Clojure code summary analysis

Hi Clojure Experts,

I am trying to do some timing tests in Clojure 1.3 and I thought I was asking a question based on an existing piece of code that solves a differential equation adapted from this blog post .

The code follows:

;; the differential equation is                                                                                            
;; dy/dt = f(y,t) = t - y                                                                                                  

(defn f [t y] (- t y))

;; solve using Euler method                                                                                              
(defn solveEuler [t0 y0 h iter]
  (if (> iter 0)
    (let [t1 (+ t0 h)
          y1 (+ y0 (* h (f t0 y0)))]
      (recur t1 y1 h (dec iter)))
    [t0 y0 h iter]))

(defn multipleSolveEuler []
  (let [steps '(1 10 100 1000 10000 100000)
        results (map #(second (solveEuler 0.0 0.0 (/ 1.0 %) %)) steps)
        errors  (map #(- (Math/exp -1) %) results)]
    (partition 3 (interleave steps results errors))))

(def *cpuspeed* 2.0)

(defmacro cyclesperit [expr its]
  `(let [start# (. System (nanoTime))
         ret# ( ~@expr (/ 1.0 ~its) ~its )
         finish# (. System (nanoTime))]
     (int (/ (* *cpuspeed* (- finish# start#)) ~its))))

(defn solveEuler-2 [t0 y0 h its]
  (let [zero (int 0)]
    (loop [t0 (double t0), y0 (double y0), h (double h), its (int its)]
      (if (> its zero)
        (let [t1 (+ t0 h)
              y1 (+ y0 (* h (- t0 y0)))]
          (recur t1 y1 h (dec its)))
        [t0 y0 h its]))))

      

        
        
        
      

    

So when I say

(time solveItEuler-2 0.0 1.0 (/ 1.0 1000000000) 1000000000))

      

        
        
        
      

    

I get 6004.184ms time on a 6 month Macbook pro. I issue the command again and I communicate at the same time. But when I run it 3 times, I get the time in the 3500ms range. I've noticed this before for other code snippets and wondered why this is the case. I guess I am expecting roughly the same execution time for sequential runs.

Don't I understand how "time" works, am I missing something or is there some caching going on under the hood?

Thank.