Constraint Constraint managementmanagement

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.

1. Identifying decision variables (e.g. quantity of a product)

2. Identifying constraints3. Solve the problem.

Graphical linear programming

1. Set up the objective function and the constraints into mathematical format.

2. Plot the constraints.3. Identify the feasible solution space.4. Plot the objective function.5. Determine the optimum solution.

1. Sliding the line of the objective function away from the origin to the farthes/closest point of the feasible solution space.

2. Enumeration approach.

Corporate system-matrix1.) 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

P1 P2 P3 P4 P5 P6 P7

R1 a11

R2 a22

R3 a32

R4 a43 a44 a45

R5 a56 a57

R6 a66 a67

Non-convertible relations Partially convertible relations

Product-mix in a pottery – corporate system

matrixJug Plate

Clay (kg/pcs) 1,0 0,5

Weel time (hrs/pcs)

0,5 1,0

Paint (kg/pcs) 0 0,1

Capacity

50 kg/week 100 HUF/kg

50 hrs/week 800 HUF/hr

10 kg/week 100 HUF/kg

Minimum (pcs/week) 10 10

Maximum (pcs/week)

100 100

Price (HUF/pcs) 700 1060

Contribution margin (HUF/pcs)

e1: 1*P1+0,5*P2 < 50e2: 0,5*P1+1*P2 < 50e3: 0,1*P2 < 10m1, m2: 10 < P1 < 100m3, m4: 10 < P2 < 100ofCM: 200 P1+200P2=MAX200 200

Objective function

refers to choosing the best element from some set of available alternatives.

X*P1 + Y*P2 = max

variables (amount of produced

goods)

weights(depends on what we want to maximize:

price, contribution margin)

Solution with linear programming

T1

T2

33,3

33,3

33 jugs and 33 plaits a per week

Contribution margin: 13 200 HUF / week

e1: 1*P1+0,5*P2 < 50e2: 0,5*P1+1*P2 < 50e3: 0,1*P2 < 10m1,m2: 10 < P1 < 100m3, m4: 10 < P2 < 100ofCM: 200 P1+200P2=MAX

e1

e2

e3ofF

100

100

What is the product-mix, that maximizes the revenues and the

contribution to profit!

  P1 P2 P3 P4 P5 P6 b (hrs/y)

R1 4           2 000

R2   2 1       3 000

R3       1     1 000

R4         2 3 6 000

R5         2 2 5 000

MIN (pcs/y) 100 200 200 200 50 100

MAX (pcs/y) 400 1100 1 000 500 1 500 2000

p (HUF/pcs) 200 270 200 30 50 150

f (HUF/pcs) 100 110 50 -10 30 20

Solution P1:

Resource constraint 2000/4 = 500 > market constraint 400

P2&P3: Which one is the better product?Rev. max.: 270/2 < 200/1 thus P3

P3=(3000-200*2)/1=2600>1000

P2=200+1600/2=1000<1100

Contr. max.: 110/2 > 50/1 thus P2

P2=(3000-200*1)/2=1400>1100

P3=200+600/1=800<1000

P4: does it worth?Revenue max.: 1000/1 > 500Contribution max.: 200

P5&P6: linear programming e1: 2*T5 + 3*T6 ≤ 6000

e2: 2*T5 + 2*T6 ≤ 5000

m1, m2: 50 ≤ T5 ≤ 1500

p3, m4: 100 ≤ T6 ≤ 2000

ofTR: 50*T5 + 150*T6 = max

ofCM: 30*T5 + 20*T6 = max

e2

e1

ofCM

ofTR

Contr. max: P5=1500, P6=1000Rev. max: P5=50, P6=1966

T5

T62000

3000

2500

2500

ExerciseExercise 1.1 1.1Set up the product-resource matrix using the following data!Set up the product-resource matrix using the following data!

RRPP coefficients: a coefficients: a1111: 10, a: 10, a2222: 20, a: 20, a2323: 30, a3: 30, a344: 10: 10 The planning period is 4 weeks (there are no holidays in it, The planning period is 4 weeks (there are no holidays in it, and no work on weekends) and no work on weekends) Work schedule: Work schedule:

RR11 and and RR22: 2 shifts, each is 8 hour long: 2 shifts, each is 8 hour long RR33: 3 shifts: 3 shifts

Homogenous machines: Homogenous machines: 1 for 1 for RR11 2 for 2 for RR22 1 for 1 for RR33

Maintenance time: only for Maintenance time: only for RR33: 5 hrs/week: 5 hrs/weekPerformance rate: Performance rate:

90% for 90% for RR11 and and RR33 80% for 80% for RR22

Solution (bSolution (bii) )

RRii = = N ∙ sN ∙ snn ∙ s ∙ shh ∙ m ∙ mnn ∙ 60 ∙ 60 ∙ ∙ N=(number of weeks) N=(number of weeks) ∙ ∙ (working days per week)(working days per week)

RR11 = 4 = 4 weeksweeks ∙ ∙ 5 5 working daysworking days ∙ ∙ 2 2 shiftsshifts ∙ ∙ 8 8 hours per shifthours per shift ∙ ∙ 60 60 minutes per hourminutes per hour ∙ ∙ 1 1 homogenous machinehomogenous machine ∙ ∙ 0,9 0,9 performance performance = = = 4 = 4 ∙ ∙ 5 5 ∙ ∙ 2 2 ∙ ∙ 8 8 ∙ ∙ 60 60 ∙ ∙ 1 1 ∙ ∙ 0,9 = 17 280 0,9 = 17 280 minutes per minutes per planning periodplanning period

RR22 = 4 = 4 ∙ ∙ 5 5 ∙ ∙ 2 2 ∙ ∙ 8 8 ∙ ∙ 60 60 ∙ ∙ 2 2 ∙ ∙ 0,8 = 38 720 0,8 = 38 720 minsmins

RR33 = (4 = (4 ∙ ∙ 5 5 ∙ ∙ 3 3 ∙ ∙ 8 8 ∙ ∙ 60 60 ∙ ∙ 1 1 ∙ ∙ 0,9) – (5 0,9) – (5 hrs per weekhrs per week maintenancemaintenance ∙ ∙ 60 60 minutes per hourminutes per hour ∙ ∙ 4 4 weeksweeks) = 25 920 – ) = 25 920 – 1200 = 24 720 1200 = 24 720 minsmins

SolutionSolution (RP matrix) (RP matrix)

   PP11 PP22 PP33 PP44 b (b (mins/ymins/y))

RR11 1010 17 28017 280

RR22 2020 3030 30 72030 720

RR33 1010 24 72024 720

ExerciseExercise 1.2 1.2 Complete the corporate system matrix with the following Complete the corporate system matrix with the following

marketing data:marketing data: There are long term contract to produce at least:There are long term contract to produce at least:

50 50 PP11 100 100 PP22 120 120 PP33 50 50 PP44

ForForeecasts says the upper limit of the market is:casts says the upper limit of the market is: 10 000 units for 10 000 units for PP11 1 500 for 1 500 for PP22 1 000 for 1 000 for PP33 3 000 for 3 000 for PP44

Unit prices: Unit prices: pp1=100, 1=100, pp2=200, 2=200, pp3=33=3330, 0, pp4=1004=100 Variable costs: Variable costs: RR1=5/min, 1=5/min, RR2=8/min, 2=8/min, RR3=11/min3=11/min

SolutionSolution (CS matrix) (CS matrix)

   PP11 PP22 PP33 PP44 b (mins/y)b (mins/y)

RR11 1010 17 28017 280

RR22 2020 3030 30 72030 720

RR33 1010 24 72024 720

MIN (MIN (pcs/ypcs/y)) 5050 100100 120120 5050

MAX (pcs/y)MAX (pcs/y)

10 10 000000 1 5001 500 1 0001 000 3 0003 000

priceprice 100100 200200 330330 100100

CMCM 5050 4040 9090 -10-10

What is the optimal product What is the optimal product mix to maximize revenues?mix to maximize revenues?

PP11= = 17 280 / 10 = 1728 < 10 00017 280 / 10 = 1728 < 10 000

PP22: : 200/20=10200/20=10

PP33: : 330/30=11330/30=11

PP44=24 720/10=2472<3000=24 720/10=2472<3000

PP22= 100= 100 PP33= = (30 720-100∙20-120∙30)/30= (30 720-100∙20-120∙30)/30=

837<MAX837<MAX

What if we want to What if we want to maximize profit?maximize profit?

The only difference is in The only difference is in the case of the case of PP44 because of its negative because of its negative contribution margin.contribution margin.

PP44=50=50

ExerciseExercise 2 2

   PP11 PP22 PP33 PP44 PP55 PP66 b (hrs/y)b (hrs/y)

RR11 66                2 0002 000

RR22    33 22          3 0003 000

RR33          44       1 0001 000

RR44             66 33 6 0006 000

RR55             11 44 5 0005 000

MIN (pcs/y)MIN (pcs/y) 00 200200 100100 250250 400400 100100

MAX (pcs/y)MAX (pcs/y) 2000020000 500500 400400 10001000 20002000 200200

p (HUF/pcs)p (HUF/pcs) 200200 100100 400400 100100 5050 100100

CM (HUF/pcs) CM (HUF/pcs) 5050 8080 4040 3030 2020 -10-10

SolutionSolution

Revenue max.Revenue max. PP11=333=333

PP22=500=500

PP33=400=400

PP44=250=250

PP55=900=900

PP66=200=200

Profit max.Profit max. PP11=333=333

PP22=500=500

PP33=400=400

PP44=250=250

PP55=966=966

PP66=100=100