Transcript Linear Programming and Distribution Planning

Introduction to Mathematical
Programming
Matthew J. Liberatore
John F. Connelly Chair in Management
Professor, Decision and Information Technologies
LINEAR PROGRAMMING
LP deals with the problem of allocating limited
resources among competing activities.
For example, consider a company that makes
tables and chairs (competing activities) using a
limited amount of large and small Legos (limited
resources).
LINEAR PROGRAMMING
The objective of LP is to select the best or optimal
solution from the set of feasible solutions (those
that satisfy all of the restrictions on the resources).
Suppose profit for each table is $20 while profit
for each chair is $16.
We may choose to identify the number of tables
and chairs to produce to maximize profit while not
using more Legos than are available.
COMPONENTS OF AN LP
Decision Variables: factors which are controlled by
the decision maker.
x1 = the number of tables produced per day
x2 = the number of chairs produced per day
Objective function: profit, cost, time, or service must
be optimized.
The objective may be to optimize profit.
COMPONENTS OF AN LP
Constraints: restrictions which limit the availability
and manner with which resources can be used to
achieve the objective.
It takes 2 large and 2 small Legos to produce a
table and 1 large and 2 smalls to produce a chair.
We may only have 6 large and 8 small Legos
available each day.
ASSUMPTIONS
Linearity: Linear objective function and linear
constraints.
This implies proportionality and additivity.
For example, it takes 2 large Legos to produce 1
table and 4 to produce 2 tables.
It takes 3 large Legos to produce 1 table and 1
chair.
ASSUMPTIONS
Divisibility: The decision variables can take on
fractional values.
The optimal solution may tell us to produce 2.5
tables each day.
Certainty: The parameters of the model are known or
can be accurately estimated.
For example, we assume that the profitability
information is accurate.
ASSUMPTIONS
Non-negativity: All decision variables must take on
positive or zero values.
LEGO PRODUCTS, INC.
Lego Products, Inc. manufactures tables and chairs.
Profit for each table is $20 while each chair generates
$16 profit.
Each table is made by assembling two large and two
small legos. Each chair requires one large and two
small legos.
Currently, Lego Products has six large and eight small
legos available each day.
LEGO EXAMPLE: Introduction
Pictures of a table and a chair are shown below.
Table
Chair
LEGO EXAMPLE: Optimal Solution
How many tables and chairs should we produce to
maximize daily profit?
producing 3 tables generates a daily profit of $60,
producing 4 chairs generates a daily profit of $64,
however,
producing 2 tables and 2 chairs generates the optimal
daily profit of $72.
Using Solver
Solver in Excel can be used to obtain the
solution and will now be demonstrated
The problem formulation is:
MAX 20*X1+16*X2
Subject to:
2*X1+1*X2<=6
2*X1+2*X2<=8
LEGO EXAMPLE: Unused legos
Do we have any unused large or small legos for all of
the solutions that you just found?
There are no unused large or small legos for the
optimal solution.
There are 2 unused small legos if 3 tables are
made.
There are 2 unused large legos if 4 chairs are made.
LEGO EXAMPLE: Slack and Surplus
The difference between the available resources and
resources used is either slack or surplus.
Slack is associated with each less than or equal to
constraint, and represents the amount of unused
resource.
Surplus is associated with each greater than or equal
to constraint, and represents the amount of excess
resource above the stated level.
LEGO EXAMPLE: Slack and Surplus
We have two slack values - one for large legos and
one for small legos - and no surplus values.
The slack or surplus section shows that both
constraints have zero slack.
Suppose we must produce at least one table
(X1>=1). The original optimal solution is still the
best. Since X1=2, we produce one surplus table.
Right Hand Side Changes
Now we are ready to illustrate the key concepts
of sensitivity analysis.
How much would you be willing to spend for
one additional large Lego?
One additional large Lego is worth $4.
Original solution: X1=2, X2=2, profit = $72;
New solution: X1=3, X2=1, profit = $76.
You would be willing to spend up to $4 (7672) for one additional large Lego.
RIGHT HAND SIDE CHANGES
How much would you be willing to spend for two
additional large legos?
Two additional large legos are worth $8.
Original solution: X1=2, X2=2, profit = $72;
New solution: X1=4, X2=0, profit = $80.
You would be willing to spend up to $8 (80-72)
for two additional large legos.
RIGHT HAND SIDE CHANGES
How much would you be willing to spend for three
more large legos?
The third large lego is not worth anything since
the optimal solution remains unchanged.
What happens if your supplier can only provide five
large legos each day?
The optimal solution is: X1=1, X2=3, profit=68,
so we lose $4.
RIGHT HAND SIDE CHANGES
What happens if your supplier can only provide four
large legos each day?
The optimal solution is: X1=0, X2=4, profit=64,
so we lose another $4.
What happens if your supplier can only provide three
large legos each day?
The optimal solution is: X1=0, X2=3, profit=48,
so we lose an additional $16, and not $4.
SHADOW PRICES
This last set of exercises enables us to determine the
shadow price (also called the shadow price) for a
resource constraint (large legos).
The shadow price, for a particular constraint, is the
amount the objective function value will increase
(decrease) if the right hand side value of that
constraint is increased (decreased) by one unit.
We found that the shadow price for large legos is $4.
SHADOW PRICES
What is the shadow price of the small legos?
With two additional small legos the new
solution:
X1=1, X2=4, profit = $84.
You would be willing to spend up to $12
(84-72) for two additional small legos, so
the shadow price is $6 (12/2).
SHADOW PRICES
In general, the shadow prices are meaningful if one
right hand side (RHS) value of a constraint is
changed, and all other parameters of the model
remain unchanged.
REDUCED COSTS
What happens if the profit of tables increases to $35?
The optimal solution is X1=3, X2=0, profit = $105.
Note that no chairs are being produced.
REDUCED COSTS
When a decision variable has an optimal value of
zero, the allowable increase for the objective
function coefficient is also called the reduced cost.
The reduced cost of a decision variable is the amount
the corresponding objective function coefficient
would have to change before the optimal value
would change from zero to some positive value.
REDUCED COSTS
The reduced cost for tables is zero in the original
formulation. Why is this the case?
We are already producing tables.
SENSITIVITY ANALYSIS PROBLEM
A manufacturing firm has discontinued production of
a certain unprofitable product line thus creating
considerable excess production capacity.
Management is considering devoting this excess
capacity to one or more of three products; call
them products 1, 2, and 3.
The available capacity on the machines that might
limit output is summarized below:
SENSITIVITY ANALYSIS PROBLEM
MACHINE TYPE
Milling machine
Lathe
Grinder
AVAILABLE TIME
(machine hours / week)
500
350
150
SENSITIVITY ANALYSIS PROBLEM
The number of machine hours required for each unit
of the respective products is:
MACHINE TYPE
Milling machine
Lathe
Grinder
P1
9
5
3
P2
3
4
0
P3
5
0
2
SENSITIVITY ANALYSIS PROBLEM
The sales department indicates that the sales potential
for products 1 and 2 exceeds the maximum
production rate and that the sales potential for
product 3 is 20 units per week.
The unit profit would be $3000, $1200, and $900,
respectively, for products 1, 2, and 3.
SENSITIVITY ANALYSIS PROBLEM
Solver is used to determine the optimal solution. .
a. What are the optimal weekly production levels for
each of the three products?
Product 1 = 45.23
Product 2 = 30.95
Product 3 = 0
SENSITIVITY ANALYSIS PROBLEM
b. What profit will be obtained if the optimal
solution is implemented?
$172,857.10 per week
c. How much unused capacity exists on the milling
machine, the lathe, and the grinder?
Milling = 0; Lathe = 0; Grinder = 14.28
(See SLACK entries)
SENSITIVITY ANALYSIS PROBLEM
d. How much would the objective function change if
the amount of available time on the grinder
increased from 150 hours per week to 250 hours?
Will the objective function increase or decrease?
Currently the grinder has 14.28 hours of slack time
so its shadow price is 0.
Increasing the available hours from 150 to 250 will
not change the total profit.
SENSITIVITY ANALYSIS PROBLEM
e. The profit for product 3 is $900 per unit and the
current production level is zero.
How much would the profit per unit have to
change before it would be profitable to produce
product 3's?
The profit per unit would have to increase by its
reduced cost of $528.57.
SENSITIVITY ANALYSIS PROBLEM
f.
The milling machine capacity can be increased at a
cost of $160 per hour. Is it economic to increase
capacity by 10 hours?
Since the shadow price per hour (285.7143) is
greater than the cost (160), it is worth increasing
milling capacity on the margin. However, since the
shadow price might change with increasing
capacity we need to rerun to see the full effect of
increasing capacity by 10 hours. Profit does
increase by 2857 (10*285.714) which is greater
than the cost increase of 1600 (10*160). Therefore
the milling capacity should be increased b y 10
hours.