Question

the gotham city maternity ward contains 2 beds. admissions are made only at the beginning of the day. each day, there is a .5 probability that a potential admission will arrive. a patient can be admitted only if there is an open bed at the beginning of the day. half of all patients are discharged after one day, and all patients that have stayed one day are discharged at the end of their second day. a what is the fraction of days where all beds are utilized? b on the average, what percentage of the beds are utilized?

Answer 1

Let's first consider the fraction of days where all beds are utilized:

a) To have both beds utilized at the same time, we need at least two patients to have stayed for two days or four patients to have stayed for one day each. The probability of having at least two patients stay for two days is the same as the probability of having two or more admissions on the first day, which can be calculated using the binomial distribution:

P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)

where X is the number of admissions on the first day. The probability of no admissions is 0.5, and the probability of one admission is 0.5 times the probability of having an open bed on the first day, which is also 0.5. Therefore:

P(X ≥ 2) = 1 - 0.5 - 0.5 × 0.5 = 0.25

Alternatively, the probability of having four patients each stay for one day is (0.5)^4 = 0.0625. The probability of having both beds utilized is the sum of these probabilities:

P(both beds utilized) = 0.25 + 0.0625 = 0.3125

Therefore, the fraction of days where all beds are utilized is 0.3125 or 31.25%.

b) To find the average percentage of beds utilized, we need to calculate the expected number of patients per day and divide by the number of beds. Let's define a random variable Y as the number of patients per day. The distribution of Y depends on the number of admissions on the first day:

If X = 0 (no admissions), Y = 0
If X = 1 (one admission), Y = 1 with probability 0.5 and 0 with probability 0.5
If X ≥ 2 (two or more admissions), Y = 2 with probability 0.25, 3 with probability 0.25, and 4 with probability 0.5

Therefore, the expected value of Y is:

E(Y) = 0 × 0.5 + 1 × 0.25 + 2 × 0.25 + 3 × 0.25 + 4 × 0.5 = 2.25

The average percentage of beds utilized is then:

2.25 / 2 = 1.125 or 112.5%

Note that this value is greater than 100% because patients can stay for more than one day.

Answer 2

Explanation:

In all Markov-chain problems, the first thing to set up is the state space; in your case the system would be described by pairs which indicate the state of the beds, say e for empty, f for full and d for discharging. So one state would be a pair, for instance {e,d} would indicate that one bed is empty and one bed contains a patient that will leave on the next day. One possible state-space describing the clinic is,

S={{e,e},{e,d},{e,f},{d,f}}.

Note that in the problem (and in the model) there is no ordering in the beds and {e,d} is the same as {d,e}; note also that the case {f,f} is impossible because two guests cannot arrive together and, in case one bed is f on one day it would become d on the following day.

Once the state space is defined, you have to define the transition function which gives the probability of the next state given the current state. In such a small problem can be done by plain enumeration of cases with their associated probabilities p

• {e,e}: no guest arrives (stay in {e,e} with p=.5) or one guest arrives (move to {e,f}, with p=.5);

• {e,f}: Either one guest arrives, and the current guest stays (move to {d,f}with p=.25); one guest arrives and the current guest leaves (stay in {e,f}with p=.2); no guest arrives and the current guest stays (move to {d,e} with p=.25); no guest arrives and the current guest leaves (move to {e,e}with p=.25)

• {e,d}: no guest arrives (move to {e,e} with p=.5); one guest arrives (move to {e,f}with p=.5)

• {d,f}: two guests leave (move to {e,e}with p=.5); one guest leaves (move to {e,d} with p=.5)

Stationary probability: Let stationary probability of a state i

is given as PI(ij).So from the lingo results, the stationary probabilities are as follows.

PI(00)=0.182

PI(01)=0.182

PI(10)=0.159

PI(11)=0.250

PI(12)=0.091

PI(20)=0.023

PI(21)=0.068

PI(22)=0.045

Here PI(00)=0.182 represents the stationary probability of state 00.

Let the fraction of days where all beds are utilized is represented as PI(2).

Here PI(12) denotes the function, where for 2 consecutive days the 1

bed was utilized for 1 day and 2 beds were in use on the consecutive day.

So, PI(2) can be calculated as:

PI(2)=PI(12/2)+PI(20/2)+PI(21/2)+PI(22)

=0.091/2+0.023/2+0.068/2+0.045

=0.0455+0.0115+0.034+0.045

=0.136

So, 13.6% of times, all the beds are utilized.