On any given day Eric is either cheerful (C), so-so (S), or glum (G). If he is c

Question

On any given day Eric is either cheerful (C), so-so (S), or glum (G). If he is cheerful today, then he will be C, S, or G tomorrow with respective probabilities 0.5, 0.3, 0.2. If he is feeling so-so today, then he will be C, S, or G tomorrow with probabilities 0.3, 0.4, 0.3. If he is glum today, then he will be C, S, or G tomorrow with probabilities 0.2, 0.2, 0.6. Generate a sequence X0, X1, ... , X49,999 of the first 50, 000 states of Eric’s mood chain and calculate the proportion of these times that the Markov chain is in each state. Use the following procedure:

(a) Generate 50, 000 independent pseudo-random numbers (uniform numbers), by computer. Print out the 40 first numbers.

(b) Use the 1st random number to chose the initial state X0 at random (1/3 probability for each state). Remember that this is equivalent to generate a discrete random variable (see the first problem of simulation). This can be considered to be Eric’s mood for day zero. Write down this number.

(c) Generate Xn recursively, n = 1, 2, ..., 49, 999, by using the successive random numbers to generate successive states. For example, if X3 = 1 then X4 should be chosen to be 0, 1, or 2 with probabilities 0.30, 0.40, and 0.30, respectively. List or print out the 40 values of the random variables (states), associating each state with the random number used to generate it.

(d) Compute the proportion of these 50, 000 days for which Eric’s mood is in each of the 3 states. Compare this with the long-run proportions of the time in each state, obtained in the preliminary problem. For this comparison, it’s enough to give the difference.

Explanation / Answer

The result of this is that the Asynchronous counter suffers from what is known as “Propagation Delay” in which the timing signal is delayed a fraction through each flip-flop.

However, with the Synchronous Counter, the external clock signal is connected to the clock input of EVERY individual flip-flop within the counter so that all of the flip-flops are clocked together simultaneously (in parallel) at the same time giving a fixed time relationship. In other words, changes in the output occur in “synchronisation” with the clock signal.

The result of this synchronisation is that all the individual output bits changing state at exactly the same time in response to the common clock signal with no ripple effect and therefore, no propagation delay.

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