2. Constraint management

3. Constraint Something that limits the performance of a process or system in achieving its goals. Categories: Market (demand side) Resources (supply side) Labour Equipment Space Material and energy Financial Supplier Competency and knowledge Policy and legal environment

4. Steps of managing constraints Identify (the most pressing ones) Maximizing the benefit, given the constraints (programming) Analyzing the other portions of the process (if they supportive or not) Explore and evaluate how to overcome the constraints (long term, strategic solution) Repeat the process

5. Linear programming

6. Linear programming… …is a quantitative management tool to obtain optimal solutions to problems that involve restrictions and limitations (called constrained optimization problems). …consists of a sequence of steps that lead to an optimal solution to linear-constrained problems, if an optimum exists.

7. Typical areas of problems Determining optimal schedules Establishing locations Identifying optimal worker-job assignments Determining optimal diet plans Identifying optimal mix of products in a factory (!!!) etc.

8. Linear programming models …are mathematical representations of constrained optimization problems. BASIC CHARACTERISTICS: Components Assumptions

9. Components of the structure of a linear programming model Objective function: a mathematical expression of the goal e. g. maximization of profits Decision variables: choices available in terms of amounts (quantities) Constraints: limitations restricting the available alternatives; define the set of feasible combinations of decision variables (feasible solutions space). Greater than or equal to Less than or equal to Equal to Parameters. Fixed values in the model

10. Assumptions of the linear programming model Linearity: the impact of decision variables is linear in constraints and the objective functions Divisibility: noninteger values are acceptable Certainty: values of parameters are known and constant Nonnegativity: negative values of decision variables are not accepted

11. Model formulation The procesess of assembling information about a problem into a model. This way the problem became solved mathematically. 1. 1. Identifying decision variables (e.g. quantity of a product) 2. 2. Identifying constraints 3. 3. Solve the problem.

12. Graphical linear programming 1. 1. Set up the objective function and the constraints into mathematical format. 2. 2. Plot the constraints. 3. 3. Identify the feasible solution space. 4. 4. Plot the objective function. 5. 5. Determine the optimum solution. 1. 1. Sliding the line of the objective function away from the origin to the farthes/closest point of the feasible solution space. 2. 2. Enumeration approach.

13. Corporate system-matrix 1.) Resource-product matrix Describes the connections between the company’s resources and products as linear and deterministic relations via coefficients of resource utilization and resource capacities. 2.) Environmental matrix (or market-matrix): Describes the minimum that we must, and maximum that we can sell on the market from each product. It also describes the conditions.

14. Contribution margin Unit Price - Variable Costs Per Unit = Contribution Margin Per Unit Contribution Margin Per Unit x Units Sold = Product’s Contribution to Profit Contributions to Profit From All Products – Firm’s Fixed Costs = Total Firm Profit

15. Resource-Product Relation types P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 R1R1 a 11 R2R2 a 22 R3R3 a 32 R4a 43 a 44 a 45 R5a 56 a 57 R6a 66 a 67 Non-convertible relations Partially convertible relations

16. Product-mix in a pottery – corporate system matrix JugPlate Clay (kg/pcs)1,00,5 Weel time (hrs/pcs) 0,51,0 Paint (kg/pcs)00,1 Capacity 50 kg/week 100 HUF/kg 50 hrs/week 800 HUF/hr 10 kg/week 100 HUF/kg Minimum (pcs/week) 10 Maximum (pcs/week) 100 Price (HUF/pcs)7001060 Contribution margin (HUF/pcs) e 1 : 1*P 1 +0,5*P 2 < 50 e 2 :0,5*P 1 +1*P 2 < 50 e 3 : 0,1*P 2 < 10 m 1, m 2 : 10 < P 1 < 100 m 3, m 4 : 10 < P 2 < 100 of CM : 200 P 1 +200P 2 =MAX 200

17. Objective function refers to choosing the best element from some set of available alternatives. X*P 1 + Y*P 2 = max variables (amount of produced goods) weights (depends on what we want to maximize: price, contribution margin)

18. Solution with linear programming T1T1 T2T2 33,3 33 jugs and 33 plaits a per week Contribution margin: 13 200 HUF / week e 1 : 1*P 1 +0,5*P 2 < 50 e 2 :0,5*P 1 +1*P 2 < 50 e 3 : 0,1*P 2 < 10 m 1,m 2 : 10 < P 1 < 100 m 3, m 4 : 10 < P 2 < 100 of CM : 200 P 1 +200P 2 =MAX e1e1 e2e2 e3e3 of F 100

19. What is the product-mix, that maximizes the revenues and the contribution to profit! P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 b (hrs/y) R1R1 4 2 000 R2R2 21 3 000 R3R3 1 1 000 R4R4 23 6 000 R5R5 22 5 000 MIN (pcs/y)100200 50100 MAX (pcs/y)4001100 1 000 500 1 500 2000 p (HUF/pcs)2002702003050150 f (HUF/pcs)10011050-103020

20. Solution P 1 : Resource constraint 2000/4 = 500 > market constraint 400 P 2 &P 3 : Which one is the better product? Rev. max.: 270/2 < 200/1thus P 3 P 3 =(3000-200*2)/1=2600>1000 P 2 =200+1600/2=1000<1100 Contr. max.: 110/2 > 50/1 thus P 2 P 2 =(3000-200*1)/2=1400>1100 P 3 =200+600/1=800<1000

21. P 4 : does it worth? Revenue max.: 1000/1 > 500 Contribution max.: 200 P 5 &P 6 : linear programming e 1 : 2*T 5 + 3*T 6 ≤ 6000 e 2 :2*T 5 + 2*T 6 ≤ 5000 m 1, m 2 :50 ≤ T 5 ≤ 1500 p 3, m 4 :100 ≤ T 6 ≤ 2000 of TR :50*T 5 + 150*T 6 = max of CM :30*T 5 + 20*T 6 = max

22. e2 e1 of CM ofTR Contr. max: P5=1500, P6=1000 Rev. max: P5=50, P6=1966 T5 T6 2000 3000 2500

23. Exercise 1.1 Set up the product-resource matrix using the following data! RP coefficients: a 11 : 10, a 22 : 20, a 23 : 30, a3 4 : 10 RP coefficients: a 11 : 10, a 22 : 20, a 23 : 30, a3 4 : 10 The planning period is 4 weeks (there are no holidays in it, and no work on weekends) The planning period is 4 weeks (there are no holidays in it, and no work on weekends) Work schedule: Work schedule: R 1 and R 2 : 2 shifts, each is 8 hour long R 1 and R 2 : 2 shifts, each is 8 hour long R 3 : 3 shifts R 3 : 3 shifts Homogenous machines: Homogenous machines: 1 for R 1 1 for R 1 2 for R 2 2 for R 2 1 for R 3 1 for R 3 Maintenance time: only for R 3 : 5 hrs/week Maintenance time: only for R 3 : 5 hrs/week Performance rate: Performance rate: 90% for R 1 and R 3 90% for R 1 and R 3 80% for R 2 80% for R 2

24. Solution (b i ) R i = N ∙ s n ∙ s h ∙ m n ∙ 60 ∙  R i = N ∙ s n ∙ s h ∙ m n ∙ 60 ∙  N=(number of weeks) ∙ (working days per week) N=(number of weeks) ∙ (working days per week) R 1 = 4 weeks ∙ 5 working days ∙ 2 shifts ∙ 8 hours per shift ∙ 60 minutes per hour ∙ 1 homogenous machine ∙ 0,9 performance = = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9 = 17 280 minutes per planning period R 1 = 4 weeks ∙ 5 working days ∙ 2 shifts ∙ 8 hours per shift ∙ 60 minutes per hour ∙ 1 homogenous machine ∙ 0,9 performance = = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9 = 17 280 minutes per planning period R 2 = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 2 ∙ 0,8 = 38 720 mins R 2 = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 2 ∙ 0,8 = 38 720 mins R 3 = (4 ∙ 5 ∙ 3 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9) – (5 hrs per week maintenance ∙ 60 minutes per hour ∙ 4 weeks ) = 25 920 – 1200 = 24 720 mins R 3 = (4 ∙ 5 ∙ 3 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9) – (5 hrs per week maintenance ∙ 60 minutes per hour ∙ 4 weeks ) = 25 920 – 1200 = 24 720 mins

25. Solution (RP matrix) Solution (RP matrix) P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 b (mins/y) R1R1R1R110 17 280 R2R2R2R22030 30 720 R3R3R3R310 24 720

26. Exercise 1.2 Complete the corporate system matrix with the following marketing data: Complete the corporate system matrix with the following marketing data: There are long term contract to produce at least: There are long term contract to produce at least: 50 P1 50 P1 100 P2 100 P2 120 P3 120 P3 50 P4 50 P4 Forecasts says the upper limit of the market is: Forecasts says the upper limit of the market is: 10 000 units for P1 10 000 units for P1 1 500 for P2 1 500 for P2 1 000 for P3 1 000 for P3 3 000 for P4 3 000 for P4 Unit prices: p1=100, p2=200, p3=330, p4=100 Unit prices: p1=100, p2=200, p3=330, p4=100 Variable costs: R1=5/min, R2=8/min, R3=11/min Variable costs: R1=5/min, R2=8/min, R3=11/min

27. Solution (CS matrix) P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 b (mins/y) R1R1R1R110 17 280 R2R2R2R22030 30 720 R3R3R3R310 24 720 MIN (pcs/y) 5010012050 MAX (pcs/y) 10 000 1 500 1 000 3 000 price100200330100 CM504090-10

28. What is the optimal product mix to maximize revenues? P 1 = 17 280 / 10 = 1728 < 10 000 P 1 = 17 280 / 10 = 1728 < 10 000 P 2 : 200/20=10 P 2 : 200/20=10 P 3 : 330/30=11 P 3 : 330/30=11 P 4 =24 720/10=2472<3000 P 4 =24 720/10=2472<3000 P 2 = 100 P 2 = 100 P 3 = (30 720-100∙20-120∙30)/30= 837<MAX P 3 = (30 720-100∙20-120∙30)/30= 837<MAX

29. What if we want to maximize profit? The only difference is in the case of P 4 because of its negative contribution margin. The only difference is in the case of P 4 because of its negative contribution margin. P 4 =50 P 4 =50

30. Exercise 2 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 b (hrs/y) R1R1R1R16 2 000 R2R2R2R2 32 3 000 R3R3R3R3 4 1 000 R4R4R4R4 636 000 R5R5R5R5 145 000 MIN (pcs/y) 0200100250400100 MAX (pcs/y) 2000050040010002000200 p (HUF/pcs) 20010040010050100 CM (HUF/pcs) 5080403020-10

31. Solution Revenue max. P 1 =333 P 1 =333 P 2 =500 P 2 =500 P 3 =400 P 3 =400 P 4 =250 P 4 =250 P 5 =900 P 5 =900 P 6 =200 P 6 =200 Profit max. P 1 =333 P 2 =500 P 3 =400 P 4 =250 P 5 =966 P 6 =100