Exponential Distribution Simulation in R

Question

I have the following graph:

I am told the following information:

(1) vertex A to vertex X is described by an exponential distribution with lambda = 4;

(2) vertex A to vertex Y is described by an exponential distribution with lambda = 2.5;

(3) vertex X to vertex Y identical to vertex Y to vertex X, and it is described by an exponential distribution with lambda = 10;

(4) vertex X to vertex B is described by an exponential distribution with lambda = 3; and, finally,

(5) vertex Y to vertex B is described by an exponential distribution with lambda = 5.

Let's assume that I'm taking the fastest path between vertices every simulation.

I now want to know the average time it takes to get from vertex A to vertex B.

My R code is as follows:

# Generate/simulate 1000 random numbers for each of the internode paths.

            AtoX <- rexp(1000, 4)
            AtoY <- rexp(1000, 2.5)
            XtoY <- rexp(1000, 10)
            XtoB <- rexp(1000, 3)
            YtoB <- rexp(1000, 5)

    # Length of path from A to X to Y and A to Y to X.

            AYX = AtoY + XtoY
            AXY = AtoX + XtoY

    # Total time of paths from A to B. 

            AXB = AtoX + XtoB
            AYB = AtoY + YtoB
            AXYB = AtoX + XtoY + YtoB
            AYXB = AtoY + XtoY + XtoB

    # Taking the fastest path of all paths. 

            minAXB = min(AXB)
            minAYB = min(AYB)
            minAXYB = min(AXYB)
            minAYXB = min(AYXB)

    # Taking an average of the fastest paths.

            averageTravelTime = 
              mean(minAXB + minAYB + minAXYB + minAYXB)

Does this look correct? I would appreciate it if people could please take the time to review my code for this problem.

Answer 1

1. It depends on the interpretation, but I'd say that you need to simulate times from X to Y and from Y to X separately, although with the same rate. If a train goes both ways and on average the speed is the same, it doesn't mean that two trains leaving from X and Y will arrive to the other point in the same time.

2. You are not using

    AYX <- AtoY + XtoY AXY <- AtoX + XtoY 

   so they are redundant.

3. Writing minAXB <- min(AXB) doesn't really make sense. You simulate 1000 travel times for each edge, AXB is a vector of 1000 times of route AXB and now you are choosing the shortest one across the time..

4. Similarly, averageTravelTime doesn't make sense because minAXB + minAYB + minAXYB + minAYXB is just a number, not a vector.

Hence, I think the code should be

set.seed(1)
AtoX <- rexp(1000, 4)
AtoY <- rexp(1000, 2.5)
XtoY <- rexp(1000, 10)
YtoX <- rexp(1000, 10) # added
XtoB <- rexp(1000, 3)
YtoB <- rexp(1000, 5)

AXB <- AtoX + XtoB
AYB <- AtoY + YtoB
AXYB <- AtoX + XtoY + YtoB
AYXB <- AtoY + YtoX + XtoB # changed XtoY to YtoX

TravelTimes <- pmin(AXB, AYB, AXYB, AYXB)
averageTravelTime <- mean(TravelTimes)

See ?pmin . For every day it chooses the fastest travel time and returns a vector of length 1000.

As a bonus, the following shows how many times which route was the fastest

table(apply(cbind(AXB, AYB, AXYB, AYXB), 1, which.min))
#   1   2   3   4 
# 317 370 240  73