Scholarship Questions
2003
2003
Question 4
• An oil company operates two refineries. The refineries
produce three types of fuel – aviation grade, regular grade
petrol and super grade petrol.
•
• Refinery 1 costs $160 000 per day to operate and Refinery
2 costs $175 000 per day to operate.
•
• The oil company has contracts to produce at least 120 000
litres of aviation fuel, 300 000 litres of regular grade petrol
and 108 000 litres of super grade petrol per month.
•
• The following table gives the daily production statistics in
litres:
• Let the number of days per month that
Refinery 1 operates be x and the number of
days per month that Refinery 2 operates be y.
•
• (a) Using graph paper where necessary,
determine the optimal number of days that
each refinery should operate in order to
minimise the total cost of operating them
both. Calculate this minimum cost.
•
•
•
•
•
•
Constraints:
10 000x + 10 000y ≥ 120 000 Aviation
20 000x + 30 000y ≥ 300 000 Regular
6 000x + 18 000y ≥ 108 000 Super
0 ≤ x ≤ 31 Restriction on days
0 ≤ y ≤ 31
Simplify the inequalities:
•
•
•
•
•
•
Constraints:
x + y ≥ 12 Aviation
2x + 3y ≥ 30 Regular
x + 3 y ≥ 18 Super
0 ≤ x ≤ 31 Restriction on days
0 ≤ y ≤ 31
• Objective function is
• C = 160 000x + 175 000y
• Gradient: -160/175 = -32/35
• Difficult to use this so look for intersection
points.
(0,12), (6,6), (12,2), (18,0),
Costs at each vertex ($000)
Optimum number of days for both refineries is 6
Minimum cost = $2 010 000
• The company is investigating possible changes
at Refinery 2 that may alter the daily cost of
operating the refinery. Assuming that the daily
cost of operating Refinery 1 remains fixed at
$160 000, determine the range of values that
the daily cost of operating Refinery 2 could lie
within and still give the production
combination in (a).
The gradient needs to stay between the gradients of the 2 lines
that form the intersection (6,6) i.e. between -1 and -2/3
-160000
2
-1 <
<m
3
160000 < m < 240000
2004
Question 4
Question 4
• To produce one bottle of POW takes 30
minutes and to produce one bottle of ZAP
takes 20 minutes, and there are 25 hours
available in total for this production-run
operation.
• x = bottles of POW
• y = bottles of ZAP
Question 4
• To produce one bottle of POW takes 30
minutes and to produce one bottle of ZAP
takes 20 minutes, and there are 25 hours
available in total for this production-run
operation.
30x + 20y £ 25 ´ 60
3x + 2y £ 150
Question 4
• Availability of a particular additive means
that the chain cannot produce more than
35 bottles of POW and a combined total of
65 bottles on any production run.
Question 4
• Availability of a particular additive means
that the chain cannot produce more than
35 bottles of POW and a combined total of
65 bottles on any production run.
x £ 35
x + y £ 65
Question 4
• As a minimum production requirement, at
least 15 bottles of each weed killer must
be produced in each run.
x ³ 15
y ³ 15
Question 4a
• The management of NAILS wants to know how
many bottles of both POW and ZAP should be
produced. The preliminary estimates of their
potential profitability are $20 per bottle of POW
and $10 per bottle of ZAP. Perform an
appropriate analysis and make a
recommendation to management about the
amount of POW and ZAP that could be
produced to maximise profit.
Question 4
• Profit
P = 20x +10 y
• Gradient is -2
80
75
70
65
60
55
50
45
40
35
30
Feasible
region
25
20
15
10
5
0
0
5
10
15
20
25
30
35
40
45
50
(15, 15) (35, 15), (35, 22.5), (20, 45), (15, 50)
55
60
65
26
25
24
23
22
21
Feasible region
20
19
18
17
16
15
14
32
33
34
Blowup of the section
35
36
26
25
24
23
22
21
Feasible region
20
19
18
17
16
15
14
32
33
34
35
36
(35, 22.5) = (35, 22) and (34, 24) as need integer values
Question 4b
• Suppose that the profit of $20 per bottle of
POW was overestimated and it was in
reality only $15 per bottle. How does that
change your production recommendation
in part (a)?
Profit = 15x + 10y
45
44
43
42
41
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
20
19
18
17
16
15
14
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
Profit = 15x + 10y
• This has the same gradient of our first
constraint
3x + 2y £ 150
• Can have any of (20, 45), (22, 42), (24,
39), (26, 36), (28, 33), (30, 30), (32, 27),
(34, 24)
• It is found that the sales of POW and ZAP are
closely related by the function y = 9 ln(x),
• where x = number of bottles of POW and y =
number of bottles of ZAP.
• Management therefore requests that the ratio
“number of bottles of POW produced : number
of bottles of ZAP produced be x : 9 ln(x)”. The
constraints need to be satisfied and the
estimated profit needs to be maximised, based
on a profit of $15 per bottle of POW and $10 per
bottle of ZAP, so as to mirror demand.
80
75
70
65
60
55
50
45
y = 9lnx
40
35
30
25
20
15
10
5
0
0
5
10
15
20
25
30
35
40
45
50
55
60
65
Equation
3x + 2 y = 150
y = 9 ln x
Þ 3x + 18 ln x -150 = 0
x + 6 ln x - 50 = 0
Solve the equation derived in part (c) and make a
recommendation to management about the amount of
POW and ZAP that could be produced to maximise
profit.
Solve the equation derived in part (c) and make a
recommendation to management about the amount of
POW and ZAP that could be produced to maximise
profit.
•
•
•
•
Use your graphics calculator:
This gives 29.661…
Integer values
29 bottles POW and 31 bottles ZAP
2005
Question 1
Question 1
• The refreshment bar sells two types of pie,
steak and mince. The number of pies
ordered each day needs
• to satisfy the following daily constraints.
At least 42 pies will be sold in total
x=mince
y = steak
x + y ³ 42
50
48
46
44
42
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
A minimum of 24 mince pies will be
sold
x ³ 24
50
48
46
44
42
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
No more than 3 mince pies will be sold for every
steak pie sold
3y ³ x
3x £ y
y £ 3x
x ³ 3y
50
48
46
44
42
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
At most, 12 more steak pies will be
sold than mince pies.
y £ x +12
50
48
46
44
42
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
50
48
46
44
42
40
38
36
34
Feasible
region
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
• The refreshment bar pays $1.30 for
each steak pie and $1.00 for each
mince pie.
Objective function
C =1.3y + x
Minimum cost
•
•
•
•
(24, 36) next point will be less
(24, 18) $47.40
(10.5, 31.5) $45.15
Cannot buy half a pie
Minimum cost
• Investigate to get minimum at
• (31, 11)
• 31 mince and 11 steak pies
• Assume the cost of steak pies remains at
$1.30 each. The cost of mince pies
increases so that the optimal solution is no
longer that obtained in (a). State the
possible changes in the cost of mince pies
that give rise to these different optimal
solution(s), and give the new optimal
solution(s) for each of these changes.
50
48
46
44
42
40
38
36
34
Feasible
region
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
50
Question 1b
• The price would have to be the same as the
steak pie, I.e. $1.30 as the gradient of the line is
-1. Then all integer points between x= 24 and 31
would give the same minimum
• I.e. (24,18), (25, 17), (26, 16), (27, 15), (28, 14),
(29, 13), (30, 12) and our result (31, 11). Once
the cost was greater than $1.30, the minimum is
achieved at (24, 18)
• Suppose the demand for the pies is
modelled by the function:
• S 2 = 66M + 10 where
• S represents the daily demand for steak
pies,
• and M represents the daily demand for
mince pies.
1c Find the optimal solution that
satisfies the daily demand
Demand enters the feasible
Region here
60
58
56
54
52
50
48
46
44
42
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
Minimum valuecheck integer values
60
58
56
54
52
50
48
46
44
42
40
38
36
34
32
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44
46
48
Solve for the intersection
y = x + 12
y = 66x + 10
2
Þ x + 24x + 144 = 66x + 10
2
Þ x - 42x + 134 = 0
2
x = 3.47, 38.52
Solve for the intersection
y = x + 12
y = 66x + 10
2
Þ x + 24x + 144 = 66x + 10
2
Þ x - 42x + 134 = 0
2
x = 3.47, 38.52
Integer answer is 39, giving y = 51
This satisfies all constraints and minimises the
cost.
2006
Note: before you even start
• All questions are based on a product,
strawberry yoghurt, produced by a
company named Fastidious Foods. This
company produces two types of
strawberry yoghurt, light and standard, in
150 mL pottles. Each type of yoghurt is
packaged in boxes containing six pottles.
Question 1
• The number of boxes of each type
produced per day needs to satisfy the
following constraints:
The total number of boxes
produced is not to exceed 400
x = number of boxes of light strawberry yoghurt
y = number of boxes of standard strawberry yoghurt
x + y £ 400
And there must be at least 200 boxes of standard
yoghurt produced.
y ³ 200
The number of boxes of standard yoghurt
produced is to be at least twice the number of
light yoghurt produced.
y ³ 2x
No more than seven boxes of standard yoghurt are to be
produced for every two boxes of light yoghurt produced.
2 y £ 7x
7
Þy£ x
2
600
550
500
450
Feasible region
400
350
300
250
200
150
100
50
0
0
50
100
150
200
250
300
350
400
450
A box of light yoghurt yields a greater profit than a box
of standard yoghurt.
P = Ax + By
A>B
-A
Gradient
< -1
B
600
550
500
450
400
350
300
250
200
150
100
50
0
0
50
100
150
200
250
300
350
400
Gradient < -1 (i.e. steeper line) makes green/red
intersection maximum i.e. (133, 267)
450
Constraints 2 and 3 in (a) are changed so that the
number of boxes of light yoghurt produced is between
33.3% and 40.0% of the number of boxes of standard
yoghurt produced. Assume that constraint 1 in (a) still
applies and that each box of light yoghurt still yields a
greater profit than a box of standard yoghurt.
Find the number of boxes of each type of yoghurt
that should be produced now to maximise the profit.
1b. New constraints
1
2
y£x£ y
3
5
y £ 3x replaces y £ 3x
5
y ³ x replaces y £ 2x
2
Optimum solution at intersection of red/green lines
as gradient < -1 ( i.e. steeper line)
Solution is (114, 286); 114 light, 286 standard
2007
Question 1
• Shedz processes two varieties of kiwifruit
– green and gold. Constraints on daily
packing are as follows:
Question 1
• Let x = number of trays of green
• and y = number of trays of gold
At most, 2 000 trays can be packed
in total.
• x + y ≤ 2000
At most, 1 200 trays of green can
be packed.
• x ≤ 1200
A minimum of 200 trays of gold
need to be packed.
• y ≥ 200
The ratio of the number of
packed trays of green to gold
must be no more than 5:2.
• 2x ≤ 5y
Constraints
•
•
•
•
x + y ≤ 2000
x ≤ 1200
y ≥ 200
2x ≤ 5y
2000
1800
1600
1400
1200
Feasible region
1000
800
600
400
200
0
0
200
400
600
800
1000
1200
1400
1600
1800
2000
The ratio of the profit per tray of the green variety to the
gold is 12:13
.
• Objective function is:
Profit = 12mx + 13my
where m is a multiple constant
Profit = 12mx + 13my
• Optimum point is x = 0 and y = 2000
• So 0 trays of green and 2000 trays of gold
should be packed daily to maximise the
profit.
The output of Shedz should reflect
market demand and maintain a
strong profit.
• Given that there is a market demand for
both green and gold varieties, suggest
what other solutions to part (a) (i) are
possible that do not involve changing
any constraints.
• Justify your suggestions fully.
1 (a) (ii)
1. It’s highly unlikely that all 2000 gold would be sold.
We know that we will sell some green due to demand.
In fact if
x = 600 and y = 1400, we would not have to sell about
50 gold for this to be a more profitable combination.
2. Objective function is almost parallel to constant
boundary line x + y = 2000. Small differences of x and y
from the optimal value don’t affect the profit
significantly.
2000
1800
1600
1400
1200
Feasible region
1000
800
600
400
200
0
0
200
400
600
800
1000
1200
1400
1600
1800
2000
2008
Question 6
• A new hotel for Statsmod Enterprises is to
have only deluxe and standard rooms. In
designing the new hotel, the following
constraints need to apply:
Define your variables
• Let x = number of standard rooms
• Let y = number of deluxe rooms
Constraints
• The maximum number of rooms is 65.
x + y £ 65
• The building cost budget for the rooms has
a maximum of $3 million. Building costs
are $45 000 for a standard room and $60
000 for a deluxe room.
45000x + 60000y £ 3000000
3x + 4y £ 200
• When the number of standard rooms (x) is
such that 10m < x ≤ 10(m + 1), then the
number of deluxe rooms (y) is such that y
≤ 5(m + 1), where m = 0, 1, 2, 3, 4.
This is a step function
• When the number of standard rooms (x) is
such that 10m < x ≤ 10(m + 1), then the
number of deluxe rooms (y) is such that y
≤ 5(m + 1), where m = 0, 1, 2, 3, 4.
This is a step function
m
x
0
1
2
3
0<x≤10
10<x≤20
20<x≤30
30<x≤40
Y≤5
Y≤10
Y≤15
y≤20
4
40<x≤50
Y≤25
Feasible
region
• The profit for deluxe rooms is 40% greater
than that for standard rooms.
P = x +1.4y
• Assuming that all rooms are occupied, find
the number of each type of room that
should be built to ensure maximum profit.
A grid is provided on page 26 of the
Answer Booklet to help you answer this
question.
Maximum occurs (40,20)
(b)
• Statsmod Enterprises wants to ensure
maximum profit and build the maximum
number of 65 rooms, but cannot do so
within the $3 million building cost budget.
• What
smallest
required
increase in
x + isy the
= 65
is still
a constraint
this budget?
We are looking at the blue line and
then calculating profit
x + y = 65
X
41
42
43
44
45
Y
24
23
22
21
20
Profit
74.6
74.2
73.8
73.4
73
P = k ( x +1.4y )
Max (41, 24)
X
41
42
43
44
45
Y
24
23
22
21
20
Profit
74.6
74.2
73.8
73.4
73
P = k ( x +1.4y )
Max (41, 24)
• Cost:
C($000) = 45 ´ 41+ 60 ´ 24 = 3285
• Increase = $285,000
• Need minimum cost to produce maximum
profit.
(c)
• Suppose the mean occupancy rate for
deluxe rooms is 60% and that all three
constraints still apply.
• While still ensuring maximum profit,
calculate the occupancy rate for standard
rooms, which will give more than one
solution for each type of room that should
be built. Clearly state all of these solutions.
p
P = 0.6 ´1.4 y +
´x
100
= 0.84 y + 0.01px
• Suppose the mean occupancy rate for
deluxe rooms is 60% and that all three
constraints still apply.
• While still ensuring maximum profit,
calculate the occupancy rate for standard
rooms, which will give more than one
solution for each type of room that should
be built. Clearly state all of these solutions.
2009
Question 6
• A proposed housing development will have
two types of home, single detached homes
and duplex units (a block of two homes).
Two of the resources available are six
hectares (60 000 m2) of land for housing
and a building budget of $7 million.
• A single detached home requires 600 m2
of land and costs $65 000 to build. A
duplex unit requires 800 m2 of land and
costs $100 000 to build.
• The profit from the sale of a duplex unit is
50% more than that from a single
detached home.
• Let
• x be the number of single detached homes
and
• y be the number of duplex units. Assume
that all homes built are sold.
A single detached home requires 600 m2 of land and costs $65
000 to build. A duplex unit requires 800 m2 of land and costs
$100 000 to build.
•
•
•
•
•
•
•
•
Let x = number of single detached homes
Let y = number of duplex units
The constraints are:
Land: 600x + 800y ≤ 60 000
simplified 3x + 4y ≤ 300
Costs: 65x + 100y ≤ 7 000
simplified 13x + 20y ≤ 1400
Non-negativity: x ≥ 0, y ≥ 0
Find the number of each type of home that should be built to
maximise the profit. A grid is provided on page 26 of the Answer
Booklet to help you answer this question.
• The profit function is
• P = k(x + 1.5y)
80
70
60
50
40
30
Feasible
region
20
10
0
0
20
40
60
80
-10
50, 37.5 need to look at points
100
120
80
70
60
50
40
30
Feasible
region
20
10
0
0
20
40
60
80
100
-10
Gradient of objective function is -2/3
120
50
Start at (40, 44) and go down 2 in y and up 3 in x
until (52, 36)
48
46
44
42
40
38
36
34
32
30
40
42
44
46
48
50
52
54
56
58
60
To ensure maximum profit
( P = 106k in each case)
•
x = 40 and y = 44 that is 40 single detached homes and 44 duplex units
should be built or
•
x = 43 and y = 42 that is 43 single detached homes and 42 duplex units
should be built or
•
x = 46 and y = 40 that is 46 single detached homes and 40 duplex units
should be built or
•
x = 49 and y = 38 that is 49 single detached homes and 38 duplex units
should be built or
•
x = 52 and y = 36 that is 52 single detached homes and 36 duplex units
should be built.
Judgement
• S: Any one of these optimal point(s).
• P: All two constraints correct OR any
constraint correct plus the correct profit
objective function
• P: Feasible region clearly shown on graph
as opposed to constraint equations.
Note:
• 1. Can use graph as evidence for the
constraints.
• 2. Profit function can take the form
• P = 1.0x + 1.5y.
Question 6b
• A new council regulation requires that
there must be at least three single
detached homes for every two duplex
units in the new development.
• Calculate the percentage reduction in
maximum profit resulting from this
regulation.
• New Council regulation: 3y ≤ 2x
Now to ensure maximum profit, x = 53 and y = 35 or
x = 56 and y = 33.
50
48
46
44
42
40
38
36
34
Feasible region
32
30
40
42
44
46
48
50
52
54
56
58
60
• For either optimal point, % reduction in
maximum profit =
• = 0.47%.
0.5k
´ 100
106k
Judgement
• S: Correct method and answer or correct
new optimal point and carried error in %
reduction in profit due to wrong answer in
(a). new optimal point.
• P: Identification of correct new optimal
point
• N: Incorrect
Note:
• 1. Only one optimal point is required.
• From previous similar developments, it has been
shown that, for social reasons, it is desirable that
the relationship between the numbers of the two
types of homes built is approximately modelled
by y = 35e-0.014x. Taking this relationship into
account and the constraint in (b), calculate the
amount of unused land and / or unused budget
when x and y are such that the profit is
maximised.
With social constraint, point where profit is
maximised is given by: x = 85 and y = 11 with profit
= $101.5k
80
70
60
50
40
30
20
10
0
0
20
40
60
80
-10
Intersection point on calculator is 86, 10.5
100
120
• Unused land = 60 000 - 800(11) - 600(85)
= 200 m2
• Unused budget = 7 000 - 100(11) - 65(85)
= $375 000
2010
Question 2b
• Two different models of Statsmobiles are
manufactured, X and Y.
• Let x be the number of model X manufactured
annually and y be the number of model Y
manufactured annually.
• Annual demand for Statsmobiles is at least
• 6 000.
x + y ³ 6000
• Annual demand for model Y is between 95%
and 170% of the annual demand for model X.
•
0.95x £ y £ 1.7x
In each of the following three cases, find the
optimal solution(s) that minimises the annual
manufacturing cost, while meeting the above
demand constraints.
The cost of producing each model Y is less than the cost of
producing each model X.
Feasible
region
The cost of producing each model Y is less than the cost of
producing each model X
C = k ( x + Ay )
Feasible
region
C = k ( x + Ay )
The cost of producing each model Y is less than the cost of
producing each model X
Feasible
region
C = k ( x + Ay )
The cost of producing each model Y is less than the cost of
producing each model X
Feasible
region
When A<1,
gradient is
steeper and so
red/blue
intersection is
a minimum i.e.
(x = 2 223 and
y = 3 777)
The cost of producing each model Y is equal to the cost of
producing each model X.
C = k ( x + Ay )
Feasible
region
When A=1,
gradient is the
same as the
blue line i.e.
where 2 223 ≤
x ≤ 3 076 )
ThThe cost of producing each model Y is greater than the cost of
producing each model X.
C = k ( x + Ay )
Feasible
region
When A>1,
gradient is less
than the blue
line i.e.
x = 3 076 and
y = 2 924
2011
Question 5
• This supplier operates out of two warehouses:
eastern and western. The supplier receives
orders from two customers, A and B, who
require sheets of plywood.
Delivery costs per sheet are:
• $0.50 from the eastern warehouse to
customer A
• $K (K is a constant) from the eastern
warehouse to customer B
• $0.40 from the western warehouse to
customer A
• $0.55 from the western warehouse to
customer B.
• Let x represent the number of sheets
delivered from the eastern warehouse to
customer A and
• y represent the number of sheets delivered
from the eastern warehouse to customer B.
• Customer A needs 50 sheets and customer B
needs 70 sheets.
Customer A needs 50 sheets and
customer B needs 70 sheets.
x £ 50, y £ 70
Customer A needs 50 sheets and
customer B needs 70 sheets.
Eastern
Western
A
x
50 - x
B
y
70 - y
• The eastern warehouse has 80 sheets in stock
and
x + y £ 80
• the western warehouse has 45 sheets in stock.
( 50 - x ) + ( 70 - y ) £ 45
x + y ³ 75
• Use K = 0.60.
• Identify all the constraints in terms of x and y.
• Draw the feasible region on the graph paper
provided on page 26 of the Answer Booklet.
• Write the overall delivery cost in terms of x
and y.
• Describe the delivery plan from each
warehouse to each customer that minimises
the overall delivery cost.
Add non-negativity
x ³ 0, y ³ 0
Feasible region is between red and
green lines
Objective Function
C = 0.5x + 0.6y + 0.4 ( 50 - x ) + 0.55 ( 70 - y )
= 0.1x + 0.05y + 58.50
Gradient of objective function is -2
Minimum at (5, 70)
Delivery plan
• Customer A needs 50 sheets
• 5 sheets from eastern warehouse
• 45 sheets from western warehouse
• Customer B needs 70 sheets
• All 70 from eastern warehouse
(ii)
• Find the value of K that gives multiple
delivery plans and also minimises the overall
delivery cost. State this minimum overall
delivery cost.
(ii)
• Find the value of K that gives multiple delivery
plans and also minimises the overall delivery cost.
State this minimum overall delivery cost.
• We need the gradient to be the same as
constraint
• i.e. -1
x + y £ 75
C = 0.1x + ( k - 0.55 ) y + 58.5
k - 0.55 = 0.1 Þ k = 0.65
• Substitute to find the minimum cost:
• Use any integer point on the line (5, 70)
C = 0.1´ 5 + 0.1´ 70 + 58.5
= $66
• O + S: Delivery Plan and minimum overall
delivery cost correct.
• S: Either (i) or (ii) correct.
2P: Both constraints correct in (i) and K correct
in (ii).
P: Constraints correct in (i) or K correct in (ii).
(b)
• This part investigates the effect of changes
made to the situation in part (a). Customer B
still needs 70 sheets and the eastern
warehouse still has 80 sheets in stock.
• For each of the following find the values of x
and y that minimise the overall delivery cost.
• Use K = 0.60, customer A still needs 50 sheets,
and the western warehouse has 40 sheets in
• stock (instead of 45).
• Use K = 0.60, customer A still needs 50 sheets,
and the western warehouse has 40 sheets in
• stock (instead of 45).
( 50 - x ) + ( 70 - y ) £ 40
x + y ³ 80
Feasible region is now the pink line
10 < x < 50
Point (10, 70) to get a cost of $66.50
Constraint changes
• Use K = 0.70, customer A needs 48 sheets
(instead of 50), and the western warehouse
has 45 sheets in stock.
x £ 48
70 - y + 48 - x £ 45
x + y ³ 73
• Use K = 0.70, customer A needs 48 sheets
(instead of 50), and the western warehouse
has 45 sheets in stock.
C = 0.5x + 0.7y + 0.4 ( 48 - x ) + 0.55 ( 70 - y )
= 0.1x + 0.15y + 57.7
Minimum at (48, 25)
2012
The gym is planning on purchasing some new treadmills,
steppers and rowing machines.
In order to meet usage requirements, there are to be a
minimum of eight treadmills, five steppers and three
rowing machines. The floor area needed for each piece of
equipment is 3 m2 per treadmill, 2 m2 per stepper and 4
m2 per rowing machine. After allowing for room between
each piece of equipment, 132 m2 is available. All of this
area must be used. Data collected on concurrent usages
in peak periods determined that there should be no more
than three steppers to every two treadmills and at least
one stepper to every treadmill.
In order to meet usage requirements, there are
to be a minimum of eight treadmills, five
steppers and three rowing machines.
T ³8
S³5
R³3
The floor area needed for each piece of
equipment is 3 m2 per treadmill, 2 m2 per
stepper and 4 m2 per rowing machine. After
allowing for room between each piece of
equipment, 132 m2 is available. All of this area
must be used.
3T + 2S + 4R =132
Data collected on concurrent usages in peak
periods determined that there should be no
more than three steppers to every two
treadmills and at least one stepper to every
treadmill.
Data collected on concurrent usages in peak
periods determined that there should be no
more than three steppers to every two
treadmills and at least one stepper to every
treadmill.
2S £ 3T
S ³T
Objective function
The cost of each piece of equipment is in the
ratio 2:3:5 for treadmills, steppers and rowing
machines respectively.
C= 0.2kT + 0.3kS + 0.5kR
Let x be the number of treadmills and
y be the number of steppers.
(a) Show that 3x +2y ≤ 120
Let z be the number of rowing machines
132 - 3x - 2y
3x + 2y + 4z = 132 Þ z =
4
x³8
y³5
132 - 3x - 2y
z ³ 3Þ
³ 3 Þ 3x + 2y £ 120
4
2y £ 3x
y³x
(b)
Find the number of treadmills, steppers and
rowing machines that should be purchased and
installed in order to minimise the cost.
GRAPH
Feasible
region
C = k ( 2x + 3y + 5z )
æ
æ 132 - 3x - 2y öö
= k ç 2x + 3y + 5 ç
÷÷
è
øø
4
è
C = k (165 -1.75x + 0.5y)
Vertices: (24, 24), (20, 30), (8,8), (8,12)
C = k (165 -1.75x + 0.5y)
(24, 24)
135k
(20, 30)
(8,8)
(8,12)
145k
Minimum cost for x = 24, y = 24
155k
157
Optimal point (x , y, z ) = (24, 24, 3) which
represents 24 treadmills, 24 steppers and 3
rowing machines.
(c)
Suppose that the cost of each piece of
equipment had been in the ratio 2:3:c for
treadmills, steppers and rowing machines
respectively.
Given that all the other constraints still apply
and that the cost is to be minimised, find the
value of c that gives multiple solutions.
New cost function
C = k ( 2x + 3y + cz )
æ
æ 132 - 3x - 2y öö
= k ç 2x + 3y + c ç
÷÷
è
øø
4
è
æ
æ 3c ö æ c öö
= k ç 33+ x ç 2 - ÷ + y ç 3- ÷÷
è
4 ø è 2 øø
è
Multiple solutions occur when the cost function
is parallel to one of the sides of the feasible
region and the cost is minimised.
Gradient of cost function
3c
- 2 3c - 8
= 4
=
c 12 - 2c
32
3c - 8
y = x, m = 1 =
Þ c = 4 Þ C = 132k
12 - 2c
-3 3c - 8
-3
3x + 2y = 120, m = =
Þc=
Þ C has no soln
2 12 - 2c
2
3x
3 3c - 8
13
y= , m= =
Þ c = Þ C = 143k
2
2 12 - 2c
3
x = 8, m is undefined Þ 12 - 2c = 0 Þ c = 6 Þ C = 218k
3c - 8
y = x, m = 1 =
Þ c = 4 Þ C = 132k
12 - 2c
-3 3c - 8
-3
3x + 2y = 120, m = =
Þc=
Þ C has no soln
2 12 - 2c
2
3x
3 3c - 8
13
y= , m= =
Þ c = Þ C = 143k
2
2 12 - 2c
3
x = 8, m is undefined Þ 12 - 2c = 0 Þ c = 6 Þ C = 218k