decision science for management-1
TRANSCRIPT
Decision Science for Management
Lecture notes are available at:http://Arash-management.blogspot.com
1Arash
Text Books…
• David R. Anderson, Dennis J. Sweeney, & Thomas A. Williams, (2003), Introduction to Management Science, Quantitative Approaches to Decision Making, 10th Edition West Publishing Company.
• Hiller F. Hiller M (2003), Introduction to Management Science: a Modeling & Case Studies approach with spreadsheets, 2nd Edition.
• Stevenson, Introduction To Management Science With Spreadsheet, Mcgraw Hill
Assessment
Assignments 30% ( 3* 10%)Midterm Examination 20% Final Exam 50%
Attendance and participation are warmly welcomed
Lecture One ( 9th June 2008)
Session Synopsis:
Linear Programming
Requirements of a linear programming problemFormulating linear programming problemsGraphical solution to linear programming : maximization problems minimization problems
Linear Programming
• Linear Programming (LP) is a mathematical procedure for determining optimal allocation of scarce resources.
• LP is a procedure that has found practical application in almost all facets of business, from advertising to production planning.
Linear Programming
Any linear program consists of four parts:
1.a set of decision variables, 2.the parameters, 3.the objective function, and 4.a set of constraints.
Formulation of Problem
In formulating a given decision problem, you should practice understanding the problem. While trying to understand the problem, ask yourself the following general questions:
1.What are the resources? 2.What is the objective? 3.What are the constraints? 4.What are the decision variables?
Requirements of a linear programming
Must seek to maximize or minimize some quantity
Presence of restrictions or constraints –
Must be alternative courses of action to choose from
Objectives and constraints must be expressible as linear equations or inequalities
Objective Function
Maximize (or Minimize) Z = C1X1 + C2X2 + ... + CnXn
• Cj is a constant that describes the rate of contribution to costs or profit of units being produced (Xj).
Z: is the total cost or profit from the given number of units being produced.
ConstraintsA11X1 + A12X2 + ... + A1nXnB1
A21X1 + A22X2 + ... + A2nXn B2
:
:
AM1X1 + AM2X2 + ... + AMnXn=BMAij are resource requirements for each of the related (Xj) decision variables.
Bi are the available resource requirements.
Note that the direction of the inequalities can be all or a combination of , , or = linear mathematical expressions.
Non-Negativity Requirement
X1,X2, …, Xn 0
• All linear programming model formulations require their decision variables to be non-negative.
• While these non-negativity requirements take the form of a constraint, they are considered a mathematical requirement to complete the formulation of an LP model.
Step 1 - Draw graph with vertical & horizontal axes (1st quadrant only)
Step 2 - Plot constraints as lines Use (X1,0), (0,X2) for line
Step 3 - Plot constraints as planes Use < or > signs
Step 4 - Find feasible region
Step 5 - Find optimal solution Objective function plotted
Step 6 – Calculate optimized value
Graphical Solution Method2 Variables
ELECTRONIC COMPANY PROBLEMHours Required to
Produce 1 Unit
DepartmentsX1
WalkmansX2
Watch-TV’sAvailable Hours
This WeekElectronics 4 3 240
Assembly 2 1 100
Profit/unit $7 $5
Constraints:4x1 + 3x2 240 (Hours of Electronic Time)2x1 + 1x2 100 (Hours of Assembly Time)
Objective: Maximize: 7x1 + 5x2
Step 1 – Draw GraphN
um
ber
of
Wat
ch-T
Vs
(X2)
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70 80
Number of Walkmans (X1)
Step 5 - Find optimal solution• Plot function line
350 5x 7x
Plot Now,
each)by divisible is that #any choose(just
350 5x 7x :Say
graph on the line get the
number toany usingfunction plot theJust
First,
5x 7x :Maximize
21
21
21
Find optimal solution (Cont’d)In This Case:
Calculate the point where both constraint lines intersect
30 X
300 3X 6X-
240 3X 4X
X of ridget to3-by Multiply Now
100 1X 2X
240 3X 4X
:zero set to XFor
1
2 1
21
2
2 1
21
2
Step 5 - Find optimal solution (Cont’d)
(30,40) Therefore,
40 X
200 2X 4X-
240 3X 4X
X of ridget to2-by Multiply Now
100 1X 2X
240 3X 4X
:zero set to XFor
2
2 1
21
1
2 1
21
1
Step 5 - Find optimal solution (Cont’d)
0
20
40
60
80
100
120
0 10 20 30 40 50 60 70 80
Number of Walkmans (X1)
Nu
mb
er o
f W
atch
-TV
s (X
2)
ElectronicsDepartment
AssemblyDepartment
Step 6 – Calculate optimized value
410 200 210 5(40) 7(30)
profit Maximized 5x 7x 21
Therefore: the best profit scenario is $410.00
Plug in values for X1 and X2
Copyright 2006 John Wiley & Sons, Inc.
LP Model Formulation
• Decision variables– mathematical symbols representing levels of activity of an operation
• Objective function– a linear relationship reflecting the objective of an operation– most frequent objective of business firms is to maximize profit– most frequent objective of individual operational units (such as a
production or packaging department) is to minimize cost• Constraint
– a linear relationship representing a restriction on decision making
Copyright 2006 John Wiley & Sons, Inc.
LP Model Formulation (cont.)
Max/min z = c1x1 + c2x2 + ... + cnxn
subject to:a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
: am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
xj = decision variablesbi = constraint levelscj = objective function coefficientsaij = constraint coefficients
Copyright 2006 John Wiley & Sons, Inc.
LP Model: Example
LaborLabor ClayClay RevenueRevenuePRODUCTPRODUCT (hr/unit)(hr/unit) (lb/unit)(lb/unit) ($/unit)($/unit)
BowlBowl 11 44 4040
MugMug 22 33 5050
There are 40 hours of labor and 120 pounds of clay There are 40 hours of labor and 120 pounds of clay available each dayavailable each day
Decision variablesDecision variables
xx11 = number of bowls to produce = number of bowls to produce
xx22 = number of mugs to produce = number of mugs to produce
RESOURCE REQUIREMENTSRESOURCE REQUIREMENTS
Copyright 2006 John Wiley & Sons, Inc.
LP Formulation: Example
Maximize Maximize ZZ = $40 = $40 xx11 + 50 + 50 xx22
Subject toSubject to
xx11 ++ 22xx22 40 hr40 hr (labor constraint)(labor constraint)
44xx11 ++ 33xx22 120 lb120 lb (clay constraint)(clay constraint)
xx1 1 , , xx22 00
Solution is Solution is xx11 = 24 bowls = 24 bowls xx2 2 = 8 mugs= 8 mugs
Revenue = $1,360Revenue = $1,360
Copyright 2006 John Wiley & Sons, Inc.
Graphical Solution Method
1.1. Plot model constraint on a set of coordinates Plot model constraint on a set of coordinates in a planein a plane
2.2. Identify the feasible solution space on the Identify the feasible solution space on the graph where all constraints are satisfied graph where all constraints are satisfied simultaneouslysimultaneously
3.3. Plot objective function to find the point on Plot objective function to find the point on boundary of this space that maximizes (or boundary of this space that maximizes (or minimizes) value of objective functionminimizes) value of objective function
Copyright 2006 John Wiley & Sons, Inc.
Graphical Solution: Example
4 4 xx11 + 3 + 3 xx2 2 120 lb120 lb
xx11 + 2 + 2 xx2 2 40 hr40 hr
Area common toArea common toboth constraintsboth constraints
50 50 –
40 40 –
30 30 –
20 20 –
10 10 –
0 0 – |1010
|6060
|5050
|2020
|3030
|4040 xx11
xx22
Copyright 2006 John Wiley & Sons, Inc.
Computing Optimal Valuesxx11 ++ 22xx22 == 4040
44xx11 ++ 33xx22 == 120120
44xx11 ++ 88xx22 == 160160
-4-4xx11 -- 33xx22 == -120-120
55xx22 == 4040
xx22 == 88
xx11 ++ 2(8)2(8) == 4040
xx11 == 2424
4 4 xx11 + 3 + 3 xx2 2 120 lb120 lb
xx11 + 2 + 2 xx2 2 40 hr40 hr
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –|1010
|2020
|3030
|4040
xx11
xx22
ZZ = $50(24) + $50(8) = $1,360 = $50(24) + $50(8) = $1,360
248
Copyright 2006 John Wiley & Sons, Inc.
Extreme Corner Points
xx11 = 224 bowls = 224 bowls
xx2 2 ==8 mugs8 mugs
ZZ = $1,360 = $1,360 xx11 = 30 bowls = 30 bowls
xx2 2 ==0 mugs0 mugs
ZZ = $1,200 = $1,200
xx11 = 0 bowls = 0 bowls
xx2 2 ==20 mugs20 mugs
ZZ = $1,000 = $1,000
AA
BBCC|
2020|3030
|4040
|1010 xx11
xx22
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-30
44xx11 + 3 + 3xx2 2 120 lb120 lb
xx11 + 2 + 2xx2 2 40 hr40 hr
40 40 –
30 30 –
20 20 –
10 10 –
0 0 –
BB
|1010
|2020
|3030
|4040 xx11
xx22
CC
AA
ZZ = 70 = 70xx11 + 20 + 20xx22
Optimal point:Optimal point:
xx11 = 30 bowls = 30 bowls
xx2 2 ==0 mugs0 mugs
ZZ = $2,100 = $2,100
Objective Function
Copyright 2006 John Wiley & Sons, Inc.
Minimization Problem
CHEMICAL CONTRIBUTIONCHEMICAL CONTRIBUTION
BrandBrand Nitrogen (lb/bag)Nitrogen (lb/bag) Phosphate (lb/bag)Phosphate (lb/bag)
Gro-plusGro-plus 22 44
Crop-fastCrop-fast 44 33
Minimize Minimize ZZ = $6x = $6x11 + $3x + $3x22
subject tosubject to
22xx11 ++ 44xx22 16 lb of nitrogen 16 lb of nitrogen
44xx11 ++ 33xx22 24 lb of phosphate 24 lb of phosphate
xx11, , xx22 0 0
Copyright 2006 John Wiley & Sons, Inc. Supplement 13-32
14 14 –
12 12 –
10 10 –
8 8 –
6 6 –
4 4 –
2 2 –
0 0 –|22
|44
|66
|88
|1010
|1212
|1414 xx11
xx22
A
B
C
Graphical Solution
x1 = 0 bags of Gro-plusx2 = 8 bags of Crop-fastZ = $24
Z = 6x1 + 3x2
Example 1: Maximization Problem
Wyndor Glass Company
Questions….