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?

Julius Vainora · Accepted Answer · 2018-03-20T12:53:36.723

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.
You are not using
```
AYX <- AtoY + XtoY
AXY <- AtoX + XtoY
```
so they are redundant.
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..
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

Ahh, I think you're right. So we should be taking the fastest route on each day, rather than the fastest route across time? — The Pointer, Mar 20 '18 at 13:04
@ThePointer, right, because of `Assuming you are taking on every day the fastest route`. And also taking the average of 1000 travel times makes more sense than of 4. — Julius Vainora, Mar 20 '18 at 13:08
Ok, thank you very much for the assistance. This was enlightening. — The Pointer, Mar 20 '18 at 13:09

Exponential distribution simulation in R

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