ACCA F5 Performance Management - 2010 - Study text

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Chapter 3: Limiting factors and linear programming

 The planned output and sales should be decided by working down through the

priority list until all the units of the limiting factor (scarce resource) have been

used.

In other words, in order to maximise profit, the aim should be to maximise the

contribution for each unit of limiting factor used.

Example

A company makes four products, A, B, C and D, using the same direct labour work

force on all the products. Budgeted data for the company is as follows:

Product A

C D

Annual sales demand (units) 4,000

5,000

8,000 4,000

$ $

Direct materials cost 3.0

6.0

5.0 6.0

Direct labour cost 6.0

12.0

3.0 9.0

Variable overhead 2.0

4.0

1.0 3.0

Fixed overhead 3.0

6.0

2.0 4.0

Full cost 14.0

28.0

11.0 22.0

Sales price 15.5

29.0

11.5 27.0

Profit per unit 1.5

1.0

0.5 5.0

The company has no inventory of finished goods. Direct labour is paid $12 per hour.

To meet the sales demand in full would require 12,000 hours of direct labour time.

However, only 6,000 direct labour hours are available during the year.

Required

Identify the quantities of production and sales of each product that would maximise

annual profit.

Answer

Direct labour hours are a scarce resource. The products should be ranked in order

of priority according to the contribution that they make per direct labour hour.

C D

$ $

Sales price/unit 15.5

29.0

11.5 27.0

Variable cost/unit 11.0

22.0

9.0 18.0

Contribution per unit 4.5

7.0

2.5 9.0

Direct labour hours/unit 0.5

1.0

0.25 0.75

Contribution/direct labour hour $9.0

$7.0

$10.0 $12.0

Priority for making and selling 3

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The products should be made and sold in the order D, C, A and then B, up to the

total sales demand for each product and until all the available direct labour hours

(limiting factor resources) are used up.

Profit-maximising budget

Product

Sales units

Direct labour

hours

Contribution

per unit

Total

contribution

$ $

D (1st) 4,000 3,000

9.0 36,000

C (2nd) 8,000 2,000

2.5 20,000

A (3rd) 2,000 (balance) 1,000

4.5 9,000

6,000

65,000

2.3 Identifying limiting factors

You might not be told by an examination question that a limiting factor exists,

although you might be told that there is a restricted supply of certain resources. You

might be expected to identify the limiting factor by calculating the budgeted

availability of each resource and the amount of the resource that is needed to meet

the available sales demand.

Example

A company manufactures and sells two products, Product X and Product Y. The two

products are manufactured on the same machines. There are two types of machine,

and the time required to make each unit of product is as follows:

Product X Product Y

Machine type 1 10 minutes per unit 6 minutes per unit

Machine type 2 5 minutes per unit 12 minutes per unit

Sales demand each year is for 12,000 units of Product X and 15,000 units of Product Y.

The contribution per unit is $7 for Product X and $5 for Product Y.

There is a limit to machine capacity, however, and in each year there are only 3,000

hours of Machine 1 time available and 4,200 hours of Machine 2 time available.

Required

Recommend what the company should make and sell in order to maximise its

annual profit.

Chapter 3: Limiting factors and linear programming

Answer

The first step is to identify whether or not there is any limiting factor other than sales

demand. Clearly, the time available on either machine type 1 or machine type 2

could be a limiting factor. To find out whether machine time on either type of

machine is a limiting factor, we calculate the required machine time to manufacture

units of Product X and Y to meet the sales demand in full. We then compare this

requirement for machine time with the actual time available.

Machine type 1

Machine type

hours

Time required

To make 12,000 units of Product X 2,000

1,000

To make 15,000 units of Product Y 1,500

3,000

Hours needed to meet sales demand 3,500

4,000

Hours available 3,000

4,200

Shortfall (500)

Machine type 1 is a limiting factor, but Machine type 2 is not. To maximise

contribution and profit, we should therefore give priority to the product that gives

the higher contribution per hour of machine type 1.

Product X Product Y

Contribution per unit $7 $5

Machine type 1 time per unit 10 minutes 6 minutes

Contribution per hour (Machine type 1)

$42 $50

Priority for making and selling 2

Profit-maximising budget

Product Sales units

Machine

type 1 hours

Contribution

per unit

Total

contribution

$ $

Y (1st) 15,000 1,500

5.0 75,000

X (2nd) - balance 9,000 1,500

7.0 63,000

3,000

138,000

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Limiting factor decisions: linear programming

 Two or more limiting factors

 Formulating a linear programming problem

 The objective function

 Formulating the constraints

3 Limiting factor decisions: linear programming

3.1 Two or more limiting factors

There may be more than one scarce resource or limiting factor. When there is more

than one limiting factor (other than sales demand for the products), the

contribution-maximising plan cannot be identified simply by ranking products in

order of contribution per unit of limiting factor, because the ranking provided by

each limiting factor could be different.

The problem is still to decide what mix of products should be made and sold in

order to maximise profits. It can be formulated and solved as a linear programming

problem.

 The problem must first be formulated.

 Having formulated the problem, it must then be solved.

3.2 Formulating a linear programming problem

A linear programming problem is formulated by:

identifying an objective function, and

formulating two or more constraints, one for each limiting factor or other

possible restriction on output and sales (such as maximum sales demand).

3.3 The objective function

The objective of a linear programming problem is to maximise or minimise the

value of something. For the purpose of your examination, it is likely to be the

objective of maximising total contribution. (The objective might possibly be

something else, such as the objective of minimising costs.)

An objective function expresses the objective, such as total contribution, as a

formula.

Chapter 3: Limiting factors and linear programming

Example

A company makes and sells two products, Product X and Product Y. The

contribution per unit is $8 for Product X and $12 for Product Y. The company wishes

to maximise profit.

If it is assumed that total fixed costs are the same at all levels of output and sales, the

objective of the company is to maximise total contribution.

Total contribution can be expressed as a formula, as follows:

Let the number of units ( made and sold) of Product X be x

Let the number of units (made and sold) of Product Y be y

The objective function is therefore to maximise:

(Total contribution): 8x + 12y.

3.4 Formulating the constraints

In a linear programming problem, there is a separate constraint for each item that

might put a limitation on the objective function.

Each constraint, like the objective function, is expressed as a formula. Each

constraint must also specify the amount of the limit or constraint.

If there is a maximum limit, the constraint must be expressed as ‘must be equal to

or less than’.

For example, if there is a maximum sales demand for Product X of 5,000 units, the

constraint for sales demand for X would be expressed as:

x ≤ 5,000.

Similarly, if there is a minimum limit, the constraint must be expressed as ‘must be

equal to or more than’. For example, if there is a requirement to supply a customer

with at least 2,000 units of Product X, this constraint would be expressed as:

x ≥ 2,000.

Constraints will often involve two (or more) variables. For example, suppose that a

company makes two products, X and Y, using the same direct labour employees. It

takes 2 hours to make one unit of X and 3 hours to make one unit of Y. In the budget

period, only 18,000 direct labour hours will be available. The constraint will be:

2x + 3y ≤ 18,000.

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Non-negativity constraints

A requirement of linear programming problems is that there should be no negative

values in the final solution. For example, it is not possible to make and sell minus

4,000 units of Product Y.

Constraints in the linear programming problem should therefore be that each

‘variable’ must be equal to or greater than 0.

For example:

(Non-negativity constraint): x, y ≥ 0.

Example

A company makes and sells two products, Product X and Product Y. The

contribution per unit is $8 for Product X and $12 for Product Y. The company wishes

to maximise profit.

The expected sales demand is for 6,000 units of Product X and 4,000 units of Product

Y. However, there are limitations to the amount of labour time and machine time

that is available in the period:

Product X Product Y

Direct labour hours per unit 3 hours 2 hours

Machine hours per unit 1 hour 2.5 hours

Total

hours

Direct labour hours available, in total 20,000

Machine hours available, in total 12,000

A linear programming problem can be formulated as follows:

Let the number of units (made and sold) of Product X be x

Let the number of units (made and sold) of Product Y be y

The objective function is to maximise total contribution: 8x + 12y.

Subject to the following constraints:

Direct labour 3x + 2y ≤ 20,000

Machine time x + 2.5y ≤ 12,000

Sales demand, X x ≤ 6,000

Sales demand, Y y ≤ 4,000

Non-negativity x, y ≥ 0

Chapter 3: Limiting factors and linear programming

If you do not like figures with fractions in a constraint, the constraint for machine

time could be expressed (by doubling it) as:

Machine time 2x + 5y ≤

24,000

Conclusion: formulating a linear programming problem

Once a linear programming problem has been formulated, it must then be solved to

decide how the objective function is maximised (or minimised).

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Linear programming: graphical solution

 The stages in a graphical solution

 Drawing the constraints on a graph

 Maximising (or minimising) the objective function

 Calculating the value for the objective function

 An alternative approach to a solution

4 Linear programming: graphical solution

When there are just two variables in a linear programming problem (x and y), the

problem can be solved by a graphical method. The solution identifies the values of x

and y that maximise (or minimise) the value of the objective function.

4.1 The stages in a graphical solution

There are three stages in solving a linear programming problem by the graphical

method:

 Step 1: Draw the constraints on a graph, to establish the feasible combinations of

values for the two variables x and y that are within all the constraints in the

problem.

 Step 2: Identify the combination of values for x and y, within this feasible area,

that maximises (or minimises) the objective function. This is the solution to the

problem.

 Step 3: Calculate the value for the objective function that this solution provides.

4.2 Drawing the constraints on a graph

The constraints in a linear programming problem can be drawn as straight lines on

a graph, provided that there are just two variables in the problem (x and y). One

axis of the graph represents values for one of the variables, and the other axis

represents values for the second variable.

The straight line for each constraint is the boundary edge of the constraint – its outer

limit in the case of maximum amounts (and inner limit, in the case of minimum

value constraints).

For example, suppose we have a constraint:

2x + 3y ≤ 600

The outer limit of this constraint is represented by a line:

2x + 3y = 600.

Combinations of values of x and y beyond this line on the graph (with higher values

for x and y) will have a value in excess of 600. These exceed the limit of the

constraint, and so cannot be feasible for a solution to the problem.

Chapter 3: Limiting factors and linear programming

The constraint is drawn as a straight line. To draw a straight line on a graph, you

need to plot just two points and join them up. The easiest points to plot are the

combinations of x and y:

 where x = 0, and

 where y = 0.

For the equation 2x + 3y = 600:

 when x = 0, y = 200 (600/3). So plot the point x = 0, y = 200 on the graph

 when y = 0, x = 300 (600/2). So plot the point y = 0, x = 300 on the graph

Join these two points, and you have a line showing the values of x and y that are the

maximum possible combined values that meet the requirements of the constraint.

Example

Suppose that we have the following linear programming problem:

The objective function is to maximise total contribution: 5x + 5y

Subject to the following constraints:

Direct labour 2x + 3y ≤ 6,000

Machine time 4x + y ≤ 4,000

Sales demand, Y y ≤ 1,800

Non-negativity x, y ≥ 0

The non-negativity constraints are represented by the lines of the x axis and y axis.

The other three constraints are drawn as follows, to produce a combination of

values for x and y that meet all three constraints. These combinations of values for x

and y represent the ‘feasible region’ on the graph for a solution to the problem.

Workings

(1) Constraint 2x + 3y = 6,000

When x = 0, y = 2,000. When y = 0, x = 3,000.

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So draw a line between 2,000 on the y axis and 3,000 on the x axis to get the

constraint line.

(2) Constraint 4x + y = 4,000

When x = 0, y = 4,000. When y = 0, x = 1,000.

So draw a line between 4,000 on the y axis and 1,000 on the x axis to get the

constraint line.

(3) Constraint y = 1,800

Draw a line parallel to the x axis from the point y = 1,800 on the y axis.

Feasible area (or feasible region) for a solution

The feasible area for a solution to the problem is shown as the shaded area OABCD.

Combinations of values for x and y within this area can be achieved within the

limits of the constraints. Combinations of values of x and y outside this area are not

possible, given the constraints that exist.

To solve the linear programming problem, we now need to identify the feasible

combination of values for x and y (the combination of x and y within the feasible

area) that maximises the objective function.

4.3 Maximising (or minimising) the objective function

As a starting point, you might recognise that the combination of values for x and y

that maximises the objective function will be a pair of values that lies somewhere

along the outer edge of the feasible area.

In the graph above, the solution to the problem will normally be the values of x and

y at one of the following points on the graph:

 A

 B

 C, or

 D

In other words, we will normally expect the solution to be the combination of values

for x and y that lies at one of the ‘corners’ of the outer edge of the feasible area.

(In some cases, the solution might be:

 any combination of values of x and y along the line AB, or

 any combination of values of x and y along the line BC, or

 any combination of values of x and y along the line CD.

However, this would be unusual.)

To identify the combination of values for x and y that are feasible (within all the

constraints) and that also maximises the objective function, we need to look at the

objective function itself.

ACCA F5 Performance Management - 2010 - Study text - Emile Woolf Publishing

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