Decision Maths

1
Decision Maths

• Lesson 9 Integer Programming

2
Ex 5a q 3

• Because the Robot is tested over 30 minutes then
  everything should be converted to output/minute.
• Let x number of minutes he will walk.
• Let y number of minutes he will run.
• The robot walks at 1.5ms-1, Therefore he will
  walk at 90m/min
• (1.5 x 60 90).
• By the same idea he will run at 240 m/min.
• Now the objective function becomes
• Distance 90x 240y
• Because he is going for half an hour we can get
  the constraint
• x y 30

3
Ex 5a q 3

• The robot consumes I unit of power/metre when
  walking and 3 units/metre when running.
• If he walks at 90 m/min then he will use power at
  90units/min.
• If he runs at 240 m/min then he will use power at
  720 units/min, because he uses 3 units/metre.
• The robot only has 9000 units of power so we can
  now obtain the second constraint.
• 90x 720y 9000
• Do not forget that x 0 and y 0.

4
Ex 5a q 4

• It is often very handy to put the information
  into a table.

• From the table it is easy to draw up the linear
  programming problem.
• Let x quantity of cloth A
• Let y quantity of cloth B
• The objective is to maximise Profit 3x 2.5y
• The first constraint is easy 2x 1y 100

5
Ex 5a q 4

• The second constraint will be
• (1/2)x (1/3)y 28
• Which we can multiply by 6 to get rid of the
  fractions
• 3x 2y 168
• The two time constraints are reasonably
  straightforward, you just need to make sure that
  everything is in the same units.
• Here the time needs to be converted in to
  minutes, so
• 5x 4y 360
• And
• 4x 5y 360
• x 0 and y 0.

6
Ex 5a q 5

• This is not an easy question, but if you think
  about it carefully and decide on your constraints
  then it is achievable.
• Let a metres cut to plan A.
• Let b metres cut to plan B.
• When the paper is sold he will make 11p profit
  from each metre of plan A and 12p profit from
  each metre of Plan B.
• Therefore he wants to maximise 11a 12b.
• However he will also make money from what is left
  at 8p a metre.
• What is left is the original length minus amount
  for plan A and plan B or 200 a b.
• So he is actually maximising
• 11a 12b 8(200 a b) 3a 4b 1600.
• As the 1600 is a fixed amount (constant) that
  will always be there. You actually just need to
  maximise 3a 4b. You can then add on the 1600
  afterwards.

7
Ex 5a q 5

• As there is 200m of paper, the first constraint
  must be
• x y 200
• The second constraint comes from the fact that
  the manufacturer does not wish to waste more than
  15 of his paper.
• This will be 15 of the area.
• the area is 200m x 0.4m 80m2 (40cm 0.4m)
• Now 15 of 80m2 0.15 x 80 12m2
• The waste with plan A is 5cm wide and the waste
  with plan B is 7cm wide. So the overall waste
  will be 0.05x 0.07y.
• The constraint will be
• 0.05x 0.07y 12

8
Ex 5a q 6

• Again this is not an easy question and It
  requires a bit of off the wall thinking.
• Let x amount of Gin
• Let y amount of Martini
• i)
• The question tells us that dryness (x 3y)/(x
  y)
• James states that the dryness must be less than
  or equal to 2.
• So (x 3y)/(x y) 2
• However x y 200
• So (x 3y)/200 2
• And (x 3y) 400

9
Ex 5a q 6

• ii)
• James has requested that his cocktail has an
  alcoholic content between 18 and 36
• Because x and y are amounts of liquid when you
  calculate the of alcohol it will also be an
  amount of liquid.
• This constraint is in capacity of drink.
• This means that out of the 200ml of drink between
  36ml and 72ml needs to be alcohol.
• 36 comes from doing 200 x 18 and 72 from 200 x
  36.
• 0.45x will tell you the alcohol provided by the
  gin and 0.15y will be the alcohol from the
  Martini. So
• 36 0.45X 0.15Y 72
• This is two constraints in one equation.

10
Ex 5a q 6

• v)
• p proportion of Gin
• q proportion of Martini
• From this we know that p q 1
• So (x 3y)/(x y) 2 can be replaced with
• (p 3q)/(p q) 2
• Which gives
• p 3q 2
• p 3(I p) 2
• 0.5 p
• Now the alcohol constraint is
• 0.18 0.45p 0.15q 0.36
• Lets make that a bit easy to work with
• 18 45p 15q 36

11
Ex 5a q 6

• 18 45p 15(1 p) 36
• 18 45p 15 15p 36
• 18 30p 15 36
• 3 30p 21
• 0.1 p 0.7
• Combine this with the result from the other
  constraint and you get
• 0.5 p 0.7
• With the method before you got the answer that
  100ml out of the 200ml or 50 should be Gin for
  the cheapest.
• You also got that 140ml out of the 200ml or 70
  should be Gin for the most expensive.
• Same answer as above.

12
Problem 1Non-integer solutions

• It is possible to have non-integer solutions that
  could be feasible in the real world.
• Example Ex 5A q 3
• It is also possible to have non-integer solutions
  that are not feasible in the real world.
• In the case of lesson 7 were we were making two
  different types of shed, it would not be sensible
  to make say 6.5 of shed A and 7.2 of shed B.
• When this occurs then you must search for the
  optimal integer solution.
• This can be found somewhere near to the optimal
  solution.

13
Problem 1-Non-integer solutions

• A company makes two types of toy, a doll and a
  bear.
• Both toys are prepared in a machine room and then
  passed on to the craft shop to be hand finished.
• The doll needs 1 minute in the machine room and
  10 minutes in the craft room.
• The bear needs 2 minutes in the machine room and
  9 minutes in the craft room.
• When sold the doll will make 4 profit and the
  bear will make 5 profit.
• Find the maximum profit that can be made from the
  company.

14
Problem 1-Non-integer solutions

• All of this can be summarised by the following
• Solve the following problem
• Maximise P 4x 5y
• Subject to the constraints
• x 2y 12
• 10x 9y 90
• x 0, y 0

15
Problem 1-Non-integer solutions

• From lesson 8 we should be confident to draw the
  feasible region and calculate the optimal
  solution.

16
Problem 1-Non-integer solutions

• You can see from the graph that the optimal
  solution will come from the solution to the
  simultaneous equations.
• x 2y 12
• 10x 9y 90
• 10x 20y 120
• 10x 9y 90
• 11y 30
• y 30/11 2.72
• x 230/11 12
• x 72/11 6.54

17
Problem 1-Non-integer solutions

• We have just seen on the previous slide that the
  optimal solution is not a feasible one when
  interpreted in the real world.

18
Problem 1-Non-integer solutions

• Lets take a look at the solution again.
• We can draw on lines to indicate where the
  integer solutions would be.

19
Problem 1-Non-integer solutions

• Hopefully you can clearly see that the feasible
  solution is bounded by the integer values x 6
  7, y 2 3.
• Lets zoom in on this area.

20
Problem 1-Non-integer solutions

• Here is the zoomed in picture of the critical
  solution.
• Remember the red lines represent integer values.

y 3
y 2
X 6
X 7
21
Problem 1-Non-integer solutions

• There are three integer solutions near to the
  optimal solution.

y 3
y 2
X 6
X 7
22
Problem 1-Non-integer solutions

• From the previous slide you can see that there
  are three integer solutions in feasible region
  close to the optimal solution.
• All you have to do is evaluate the profit
  function at these co-ordinates.
• From the table we can see that the optimal
  solution comes from making 6 dolls and 3 bears

23
Problem 1-Non-integer solutions

• We can support this result by using the ruler
  method.

y 3
y 2
X 6
X 7
24
Problem 1-Non-integer solutions

• You can see that the furthest the ruler can move
  and still be on a feasible integer solution is x
  6, y 3.

y 3
y 2
X 6
X 7
25
Problem 2Multiple Solutions

• Sketch the feasible region for the following
  linear programming problem.
• Maximise P 5x 5y
• Subject to the following constraints
• Y 7, X 7
• X Y 10
• X 0, Y 0

26
Problem 2-Multiple Solutions

• The feasible region should look like this.

27
Problem 2-Multiple Solutions

• Now I can draw the objective function and find
  the optimal solution.

28
Problem 2-Multiple Solutions

• As you can see in this particular case there are
  many solutions each one will give you the same
  answer.

29
Problem 2-Multiple Solutions

• We can illustrate this further if we consider a
  table of the possible feasible values.

30
Problem 3Minimisation

• So far all of the examples we have looked at have
  been Maximisation problems.
• Minimisation is no different, you still draw the
  graphs and calculate values in the tables, its
  just you are looking for the smallest answer and
  not the largest.
• With the ruler method the aim is bring the ruler
  as close to the origin as you can.
• Be careful when you plot your constraints, check
  that you shade in the correct side of the line.
• Try this example for practice.????

31
Work

• You should now be able to complete the rest of
  chapter 5.
• Ex 5b pg 150
• Ex 5c pg 158