Q#1:

Let unit time be 1 minnute. Then,
        lambda1 = 1,
        lambda2 = 1.5,
        mu = 2.

(a)
                   1
        Ws1 = ------------ = 1 (minute).
              mu - lambda1

(b)
               lambda2 
        rho2 = ------- = 0.75,
                 mu

                 rho2
        Wq2 = ------------ = 1.5 (minutes).
              mu - lambda2


Q#2:

Let unit time be 1 hour. Then,
        lambda = 60,
        mu = 30.
The cost factors are
        Cb = $18 per booth per hour, and
        Cd = $15 per booth per hour.
Let the # of booths in operation be
        n.

(a)
        No,
        because if n = 2, then

              lambda
        rho = ------- = 1,  
               n*mu 

        which means the average length of queue will be infinity.

(b)
              lambda
        rho = -------.
               n*mu

Given "n" and "rho", find "Wqmu" by using the "Wqmu - n" chart. Then,
        
             Wqmu
        Wq = -----.
              mu

The total system cost is
        TC(n)
        = n*Cb + Ws*lambda*Cd
        = n*Cb + (Wq + 1/mu)*lambda*Cd
        = 18*n + 30*Wqmu + 30

       ---------------------------------
        n       rho     Wqmu    TC(n)
        3       2/3     0.4     96
        4       2/4     0.08    104.4

(c)
In terms of total system cost, the optimal number of booths in operation
should be n=3.

We assume that the total system cost is a convex function of "n",
and actually, it really is.  Thus, from (b), we know TC(n) is
increasing in "n" when "n" is greater than 3.  Hence, the optimal number
is less than or equal to 3.  When "n" is less than 3, the average length
of queue is infinity, so is TC(n).  Therefore, the optimal number is
exactly 3.