My home uses two light bulbs. On average, a light bulb last for 22 days (expo-...

 

Transcribed Image Text:

My home uses two light bulbs. On average, a light bulb last for 22 days (expo- nentially distributed). When a light bulb burns out, it takes and average of 2 days (exponentially distributed) before I replace the bulb. (a) Formulate a three-state birth and death process of this situation. (b) Determine the fraction of the tme that both light bulbs are working. (c) Determine the fraction of the time that no light bulbs are working. (d) Suppose that whenever a bulb burns out, a fixed failure cost of 5 dollars is incurred. Moreover, when any of the bulb is not working, an inconvenience cost of 1 dollar per bulb per day is incurred. Determine the long run cost per day for this system. 2 My home uses two light bulbs. On average, a light bulb last for 22 days (expo- nentially distributed). When a light bulb burns out, it takes and average of 2 days (exponentially distributed) before I replace the bulb. (a) Formulate a three-state birth and death process of this situation. (b) Determine the fraction of the tme that both light bulbs are working. (c) Determine the fraction of the time that no light bulbs are working. (d) Suppose that whenever a bulb burns out, a fixed failure cost of 5 dollars is incurred. Moreover, when any of the bulb is not working, an inconvenience cost of 1 dollar per bulb per day is incurred. Determine the long run cost per day for this system. 2

Answer rating: 100% (QA)

Solution a The three states in this birth and death process are State 0 Both light bulbs are working State 1 One light bulb is working State 2 Both li... View the full answer