Linear programming
An area of relatively recent development in Mathematics, and one which
has a financially important application, is that of linear programming
It is a method for maximising or minimising variables, subject to constraints.
This can be applied to maximising profit, given constraints on time, material & money
Or applied to minimising costs, given constraints
Eg Al is buying some cows and sheep for his farm.
He buys c cows at £120 each
He buys s sheep at £200 each.
c  s  10
He wants at least 10 animals in total.
He wants more sheep than cows.
He has a maximum of £1800 to spend.
s c
120c  200s  1800
Write down three inequalities involving c and s
By the end of the module you will know how to find a possible
combination of cows and sheep that satisfies all 3 inequalities.
Linear programming
We will look in more detail about how to sketch inequalities, but for now,
just to give an overall sense of the purpose of the method…
Eg from previously
Al is buying c cows and s sheep for his farm.
Three inequalities involving c and s are:
c  s  10
s c
120c  200s  1800
What possible combinations could Al buy?
Al can sell the produce of each cow for £150
and the produce of each sheep for £450.
How many of each should he buy to
maximise his profit?
Profit  150c  450 s
Cows
Sheep
Profit
3
7
£3600
4
6
£3300
5
6
£3450
Sketching regions
Eg find the region that satisfies the inequalities:
2 x  y  30
y  2x
y6
2 x  3 y  36
0  0 the
36line
Eliminate
below
the
line
0  0  30 so eliminate
above
line
y  6above
0
10
If
we
want
eliminate
below
sothe
eliminate
theas
line
2 x  y  30
2 x  y  30
x  0  y  30  (0,30)
Then plot these lines using:
y  0  x  15  (15,0)
y  2x
First identify
Test a the
coordinate
lines of the
notboundaries
on the line to
of each
Either use intercept and gradient to plot
region by
see
replacing
which side
inequality
satisfies
signs
the with
signs
yinequality
 2equals
x
Or find where each line intersects the axes
Gradient = 2, Intercept = 0
2 x  3 y  36
y6
Horizontal line through 6
y6
Test
Test aa coordinate
coordinate
not on
on the
the line
line
Test a not
coordinate
not on the line
2 x  3 y  36
x  0  y  12  (0,12)
y  0  x  18  (18,0)
Eg find the coordinates of the vertices of the feasible region below
y  2x
2 x  y  30
A is at the intersection of:
y  2 x (1)
2 x  y  30 (2)
A
Sub (1) in (2): 4 x  30  x  7.5
Sub x = 7.5 in (1):  y  15
R
So A has coordinates (7.5,15)
y6
2 x  3 y  36
Eg find the coordinates of the vertices of the feasible region below
y  2x
2 x  y  30
B is at the intersection of:
y  6 (1)
2 x  y  30 (2)
Sub (1) in (2): 2 x  6  30
 x  12
R
So B has coordinates (12,6)
B
y6
2 x  3 y  36
Eg find the coordinates of the vertices of the feasible region below
y  2x
2 x  y  30
C is at the intersection of:
y  6 (1)
2 x  3 y  36 (2)
Sub (1) in (2): 2 x  18  36
 x9
R
So C has coordinates (9,6)
C
y6
2 x  3 y  36
Eg find the coordinates of the vertices of the feasible region below
y  2x
2 x  y  30
D is at the intersection of:
y  2 x (1)
2 x  3 y  36 (2)
Sub (1) in (2): 8 x  36  x  4.5
D
Sub x = 4.5 in (1):  y  9
R
So D has coordinates (4.5,9)
y6
2 x  3 y  36
In linear programming, you need to be able to draw the line of an equation
where, effectively, the intercept is a variable. Consider:
3 x  5 y  c  5 y  c  3x  y 
c 3
 x
5 5
c
3
so the gradient is 
and the intercept is
5
5
ie you know the gradient but not the intercept
The intercept is a function of c
As c is varied, so will the intercept, but not the gradient
Watch the demo!
Whatever the value of c, you get parallel lines
So to draw 3 x  5 y  c, draw any line with a gradient of 
3
5
If ax  by  c then y 
This means
a
c a
 x and the line will have gradient 
b b
b
a
b
Eg 3 x  5 y  c has gradient 
3
5
When drawing this, we could imagine
3
triangles repeatedly,
5
or use a larger similar triangle like:
12
or
20
300
500
Depending on the scale of the axes, this may be a lot faster for accurately
sketching a line
Eg show a suitable line, for each axes, of the equation 3 x  5 y  c
3
use
5
use 30
3
6
10
5
50
Eg John makes £1 profit for every chocolate brownie he sells and £2
for every muffin. Using x as the number of brownies he sells and y as
the number of muffins he sells, write an equation for his total profit P
x  2y  P
If To
he increase
needs toprofit,
make move
a profit
ofobjective
£240,
combination
he sell?
£120, what
£180,
the
line
away fromcould
the origin
x  2 y  180
240
120
(0,60), he sells no brownies and 60 muffins
(20,50), he sells 20 brownies and 50 muffins
x  2(40,40),
y  240 he sells 40 brownies and 40 muffins
(60,30), he sells 60 brownies and 30 muffins
(80,20), he sells 80 brownies and 20 muffins
x  2 y  180
(100,10), he sells 100 brownies and 10 muffins
(120,0), he sells 120 brownies and no muffins
x  2 y  120
Eg The feasibility region of a linear programming problem is given
below. Use the objective line method to identify the optimum solution
if the aim is to:
Maximise P  x  2 y
The optimum
solution
is:
Increase
the
profit
byline
Draw
a
‘suitable’
with
The
optimum
point
isimagining
the
last
the
line
sliding
away
from
the
the
gradient
value
x =correct
10,
within
y =the
10,critical
P = 30region
origin, keeping it parallel
1
use 10
2
20
Eg The feasibility region of a linear programming problem is given
below. Use the objective line method to identify the optimum solution
if the aim is to:
Maximise N  3 x  y
Draw
aoptimum
‘suitable’
line
withis:
Increase
the
profit
by
imagining
The
solution
The
optimum
point is the last
the
correct
gradient
the
linewithin
sliding
away
from
the
value
the
critical
region
x = keeping
20, y = 0,it N
= 60
origin,
parallel
3
use 12
1
4
Eg The feasibility region of a linear programming problem is given
below. Use the objective line method to identify the optimum solution
if the aim is to:
Minimise M  200 x  200 y
The optimum solution is
Minimise
bypoint
imagining
the
The
not
optimum
a whole
number
is
the
and
last
Draw
a ‘suitable’
line
with
line
sliding
the
value
we
within
totowards
the
solve
critical
region
theneed
correct
gradient
origin,
keeping itequations
parallel
simultaneous
to find their values
200
use 12
200
12
The fact that the optimal point is always a vertex of the feasibility region gives
rise to a second method of identifying the optimal point called vertex testing
With this method, we find the value of the objective function at every vertex.
For maximise problems, we then identify the vertex that gives the largest value
For minimise problems, we then identify the vertex that gives the smallest value
Eg previously used
Identify the optimum point and value if the aim is to:
a) Maximise M = x + y x  7.5, y  15, M  22.5
x  9, y  6, N  27
b) Minimise N = x + 3y
A(7.5,15)
D(4.5,9)
C(9,6)
B(12,6)
Vertex
M=x+y
N = x + 3y
A(7.5,15)
22.5
52.5
B(12,6)
18
30
C(9,6)
15
27
D(4.5,9)
13.5
31.5
Ex 6C, Q1
Find the optimum point and the optimum value, using:
a) the objective line method, with the
objective ‘maximise M  2 x  y ‘
4 x  y  1400
M  2x  y
use
2
1
400
200
3 x  2 y  1200 (1)
4 x  y  1400 (2)
2  (2)  8 x  2 y  2800 (3)
x  3 y  1200 (3)  (1) 
5 x  1600  x  320
x  320  (2) 1280  y  1400
 y  120
R
Solution x = 320, y = 120
3 x  2 y  1200
 M  2  320  120
 760
Ex 6C, Q1
Find the optimum point and the optimum value, using:
a) the objective line method, with the
objective ‘maximise N  x  4 y ‘
4 x  y  1400
N  x  4y
use
1
4
100
400
x  3 y  1200 (1)
x0
(2)
x  0  y  400
x  3 y  1200
Solution x = 0, y = 400
 N  0  4  400
R
 1600
3 x  2 y  1200
Integer value solutions
Often, solutions to linear programming problems require integer values.
There are 2 methods for doing this, depending on the scale of the graph:
Maximise P  x  2 y
Minimise C  x  y
5 x  2 y  1000
Objective line
3 x  5 y  1500
The closest point is obviously (3,6)
The closest point is NOT obvious…
5 x  2 y  1000 (1)
3 x  5 y  1500 (2)
5 x  2 y  1000
5  (1)  25 x  10 y  5000 (3)
2  (2)  6 x  10 y  3000 (4)
(3)  (4) 19 x  2000  x  105 195  y  236 16
19
Objective line C  x  y
Test 4 nearest points if integer solutions required
106,237
105,237
105 165 ,236 1619 
3 x  5 y  1500
106,236
105,236
Point
105,236
105,237
106,236
106,237
5 x  2 y  1000 3 x  5 y  1500
997  1000
In R?
C  x y
999  1000
1002  1000
1498  1500
1004  1000
1503  1500
Yes
343
WB11 A manager wishes to purchase seats for a new cinema.
He wishes to buy three types of seat; standard, deluxe and majestic.
Let the number of standard, deluxe and majestic seats to be bought be x, y and z
Standard, deluxe and majestic seats each cost £20, £26 and £36, respectively.
The manager wishes to minimise the total cost, £C, of the seats.
Constraints
He decides that the total number of deluxe
x
and majestic seats should be at most half
y  z   2 y  2z  x
2
of the number of standard seats.
y
 0.1 10 y  x  y  z  9 y  x  z
x yz
y
 0.2  5 y  x  y  z  4 y  x  z
x yz
y
The number of majestic seats should be at
z   2z  y
least half of the number of deluxe seats.
2
The number of deluxe
seats should be at least
10% and at most 20% of
the total number of seats.
The total number of seats should be at least 250.
x  y  z  250
Non-negativity x, y, z  0
Objective function
Minimise C  20 x  26 y  36 z
Formulate this situation as a linear programming problem,
simplifying your inequalities so that all coefficients are integers.
WB12. A company produces two types of
self-assembly wooden bedroom suites, the
Oxford
York
‘Oxford’ and the ‘York’. After the pieces of
Cutting
4
6
wood have been cut and finished, all the
materials have to be packaged.
Finishing
3.5
4
The table below shows the time, in hours,
Packaging
2
4
needed to complete each stage of the
Profit (£)
300
500
process and the profit made, in pounds, on
each type of suite.
The times available each week for cutting, finishing and packaging are 66, 56 and 40
hours respectively.
The company wishes to maximise its profit.
Let x be the number of Oxford, and y be the number of York suites made each week.
(a) Write down the objective function.
Maximise P  300 x  500 y
(b) In addition to
2x  3 y  33
x0
y0
find two further inequalities to model the company’s situation.
Cutting: 4 x  6 y  66
 2 x  3 y  33 given
Finishing: 3.5 x  4 y  56  7 x  8 y  112
Packaging: 2 x  4 y  40  x  2 y  20
(c) On the grid below, illustrate all the inequalities, indicating clearly the feasible region
2x  3 y  33
x  0  y  11  (0,11)
y  0  x  16.5  (16.5,0)
14
12
7 x  8 y  112
x  0  y  14  (0,14)
y  0  x  16  (16,0)
10
x  2 y  20
x  0  y  10  (0,10)
y  0  x  20  (20,0)
8
6
4
xmax  20
R
ymax  14
P  300 x  500 y
2
use
300
0
0
2
4
6
8
10
12
14
16
18
20
500
6
10
(d) Explain how you would locate the optimal point
P  300 x  500 y
14
use
300
10
500
12
6
Profit line: draw profit line,
then select point on profit
line furthest from the origin
Furthest point (6,7)
Make 6 Oxford and 7 York
10
Profit  300x  500 y = £5300
8
Point testing: test corner
points in feasible region,
find profit at each and select
the point yielding maximum
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
Vertex
Profit
(0,10)
5000
(16,0)
4800
(6,7)
5300
(14,1)
4700
It is noticed that when the optimal solution is adopted, the time needed for one
of the three stages of the process is less than that available.
(f) Identify this stage and state by how many hours the time may be reduced
2
14
Cutting: 4 x  6 y  66 1
Finishing: 3.5 x  4 y  56 2
12
Packaging: 2 x  4 y  40 3 
1
10
The maximum time possible
for finishing is not used in
the optimum solution
8
x  6, y  7  3.5 x  4 y  49
6
4
So only 49 hours of
finishing is required
2
Reduce the amount
of time by 7 hours
3
0
0
2
4
6
8
10
12
14
16
18
20