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What is Operations Research?
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Meaning Of OR:
According to T.L.Saaty “OR is the art of giving badresults to problems to which otherwise worseresults are given”.
According to Churchman “OR is the application of
scientific methods, techniques and tools toproblems involving the operations of systems soas to provide these in control of the applicationwith optimum solutions to the problem”.
According to H.M.Wagner “ OR is the scientificapproach to problem solving for executivemanagement”.
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Scope of OR:
• In Agriculture
• In Finance
• In Industry• In Marketing
• In Production Management
• In Personnel Management• In Production Management
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Methodology Of OR:
Define theProblem
Develop theModel
Obtain InputData
Solve the Model
Model Validation
Analyze the results
Implement the Solution
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What is Linear Programming Problem?(LPP)
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Formulation of LPP:
Step 1:Determine the objective function as alinear function of the variables.
Step 2: Formulate the other conditions of the problem such as resources limitationsas linear equations or inequations intermsof the variables.
Step 3: Add the non-negativity constraints.
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Production Allocation Problem:
A manufacturer produces two types of models A &B. Each model A requires 4 hours of grinding
and 2 hours of polishing, where as model Brequires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grinders and3 polishers. Each grinder works for 40 hours a
week and each polisher works for 60 hours aweek. Profit on model A is Rs.3 and on model Bis Rs.4. Whatever is produced in a week is soldin the market. How should the manufacturer
allocate his production capacity to the types of models so that he may make the maximum profitin week?
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2. A manufacturing firm needs 5 component parts.
Due to inadequate resources, the firm is unable
to manufacture all its requirements. So themanagement is interested in determining as to
how many , if any , units of each component
should be purchased from the outside and how
many should be produced internally. Therelevant data are given below.
Component M A T TR PP PC
C1 4 1 1.5 20 48 30
C2 3 3 2 50 80 52
C3 1 1 0 45 24 18
C4 3 1 0.5 70 42 31
C5 2 0 0.5 40 28 16
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Where
M: Per unit milling time in hours
A: Per unit assembly time in hours
T: Per unit testing time in hoursTR: Total requirements in units
PP: Price per unit quoted in the market
PC: Per unit direct costs
Resources available are as follows:Milling hours: 300
Assembly hours: 160
Testing hours: 150
Formulate this as an LPP, taking the objective function asmaximization of saving by producing the componentsinternally.
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3. The marketing department of Everest company
has collected information on the problem of
advertising for its product. This relates to theadvertising media available, the number of
families expected to be reached with each
alternative, cost per advertisement, the
maximum availability of each medium and theexpected exposure of each one. The information
is as given below:
Advertising Media
No.of families
expected tocover Cost perAd (Rs)
Maximum
availability (No.oftimes)
Expected
exposure(units)
TV(30 sec) 3000 8000 8 80
Radio(15 sec) 7000 3000 30 20
Sunday edition of a daily (1/4page) 5000 4000 4 50
Magazine(1 page) 2000 3000 2 60
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Other information and requirements:
a) The advertising budget is Rs 70,000
b) At least 40,000 families should be
covered.
c) At least 2 insertions be given in Sunday
edition of a daily.
Formulate this as LPP to maximize the
expected exposure.
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Graphical Method of solving a GLPP:
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Step 1: Consider each inequality asequation.
Step 2: Plot each equation on the graph.Step 3: Identify the feasible solution or
common region satisfying all theconstraints simultaneously.
Step 4: Determine the corner points of thefeasible region.
Step 5: Find the optimal solution by
substituting the corner points in theobjective function.
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Two products: Chairs and Tables
Decision: How many of each to make this
month?
Objective: Maximize profit
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Flair Furniture Co. DataTables
(per table)
Chairs
(per chair)
Hours
Available
Profit
Contribution7 5
Carpentry 3 hrs 4 hrs 2400
Painting 2 hrs 1 hr 1000
Other Limitations:
• Make no more than 450 chairs
• Make at least 100 tables
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Decision Variables:
T = No. of tables to makeC = No. of chairs to make
Objective Function: Maximize Profit
Z = 7 T + 5 C
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Constraints:
• Have 2400 hours of carpentry time
available
3 T + 4 C < 2400 (hours)
• Have 1000 hours of painting time available
2 T + 1 C < 1000 (hours)
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More Constraints:
• Make no more than 450 chairs
C < 450 (no. chairs)• Make at least 100 tables
T > 100 (no. tables)
Non-negativity:
Cannot make a negative number of chairs or
tables
T > 0
C > 0
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Max Z = 7T + 5C (Profit)
Subject to the constraints:
3T + 4C < 2400 (carpentry hrs)
2T + 1C < 1000 (painting hrs)
C < 450 (max no.of chairs)
T > 100 (min no.of tables)
T, C > 0 (non-negativity)
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Carpentry
Constraint Line
3T + 4C = 2400
Intercepts
(T = 0, C = 600)
(T = 800, C = 0)0 800 T
C
600
0
Feasible
< 2400 hrs
Infeasible
> 2400 hrs
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Painting
Constraint Line
2T + 1C = 1000
Intercepts
(T = 0, C = 1000)
(T = 500, C = 0)0 500 800 T
C
1000
600
0
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0 100 500 800 T
C
1000
600
450
0
Max Chair Line
C = 450
(T=0, C=450)
Min Table Line
T = 100
(T=100, C=0)
Feasible
Region
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The corner points of the feasible region are:
A(200,0)
B(500,0)
C(320,360)
D(200,450)E(450,100)
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Z value at A(100,0) = 700
Z value at B(500,0) =3500Z value at C(320,360) =4040
Z value at D(200,450) =3650
Z value at E(100,450) =2950
Z value maximum at C(320,360) hence an
optimum solution is given byT=320 & C=360 and max profit is Rs.4040
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2. Solve graphically:
0x,0x18x2x
84x73x
60x45x Subject to
x146xZMinimize
21
21
21
21
21
and
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3. Solve graphically
0x,0x
3xx
1xxSubject to
x23xZMinimize
21
21
21
21
and
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Simplex method of solving GLPP
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Standard from of LPP:
1.Check whether the objective function is to
be maximized or minimized. If minimizedconvert it into maximization form
Min Z= -{Max(-Z)}
EX:Min Z= 2x-3y+7z
Min Z= -{Max(-[2x-3y+7z ])}
Min Z= -{Max(-2x+3y-7z)}
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2. The right hand side constant of each
constraint should be non-negative. If not made
it positive by multiplying it by-1(one).
Ex:
0y)(x,and
7y-2x-7))(1(y]-(-1)[2x
265y-8xs.t
5y2xMax Z
0y)(x,and
-7y-2x
265y-8xs.t
5y2xMax Z
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3. All the decision variables should be non-
negatively restricted i.e if the variable x is
unrestricted in sign then such a variable isexpressed as the difference between two
variables which are non-negative
0,0where x x x x x
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0y0,y0,x
7 yy-3x
7 )y-y(-3x
12y9-y92x
12)y-y9(2x s.t
y7y7-2x)y-y7(-2xMax Z
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4.Convert all the inequalities into equations by
introducing non-negative variables on the left hand
side of each constraint called slack or surplusvariables.
.constrainteachof
sidehandlefton thevariablesurplusasubstractthenof is
inequalityof typetheIf .constrainteachof sidehandlefttheonableslack variaaddthenof is inequalityof typetheIf
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0y),(xand254y7x
842yx-
60y2xs.t
3y2xMax Z
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0y),(xand254y7x
84s-2yx-
60sy2xs.t
3y2xMax Z
2
1
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Simplex method of solving LPP
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• Solve by simplex method
0)x,(xand
2x
1xx- 62xx
244x6x .x45xMax Z
21
2
21
21
21
21
t s
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Step 1: Convert the LPP into standard form.
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)0s0,,0,0,0x,0(xand2s x
1 xx- 6 2xx
244x6x .x45xMax Z
432121
42
321
221
121
21
sss
s
s
st s
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)0s0,,0,0,0x,0(xand2s x
1 xx-
6 2xx
244x6x .
.0.0.0.0x45xMax Z
432121
42
321
221
121
432121
sss
s
s
st s
ssss
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Step 2: Prepare simplex table
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j j
j
s).(zwhere
computeNow
or x BC
c z
j
j j
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5
(I.V)
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/x Ratios(Sol 1 )
6
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Ratios
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Step 5: Mark the element in the incoming variable
corresponding to the outgoing variable called
key element or pivotal element .
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Step 6: Divide each element of the Key row
(including bi) by the key element to get thecorresponding values in the new table.
1st row (new)
0 0 0 1/6 4/6 1 4
s s s s xxbSol 432121i
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For each row other than the key row,
valuerowtreplacemeningCorrespond
columnkeytheinelementRow
-elementrowOld
elementrowNew
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Corresponding replacement row value:
2 nd old row:
Old row element- Row element in key column* corresponding replacementvalue
6 1 4 =2
1 1 1 =0
2 1 4/6 =8/6
0 1 1/6 =-1/6
1 1 0 =1
0 1 0 =0
0 1 0 =0
0 0 0 1/6 4/6 1 4
s s s s xxbSol 432121i
0 0 1 0 2 1 6
s s s s xxbSol 432121i
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2 nd row (new):
0 0 1 1/6- 8/6 0 2
s s s s x x bSol432121i
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Corresponding replacement row value:
3 rd old row:
Old row element- Row element in key column* corresponding replacementvalue
1 -1 4 =5
-1 -1 1 =0
1 -1 4/6 =10/6
0 -1 1/6 =1/6
0 -1 0 =0
1 -1 0 =1
0 -1 0 =0
0 0 0 1/6 4/6 1 4
s s s s xxbSol 432121i
0 1 0 0 1 1- 1
s s s s xxbSol 432121i
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3 rd row (new):
0 1 0 1/6 10/6 0 5
s s s s x x bSol432121i
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Corresponding replacement row value:
4 th old row:
Old row element- Row element in key column* corresponding replacementvalue
2 0 4 =20 0 1 =01 0 4/6 =1
0 0 1/6 =00 0 0 =00 0 0 =01 0 0 =1
0 0 0 1/6 4/6 1 4
s s s s xxbSol 432121i
1 0 0 0 1 0 2
s s s s xxbSol 432121i
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4 th row (new)
1 0 0 0 1 0 2
s s s s x x bSol432121i
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Now compute again ΔJ
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0 0 0 6
5 6
4- 0 j
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0 0 0 6
5 6
4 - 0 j
sol/ Ratio
6 6 4
4
8
12
6 8
2
3
( ) .
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• The Omega manufacturing company has
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• The Omega manufacturing company has
discontinued the production of a certain
unprofitable product lin. This act created
considerable excess production capacity.
Management is considering devoting this
excess capacity to one or more of the
three products; call them products 1,2 and3. The available capacity on the machines
that might limit output is summarized in the
following table:
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The number of machine hours required for eachunit of the respective product is:
Machine Type Available Time(Machine Hrs perweek)
Milling machine 500Lathe 350
Grinder 150
Machine Type Product 1 Product 2 Product 3
Milling machine 9 3 5
Lathe 5 4 0
Grinder 3 0 2
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The sales department indicates that the
sales potential for product 3 is 20 units per
week. The unit profit would be Rs. 50,
Rs.20, and Rs.25 respectively on products1,2 and 3. the objective is to determine
how much of each product Omega should
produce to maximize the profit.
• A firm produces three products A B and C
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• A firm produces three products A,B and C,
each of which passes through three
departments: Fabrication, Finishing and
Packaging. Each unit of product A requires
3,4 and 2; a unit of product B requires 5, 4
and 4, while each unit of product C requires
2, 4 and 5 hours respectively in threedepartments. Everyday, 60 hours are
available in the fabrication department, 72
hours in the finishing department and 100hours in the packaging department. The unit
contribution of product A is Rs.5, of product