1.  u_i = i(g-i).  Problem incorrectly stated mu_i.

2.  

varTimeToAbsorb <- function(g)
{
   iVals <- 1:(g-2)
   rates <- iVals * (g - iVals)
   vars <- 1 / rates^2
   sum(vars)
}

print(varTimeToAbsorb(10))

3.  

# A buses arrive at the rate 1/10 = 0.10, and B buses arrive at the rate
# of 1/20 = 0.05.  So buses, collectively of both lines, arrive at the
# rate of 0.15, with a mean time between buses of 1/0.15, a little less
# than 7 minutes.  By the Markov property, that is also the mean time I
# will wait until some bus arrives.  And again, since "time starts over"
# when I arrive, our theorem about the min of 2 independent expons applies

nextBus <- function()
{
   rateA <- 1/10
   rateB <- 1/20
   rateBoth <- rateA + rateB
   meanTimeBetweenBuses <- 1/rateBoth
   meanTimetoNextBus <- meanTimeBetweenBuses
   probBusA <- rateA / rateBoth
   c(probBusA,meanTimetoNextBus)
}

print(nextBus())

4.

In Eqns. (11.10)-(11.12), the original rates in and out were:

state   rate out           rate in

0       pi0 (1/8+1/8)      pi1 (1/20)
1       pi1 (1/20+1/8)     pi0 (1/8+1/8) + pi2 (1/25+1/25)
2       pi2 (1/25+1/25)    pi1 (1/8)

But now, with just one repairperson, each instance of 1/8+1/8 becomes
1/8.  So our new rates in and out are.

state   rate out           rate in

0       pi0 (1/8)          pi1 (1/20)
1       pi1 (1/20+1/8)     pi0 (1/8) + pi2 (1/25+1/25)
2       pi2 (1/25+1/25)    pi1 (1/8)


findQ <- function() 
{
   q <- matrix(0,nrow=3,ncol=3)
   q[1,1] <- -0.125
   q[1,2] <- 0.05
   q[2,1] <- 0.125
   q[2,2] <- -0.175
   q[2,3] <- 0.08
   q[3,2] <- 0.125
   q[3,3] <- -0.08

   q

}

print(findQ())

> print(findQ())
       [,1]   [,2]  [,3]
[1,] -0.125  0.050  0.00
[2,]  0.125 -0.175  0.08
[3,]  0.000  0.125 -0.08
 

5.

# EX = c/2

# can use the integrate() function if you wish
# EY = integral(0,c) t f_Y(t) dt 
#    = integral(0,c) t t (1/c) / EX dt 
#    = (c^3/3) (1/c) (2/c)
#    = 2/3 c
# 
# EY / EX = 2/3 c / (c/2) = 4/3

print(4/3)