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Unformatted text preview: 540:311 DETERMINISTIC MODELS IN OPERATIONS RESEARCH Lecture 8: Chapters 3
4 Class MeeBng: Thu Feb 24th 10:20
11:40am Prof. W. Art Chaovalitwongse Key to Success From the problem statement, IDENTIFY • Decision variables: – the quanEEes that can vary; • ObjecBve funcBon: – the expression that is being minimized or maximized; min f(x) = max
f(x) • Constraints: – equaEons and inequaliEes that the decision variables must saEsfy Diet Problem • A farmer is planning to buy food for his cows. He can choose between two diﬀerent brands, A and B. A costs 30 cent/kg and B costs 12 cent/kg. • The brands contains the following units (per kg) of the three nutrients n1, n2, and n3. • The farmer needs at lest 60 units of n1, 84 units of n2 and 72 units of n3. • Determine the cheapest way to fulﬁll these requirements, i.e., how much he should buy of brand A and of brand B to minimize the cost! Problem FormulaEon how much he should buy of brand A and of brand B to minimize the cost! Decision Variables – A = quanEty of brand A to buy – B = quanEty of brand B to buy A costs 30 cent/kg and B costs 12 cent/kg. ObjecBve FuncBon – Minimize Cost = 30*A + 12*B The farmer needs at lest 60 units of n1, 84 units of n2 and 72 units of n3. Constraints – n1: 3A + 2B ≥ 60 – n2: 7A + 2B ≥ 84 – n3: 3A + 6B ≥ 72 A≥0
B≥0 Problems 1 – in text book • Farmer Jones must determine how many acres of corn and wheat to plant this year. • An acre of wheat yields 25 Bushels of wheat and requires 10 hours of labor per week. • An acre of corn yields 10 bushels of corn and requires 4 hours of labor per week. • All wheat can be sold at $4 a bushel. • All corn can be sold at $3 a bushel. • 7 acres of land and 40 hours per week of labor are available. • Government regulaEons require that at least 30 bushels of corn be produced during the current year. • Maximize the total revenue from wheat and corn! Problem FormulaEon determine how many acres of corn and wheat to plant this year • Decision Variables – C = acres of corn to plant – W = acres of wheat to plant Maximize the total revenue from wheat and corn! An acre of wheat yields 25 Bushels of wheat…. An acre of corn yields 10 bushels of corn…. All wheat can be sold at $4 a bushel. All corn can be sold at $3 a bushel. • ObjecBve FuncBon – Maximize Revenue = 25*4*W + 10*3*C = 100W + 30C An acre of wheat…requires 10 hours of labor per week… An acre of corn … requires 4 hours of labor per week….7 acres of land and 40 hours per week of labor are available. • Constraints – land: W + C ≤ 7 W≥0
– labor: 10W + 4C ≤ 40 C≥0
– Government: 10C ≥ 30 ⇒ C ≥ 3 2
D Graphical RepresentaEon W+C≤7 W C≥3 Objective function:
100W + 30C Feasible Region C
10W + 4C ≤ 40 Binding and Non
binding constraints • When the opEmal values of the decision variables are subsEtuted into the constraint – A constraint is binding if the le=
hand side and the right
hand side of the constraint are equal – A constraint is non
binding if the le=
hand side and the right
hand side of the constraint are unequal Example: GTC Problem Want to determine the number of wrenches and pliers to produce given the
available raw materials, machine hours and demand. MIT and James Orlin © 2003 FormulaEng the GTC Problem
P = number of pliers manufactured W = number of wrenches manufactured Maximize Proﬁt = .4 W + .3 P
Steel:
Molding:
Assembly:
Wrench Demand:
Pliers Demand:
Nonnegativity: 1.5 W + P ≤ 15,000 W+ P ≤ 12,000 0.4 W + 0.5 P
W 8,000 ≤
P 5,000 ≤ ≤ 10,000 P,W ≥ 0
MIT and James Orlin © 2003 ReformulaEon
P = number of 1000s of pliers manufactured W = number of 1000s of wrenches manufactured Maximize Proﬁt = 400 W + 300 P
Steel:
Molding:
Assembly:
Wrench Demand:
Pliers Demand:
Nonnegativity: 1.5 W + P ≤ 15 W+ P ≤ 12 0.4 W + 0.5 P
W 8 ≤
P 5 ≤ ≤ 10 P,W ≥ 0
MIT and James Orlin © 2003 Graphing the Feasible Region
P 14 We will construct and shade the
feasible region one or two
constraints at a time. 12
10
8
6
4
2
W
2 4 6 8 10 12 14 Graphing the Feasible Region
P 14
12
10 Graph the Constraint:
1.5 W + P 8 ≤ 15 6
4
2
W
2 4 6 8 10 12 14 Graphing the Feasible Region
P 14
12
10 Graph the Constraint: 8 W+ P ≤ 12 6
4
2
W
2 4 6 8 10 12 14 Graphing the Feasible Region
P 14
12
10 Graph the Constraint:
0.4 W + 0.5 P ≤ 5 8
6 What happened to the constraint :
W + P ≤ 12? 4
2
W
2 4 6 8 10 12 14 Graphing the Feasible Region
P 14
12
10
8
Graph the Constraints:
6 W ≤ 8 4
P
2 ≤ 10 W
2 4 6 8 10 12 14 How do we ﬁnd an opEmal soluEon?
P 14
12
10 Maximize z = 400W + 300P 8
6 It is the largest value of q such that
400W + 300P = q has a feasible
solution 4
2
W
2 4 6 8 10 12 14 How do we ﬁnd an opEmal soluEon?
P 14
12
10 Maximize z = 400W + 300P 8
6 Is there a feasible solution with
z = 400W + 300P = 1200? 4
2 z=1200 W
2 4 6 8 10 12 14 How do we ﬁnd an opEmal soluEon?
P 14
12 Maximize z = 400W + 300P 10
8 z=3600 Is there a feasible solution with z =
2400? 6 z =2400
Is there a feasible solution with z
= 3600? 4
2 W
2 4 6 8 10 12 14 How do we ﬁnd an opEmal soluEon?
P 14
12 Maximize z = 400W + 300P 10
8 z=
3600
z=
2400 Can you see what the optimal
solution will be? 6
4
2 W
2 4 6 8 10 12 14 How do we ﬁnd an opEmal soluEon?
P
Maximize z = 400W + 300P
14
12 What characterizes the optimal
solution? 10
8 What is the optimal solution vector? 6 W=? P=?
What is its solution value? 4
2
W
2 4 6 8 10 12 14 z=? OpEmal SoluEon Structure
P Binding constraints 14 Maximize z = 400W + 300P
1.5 W + P ≤ 15
.4W + .5P ≤ 5 12 plus other constraints 10
8
6 A constraint is said to be binding if it
holds with equality at the optimum
solution. 4 Other constraints are nonbinding 2
W
2 4 6 8 10 12 14 How do we ﬁnd an opEmal soluEon?
P 14 Optimal solutions occur at
corner points. In two
dimensions, this is the
intersection of 2 lines. 12 Maximize z = 400W + 300P 10 1.5W + P = 15 8 .4W + .5P = 5 6 Solution:
.7W = 5, W = 50/7 4 P = 15  75/7 = 30/7 2 z = 29,000/7 = 4,142 6/7
W
2 4 6 8 10 12 14 LP SoluEons: Four Cases • Case 1: The LP has a unique soluEon. • Case 2: The LP has more than one (actually inﬁnite number of) opEmal soluEons. This is the case of alternaEve opEmal soluEons. Graphically, we recognize this case when the isoproﬁt line last hits an enEre line segment before leaving the feasible region. • Case3: The LP is infeasible (has no feasible soluEon). • Case 4: The LP is unbounded. This means (in a max problem) that there are points in the feasible region with arbitrarily large objecEve funcEon values. Graphically, we recognize this case by the fact that when we move parallel to an isoproﬁt line in the direcEon of increasing the objecEve funcEon, we never lose contact with the LP s feasible region. 3.5 Work Scheduling Problem • Each postal worker works for 5 consecuEve days, followed by 2 days oﬀ, repeated weekly. • Minimize the number of postal workers (for the time
being, we will permit fractional workers on each day.)
• Select the decision variables – Let x1 be the number of workers who start working on Monday, and work Ell Friday – Let x2 be the number of workers who start on Tuesday … – Let x3, x4, …, x7 be deﬁned similarly. The linear program Minimize z = x1 + x2 + x3 + x4 + x5 + x6 + x7
subject to x1 + x4 + x5 + x6 + x7 ≥ 17 x1 + x2 + x5 + x6 + x7 ≥ 13 x1 + x2 + x3 +
x1 + x2 + x3 + x4 + x6 + x7 ≥ 15
x7 ≥ 19 x1 + x2 + x3 + x4 + x5
x2 + x 3 + x 4 + x 5 + x 6 ≥ 14
≥ 16 x3 + x4 + x5 + x6 + x7 ≥ 11
xj ≥ 0 for j = 1 to 7 Some Enhancements of the Model
• Suppose that there was a pay diﬀerenEal. The cost of workers who start work on day j is cj per worker. • Minimize z = c1x1 + c2x2 + c3x3 + … + c7x7 • Suppose that one can hire part Eme workers (one day at a Eme), and that the cost of a part Eme worker on day j is PTj. • Let yj = number of part Eme workers on day j What is the Revised Linear Program? Minimize z = x1 + x2 + x3 + x4 + x5 + x6 + x7
+ PT1 y1 + PT2 y2 + … + PT7 y7
subject to x1 + x4 + x5 + x6 + x7 + y1 ≥ 17 x1 + x2 + x 5 + x 6 + x 7 + y 2 ≥ 13 x1 + x2 + x3 +
x1 + x2 + x3 + x4 + x 6 + x 7 + y 3 ≥ 15
x 7 + y 4 ≥ 19 x1 + x2 + x3 + x4 + x5 +
x2 + x 3 + x 4 + x 5 + x 6 y 5 ≥ 14
+ y 6 ≥ 16 x3 + x4 + x5 + x6 + x7 + y7 ≥ 11
xj ≥ 0 , yj ≥ 0 for j = 1 to 7 Non
linear objecEve: Maximum
• Suppose that one wants to minimize the maximum of the slacks, that is • minimize z = max (s1, s2, …, s7). • This is a non
linear objecEve but we can transform it, so the problem becomes an LP. The new constraint ensures that
z ≥ max (s1, …, s7)
The objective ensures that z = sj
for some j. Nonlinear objective: Absolute Value
• Suppose that the goal is to have EXACTLY dj workers on day j. Let yj be the number of workers on day j. For each addiEonal or less worker on day j, it incurs an extra cost of pj. • The objecEve is minimize Σi pj  yj – dj 
The new constraints ensure that
zj ≥  yj – dj  for each j.
The objective ensures that zj = 
yj – dj  for each j. Nonlinear function: Ratio Constraint
Suppose that we need to ensure that at least 30% of the
workers have Sunday off.
How do we model this?
(x1 + x2 )/x1 + x2 + x3 + x4 + x5 + x6 + x7 ≥ .3
(x1 + x2 ) ≥ .3 x1 + .3 x2 + .3 x3 + .3 x4 + .3 x5 + .3 x6 + .3 x7
.7 x1  .7 x2 + .3 x3 + .3 x4 + .3 x5 + .3 x6 + .3 x7 <= 0 3.8 Oil Blending Problem
•
•
•
• Sunco Oil manufactures 3 types of gasoline (gas 1, gas 2, and gas 3). Each type is produced by blending 3 types of crude oil (crude 1,2,3). Prices: Crude 1 = $45/Barrel; Crude 2 = $35 Barrel; Crude 3 = $25/Bar Prices: Gas 1 = $70/Bar; Gas 2 = $60 Bar; Gas 3 = $50/Bar • Gas 1 must have an average octane raEng of at least 10% and contain at most 1% sulfer. Gas 2 have an average octane raEng of at least 8% and contain at most 2% sulfer. Gas 3 have an average octane raEng of at least 6% and contain at most 1% sulfer. •
• •
•
• Crude 1 has: Octane RaEng = 12, Sulfer = 0.5% Crude 2 has: Octane RaEng = 6, Sulfer = 2.0% Crude 3 has: Octane RaEng = 8, Sulfer = 3.0% • It costs $4 to transform one barrel of oil into one barrel of gasoline and Sunco s reﬁnery can produce up to 14,000 barrels of gasoline daily. 3.8 Oil Blending Problem (Cont d)
• • Customers require: – Gas 1 – 3,000 barrels per day – Gas 2 – 2,000 barrels per day – Gas 3 – 1,000 barrels per day Sunco has the opEon of adverEsing to sEmulate the demand. Each dollar spent daily in adverEsing a parEcular type of gas increases the daily demand for that type of gas by 10 barrels. • Formulate an LP to maximize daily proﬁts (proﬁts = revenues – costs). • Decisions: – How much to spend on adverEsing – How to blend each type of gasoline Decision Variables – aj = dollars spent daily on adverEsing gas j – xij = barrels of crude oil i used daily to produce gas j • CRUDE 1
Octane = 12, Sulfer = 0.5%
x11 x13 CRUDE 2
CRUDE 3
Octane = 6, Sulfer = 2.0% Octane = 8, Sulfer = 3.0%
x21 x12 x23 x31 x22 x33
x32 REFINERY x11 x21
x31 GAS 1
octane ≥ 10%, sulfer ≤ 1% x11 x21
x31 GAS 2
octane ≥ 8%, sulfer ≤ 1% x11 x21
x31 GAS 3
octane ≥ 6%, sulfer ≤ 1% Problem FormulaEon
Maximize the total daily proﬁts (proﬁts = revenues – costs) • ObjecBve FuncBon: – Revenues = Price1*Gas1 + Price2*Gas2 + Price3*Gas3 – = 70(x11+x21+x31) + 60(x12+x22+x32) + 50(x13+x23+x33) – Costs of purchasing oil = 45(x11+x12+x13) + 35(x21+x22+x23) + 25(x31+x32+x33) – Daily adverEsing costs = a1 + a2 + a3 – Daily producEon costs = 4(x11+x21+x31 + x12+x22+x32 + x13+x23+x33 ) – Daily Proﬁt z = 21x11+11x12+x13+31x21+21x22+11x23+41x31+31x32+21x33
a1
a2
a3 Problem FormulaEon
• Constraints: – Constraint 1: Gas 1 produced daily should equal its daily demand x11+x21+x31 = 3,000 + 10a1 – Constraint 2: Gas 2 produced daily should equal its daily demand x12+x22+x32 = 2,000 + 10a2 – Constraint 3: Gas 3 produced daily should equal its daily demand x13+x23+x33  1,000 + 10a2 = – Constraint 4: At most 5,000 barrels of crude 1 can be purchased daily x11+x12+x13 ≤ 5,000 – Constraint 5: At most 5,000 barrels of crude 2 can be purchased daily x21+x22+x23 ≤ 5,000 – Constraint 6: At most 5,000 barrels of crude 3 can be purchased daily x31+x32+x33 ≤ 5,000 – Constraint 7: LimitaBon of reﬁnery capacity of at most 14,000 barrels daily x11+x21+x31+x12+x22+x32+x13+x23+x33 ≤ 14,000 Problem FormulaEon
• Constraints: – Constraint 8: Gas 1 must have an average octane level of at least 10% Average octane level = (Total Octane value in mixture)/(number of barrels in mixture)
(12x11+6x21+8x31)/(x11+x21+x31) ≥ 10 ⇒ 2x11
4x21
2x31 ≥ 0 – Constraint 9: Gas 1 must have an average octane level of at least 8% 4x12
2x22 ≥ 0 – Constraint 10: Gas 1 must have an average octane level of at least 6% 6x13+2x33 ≥ 0 – Constraint 11: Gas 1 must contain at most 1% sulfur
0.005x11+0.01x21+0.02x31 ≤ 0 – Constraint 12: Gas 2 must contain at most 2% sulfur
0.015x12+0.01x32 ≤ 0 – Constraint 11: Gas 3 must contain at most 1% sulfur
0.005x13+0.01x23+0.02x33 ≤ 0 – NonnegaBvity Constraints 4.1 – How to Convert an LP to Standard Form Before the simplex algorithm can be used to solve an LP, the
LP must be converted into a problem where all the
constraints are equations and all variables are nonnegative.
We say that a linear program is in standard form if the following
are all true:
1. Nonnegativity constraints for all variables.
2. All remaining constraints are expressed as equality
constraints.
3. The right hand side vector, b, is nonnegative. 4.1 – How to Convert an LP to Standard Form If an LP has both ≤ and ≥ constraints, apply the previous procedures to the individual constraints. Consider the example below. Nonstandard Form Standard Form max z = 20x1 + 15x2 max z = 20x1 + 15x2 s.t. x1 ≤ 100 s.t. x1 + s1 x2 ≤ 100 = 100 x2 + s2 = 100 50x1 + 35x2 ≤ 6000 50x1 + 35x2 + s3 = 6000 20x1 + 15x2 ≥ 2000 20x1 + 15x2 x1, x2 > 0
e4 = 2000 x1, x2, s1, s2, s3, e4 > 0 4.2 – Preview of the Simplex Algorithm The relaEonship between extreme points and basic feasible soluEons outlined in Theorem 2, is seen in the Leather Limited problem. The LP (with slack variables) was:
max z = 4x1 + 3x2 s.t. x1 + x2 + s1 = 40 2x1 + x2 + s2 = 60 x1, x2, s1, s2 ≥ 0
Both inequaliEes are saEsﬁed in the shaded area. The extreme points are of the feasible region are B, C, E, and F. 4.2 – Preview of the Simplex Algorithm The table above shows the correspondence between the basic feasible soluEons to the LP and the extreme points of the feasible region. The basic feasible soluEons to the standard form of the LP correspond in a natural fashion to the LP s extreme points. 4.3 – The Simplex Algorithm (max LPs) The Simplex Algorithm Procedure for maximizaBon LPs Step 1 Convert the LP to standard form Step 2 Obtain a bfs (if possible) from the standard form Step 3 Determine whether the current bfs is opEmal Step 4 If the current bfs is not opEmal, determine which nonbasic variable should be come a basic variable and which basic variable should become a nonbasic variable to ﬁnd a bfs with a beter objecEve funcEon value. Step 5 Use ero s to ﬁnd a new bfs with a beter objecEve funcEon value. Go back to Step 3. In performing the simplex algorithm, write the objecEve funcEon in the form: z – c1x1 – c2x2
…
cnxn = 0 We call this format the row 0 version of the objecEve funcEon (row 0 for short). Example Max z = 3x1 + 2x2
s.t.
2 x1 + x2 ≤ 100
x1 + x2 ≤ 80
x1 ≤ 40 4.3 – The Simplex Algorithm (max LPs) Deﬁne: x1 = number of desks produced x2 = number of tables produced x3 = number of chairs produced.
The LP is: max z = 60x1 + 30x2 + 20x3 s.t. 8x1 + 6x2 + x3 ≤ 48 (lumber constraint) 4x1 + 2x2 + 1.5x3 ≤ 20 (ﬁnishing constraint) 2x1 + 1.5x2 + 0.5x3 ≤ 8 (carpentry constraint) x2 ≤ 5 (table demand constraint) x1, x2, x3 ≥ 0 4.3 – The Simplex Algorithm (max LPs) 4.6 – The Simplex Algorithm (min LPs) Two diﬀerent ways the simplex method can be used to solve minimizaEon problems. Consider the LP shown to the right: min z = 2x1 – 3x2
s.t. x1 + x2 ≤ 4
x1 – x2 ≤ 6
x1, x2 ≥ 0 Method 1
The optimal solution is the point
(x1,x2) that makes z = 2x1 – 3x2 the
smallest. Equivalently, this point
makes max z =  2x1 + 3x2 the
largest. This means we can find the
optimal solution to the LP by solving
the LP shown to the right. max z = 2x1 + 3x2
s.t. x1 + x2 ≤ 4
x1 – x2 ≤ 6
x1, x2 ≥ 0 4.6 – The Simplex Algorithm (min LPs) Method 2 A simple modification of the simplex algorithm can be
used to solve min problems directly.
Modify Step 3 of the simplex algorithm as follows:
• If all nonbasic variables (NBV) in row 0 have
nonpositive coefficients, the current bfs is optimal.
• If any nonbasic variable has a positive coefficient,
choose the variable with the most positive
coefficient in row 0 as the entering variable.
This modification of the simplex algorithm works
because increasing a nonbasic variable (NBV) with a
positive coefficient in row 0 will decrease z. 4.7 – Alternate OpEmal soluEons For some LPs, more than one extreme point is opEmal. If an LP has more than one opEmal soluEon, it has mulEple opEmal soluEons. If there is no nonbasic variable (NBV) with a zero
coefficient in row 0 of the optimal tableau, the LP has
a unique optimal solution.
Even if there is a nonbasic variable with a zero
coefficient in row 0 of the optimal tableau, it is possible
that the LP may not have alternative optimal solutions. 4.8 – Unbounded LPs For some LPs, there exist points in the feasible region for which z assumes arbitrarily large (in max problems) or arbitrarily small (in min problems) values. When this occurs, we say the LP is unbounded. An unbounded LP occurs in a max problem if there is a nonbasic variable with a negaEve coeﬃcient in row 0 and there is no constraint that limits how large we can make this NBV. Specifically, an unbounded LP for a max problem occurs
when a variable with a negative coefficient in row 0 has a
non positive coefficient in each constraint 4.11 Degeneracy and the Convergence of the Simplex Algorithm
• TheoreEcally, the simplex algorithm can fail to ﬁnd an opEmal soluEon to an LP. • However, LPs arising from actual applicaEons seldom exhibit this unpleasant behavior. • The following are facts: – If (value of entering variable in new bfs) > 0, then (z
value for new bfs) > (z
value for current bfs). – If (value of entering variable in new bfs) = 0, then (z
value for new bfs) = (z
value for current bfs). • Assume that in each of the LP’s basic feasible soluEons all basic variables are posiEve. 4.11 Degeneracy and the Convergence of the Simplex Algorithm
• An LP with this property is a nondegenerate LP. • An LP is degenerate if it has at least one bfs in which a basic variable is equal to zero. • Any bfs that has at least one basic variable equal to zero is a degenerate bfs. • When the same bfs is encountered twice it is called cycling. – If cycling occurs, then we will loop, or cycle, forever among a set of basic feasible soluEons and never get to an opEmal
soluEon. • Some degenerate LPs have a special structure that enables us to solve them by methods other than the simplex. 4.11 Problem 1
• Solve the following LP: Max z = 5x1 + 3x2 st. 4x1 + 2x2 ≤ 12 4x1 + x2 ≤ 10 x1 + x2 ≤ 4 All variables ≥ 0 Simplex Weaknesses: DegeneraEon: A Graphic Example
Simplex Weaknesses:
DegeneraEon: Summary
• How does the degeneracy of this problem impact the graphical soluEon? – Degenerate soluEons express the same vertex in a diﬀerent way. • How have we dealt with degeneracy previously? 4.12 – The Big M Method
• The simplex method algorithm requires a starEng bfs. • Previous problems have found starEng bfs by using the slack variables as our basic variables. – If an LP have ≥ or = constraints, however, a starEng bfs may not be readily apparent. • In such a case, the Big M method may be used to solve the problem. • Idea: – Introduce arEﬁcial variables to start the simplex with a bfs – Use the simplex method and perform ero’s unEl the opEmal soluEon found, infeasible, or unbounded – If opEmal soluEon is found and the arEﬁcial variable(s) is a BVs, the problem is infeasible. 4.12 – The Big M Method
Letting x1 = number of ounces of orange soda in a bottle of Oranj
x2 = number of ounces of orange juice in a bottle of Oranj
The LP is:
min z = 2x1 + 3x2
st 0.5x1 + 0.25x2 ≤ 4
x1 + 3x2 ≥ 20 x1 + x 2 = 10 (sugar constraint)
(Vitamin C constraint)
(10 oz in 1 bottle of Oranj) x 1 , x2 , > 0
The LP in standard form is shown on the next slide. 4.12 – The Big M Method
The LP in standard form has z
and s1 which could be used
for BVs but row 2 would
violate sign restrictions and
row 3 no readily apparent
basic variable. Row 1: z  2x1 Row 2: 3x2 0.5x1 + 0.25x2 + s1 =0
=4 Row 3: x1 + 3x2  e 2 = 20 Row 4: x1 + x2 = 10 In order to use the simplex method, a bfs is needed. To remedy the
predicament, artificial variables are created. The variables will be
labeled according to the row in which they are used as seen below.
Row 1: z  2x1 Row 2: 3x2 =0 0.5x1 + 0.25x2 + s1 Row 3: x1 + 3x2 Row 4: x1 + x2 =4
 e 2 + a2 = 20
+ a 3 = 10 4.12 – The Big M Method
In the optimal solution, all artificial variables must be set equal to zero.
To accomplish this, in a min LP, a term Mai is added to the objective
function for each artificial variable ai. For a max LP, the term –Mai is
added to the objective function for each ai. M represents some very
large number. The modified Bevco LP in standard form then becomes:
Row 1: z  2x1 Row 2: 3x2 Ma2  Ma3 0.5x1 + 0.25x2 + s1 =0
=4 Row 3: x1 + 3x2  e2 + Row 4: x1 + x2 + a2 = 20
a3 = 10 Modifying the objective function this way makes it extremely costly for
an artificial variable to be positive. The optimal solution should force
a2 = a3 =0. 4.12 – The Big M Method
DescripEon of the Big M Method 1. Modify the constraints so that the rhs of each constraint is
nonnegative. Identify each constraint that is now an = or ≥ constraint.
2. Convert each inequality constraint to standard form (add a slack
variable for ≤ constraints, add an excess variable for ≥ constraints).
3. For each ≥ or = constraint, add artificial variables. Add sign restriction
ai ≥ 0.
4. Let M denote a very large positive number. Add (for each artificial
variable) Mai to min problem objective functions or Mai to max
problem objective functions.
5. Since each artificial variable will be in the starting basis, all artificial
variables must be eliminated from row 0 before beginning the simplex.
Remembering M represents a very large number, solve the
transformed problem by the simplex. 4.12 – The Big M Method
If all artificial variables in the optimal solution equal zero, the
solution is optimal. If any artificial variables are positive in
the optimal solution, the problem is infeasible.
The Bevco example continued:
Initial Tableau 4.12 – The Big M Method
4.12 – The Big M Method
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