Equitable division


An equitable division is a division of a heterogeneous object, in which the subjective value of all partners is the same, i.e., each partner is equally happy with his/her share. Mathematically, that means that for all partners and :
Where:
The following table illustrates the difference. In all examples there are two partners, Alice and Bob. Alice receives the left part and Bob receives the right part.
Note that the table has only 6 rows, because 2 combinations are impossible: an EX+EF division must be EQ, and an EX+EQ division must be EF.

Finding an equitable division for two partners

One cut - full revelation

When there are 2 partners, it is possible to get an EQ division with a single cut, but it requires full knowledge of the partners' valuations. Assume that the cake is the interval . For each, calculate and and plot them on the same graph. Note that the first graph is increasing from 0 to 1 and the second graph is decreasing from 1 to 0, so they have an intersection point. Cutting the cake at that point yields an equitable division. This division has several additional properties:
The same procedure can be used for dividing chores.

Proportional-equitability variant

The full revelation procedure has a variant which satisfies a weaker kind of equitability and a stronger kind of truthfulness. The procedure first finds the median points of each partner. Suppose the median point of partner A is and of partner B is, with. Then, A receives and B receives. Now there is a surplus -. The surplus is divided between the partners in equal proportions. So, for example, if A values the surplus as 0.4 and B values the surplus as 0.2, then A will receive twice more value from than B. So this protocol is not equitable, but it is still EF. It is weakly-truthful in the following sense: a risk-averse player has an incentive to report his true valuation, because reporting an untrue valuation might leave him with a smaller value.

Two cuts - moving knife

gives each of the two partners a piece with a subjective value of exactly 1/2. Thus the division is EQ, EX and EF. It requires 2 cuts, and gives one of the partners two disconnected pieces.

Many cuts - full revelation

When more than two cuts are allowed, it is possible to achieve a division which is not only EQ but also EF and PE. Some authors call such a division "perfect".
The minimum number of cuts required for a PE-EF-EQ division depends on the valuations of the partners. In most practical cases the number of required cuts is finite. In these cases, it is possible to both find the optimal number of cuts and their exact locations. The algorithm requires full knowledge of the partners' valuations.

Run-time

All the above procedures are continuous: the second requires a knife that moves continuously, the others requires a continuous plot of the two value measures. Therefore, they cannot be carried out in a finite number of discrete steps.
This infinity property is characteristic of division problems which require an exact result. See Exact division#Impossibility.

One cut - near-equitable division

A near-equitable division is a division in which the partners' values differ by at most, for any given. A near-equitable division for two partners can be found in finite time and with a single cut.

Finding an equitable division for three or more partners

Moving knife procedure

. It gives each partner a piece with a subjective value of exactly. This division is EQ, but not necessarily EX or EF or PE.

Connected pieces - full revelation

Jones' full revelation procedure can be extended to partners in the following way:
Note that the maximal equitable value must be at least, because we already know that a proportional division is possible.
If the value measures of the partners are absolutely continuous with respect to each other, then any attempt to increase the value of a partner must decrease the value of another partner. This means that the solution is PE among the solutions which give connected pieces.

Impossibility results

Brams, Jones and Klamler study a division which is EQ, PE and EF.
They first prove that, for 3 partners that must get connected pieces, an EQ+EF division may not exist. They do this by describing 3 specific value measures on a 1-dimensional cake, in which every EQ allocation with 2 cuts is not EF.
Then they prove that, for 3 or more partners, a PE+EF+EQ division may not exist even with disconnected pieces. They do this by describing 3 specific value measures on a 1-dimensional cake, with the following properties:
A pie is a cake in the shape of a 1-dimensional circle.
Barbanel, Brams and Stromquist study the existence of divisions of a pie, which are both EQ and EF. The following existence results are proved without providing a specific division algorithm:
The adjusted winner procedure calculates an equitable, envy-free and efficient division of a set of divisible goods between two partners.

Summary table

NameType# partners# cutsProperties
JonesFull-revelation proc21 EQ, EF, 1-PE
Brams-Jones-KlamlerFull-revelation procnn−1 EQ, -PE
AustinMoving-knife proc22EQ, EF, EX
AustinMoving-knife procnmanyEQ
Barbanel-BramsFull-revelation proc2manyEQ, EF, PE
Cechlárová-PillárováDiscrete approximation proc21 near-EQ