Answer to Question #234592 in Statistics and Probability for peterclain

Question

A 3D printer produces one item every hour. There is a probability of 0.2 that a produced item is defective. If an item is defective, it is discarded immediately, else it is stored in a buffer. Right after inspection, if the item is non-defective, a robot is called upon to pick the item up. However, there is only a 50% chance the robot is available whenever it is called. If the robot is available, it come instantaneously and picks exactly one item. If the robot is unavailable, only at the end of the next hour the robot is called. Further, the buffer can only hold at most 3 items. So. if the buffer is full, the 3D printer does not produce during that hour. Let X„ denote the number of items in the buffer at the beginning of the nth hour (if the buffer is not empty. this is when the 3D printer starts to produce). Then IX.n, it > 01 is a DTMC. 

Answer 1

We are given here that:

P(D) = 0.2 is the probability of the produced item to be defective.

We are given here that:

P( discarded ) = 0.2

P( buffer ) = 1 - 0.2 = 0.8

P( picked | non defective ) = 0.5

Buffer can only hold 3 items at a moment. Therefore, there are 4 possible states here: 0, 1, 2 or 3 which represents the number of items in the buffer at any moment here.

P(0 --> 0) = 0.2 + 0.8*0.5 = 0.6

P(0 --> 1) = 0.8*0.5 = 0.4 when the robot is able to arrive. These are the probabilities for row 1.

P(1 --> 0) = 0.2*0.5 = 0.1

P(1 --> 1) = 0.8*0.5 + 0.2*0.5 = 0.5

P(1 --> 2) = 0.8*0.5 = 0.4

P(2 --> 0) = 0

P(2 --> 1) = 0.2*0.5 = 0.1

P(2 --> 2) = 0.8*0.5 + 0.2*0.5 = 0.5

P(2 --> 3) = 0.4

P(3 --> 2) = 0.2*0.5 = 0.1

P(3 --> 3) = 0.8*0.5 + 0.2*0.5 + 0.4 = 0.9

Therefore, the transition probability matirx here is given as: