I have this problems to solve. There are 4 areas. Each area has 4 switches.

For example, here are areas:

A1 A2 A3 A4

B1 B2 B3 B4

C1 C2 C3 C4

D1 D2 D3 D4

Conditions:

1) If one switch in one area is down, the other areas can handle the corresponding load.

2) Even if one full area is down BUT all other 3 areas are fully up, they can handle the load.

3) If two corresponding switches in two independent areas go down, then the whole grid will go down. Which means, if A1 and B1 go down, the full grid is down. But if A1 and B2 go down, then the grid will not go down.

The question is to have the uptime of 99.999% how much time each single switch should be up in the year?

I solved it by doing this:

Let’s say, available = average uptime percentage of each switch

Unavailable would be = (1 – available)

When 3 switches are available = Given one is available * seconds is available * 3rd = (available)^3

So, the availability target = When one library is down the other three libraries are still available

0.99999 = available + (1 - available) * (available)^3

available = 0.998173

(1 - 0.998173) * one year = 16 hours downtime per switch (given they go down independently)

Now what I am confused about is that in the equation do I need to consider 1 choose 4 since each area has 4 switches or is that information irrelevant as probability of available will include that? Also, I used union since a switch down in one area seems to be independent of other areas. Not sure if I should use intersection?

If you would have done it, what would have you used? Thanks in advance.