ACCA F5 Performance Management - 2010 - Study text

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Chapter 3: Limiting factors and linear programming

Iso-contribution lines

We do not know the maximum value (or minimum value) for the objective function.

However, we can draw a line that shows all the combinations of x and y that

provide the same total value for the objective function.

For example, suppose that the objective function is to maximise contribution 4x +

3y. We can draw a line on a graph that shows combinations of values for x and y

that give the same total contribution, when x has a contribution of 4 and y has a

contribution of 3. Any total contribution figure can be chosen, but a convenient

multiple of 4 and 3 is simplest and easiest.

 For example, we could select a total contribution value of 4x + 3y = 12,000. This

contribution line could be found by joining the points on the graph x = 0, y =

4,000 and y = 0, x = 3,000.

 Instead, we might select a total contribution value of 4x + 3y = 24,000. This

contribution line could be found by joining the points on the graph x = 0, y =

8,000 and y = 0, x = 6,000.

A line showing combinations of x and y that would earn the same total contribution

is sometimes called an iso-contribution line. (‘Iso’ means ‘the same’.)

If you draw both of these iso-contribution lines on a graph, you will find that:

 the two lines are parallel to each other on the graph, and

 the line with the higher total contribution value for values of x and y (24,000) is

further away from the origin of the graph (point 0).

This can be used to identify the solution to a linear programming problem.

 On the graph showing the feasible area, draw an iso-contribution line. Choose

any iso-contribution line that is easy to draw on the graph. The total value of the

line doesn’t matter, because all you are trying to find out is the slope of the line,

and the slope is the same for every iso-contribution line.

8,000

4,000

3,000 6,000

4x+3y=24,000 4x+3y=12,000

ISO – contribution lines

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 Having established the slope of an iso-contribution line, the next step is to

identify the point in the feasible area on the graph where an iso-contribution line

can be drawn which is as far from the origin of the graph as possible.

 Look at the slope of the contribution line, and (using a ruler if necessary) identify

which combination of values of x and y within the feasible area for the

constraints is furthest away from the origin of the graph, through which you

could draw an iso-contribution line. This is usually the combination of values for

x and y at one corner of the feasible area.

 In other words, you can use the iso-contribution line you have drawn to identify

the point in the feasible area that will maximise total contribution.

This is the combination of values of x and y that provides the solution to the linear

programming problem.

Example

Returning to the previous example, we can draw an iso-contribution line for, say, 5x

+ 5y = 10,000, by joining the points x = 0, y = 2,000 and y = 0, x = 2,000. (The value

10,000 is chosen here as a convenient multiple of the values 5 and 5 that can be

drawn clearly on the graph.)

This should show that the combination of values of x and y that will maximise total

contribution lies at point C on the graph. An iso-contribution line going through

point C would be further from the origin than an iso-contribution line going through

point B (and is obviously further from the origin than an iso-contribution lines going

through point A and point D).

The combination of x and y at point C is therefore the solution to the linear

programming problem.

Chapter 3: Limiting factors and linear programming

4.4 Calculating the value for the objective function

Having identified which combination of values for x and y provides the solution to

the linear programming problem, the final step is to:

 establish the exact values for x and y at this point, and

 use these values to calculate the value of the objective function at this point.

It might be possible to read the value of x and y from the graph, but it is sensible to

measure the values of x and y exactly by solving two simultaneous equations for the

values of x and y.

Simultaneous equations are equations that are both correct at the same time.

The simultaneous equations in the solution to a linear programming problem are

the equations for the two constraint lines at the point on the graph that provides the

solution. In this example, these are the equations at point C on the graph.

Example

At point C, we have the simultaneous equations:

2x + 3y = 6,000

4x + y = 4,000

You should know how to solve simultaneous equations. However, if you are not

sure, the technique is to multiply one or both of the simultaneous equations so that

you obtain two equations where the coefficient for either x or y is the same. You can

then subtract one equation from the other (or possible add them together, if minus

values for coefficients are involved). This will allow you to calculate the value for

wither x or y.

Having obtained a value for x or y, you can then substitute this value in any of the

simultaneous equations to obtain the value for the other variable.

In this example the simultaneous equations are solved as follows:

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

(2) 4x + y = 4,000

Multiply (1) by 2:

(3) 4x + 6y = 12,000

The coefficient for x is 4 in both (2) and (3)

Subtract (2) from (3):

5y = 8,000

Therefore y = 1,600

Substitute in equation (2)

4x + 1,600 = 4,000

4x = 2,400

x = 600

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The total contribution is maximised by producing 600 units of X and 1,600 units of

The objective in this problem is to maximise 5x + 5y.

The total contribution where x = 600 and y = 1,600 is:

5 (600) + 5 (1,600) = $11,000.

This is the amount of the maximum achievable contribution.

4.4 An alternative approach to a solution

If you find the iso-contribution line method confusing, there is an alternative

method of identifying the solution to a linear programming problem.

When you have drawn the feasible area on a graph, and you have identified the

feasible combinations of values for x and y, you will know that the solution lies at

one of the corners of the feasible area.

In the previous example, the solution has to be at points A, B, C or D.

You can calculate the values of x and y at each of these points, using simultaneous

equations if necessary to calculate the x and y values. Having established the values

of x and y at each of the points, calculate the value of the objective function (total

contribution) for each.

The solution is the combination of values for x and y at the point where the total

contribution is highest.

However, you must learn how to solve a linear programming problem using the

graphical method with iso-contribution lines, because this might be a specific

requirement of an exam question.

Chapter 3: Limiting factors and linear programming

Linear programming: shadow prices and slack

 Shadow prices

 Calculating a shadow price

 Implications of shadow prices

 Slack

5 Linear programming: shadow prices and slack

5.1 Shadow prices

In a linear programming problem, every constraint has a shadow price. (A shadow

price is also called a dual price.)

A shadow price is the amount by which the value of the objective function would

change in the optimal solution if the constraint was one unit extra (or one unit less).

 For example, suppose that direct labour hours are a constraint in a linear

programming problem, and the maximum labour time available is restricted to

20,000 hours. If the objective function is to maximise contribution, the shadow

price of direct labour is the amount by which total contribution could be

increased in 20,001 labour hours were available (and the amount by which total

contribution would be reduced if only 19,999 labour hours were available).

 Similarly suppose that in a linear programming problem there is a maximum

sales demand of 1,800 units for Product Y. The shadow price of sales demand for

Product Y is the amount by which total contribution in the optimal solution

would increase if the maximum sales demand were 1,801 units (of the amount

by which total contribution would fall if maximum sales demand were 1,799

units).

5.2 Calculating a shadow price

The shadow price of an item cannot be a negative value. However, if it is not a

limiting factor in the optimal solution to the linear programming problem, the

shadow price of an item will be $0.

When an item is an actual limiting factor in the solution to a linear programming

problem, this means that in the optimal solution, the limit of the constraint has been

reached. The item therefore has a shadow price with a positive value.

When the graphical method of solving a linear programming problem is used, the

shadow prices of the constraining items (effective limiting factors) are calculated as

follows.

 The shadow price for each constraining item (effective limiting factor) should be

calculated one at a time.

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 Identify the constraints in the optimal solution to the linear programming

problem.

 Take one of the constraints in that solution and increase the limit of the

constraint by one unit.

 Calculate the optimal solution when this constraint has been increased. The

optimal solution will be at the point on the graph where the same constraint

lines intersect, so calculate the solution using simultaneous equations.

 Compare the value of the objective function with the new constraint with the

value of the objective function in the original linear programming problem. The

amount by which the total value of the objective function (total contribution) has

increased is the shadow price of the item.

The same technique can be used to calculate the shadow price of any other effective

limiting factor in the original solution.

Example

A company makes two products B and C. Product B earns a contribution of $10 per

unit and a unit of Product C earns a contribution of $30.

The following resources are used to make one unit of each product:

Product B Product C

Maximum

resource available

Machine type 1 30 minutes 10 minutes 15 hours per day

Machine type 2 40 minutes 45 minutes 30 hours per day

Skilled labour 20 minutes 30 minutes 15 hours per hay

The maximum sales demand for C is 20 units per day. The sales demand for Product

B is unlimited.

Required

Calculate the optimum production plan for the baker and the shadow prices of

resources.

Answer

Let x = the number of units of Product B per day

Let y = the number of units of Product C per day

The objective function is to maximise:

(Total contribution): 10x + 30y.

Subject to the constraints:

Machine type 1 30x + 10y ≤ 900

Machine type 2 40x + 45y ≤ 1,800

Skilled labour 20x + 30y ≤ 900

Chapter 3: Limiting factors and linear programming

Demand for C y ≤ 20

Minimum production x ≥ 0, y ≥ 0

A graph can be drawn as follows to identify the feasible area for a solution. This is

shown below. The feasible area is 0ABCD, and the optimal solution is at point A,

point B, point C or point D.

An iso-contribution line 10x + 30y = 300 has been drawn to establish the slope of the

objective function. This can be used to identify the optimal solution, which is at

point B.

At point B, the effective constraints are sales demand (y = 20) and skilled labour

(20x + 30y = 900).

Since y = 20: 20x + (30 × 20) = 900

Therefore 20x = 300 and x = 15.

Total contribution at the optimal solution is therefore (15 × $10) + (20 × $30) = $750

per day.

Shadow price of skilled labour

If the available skilled labour could be increased from 900 minutes to 901 minutes

per day, the optimal solution would be at a point where:

(1) y = 20

(2) 20x + 30y = 901

40x+45y=1,800

30x+10y=900

20x+30y=900

10x+30y=300

(ISO – Contribution Line)

y=20

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Multiply (1) by 30:

(3) 30y = 600

Subtract (3) from (2) 20x = 301

x = 15.05

Total contribution with new labour constraint:

Product C: 20 × $30

600.0

Product B: 15.05 × $10

150.5

Total contribution 750.0

Total contribution with original constraint for labour 750.0

Shadow price of skilled labour, per minute 0.5

The shadow price of skilled labour is therefore $0.50 per minute, or $30 per hour.

Shadow price of sales demand for Product C

If the maximum demand for Product C could be increased to 21 units per day, the

optimal solution would be at the point where:

(1) y = 21

(2) 20x + 30y = 900

Multiply (1) by 30:

(3) 30y = 630

Subtract (3) from (2) 20x = 270

x = 13.5

Total contribution with new labour constraint:

Product C: 21 × $30

630.0

Product B: 13.5 × $10

135.0

Total contribution 765.0

Total contribution with original constraint for labour 750.0

Shadow price of skilled labour, per minute 15.0

The shadow price of sales demand for Product is therefore 15 per unit of demand.

Shadow price of time on machine type 1 and machine type 2

In the optimal solution, the total available time on both machine type 1 and machine

type 2 is not fully used. This means that there is spare capacity on both types of

machine and the shadow price for both types of machine is $0.

5.3 Implications of shadow prices

An implication of shadow prices is that the value of the objective function (total

contribution) could be increased if more units of the limiting factor could be made

available. In the previous example:

Chapter 3: Limiting factors and linear programming

 Total contribution could be increased by $30 per skilled labour hour if more

hours of skilled labour could be obtained at its normal variable cost per hour.

 Total contribution would be increased even if extra hours of skilled labour could

be made available at a cost above its normal variable cost, provided that the

additional cost per hour was less than $30.

 For example, if skilled labour is paid $18 per hour, total contribution would be

increased even if the company had to pay an overtime premium of 50% to get

skilled employees to work some overtime. The overtime premium would be $9

per hour, so total contribution could still be increased by $21 (= $30 - $9) for each

extra hour of overtime worked.

The point where the shadow price falls to $0

The shadow price of a limiting resource applies only to additional quantities of the

resource for as long as it remains a limiting factor.

Eventually, if more and more units of a scarce resource are made available, a point

will be reached when it ceases to be an effective limiting factor, and a different

constraint becomes a limiting factor instead. When this point is reached, the shadow

price for additional units of the resource will become $0.

In the above example, it is difficult to identify clearly where this point would be

reached for skilled labour. As more and more hours of skilled labour are made

available, the constraint line for skilled labour will move out from the origin of the

graph, and skilled labour will cease to be a limiting factor when its constraint line

moves beyond the point on the graph were the lines y = 20 and 40x + 45y = 1,800

intersect. (Workings are not shown.)

At this point:

(1) y = 20

(2) 40x + 45y = 1,800

Multiply (1) by 45:

(3) 45y = 900

Subtract (3) from (2) 40x = 900

x = 22.5

The skilled labour hours required to produce 20 units of y (Product C) and 22.5

units of x (Product B) would be (20 × 30) + (22.5 × 20) = 1,050 minutes per day.

The original constraint for skilled labour was 900 minutes per day.

The shadow price of $0.50 per minute or $30 per hour for skilled labour therefore

applies only for an additional 150 minutes or 2.5 hours per day.

Summary: shadow prices

Important implications of shadow prices for management can be summarised as

follows.

Paper F5: Performance management

 Management need to be aware that for each limiting factor it might be profitable

to acquire additional units of the limiting factor even if they cost more than the

normal variable cost.

 However, it is also necessary to know the limit to the additional units of the

scarce resource that would add to total contribution. Beyond this point, the

shadow price of the resource falls to $0.

 Although it is assumed that the aim is always to maximise the value of the

objective function (total contribution), non-financial considerations might

influence the decision. For example if skilled labour is a limiting factor and

management is considering whether to pay workers a premium to work

overtime, non-financial factors could be:

- The amount of premium paid for overtime working: if this is set at a high

level, it might cause discontent amongst the rest of the work force

- The number of overtime hours to be worked per employee. If individuals

work excessive hours of overtime, they could get tired and the quality and

efficiency of their work might decline.

5.4 Slack

Slack refers to the amount of a constraint that is not used in the optimal solution to a

linear programming problem. In the previous example, there was slack for both

machine type 1 and machine type 2, because the total available time was not fully

used in the optimal solution.

In the original solution where x = 15 and y = 20, the total time required per day on

machine type 1 was (15 × 30 minutes) + (20 × 10 minutes) = 650 minutes, and up to

900 minutes were available.

The total time required per day on machine type 2 was (15 × 40 minutes) + (20 × 45

minutes) = 1,500 minutes, and up to 1,800 minutes were available.

Implications of slack

When there is some slack, management should consider possible implications.

If there is some slack for a production resource, such as machine time, it might be

worth considering whether a reduction in production capacity would be advisable.

If the slack is long-term, there will be over-capacity in production. Machines might

be sold off, or the spare capacity might be used for other purposes that might add to

contribution and profit.

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

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