DMOR
Linear Programming 2
Solving LPs – general case
• Till now we talked about the standard form
–
–
–
–
Maximization
Constraints of the form ≤
RHS of the constraints ≥ 0
All variables ≥ 0
• In this case, point (0,0,…,0) is always a feasible cornerpoint
solution
• What if the problem is NOT in the standard form?
Minimization
1. Transform into maximization:
– min Z = 12x1 + 5x2 – 7x3
– max (–Z) = –12x1 – 5x2 + 7x3
• Put -1 in column Z in the simplex tableau to remember
• After solving an LP as a maximization problem:
– Multiply the objective function value by -1
– Decision variables values taken from the tableau are the
solution to the original problem
2. Another method – less popular – solve the simplex
tableau as minimization problem with appropriately
modified rules.
Equality constraints
• Bicycle company problem:
• Introduce an artificial variable a1 [artificial variable]
• An artificial variable differs from a slack variable – we have to make
it zero, because point (x1,x2,s1,s2,a1) = (0,0,2,3,4) is not feasible.
2-phase simplex algorithm
• To do it we use 2-phase simplex algorithm
– Phase 1: minimize all artificial variables (we only
have one: a1); if the minimum is zero, then we
have a feasible cornerpoint solution…
– Phase 2: … from which we start to search for the
optimal solution the same way as in basic simplex
algorithm introduced in the last lecture
Phase 1
• Transform
– min W = a1 + a2 + a3 + …
– max –W + a1 + a2 + a3 + … = 0
• The original objective function is written along with the objective function
used in phase 1 of the algorithm
• The first simplex tableau in phase 1 is never in the proper form, because
artificial variables, which are basic variables first „appear” two times in
the given column.
• We need some preliminary transformation
Phase 1
• We make it into the proper form
• 4 units of constraint violation
• d
Phase 1
• When solving, we have to remember to modify the
original objective function as well
• We finished phase 1. We have to continue into phase 2, because
there is a negative coefficient in the original objective function raw
Phase 2
• Solution: (x1,x2,s1,s2) = (2,2,0,1); Z=50
What happened?
• We start in the cornerpoint which is not feasible for phase
2, but is feasible for phase 1
• We finish phase 1 in the cornerpoint, which is feasible both
for phase 1 and phase 2
Infeasible problem
• If phase 1 ends and value of W is still positive,
the problem is infeasible.
• How to check why:
– IIS – irreducible infeasible set – commonly a very
small subset of all constraints, which taken
together are infeasible, but
– Throwing away one or more constraints from this
set results in a feasible problem
Constraints of the type ≥ with positive
RHS
• These constraints mean that the origin of the coordinate
system is infeasible
• We cannot simply multiply eveything by -1, because the
RHS will then be negative.
• Solution: we introduce a surplus variable
3x1 + 5x2 ≥ 20 ⇒ 3x1 + 5x2 − s1 = 20.
• Surplus variable may not be used as a basic variable for the
constraint, in which it occurs, because it has a coefficient -1
instead of +1
• But the constraint is now the equality constraint, so we can
treat it the same way as other equality constraints:
3x1 + 5x2 − s1 + a1 = 20
Negative RHS of the constraint
• Problem – basic variable for this constraint
would have to be negative and it cannot
• Solution: multiply by -1 and treat the resulting
constraint of the type ≥, =, ≤ respectively.
Example conversion
min Z = 15x1 + 10x2 ← minimiation instead of maximization
x1 ≤ 2
x2 = 3
← equality instead of ≤
x1 + x2 ≥ 4
← ≥ instead of ≤
max (-W) + a1 + a2 = 0
max (-Z) + 15x1 + 10x2 =0
x1 + s1 = 2
x2 + a1 = 3
x1 + x2 – s2 + a2 = 4
[phase 1 objective function]
[phase 2 objective function]
Variables that may be negative
• Two possibilities:
1. Lower bound
•
•
E.g. production level can fall no more than Lj from the
present level xj ≥ Lj
Solution – change of variables xj = xj′ + Lj
2. No bounds at all: next slide
No bounds on a variable
• Unbounded variable xj is replaced by the difference between two
nonnegative variables xj = xj+ – xj• Simplex will automatically assign one of them a nonzero value.
– Columns for these variables are a mirror image of one another (+1 in
one and -1 in another), so both variables cannot be basic at the same
time
– If xj+ > 0
then xj = xj+
– If xj- > 0
then xj = - xj-,
– If xj+ =xj- = 0 then xj = 0.
• If we have a lot of unbounded variables we use for each of them
one variable xj-. In the optimal solution xj- will be the absolute value
of the most negative variable xj . Variables xj+ tell us then by how
much a given variable exceeds this most negative variable
Tropical Treats Kissable Lip shine– Jessica
Simpson & Urban Chic Boutique
• Dear Students of DMOR class at WSE,
My name is Jessica Simpson. I have been going through some
tough times recently and am having a real problem with one of my
cosmetic lines. The info for the line is on the next page. Recently
though costs are changing based on market demand in addition to
highly fluctuating resource costs. My problem is this we currently
have an LP that we solve to find the optimal amount to produce of
each product. However, every time a parameter changes, I am
always forced to resolve the LP and this takes too long. I was
hoping you guys could find a better way. Lately I have just been out
of it. For example, Nick and I decided to split our Hummer in half,
and now I need to buy a new one. Oh yeah, about the LP it seems
to have been misplaced when I was moving out of my Malibu
house.
Please Help!
-Jessica
http://www.urbanchic-boutique.com/dhop/jessica-simpsondessert-treats/jessica-simpson-tropical-treats-kissable-lipshine/prod_89.html
Problem data
•
•
•
•
•
•
Peach
Pineapple
Orange
Mango
Raw material
2
3
4
7
Hours of labor
3
4
5
6
Sale price ($)
4
6
7
8
Jessica sells 4 kinds of tropical lip shine.
Exactly 950 total units must be produced.
Customers demand that at least 400 units of Mango lip shine be produced.
Formulate an LP to maximize profit.
Raw Materials Available <=4600
Labor Available <=5000
Formulating an LP
• Amount of each lip shine
– X1: peach
– X2: pineapple
– X3: orange
– X4: mango
max
Our results
Decision variables
Objective function
0
400
150
400
6650
• What may change?:
– Based on market demand, prices can change
– Cost of resources and labor change constantly
– Amount of resources can change
– New products can be introduced
• Pytanie: How/when can I determine if my current solution (or basis) is still
optimal given the change without having to resolve the LP?
Sensitivity report
• Should it look like this?
Microsoft Excel 12.0 Raport wrażliwości
Arkusz: [example.xlsx]Problem
Raport utworzony: 2013-03-12 12:07:53
Komórki decyzyjne
Komórka
$B$6
$C$6
$D$6
$E$6
Nazwa
Decision variables Peach
Decision variables Pineapple
Decision variables Orange
Decision variables Mango
Wartość
Przyrost
końcowa marginalny
0 -1,000001907
400
0
150
0
400
0
Warunki ograniczające
Komórka
$H$2
$H$3
$H$4
$H$5
Nazwa
Raw Material
Hours of labor
Quantity all
Qunatity mango
Wartość
Mnożnik
końcowa Lagrange'a
4600
1
4750
0
950
3
400
-2
Common errors
Forgetting to Check
• Assume Non-Negativity
• Forgetting to Check Assume Linear Model
• We will assume for all conditions derived that
degeneracy is not present
• At the end of the lecture we will show how to
modify the conditions when degeneracy is
present.
Sensitivity report for an LP
Decision variables
0
Objective function
400
150
400
6650
Microsoft Excel 12.0 Raport wrażliwości
Arkusz: [example.xlsx]Problem
Raport utworzony: 2013-03-12 12:07:18
Komórki decyzyjne
Wartość Przyrost Współczynnik Dopuszczalny Dopuszczalny
Komórka
Nazwa
końcowa krańcowy funkcji celu
wzrost
spadek
$B$6
Decision variables Peach
0
-1
4
1
1E+30
$C$6
Decision variables Pineapple
400
0
6 0,666666667
0,5
$D$6
Decision variables Orange
150
0
7 1,000000001
0,5
$E$6
Decision variables Mango
400
0
8
2
1E+30
Warunki ograniczające
Komórka
Nazwa
$H$2
Raw Material
$H$3
Hours of labor
$H$4
Quantity all
$H$5
Qunatity mango
Wartość
Cena Prawa strona Dopuszczalny Dopuszczalny
końcowa dualna
w. o.
wzrost
spadek
4600
1
4600
250
150
4750
0
5000
1E+30
250
950
3
950
50
100
400
-2
400
37,5
125
Type of Change 1: Changing the Cost
Coefficient of a Basic Variable
“Hi guys I was at the market and noticed the price of
pineapple went up by 50 cents. That means if I raise the
price of the pineapple lip shine by 50 cents I will make more
money. Right?” Jessica
Solution:
Be careful!! The next steps applies only to basic variables.
Steps to Take:
• Step 1: Check if the change of the objective coefficient is
within the allowable range?
• Step 2: If so, the optimal basic feasible solution will not
change. Calculate the change in profit.
Microsoft Excel 12.0 Raport wrażliwości
Arkusz: [example.xlsx]Problem
Raport utworzony: 2013-03-12 12:07:18
Komórki decyzyjne
Wartość Przyrost Współczynnik Dopuszczalny Dopuszczalny
Komórka
Nazwa
końcowa krańcowy funkcji celu
wzrost
spadek
$B$6
Decision variables Peach
0
-1
4
1
1E+30
$C$6
Decision variables Pineapple
400
0
6 0,666666667
0,5
$D$6
Decision variables Orange
150
0
7 1,000000001
0,5
$E$6
Decision variables Mango
400
0
8
2
1E+30
Warunki ograniczające
Komórka
Nazwa
$H$2
Raw Material
$H$3
Hours of labor
$H$4
Quantity all
$H$5
Qunatity mango
Wartość
Cena
Prawa strona Dopuszczalny Dopuszczalny
końcowa dualna
w. o.
wzrost
spadek
4600
1
4600
250
150
4750
0
5000
1E+30
250
950
3
950
50
100
400
-2
400
37,5
125
Exercise
Problem A
• Suppose the price of x1 is increased by 60 cents.
What is the new optimal solution and change in
profit?
Problem B
• Suppose the price of x3 is decreased by 60 cents.
What is the new optimal solution and change in
profit?
Type of Change 2: Changing the Cost
Coefficient of a Non-Basic Variable
“Hello Class, I went to the store to buy some of my peach gloss
and found out none of it was being produced because it
wasn’t profitable. What should I charge to make them in
the optimal mix?”, Jessica (with help of agent)
In order to answer this, we need to look at the “reduced
costs”
• If the reduced cost of a non basic variable xi is – ri, it
means that increasing the “cost” of the variable by ri will
lead to an optimum basis that includes xi.
• What is the relationship between the reduced cost and the
z row coefficient?
Microsoft Excel 12.0 Raport wrażliwości
Arkusz: [example.xlsx]Problem
Raport utworzony: 2013-03-12 12:07:18
Komórki decyzyjne
Wartość Przyrost Współczynnik Dopuszczalny Dopuszczalny
Komórka
Nazwa
końcowa krańcowy funkcji celu
wzrost
spadek
$B$6
Decision variables Peach
0
-1
4
1
1E+30
$C$6
Decision variables Pineapple
400
0
6 0,666666667
0,5
$D$6
Decision variables Orange
150
0
7 1,000000001
0,5
$E$6
Decision variables Mango
400
0
8
2
1E+30
Warunki ograniczające
Komórka
Nazwa
$H$2
Raw Material
$H$3
Hours of labor
$H$4
Quantity all
$H$5
Qunatity mango
Wartość
Cena
Prawa strona Dopuszczalny Dopuszczalny
końcowa dualna
w. o.
wzrost
spadek
4600
1
4600
250
150
4750
0
5000
1E+30
250
950
3
950
50
100
400
-2
400
37,5
125
Exercise
Problem C:
• What property exists after we increase the price of the peach gloss to exactly $5?.
Problem D:
• What is the reduced cost of a basic variable? Explain!
Problem E: (Assume max problem!)
• Here is an optimal simplex tableau: determine the reduced cost of variable c:
Problem C: If we increase the price of cream by 1, then we are indifferent about pivoting
it into the basis. Thus there exist multiple optimal solutions.
Problem D: Reduced costs for max problems are non-positive. They tell us how much we
need to increase the price of a product by before we start producing it. If we are already
producing it (i.e., it is basic) it has a reduced cost of zero.
Problem E : The reduced cost of c is -2 .
Type of Change 3: Changing a Right
Hand Side Coefficient
“Uggh! You won’t believe this. After seeing me
on Newlyweds, MTV decided it would be
profitable to make a reality show where
instead of having 4600 of raw materials, I
have only 4499 .What should I do (that is,
what happens to the optimal solution)?”,
Jessica from Maui
Solution:
• Step 1: Determine if the right hand side
change is within the allowable range
• Step 2: If so the optimal basis will NOT
change, and we can use the shadow price of
the constraint to determine the change in
the optimal objective value.
Microsoft Excel 12.0 Raport wrażliwości
Arkusz: [example.xlsx]Problem
Raport utworzony: 2013-03-12 12:07:18
Komórki decyzyjne
Wartość Przyrost Współczynnik Dopuszczalny Dopuszczalny
Komórka
Nazwa
końcowa krańcowy funkcji celu
wzrost
spadek
$B$6
Decision variables Peach
0
-1
4
1
1E+30
$C$6
Decision variables Pineapple
400
0
6 0,666666667
0,5
$D$6
Decision variables Orange
150
0
7 1,000000001
0,5
$E$6
Decision variables Mango
400
0
8
2
1E+30
Warunki ograniczające
Komórka
Nazwa
$H$2
Raw Material
$H$3
Hours of labor
$H$4
Quantity all
$H$5
Qunatity mango
Wartość
Cena
Prawa strona Dopuszczalny Dopuszczalny
końcowa dualna
w. o.
wzrost
spadek
4600
1
4600
250
150
4750
0
5000
1E+30
250
950
3
950
50
100
400
-2
400
37,5
125
Definition: . The shadow price of the i-th
constraint is the amount by which the optimal
Z-Value of the LP is improved if the RHS is
increased by one unit.
VERY IMPORTANT: The Shadow Price of the i-th
constraint is ONLY valid within the RHS range
of the i-th constraint.
Exercise
Problem F: What is the change in the objective function if the number
of available labor hours changes to 4800? What if this number is
4700?
Problem G: . What can you tell me about the shadow price of a “≥
constraint”? How about an “= constraint”?
Problem F:
Step 1: If the number is 4800, then the change is within the allowable decrease of
250, and the current basis remains optimal.
Step 2: The shadow price of the labor constraint is zero. Why? Thus the change in the
objective is 0 × 200 = 0.
If the number if 4700, this is a decrease of 300 and we are outside the allowable range.
Can you say anything about this case?
Problem G:
A “≥ constraint” for a maximization problem always has a nonpositive shadow price.
Intuitively, if we increase the RHS, then we further restrict the feasible region, which
can not make us better off! We can say nothing about the sign of an “= constraint”. It
could be positive, negative, or 0.
Type of Change 4: Purchasing Extra
Resources
“Guys, My sister Ashlee just lost her recording contract. I
know, it’s shocking. Anyway, she needs a job; she is
willing to work for 1 hour. She also said she could convert
her unit of talent into a unit of raw material, whatever
that means. What is the most I should pay for the unit of
raw materials and for her?”, Jessica
Solution:
Each increases is within the range. So, we can use the
shadow prices to determine the change in objective if
either change takes place. But we cannot assume that
changing both is OK.
• The Raw Material SP is 1, corresponding to a revenue
increase of $1 if we accept the offer. Thus to breakeven
or profit we should pay no more then $1 for the unit
• The Labor SP is 0, resulting in no additional revenue if we
hire Ashley. And you say our models don’t reflect
reality!!!
Microsoft Excel 12.0 Raport wrażliwości
Arkusz: [example.xlsx]Problem
Raport utworzony: 2013-03-12 12:07:18
Komórki decyzyjne
Wartość Przyrost Współczynnik Dopuszczalny Dopuszczalny
Komórka
Nazwa
końcowa krańcowy funkcji celu
wzrost
spadek
$B$6
Decision variables Peach
0
-1
4
1
1E+30
$C$6
Decision variables Pineapple
400
0
6 0,666666667
0,5
$D$6
Decision variables Orange
150
0
7 1,000000001
0,5
$E$6
Decision variables Mango
400
0
8
2
1E+30
Warunki ograniczające
Komórka
Nazwa
$H$2
Raw Material
$H$3
Hours of labor
$H$4
Quantity all
$H$5
Qunatity mango
Wartość
Cena
Prawa strona Dopuszczalny Dopuszczalny
końcowa dualna
w. o.
wzrost
spadek
4600
1
4600
250
150
4750
0
5000
1E+30
250
950
3
950
50
100
400
-2
400
37,5
125
Exercise
Problem H
• Suppose now that Raw Materials cost $5 per unit . Suppose
Johnny Knoxville offers to sell you an addition unit for $.50
Should you take the deal? What is the break-even price?
Problem I
Same for labor as problem H
Problem H
The key is to interpret the shadow price as follows: If Jessica can buy one more unit
at $4, then profits increase by $1 since the shadow price is 1. Thus Jessica could pay
4+1=5 and profits will increase by 1-1=0. Thus the most Jessica should pay is 5.
Since 4.5 is less then 5, she should take the deal.
Problem I T
The shadow price here is 0. We are not using all of the labor we are given.
Additional labor is worthless, and we should not pay for it.
Type of Change 5: A New Product
•
“Hey Yall, I just got the best idea for a new flavor of Lip Gloss called Limonette.
To make some, 8 units of raw material are needed and 7 hours of labor are
needed. If I sell it for $7, will any be produced?”, Jessica
Solution:
• Step 1: Determine if the right hand side change is within the allowable range.
• Step 2: Price out to calculate the reduced cost.
Peach
Pineapple
Orange
Raw material
2
3
4
7
8
Hours of labor
3
4
5
6
7
Sale price ($)
4
6
7
8
7
Microsoft Excel 12.0 Raport wrażliwości
Arkusz: [example.xlsx]Problem
Raport utworzony: 2013-03-12 12:07:18
Mango Limonette
“reduced cost”: 7 - [ 3 * 1 – 2 * 0
+ 1 * 8 + 0 * 7 ] = -4
One more
altogether
Komórki decyzyjne
Wartość Przyrost Współczynnik Dopuszczalny Dopuszczalny
Komórka
Nazwa
końcowa krańcowy funkcji celu
wzrost
spadek
$B$6
Decision variables Peach
0
-1
4
1
1E+30
$C$6
Decision variables Pineapple
400
0
6 0,666666667
0,5
$D$6
Decision variables Orange
150
0
7 1,000000001
0,5
$E$6
Decision variables Mango
400
0
8
2
1E+30
Warunki ograniczające
Komórka
Nazwa
$H$2
Raw Material
$H$3
Hours of labor
$H$4
Quantity all
$H$5
Qunatity mango
Mango
does not
increase
Wartość
Cena
Prawa strona Dopuszczalny Dopuszczalny
końcowa dualna
w. o.
wzrost
spadek
4600
1
4600
250
150
4750
0
5000
1E+30
250
950
3
950
50
100
400
-2
400
37,5
125
Type of Change 6: Parametrics
“Guys, what would a graph of the optimal objective value look like that
used the amount of available raw materials as a parameter?”
Definitely not Jessica Simpson
Solution:
• In the allowable range (4450,4850), profit is linear in raw materials
and the slope is equal to the shadow price value (1)
• We do the same for values outside this interval.
• In the respective intervals, the function is linear. In general it is
concave.
3000
3050
3100
3150
3200
3250
3300
3350
3400
3450
3500
3550
3600
3650
3700
3750
3800
3850
3900
3940
3990
4040
4090
4140
4190
4240
4290
4340
4390
4440
4490
4540
4590
4640
4690
4740
4790
4840
4890
4940
4990
5040
5090
5140
5190
Profit as a function of raw materials
8000
7000
6000
5000
4000
3000
2000
1000
0
Exercise
Problem I
• What type of function is our graph?
Problem J
• Is it possible for an optimal BFS to have more than one
shadow price correspond to a constraint?
Problem N: It is a piecewise linear concave graph
Problem O: Yes it is. When an LP has multiple optimal solutions multiple
shadow prices are possible. This occurred at the breakpoints of our graph
Degenerate solutions
Up until now we have assumed all BFS’s were non degenerate. What happens
if a basis is degenerate?
Three Oddities Occur When a BFS is degenerate
• Oddity 1: In the RANGE IN WHICH THE BASIS IS UNCHANGED at least one
constraint will have a 0 AI or AD. This means that for at least one
constraint, the SHADOW PRICE can tell us about the new z-value for either
an increase or decrease in the RHS, but not both.
• Oddity 2: For a nonbasic variable to become positive, a nonbasic
variable’s objective function coefficient may have to be improved by
more than its REDUCED COST.
• Oddity 3: Increasing a variable’s objective function coefficient by more
than its AI or decreasing it by more than its AD may leave the optimal
solution the same.
Remember when performing analysis, always ask first if the current BFS is
degenerate. If so follow the rules on this slide
Lecture summary
• Using Excel to determine information
• Shadow prices
• Determining upper and lower bounds so that
the shadow price remains valid.
• Changes in cost coefficients.
• Key Idea: never go into business with Jessica
Simpson