr. Farouk Sha'ban 30

To determine the direction in which the profit function increases we assign arbitrary increasing values of 10 and 15 5 X 1 + 4 X 2=10 And 5 X 1 + 4 X 2=15 The optimum solution is mixture of 3 tons of exterior and 1. 5 tons of interior paints will yield a daily profit of 21000$. Dr. Farouk Sha'ban 31

Example 2: The Diet Problem Farm uses at least 800 lb of special feed daily. The special feed is a mixture of corn and soybean with the following composition. The food mixture must contain at least 30% protein and at most 5% fiber. They want to determine the daily minimum cost feed mix. Dr. Farouk Sha'ban 32

Problem Formulation Decision variables X 1= lb of corn in the daily mix. X 2= lb of soybean in the daily mix. Objective Function Minimize Z= 0. 3 X 1 + 0. 9 X 2 Constraints X 1 + X 2 >= 800 0. 09 X 1+0. 6 X 2 >= 0. 3(X 1+X 2) = -0. 21 X 1+0. 3 X 2 >= 0 0. 02 X 1+0. 06 X 2 <= 0. 05(X 1+X 2) = 0. 03 X 1 -0. 01 X 2 >= 0 X 1, X 2 >= 0 Dr. Farouk Sha'ban 33

Dr. Farouk Sha'ban 34

Slack, Surplus, and Unrestricted variables Slack Variable: For constraints of the type (<=) the R. H. S normally represents the limit on the availability of a resources and the L. H. S represents the usage of this limited resource by the different activities (variables) of the model. A slack represents the amount by which the available amount of the resource exceeds its usage by the activities. For example (6 X 1+4 X 2<= 24) is equivalent to (6 X 1+4 X 2 + S 1= 24) provided that S 1>=0. The slack variable S 1=24 -6 X 1 -4 X 2 represents the unused amount of raw material M 1. Dr. Farouk Sha'ban 35

Surplus Variable: It is used in the constraints of type (>=) normally set minimum specification requirements. Surplus represents the excess of the L. H. S over the minimum requirement. For example (X 1+X 2 >= 800) is equivalent to (X 1+X 2 - S 1= 800) provided that S 1>=0, this signifies that a surplus amount of feed over the minimum requirement will be produced. Unrestricted Variable: The variable which can be positive or negative. Dr. Farouk Sha'ban 36

The Simplex Method Introduction It is a general algebraic method to solve a set of linear equations. We use simplex method to get extreme (or corner) point solution. We must first convert the model into the standard LP form by using slack or surplus variables to convert the inequality constraints into equations. Our interest in the standard LP form lies in the basic solutions of the simultaneous linear equations. Dr. Farouk Sha'ban 37

Standard LP Form and its Basic Solution Example 1: Express the following LP model in standard form. Maximize Z= 2 X 1 + 3 X 2 + 5 X 3 S. T X 1 + X 2 - X 3 >= -5 -6 X 1 + 7 X 2 - 9 X 3 <= 4 X 1 + X 2 + 4 X 3 = 10 X 1, X 2 >= 0 X 3 unrestricted = X+3 - X-3 where X+3 , X-3 Standard LP form Maximize Z= 2 X 1 + 3 X 2 + 5 X+3 - 5 X-3 S. T -X 1 - X 2 + 3 X+3 - 3 X-3 + X 4 =5 -6 X 1 + 7 X 2 - 9 X+3 + 9 X-3 + X 5 = 4 X 1 + X 2 + 4 X+3 - 4 X-3 = 10 X 1, X 2, X+3, X-3, X 4, X 5 >=0 Dr. Farouk Sha'ban 38

Determination of basic solutions The standard LP form includes m simultaneous linear equations in n unknowns or variables (m<n). We divide the n variables into 2 sets: 1. (n-m) variables to which we assign zero values which are called non-basic variables. 1. Remaining m variables whose values are determined by solving the resulting m equations which is called basic variables. The resulting solution is Basic Solution(BS). If all values are satisfying non- negativity then resulting BS is Feasible Basic Solution(BFS), otherwise, it is infeasible. The maximum number of possible basic solution for m equations in n unknowns is Dr. Farouk Sha'ban 39

Example 2: Maximize Z= 5 X 1 + 6 X 2 S. T 2 X 1 + 3 X 2 2 X 1 +X 2 3 X 1 + 3 X 2 X 1, X 2 >= 18 <= 12 = 24 >=0 Standard LP form Maximize Z=5 X 1 + 6 X 2 S. T 2 X 1 +3 X 2 + X 3 2 X 1 + X 2 3 X 1 + 3 X 2 X 1, X 2, X 3, X 4, X 5 Dr. Farouk Sha'ban + X 4 + X 5 = 18 = 12 = 24 >=0 40

The Simplex Algorithm We solve the Reddy Mikks model, where X 1= Tons produced daily of exterior paint. X 2= Tons produced daily of interior paint. Exterior and interior paints are produced from two types of raw materials M 1 and M 2 Maximize Z- 5 X 1 - 4 X 2 + 0 X 3 + 0 X 4 + 0 X 5 +0 X 6 S. T. 6 X 1 +4 X 2 + X 3 X 1 +2 X 2 + X 4 - X 1 + X 2 + X 5 X 2 + X 6 X 1, X 2, X 3, X 4, X 5, X 6 = 0 <= 24 <= 6 <= 1 <=2 >=0 The variables X 3, X 4, X 5, X 6 are the slacks associated with the four (<=) constraints. Dr. Farouk Sha'ban 41

The Starting Basic Feasible Solution is Is the starting solution optimal? No, since the coefficients of X 1 and X 2 are still negative so they can increase the profit of Z. We choose X 1 with more negative (-5) i. e. ; X 1 is the entering variable. Dr. Farouk Sha'ban 42

Since none of the Z-row coefficients associated with the non-basic variables X 3, X 4 is negative the last table is optimal. Dr. Farouk Sha'ban 43

The rules for selecting the entering and leaving variables are referred to as the optimality and feasibility conditions (a) Optimality conditions: The entering variable in maximization (minimization) problem is the non-basic variable having the most negative (positive) coefficient in the Zrow. Ties are broken arbitrarily. The optimum is reached at the iteration where all the Z-row coefficients of the non-basic variables are non-negative (non-positive) (b) Feasibility Condition: for both the maximum and minimum problems the leaving variables is the basic variables associated with the smallest non-negative ratio. Ties are broken arbitrarily. The steps of the simplex method are: 1. Determine a starting basic feasible solution 2. Select an entering variable using the optimality condition. Stop if there is no entering variable. 3. Select a leaving variable using the feasibility condition. 4. Determine the new basic solution by using the appropriate Gauss. Jordan computation. Dr. Farouk Sha'ban 44 5. Go to step 1

Special Cases in Simplex Method 1. 2. 3. 4. Degeneracy Alternative optima (Infinity of Solution) Unbounded Solution Non-existing or Infeasible Solution 1) Degeneracy one or more basic variable(s) has zero value. If you find more than one leaving variable (i. e. ; we have two or more variables having the same ratio in the R. H. S) so one or more of the basic variable(s)will be equal zero in the next iteration. This condition indicates that the model has at least one redundant constraint. Dr. Farouk Sha'ban 45

Example 1: Maximize Z= 3 X 1 + 9 X 2 X 1 +4 X 2 <= 8 X 1 +2 X 2 <= 4 X 1, X 2 >=0 Standard LP form Maximize Z=3 X 1 + 9 X 2 S. T X 1 +4 X 2 + X 3 X 1 + 2 X 2 X 1, X 2, X 3, X 4 + X 4 Dr. Farouk Sha'ban =8 =4 >=0 46

X 1=0, X 2 =2, X 3 =0, X 4=0, Z=18 Is it possible to stop at the second iteration (when degeneracy first appears) even though it is not optimum? The answer is NO, because the solution may be temporarily degenerate. Dr. Farouk Sha'ban 47

2) Alternative optima (Infinity of. Solution): When the objective function is parallel to a binding constraint. So the objective function (Z) will have the same optimal value at more than one solution point. Example 2: Maximize Z= 2 X 1 + 4 X 2 X 1 + 2 X 2 X 1 + X 2 X 1, X 2 <= 5 <= 4 >=0 Standard LP form Maximize Z=2 X 1 + 4 X 2 S. T X 1 +2 X 2 + X 3 =8 X 1 + X 2 + X 4 = 4 X 1, X 2, X 3, X 4 >=0 Dr. Farouk Sha'ban 48

First Optimum Solution: X 1*=0, X 2*=5/2, X 3*=0, X 4*=3/2, Z*=10 Second Optimum Solution: X 1*=3, X 2*=1, X 3*=0, X 4*=0, Z*=10 In the second table: X 1=0 , X 2= 5/2 In the third table: X 1= 3 , X 2= 1 The general form: X 1 = α (0) + (1 -α) (3) = 3 -3α X 2 = α (5/2) + (1 -α) (1) = 1+3/2α α=0→ 1 Dr. Farouk Sha'ban 49

3) Unbounded Solution: In which the objective function (Z) can be increased indefinitely without violating any of the constraints i. e. ; the solution space is unbounded in at least one direction. Example 3: Maximize Z= 2 X 1+ X 2 S. T X 1 - X 2 <= 10 2 X 1 <= 40 X 1, X 2 >=0 Dr. Farouk Sha'ban 50

Standard LP form Maximize Z=2 X 1 +X 2 S. T X 1 -X 2 + X 3 = 10 2 X 1 + X 4= 40 X 1, X 2, X 3, X 4 >=0 Both X 1, X 2 are candidates for entering the solution. But all the constraint coefficient under X 2 are negative or zero, meaning that X 2 can be increased indefinitely without violating any of the constraints. Any increase in X 2 will increase Z so it can be increased ∞ as X 2 → ∞. Thus the problem has no bounded solution. Dr. Farouk Sha'ban 51

4) Non-existing or Infeasible Solution: If the constraints are not satisfied simultaneously, the model has no feasible solution. This situation can never occur if all the constraints are of the type <= (assuming nonnegative R. H. S constraints) because the slacks provide a feasible solution. Example 4: Maximize Z= 3 X 1 +2 X 2 S. T 2 X 1 + X 2 <= 2 3 X 1 +4 X 2 >= 12 X 1, X 2 >=0 Dr. Farouk Sha'ban 52

Transportation Model and Its Variants Definition of the transportation model The transportation model is a special class of the LP problem. It deals with the situation in which a commodity is shipped from source (e. g. factories) to destinations (e. g. warehouses). The objective is to determine the amounts shipped from each source to each destination that minimize the total shipping cost while satisfying both the supply limits and the demand requirements. Dr. Farouk Sha'ban 53

Example 1: There are three plants A, B, C and two destinations 1 and 2, the distance between them is as follows The mile costs 8 cents find the cost/unit round to $ We define X 11 as the amount shipped from A to 1 X 12 as the amount shipped from A to 2 X 21 as the amount shipped from B to 1 X 22 as the amount shipped from B to 2 X 31 as the amount shipped from C to 1 X 32 as the amount shipped from C to 2 The plants A, B, and C produces 1000, 1500, and 1200 cars respectively. The destination 1&2 demands are 2300 and 1400 cars. Dr. Farouk Sha'ban 54

The LP model is Minimize Z=80 X 11+215 X 12 +100 X 21 +108 X 22 +102 X 31 +68 X 32 S. T X 11 + X 12 = 1000 ………. …… (A) X 21 + X 22 = 1500 ……………. (B) X 31 + X 32 = 1200 ……………. (C) X 11 + X 21+ X 31 = 2300 ……………. (1) X 12 + X 22+ X 32 = 1400 …………. . . (2) Xij >=0 i=1, 2, 3 j=1, 2 These constraints are all equations because the total supply from the three plants = 1000 + 1500 + 1200 = 3700 cars, equals the total demands at the two destinations = 2300 + 1400 = 3700 cars. The LP model can be solved by the simplex method. However, the special structure of the constraints allows us to solve the problem more conveniently using the transportation tableau. When the total supply does not equal the total demand, the transportation model is said to be unbalanced. We must add a dummy source or destination. Dr. Farouk Sha'ban 55

Example 2: In example (1) suppose that plant B capacity is 1300 cars (instead of 1500). The total supply =3500 cars and the total demand = 3700 cars. So the total supply < the total demand. Therefore, we have to add dummy source with capacity 200 cars. Dr. Farouk Sha'ban 56

Example 3: In example (1) suppose that the demand of the destination 1 is 1900 cars (instead of 2300). The total supply =3700 cars and the total demand = 3300 cars. So the total supply > the total demand. Therefore, we have to add dummy destination with 400 cars. Dr. Farouk Sha'ban 57

The Transportation Algorithm Step (1): Determine an Initial Basic Feasible Solution (IBFS) by using one of the following methods. 1. The North-West Corner Method OR, 2. The Least-Cost Method OR, 3. Vogel Approximation Method. A general T. M. with m sources and n destinations has m + n constraints, one for each source and each destination. However, because the T. M. is always balanced, one of these equation is redundant. Thus, the model has m + n - 1 independent constraint equations, which means that the starting basic solution consists of m + n – 1 basic variables. Step (2): Improve the IBFS • Determine an entering variable. • Determine a leaving variable. • Obtain a new BFS • If the solution is optimal stop. Otherwise; repeat step (2). Dr. Farouk Sha'ban 58

Steps of North-West Method 1. Allocate as much as possible to a selected cell X 11 2. Cross out the row or column with zero supply or demand (if both are satisfied at the same time, only one should be crossed out). 3. If exactly one row or column is left uncrossed out stop. Otherwise go to step(1) Dr. Farouk Sha'ban 59

Example 3: Solve the following transportation problem. Solution Step (1): finding the IBFS • Using the North-West Corner Method Transportation Cost = 5(10) +10(2) +5(7) +15(9) +5(20) +10(18) =520 Dr. Farouk Sha'ban 60

The Least-Cost Method: This method gives a better starting solution because it takes the cost into consideration Transportation Cost = 15(2) +15(9) +0(7) +10(20) +5(4) +5(18) =475 Dr. Farouk Sha'ban 61

Steps of Least-Cost Method 1. Assign as much as possible to the variable with the smallest unit cost cell. 2. Cross out the row or column with zero supply of demand (if both are satisfied at the same time, only one should be crossed out). 3. If exactly one row or column is left uncrossed out stop. Otherwise go to step(1) Dr. Farouk Sha'ban 62

The Vogel Approximation Method: This method provides a better starting solution (optimal or closed to the optimal solution) Steps of Vogel Approximation Method 1. In each row and column, subtract the lowest two costs from each other. 2. Determine the row or column with the largest difference. 3. Allocate as much as possible to the variable with the least cost in the selected row or column. Any row or column with zero supply or demand would not be used in computing the following step. 4. If exactly one row or column remains uncrossed out stop. Otherwise, go to step (1). 5. If only one row or column with positive supply or demand remains uncrossed out determine the basic variable by the least-cost method. 6. If all uncrossed out rows and columns have assigned zero supply and demand, determine the zero basic variable by the least-cost method. Note: In all the three methods, if you have a tie you can break the tie arbitrarily. Dr. Farouk Sha'ban 63

X 31=5 X 12=15 X 23=15 X 14=0, X 34=5, X 24=10 Dr. Farouk Sha'ban Transportation Cost = 15(2) + 15(9) +0(4) +10(20) +5(4) +5(18) = 475 It is expected to produce better starting solution than least-cost method. 64

Improving the IBFS Determine an entering variable using method of multipliers (U, V method), In this method associate the multipliers with row I and column j of the transportation tableau. • For each current basic variable Ui + Vj = Cij for each basic Xij • For non-basic variable C'ij = Ui + Vj - Cij for each non-basic Xij Choose the non-basic variable with the largest positive coefficient (C'ij) to be the entering variable. The solution is optimal if and only if all C'ij are nonpositive. Dr. Farouk Sha'ban 65

Improving the IBFS Determine an entering variable using method of multipliers (U, V) method, In this method associate the multipliers with row I and column j of the transportation tableau. For each current basic variable Ui + Vj = Cij for each basic Xij For non-basic variable C’ij = Ui + Vj - Cij for each non-basic Xij Choose the non-basic variable with the largest positive coefficient (C’ij) to be the entering variable. The solution is optimal if and only if all C’ij are non-positive. Consider the same example: Dr. Farouk Sha'ban 66

Dr. Farouk Sha'ban 67

The entering variable must have the maximum C'ij which is X 31 = 9 To determine the leaving variable make a closed loop of basic variable starting with maximum C'ij of non-basic variable keeping: 1. Supply and demand requirements remain satisfied. 2. No negative shipments are allowed through any of the routes. Therefore, θ=5 and X 31 will be basic with value 5 and X 22 is a basic variable but with value 0 and X 11 will be non-basic (leaving variable) The cost of the IBFS by Northwest corner method was 520$ The first iteration cost is 475$ (less which is better) In the second iteration Get the entering variable with maximum C'ij = C'14 = 4 X 14 = 10 The Leaving variable is X 24 =10 and the new solution is: Dr. Farouk Sha'ban 68

All C'ij is negative for all non-basic variables, so the solution is optimal which is: X 12 =5, X 14 =10, X 22 =10, X 23 =15, X 31 =5, X 34 =5 With optimum cost = 5*2 +10*11 + 10*7 + 15*9 +5*4 + 5*18 = 10+110+70+135+20+90 = 435 Dr. Farouk Sha'ban 69

The Assignment Model There are n machines M 1, M 2, M 3…, Mn. And n different jobs J 1, J 2, …. , Jn. The jobs are to be assigned to these machines. The machine cost for each job depends on the machine to which it is assigned. Each machine can work only on one job. The problem is to assign the jobs to the machines to minimize the total cost of machining. Number of possible assignment = n! If n=5, then 5! = 120 ways Dr. Farouk Sha'ban 70

The Linear Programming Formulation of the Problem Decision variables Xij job i is assigned to machine j Objective Function Minimize Z= Σnj=1 Σni=1 Cij Xij Constraints Σnj=1 Xij =1 Σni=1 Xij =0 i= 1, 2… n j= 1, 2… n if the job i is assigned to the machine j other wise The Assignment Algorithm (Hungarian Algorithm) It is the most efficient way or solving the assignment problem. It is based on the idea that if a constant is added or subtracted from any row or column the optimal assignment is not affected. Dr. Farouk Sha'ban 71

The Algorithm Steps 1. Subtract the smallest value in each column from all the values in that column. 2. Subtract the smallest value in each row from all the values in that row. 3. Note: step 1 & 2 will produce a cost matrix containing at least one zero element in each row or column. 4. Determine a feasible assignment using the cells with zero cost. If this is possible stop, as optimal solution is found. This step is done as follows: a) Determine the number of lines required to cover all zeros. b) If the number of lines is equal to n, stop (This is the optimal assignment). c) If the number of lines is less than n then go to (4). 5. Subtract the smallest value not covered with lines from all uncovered values and add it to the value at the intersection of the lines then go to (3). Note: Obtain the minimum cost from the original table. Dr. Farouk Sha'ban 72

Example 1: Solve the following assignment problem. Jobs Person Solution: The optimal assignment is X 12, X 21, X 33 Job 1 is assigned to machine 2 Job 2 is assigned to machine 1 Job 3 is assigned to machine 3 The cost = 10 + 9 + 8 = 27$ Dr. Farouk Sha'ban 73

Example 2: Solve the following assignment problem. Solution: The optimal assignment is X 15, X 22, X 33, X 44, X 51 Job 1 is assigned to machine 5 Job 2 is assigned to machine 2 Job 3 is assigned to machine 3 Job 4 is assigned to machine 4 Job 5 is assigned to machine 1 The cost = 3 + 4 + 2 + 5 + 3 = 17$ Dr. Farouk Sha'ban 74

6. Network Models 6. 1 Introduction There is a multitude of operations research situations that can be modeled and solved as networks (nodes connected with branches) We have 5 network algorithms such as: 1. Minimal spanning tree. 2. Shortest route algorithm. 3. Minimum-cost capacitated network algorithm. 4. Critical Path Method (CPM). 5. Program Evaluation and Review Technique (PERT). 6. 7 Project planning and control with PERT-CPM The successful management of large-scale projects requires careful planning, scheduling and control of numerous interrelated activities especially when these activities have to be performed in a specified technological sequence, to aid in these tasks special procedures based on the use of network techniques were developed. The most important of these procedures are: Critical Path Method (CPM). Dr. Farouk Sha'ban 75 Program Evaluation and Review Technique (PERT).

Applications of PERT & CPM 1. Construction projects ( buildings, highways, houses, bridges) 2. Preparations of bids and proposals for large projects. 3. Maintenance planning of oil refineries, ship repair and other large operations. 4. Manufacture and assembly of large items as aero plans and computers. 5. Development of new weapons system. 6. Simple projects such as cleaning and painting. Dr. Farouk Sha'ban 76

Basic Terminology 1. A Project: A project defines a combination of interrelated activities that must be done in a certain order before the entire task can be completed. 2. An Activity: It is an effort required to complete a plan of the project. Most activities can not start until some others are completed. 3. Real Activity: Consumes resources and time. 4. Waiting Activity: Consumes time. 5. Dummy Activity: Consumes no time and no resources. It represents the dependence of one activity upon another. 6. An Event: Represents a point in time. It signifies the beginning or ending of Dr. Farouk Sha'ban some activities. 77

Network Representation 1. An activity is represented by a line or an arrow (usually with time estimate written along it). The arrow begins and ends with an event which is represented by a node. 1. A dummy activity is represented by dashed line arrow or solid arrow with zero time estimate. 2. The arrow must be oriented from left to right. 3. The network must have one initial node and one ending node. 4. Any two events may not be connected by more than one activity. 5. Event numbers must not be duplicated in the network. 6. Each activity is represented by one and only one arrow. No single activity can be represented twice in the network. Dr. Farouk Sha'ban 78

Numbering a network 1. An initial event is one which has arrows coming out of it only. Number it (1). 2. Delete all arrows emerging from event (1). This will create at least one more initial event, number these new events(s) as (2, 3…). 3. Delete all emerging arrows from these numbered events which will create new initial events and continue until the last event which has no arrows emerging from it. Dr. Farouk Sha'ban 79

Solution by Linear Programming Formulation Each node represents a variable. Each arc represents a constraint. Example 1: Consider the following Network problem. Decision variables Let ti represents the time at which event (i) occurs i= 1, 2, 3, 4, 5. t 4: time at which jobs (C) & (D) are completed t 5: time at which the project is completed Time of completion of the whole project = t 5 – t 1 Objective Function Constraints Minimize Z= t 5 – t 1 t 2 – t 1 ≥ 3 , t 3 – t 1 ≥ 1 , t 3 – t 2 ≥ 0 t 4 – t 3 ≥ 2 , t 4 – t 2 ≥ 4 , t 5 – t 4 ≥ 5 ti ≥ 0 i= 1, 2, 3, 4, 5 This problem can be solved by simplex method Z* is the minimum time required to complete the project. Dr. Farouk Sha'ban 80

Solution by Network Analysis CPM Calculations: Let Dij = Duration (completion time) of activity ij ESi = The earliest start time at which event (i) occurs LFi = The latest finish time at which event (i) occurs without delaying the project beyond its earliest time. ESi ………. Forward ESi = max [ESi+ Dij] for all (i, j) ESi = max [(2+8), (1+6), (3+5)] Dr. Farouk Sha'ban 81

Start from the terminal node and move backward to the initial node for terminal node LFn = ESn = max [10, 7, 8] =10 Start from the initial node and move to the terminal node for initial node ES 1= 0 LFi ………. Backward ESi = min [LFj - Dij] for all (i, j) LFi = min [(10 -3), (9 -5), (13 -4)] = min [7, 4, 9] =10 Dr. Farouk Sha'ban 82

Example 1: Determine the critical path for the project network given below (all the durations are in days) Solution The critical path is defined by 1 -2 -4 -5 -6 The duration of the project = 25 days Dr. Farouk Sha'ban 83