Use the three-person, envy-free chore-division procedure from problem 30 to divide the f...

Use the three-person, envy-free chore-division procedure from problem 30 to divide the following chores. Each boy’s estimate of the burdensomeness of the chores is given in the following table. For the final division of chores, explain why it is envy-free.

Problem 30:

Although we have given less attention to dividing chores and undesirable items in this chapter than we have to dividing desirable items, we will now look at a three-person, envy-free chore division procedure created by Elisha Peterson and Francis Su. It is a moving-knife procedure similar to the one described in problem 29. The procedure’s aim is to divide the chores into six pieces and assign to each player two of the pieces that he or she feels are as small as or smaller than each pair of pieces assigned to the other players, as explained below.

STEP 1: The three chores must be divided into three portions using any three-person, envy-free fair division scheme, such as the one developed in Section 4.3. Thus, each player has a piece he or she feels is the largest.

STEP 2: Player 1 divides portion 1 into two pieces that he or she feels is exactly half and then assign those pieces to the other two players in such a way that each feels he or she has received no more than half of portion 1. (The players can achieve this result using Austin’s moving knife procedure from the previous problem. For example, player 1 and player 2 can divide the chore using Austin’s procedure and then player 3 can choose the half he or she thinks is the smallest. The remaining half goes to player 2.)

STEP 3: Repeat step 2 for each of the other players and then end the procedure. At this point, each player has two out of six total pieces such that each feels his or her share is the smallest.

a. Explain why a player will not envy the two pieces from another player.

b. What is the largest number of cuts required by this procedure?

c. This method assumes that a chore is infinitely divisible. Is this a reasonable assumption? Explain.