How long does it take you to load the dishwasher? Placing all those tricky-shaped objects in position to maintain the perfect balance between getting the maximum numbers of objects into the machine and placing them so that they clean optimally.
Most likely, less time than it takes to load ours. This isn’t because it’s a particularly difficult machine to load (and we don’t have to worry about it cleaning optimally – because it doesn’t – if we want something really clean then it goes in the washing up bowl). No, it’s because the baby is getting rather adept at unloading it. That’s his latest bit of amusement. Pull out all the items of cutlery, one by one, and scatter them over the kitchen floor.
One can construct a simple theoretical model of the process, to see what happens. Suppose I can place objects into the dishwasher at a rate m (for me, Marcus). Suppose baby can unload them at a rate b (for baby, Benjamin). Then the rate of increase of objects is given by m – b. Simple enough. If m is greater than b, I eventually win and the dishwasher ends up fully loaded and I can shut the door. If b is greater than m we end up with all our dirty dishes on the kitchen floor. If m equals b, we will go on forever.
But it’s slightly more complicated. The rate at which I can load them diminishes as more objects are in the dishwasher. This is because 1. there are fewer dirty objects in the kitchen to grab, so I have to take longer hunting down the next bit of washing-up, and 2. the dishwasher is more crowded so it takes longer to negotiate the tangle of dishes to place the plate into position. So the rate m depends on the number of objects in the dishwasher, N. We might suggest that m reduces linearly with N, so that m = c – kN, where c and k are constants. The maximum number of objects that one can load is then c/k, because when N=c/k, m will be zero.
But it’s worse than that. The more objects in the dishwasher the more quickly baby can grab the next and remove it. So b grows with N. Let’s assume it grows approximately linearly, so b = a N, where a is another constant.
What will happen now? Let’s look at the difference between m and b? m minus b is now (c-kN) – a N, which equals c – (k+a)N. At the start, N is low, and I have the upper hand. More objects go in, so N increases. But as N gets larger, the game swings in favour of Benji. We can see that when c = (k+a)N, that is N = c/(k+a), m – b is exactly zero. Then we have reached a dynamic equilibrium – the rate at which I can load the dishwasher is exactly balanced by the rate Benjamin unloads it. It’s then time to give up, shut the door, and start the machine going, with c/(k+a) objects inside.
So, if we want to maximize our use of the dishwasher, we simply need to increase ‘c’ (practise grabbing objects quickly), reduce ‘k’ (practise manoeuvering objects into the last remaining slot at the back) and reduce ‘a’ (distracting the bubble with something more interesting, like emptying the entire contents of our cereal packets onto the floor).
Or we could just wait until the baby’s in bed.