1. Linear Programming - Graphical Method Flashcards
(29 cards)
What table should be constructed during the ‘construct a model’ stage of Linear Programming?
A table of R1, R2 and Profit (P) on the rows against x, y and max resources on the columns
What is the resource equation, R1, equal to?
Suppose Commodity 1 = 3, Commodity 2 = 2 and the max resources for R1 = 7, define the resource equation, R1.
R1: Commodity 1 + Commodity 2 ≤ Max Resource
R1 = 3x + 2Y ≤ 7
What is the resource equation, R2, equal to?
Suppose Commodity 1 = 2, Commodity 2 = 4 and the max resources for R2 = 10, define the resource equation, R2.
The same process is used as with R1.
R2 = 2x + 4y ≤ 10
What is the profit equation equal to?
Assume the cost of Commodity 1 (x) = £2 and the cost of Commodity 2 (y) = £3
P = 2x + 3y
The least amount of commodities that can be produced is zero, therefore only positive production is possible. How is this represented mathematically?
x=0 and y=0
Once the resource and profit equations are formed, how are the production constraints produced?
Find the limits of each resource equation.
i.e. when x=0, y=? and when x=?, y=0 for each resource equation
If the resource equation, R1, is as follows, identify the limits of this equation.
R1: 3x + 2y = 7
When x=0, y=3.5
When y=0, x=2.33
What should be done with the limits of the resource equations?
They should be plotted on a graph, the area beneath the line is referred to as the ‘Feasible Domain of Production’.
Why is the area below the limits of the resource equation referred to as the ‘Feasible Domain of Production’?
Only positive production is allowed, so anywhere where x and y are greater than 0 sits above the line.
What should be plotted on the graph following the feasible domain for R1?
The feasible domain for R2
When both R1 and R2 are plotted, where is the feasible domain of production located for both maximum and minimum problems.
In the intersecting area that falls beneath BOTH lines, not one singular line for maximum, and above both lines for minimum.
What technique should be used to determine the maximum profit within the system?
Corner point method
How do you use the Corner Point method? (4)
Identify the x,y coordinates for all corners of the feasible domain of production.
Draw up a table of corners of shaded region (x,y) and Profit (p)
Use the profit equation to calculate the profit using each value of x,y
Select the option with the highest profit
How do you test your solution?
Put the values of x,y that returned the highest profit into each of your resource equations and check they achieve 100% usage. Also put these values into your constraints formula (positive production) and profit equation.
What is different about the resource equations when considering a minimisation problem rather than a maximisation problem?
The operator in the resource question is ‘greater than’ in minimisation questions (≥) and ‘less than’ in maximisation questions (≤)
What is different about the feasible domain of the system when considering a minimisation problem rather than a maximisation problem?
The feasible domain of the system is above the resource equation lines on the graph, rather than below.
What is different about calculating the profit/cost in the corner point method when considering a minimisation problem rather than a maximisation problem?
You select the solution with the lowest profit, or cost.
If a supply constraint (typically an ii.) question) is put on the product in a minimisation question, how is this calculated?
E.g. 3<x<8 and 0<y<4
The constraint range for x (i.e. 3< x <8) and the constraint range for y (i.e. 0<y<4) are drawn on the pre-existing graph as a series of horizontal and vertical lines, the feasible domain of production then becomes the area above the resource lines but within the supply constraint lines. The corner point method can then be applied.
How is the constraint ‘the company sells twice as many of item ‘x’ compared to item ‘y’’ formulated?
We need to produce 2 X for every Y, therefore at Y=1, X=2
x=2y -> x - 2y = 0
How is the constraint of ‘x - 2y = 0’ added to a graph and interpreted?
x-2y=0 is also equal to x=2y, therefore a straight grade is drawn using this line and anything on this line is compared to the feasible domain of production in reference to the original resource lines.
How is a profit equation made for more complex situations?
Profit = revenue - cost
How would you deal with the following additional constraint?
‘x’ takes up twice the amount of storage as ‘y’. There is currently enough space for 70 of item ‘x’. Determine the optimal level of production needed to maximise profit with this additional constraint.
This constraint means we can store either 70x,0y or 0x,140y (eq. x + 0.5y = 70)
The limits are plotted on the graph and the corner point method is used to identify the area of highest profit.
How would you discuss what effect the minimum or maximum profit solution has on the labour capacity or maximum resources involved?
Using the x and y coordinates for the optimal solution, workout the total time or resources used to deliver the optimal solution, then comment on whether or not there is any spare capacity.
Determine the time, demand and storage constraints based on the following information:
- It takes 6 hours to make a slab (s) and 3 hours to make a beam (b). The manufacturer has up to 40 hours available a week.
- Customer demand dictates the company must make at least three times as many beams (b) as slabs (s)
- Slabs (s) require 4 times the storage space compared to beams (b), there is only room for the equivalent of 4 slabs (s) in any week.
Time: 6s +3b < 40
Demand: 3s < b (line = b=3s, constraint is below this line)
Storage: s + 0.25b <4
