Ex-3.40

(define x 10)
(parallel-execute (lambda () (set! x (* x x)))
                  (lambda () (set! x (* x x x))))

Let P, Q represent both procedures and Pk/Qk denotes accessing value of x the kth time then following possibilities are possible..

x = 1000000: P1, P2, P sets x, Q1, Q2, Q3, Q sets x
x = 1000000: Q1, Q2, Q3, Q sets x, P1, P2, P sets x
x = 10000: P1, Q1, Q2, Q3, Q sets x, P2, P sets x
x = 100: P1, P2, Q1, Q2, Q3, Q sets x, P sets x
x = 100000: Q1, P1, P2 ,P sets x, Q2, Q3, Q sets x
x = 10000: Q1, Q2, P1, P2, P sets x, Q3, Q sets x
x = 1000: Q1, Q2, Q3, P1, P2, P sets X, Q sets x

.... There are other execution sequences possible also, but all the possible values that x can take after the execution are {100, 1000, 10000, 100000, 1000000}

In the next case when P, Q are serialized. After the execution, x can only be 1000000.

In general, This phenomenon when multiple threads/processes are modifying the same resource and final value of the resource could be different depending upon the interleaving is called "Race Condition"