Figure 9.2

9.3 Tax on cigarette smoking

Context

An increase in the level of tax on cigarettes will bring in extra tax revenue, however, it could lead to a reduction in the number of cigarettes sold, which could mean less revenue.

Problem statement

If the tax is increased by x pence per packet, what is the limit on the value of x if the increase is to be worthwhile from the point of view of the Chancellor of the Exchequer?

Formulate a mathematical model

Our list of factors is quite brief and is shown in Table 9.2.

Table 9.2

Assumptions

1 At present, N packets of cigarettes are sold every day and there is a tax t on each packet.

2(a) The reduction in the number of cigarettes smoked is proportional to the increase in price.

Obtain the mathematical solution

Using assumptions 1 and 2(a), before the tax rise we have N packets sold every day with a tax t on each. After the rise the tax on each packet changes to t + x

and the number of packets sold every day changes to N − kx where k is some constant (we are

assuming that the price rise is all tax). The difference in tax revenue every day is therefore

which will be positive if N > k ( t + x ). This is equivalent to the condition x < N / k − t. We could consider an alternative to assumption 2(a):

2(b) There is a constant percentage decrease in cigarette smoking for each penny rise in the price of a packet.

Using assumption 2(b) instead of 2(a), let r % be the percentage reduction. Note that this does not mean that if the tax is increased by, say 10 pence, the reduction will be 10 r %. The correct expression for the number of packets sold when the tax is increased by x pence is (1 − r /100) x N. The effects of the two assumptions 2(a) and 2(b) are compared in Figure 9.3, which shows how the number of packets sold daily decreases when the tax is increased. We note that for a certain level of increase x the assumption 2(a) predicts that the number of packets sold will become zero. This is probably less realistic than the gradual diminution predicted by model 2(b).

After the tax rise we now have (1 − r /100) x N packets sold daily with tax ( t + x ) on each so the net difference in the daily tax revenue is (1 − r /100) x N ( t + x ) − Nt = {(1 − r /100) x ( t + x ) − t } . This will be > 0 if (1 − r /100) x ( t + x ) > t. The graph of the function (1 − r /100) x ( t + x ) is shown in Figure 9.4. For the case r = 10, t = 2, it seems to indicate that x could go up to 25 before the net revenue drops.

Figure 9.3