{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# sol11 - STA3007 Applied probability(08-09 Chapter 6...

This preview shows pages 1–3. Sign up to view the full content.

STA3007 Applied probability (08-09) Solution to Assignment 11 Chapter 6 Problem 4.1 If there is single repairman. As N = M = 5 , R = 1 , λ = 2 , μ = 1, we have λ k = λ = 2 , k = 0 , 1 , 2 , 3 , 4 , λ 5 = 0 , μ k = = k, k = 0 , 1 , 2 , 3 , 4 , 5 , and θ 0 = 1 , θ 1 = 2 , θ 2 = 2 , θ 3 = 4 3 , θ 4 = 2 3 , θ 5 = 4 15 . Therefore, the limiting distribution is π i = 15 109 θ i , i = 0 , 1 , 2 , 3 , 4 , 5. i.e. π 0 = 15 109 , π 1 = 30 109 , π 2 = 30 109 , π 3 = 20 109 , π 4 = 10 109 , π 5 = 4 109 . (a) The average number of machines operating is 5 k =0 k = 210 109 = 1 . 9266 . (b) The equipment utilization is 5 k =0 k /M = 210 109 / 5 = 0 . 3853 . (c) The average idle repair capacity is π 5 = 4 109 = 0 . 0367 . If there are two repairmen. As N = M = 5 , R = 2 , λ = 2 , μ = 1, we have λ k = = 4 , k = 0 , 1 , 2 , 3 , λ 4 = λ = 2 , λ 5 = 0 , μ k = = k, k = 0 , 1 , 2 , 3 , 4 , 5 , and θ 0 = 1 , θ 1 = 4 , θ 2 = 8 , θ 3 = 32 3 , θ 4 = 32 3 , θ 5 = 64 15 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Therefore, the limiting distribution is π i = 15 579 θ i , i = 0 , 1 , 2 , 3 , 4 , 5. i.e. π 0 = 15 579 , π 1 = 60 579 , π 2 = 120 579 , π 3 = 160 579 , π 4 = 160 579 , π 5 = 64 579 . (a) The average number of machines operating is 5 k =0 k = 1740 579 = 3 . 0052 . (b) The equipment utilization is 5 k =0 k /M = 1740 579 / 5 = 0 . 6010 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}