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.
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Matthew J. Liberatore
John F. Connelly Chair in Management
Professor, Decision and Information Technologies
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).
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.
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.
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.
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.
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.
Non-negativity: All decision variables must take on positive or zero values.
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.
Pictures of a table and a chair are shown below.
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.
Solver in Excel can be used to obtain the solution and will now be demonstrated
The problem formulation is:
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.
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.
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.
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 (76-72) for one additional large Lego.
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.
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.
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.
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.
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).
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.
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.
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.
The reduced cost for tables is zero in the original formulation. Why is this the case?
We are already producing tables.
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:
MACHINE TYPE(machine hours / week)
Milling machine 500
The number of machine hours required for each unit of the respective products is:
MACHINE TYPE P1 P2 P3
Milling machine 9 3 5
Lathe 5 4 0
Grinder 3 0 2
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.
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
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)
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.
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.
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.