Suppose you just bought several pairs of socks. Would it be better for you to wear out one pair of socks first before wearing the next pair or would it be better for you to rotate the pairs on a regular basis? Is there even any difference?
It is quite intuitive to realize that rotation wears out the pairs evenly. But what about the first strategy of wearing out each pair one at a time? Are your socks newer on an average day? Do your socks last longer or shorter?
We can examine this problem by using a simplified model below:
1. One of pair of socks is worn per day.
2. Let us assume each pair of socks has a lifespan of 8n units. If a pair of socks is left unused for a day, 1 unit is consumed. If a pair of socks is worn for a day, 2 units are consumed. Essentially, when a pair of socks is being used, it wears out twice as fast as a pair that is not.
3. The age of a pair of socks is equal to x when the pair of socks has used up x units of life by the end of the previous day.
4. A pair of socks has age = 0 on the first day of the experiment and age = 8n-2 on the last day of its life if it is being used on that day.
5. The objective is to minimize the average age of the pairs of socks worn per day and to maximize the number of days the whole batch of socks will last.
Now, suppose you just bought 3 pairs of socks.
Strategy 1: Wear them out one pair at a time Calculate the total sum of the age of the socks worn on each day. Total age = 0 + 2 + 4 + ... + (8n-2) 1st pair worn out after 4n days + 4n + (4n+2) + (4n+4) + ... + (4n+(4n-2)) 2nd pair worn out after 2n days + 6n + (6n+2) + (6n+4) + ... + (6n+(2n-2)) 3rd pair worn out after n days = 7n(5n - 1)
Total number of days the socks last = 4n + 2n + n = 7n Average age per day = 7n(5n - 1)/7n = 5n - 1
Strategy 2: 3-day cycle rotation Calculate the total sum of the age of the socks worn on each day. Total age = (0+1+2) + (4+5+6) + (8+9+10) + ... + ((8n-4)+(8n-3)+(8n-2)) = 6n(4n - 1)
Total number of days the socks last = (2n cycles)(3 day per cycle) = 6n Average age per day = 6n(4n - 1)/6n = 4n - 1
Conclusion Clearly, strategy 2 allows us to appear to wear socks that aren't that old on average while strategy 1 allows the socks to last longer. The reason is rather simple. Strategy 1 is able to use the socks for more days because it tries to minimize the amount of time the socks are rotting away in the dresser. Of course, the price to pay is that you end up wearing really old socks on many days. A strategy that randomly chooses the pair of socks to be worn will probably achieve a result between these two extremes.
