Create a linprog program based on
max 6π₯ β π¦
s. t. 4π¦ β π₯ β₯ 10
2π₯ + π¦ β€ 10
2π¦ β π₯ β€ 5
π₯, π¦ β β
f = [-6 1]
A = [1 -4;2 1;-1 2]
b = [-10 10 5]
linprog(f,A,b)
Probability shop is empty
is just p0
probability queue is empty
all shop employees are used
aka if 4 shop employies: p0+p1+p2+p3
aka p0(1+c1+c2+c3+c4)
newtons step
minimise π(π‘) = π(π₯0 β π‘πβ²(π₯0))
πβ²(π‘) = πβ²(π₯0 β π‘πβ²(π₯0))(βπβ²(π₯0)) = 0
solve for t
π‘ = (π₯0 β ππ)/πβ²(π₯0)
find the smallest positive optimal step size for k
π₯1 = π₯0 β tπβ²(π₯0) = π₯0 β [(π₯0 β π)/πβ²(π₯0)]πβ²(π₯0)
mean service
times
W-Wq