Constraint management

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

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

Linear programming

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.

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.

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

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

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

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

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

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.

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

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

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) 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 P 2 =MAX 200

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)

Solution with linear programming T1T1 T2T2 33,3 33 jugs and 33 plaits a per week Contribution margin: 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 P 2 =MAX e1e1 e2e2 e3e3 of F 100

What is the product-mix, that maximizes the revenues and the contribution to profit! P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 b (hrs/y) R1R R2R R3R R4R R5R MIN (pcs/y) MAX (pcs/y) p (HUF/pcs) f (HUF/pcs)

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 =( *2)/1=2600>1000 P 2 = /2=1000<1100 Contr. max.: 110/2 > 50/1 thus P 2 P 2 =( *1)/2=1400>1100 P 3 = /1=800<1000

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 *T 6 = max of CM :30*T *T 6 = max

e2 e1 of CM ofTR Contr. max: P5=1500, P6=1000 Rev. max: P5=50, P6=1966 T5 T

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

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 = 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 = minutes per planning period R 2 = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 2 ∙ 0,8 = mins R 2 = 4 ∙ 5 ∙ 2 ∙ 8 ∙ 60 ∙ 2 ∙ 0,8 = mins R 3 = (4 ∙ 5 ∙ 3 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9) – (5 hrs per week maintenance ∙ 60 minutes per hour ∙ 4 weeks ) = – 1200 = mins R 3 = (4 ∙ 5 ∙ 3 ∙ 8 ∙ 60 ∙ 1 ∙ 0,9) – (5 hrs per week maintenance ∙ 60 minutes per hour ∙ 4 weeks ) = – 1200 = mins

Solution (RP matrix) Solution (RP matrix) P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 b (mins/y) R1R1R1R R2R2R2R R3R3R3R

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: units for P units for P for P for P for P for P for P 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

Solution (CS matrix) P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 b (mins/y) R1R1R1R R2R2R2R R3R3R3R MIN (pcs/y) MAX (pcs/y) price CM

What is the optimal product mix to maximize revenues? P 1 = / 10 = 1728 < P 1 = / 10 = 1728 < 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 = ( ∙20-120∙30)/30= 837<MAX P 3 = ( ∙20-120∙30)/30= 837<MAX

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

Exercise 2 P1P1P1P1 P2P2P2P2 P3P3P3P3 P4P4P4P4 P5P5P5P5 P6P6P6P6 b (hrs/y) R1R1R1R R2R2R2R R3R3R3R R4R4R4R R5R5R5R MIN (pcs/y) MAX (pcs/y) p (HUF/pcs) CM (HUF/pcs)

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