Assume that the scheduling logic will issue an instruction to the execution unit when all of its register operands are ready. Further assume that registers codice_10 and codice_11 are ready: the values in codice_10 and codice_11 were computed a long time ago and have not changed. However, assume codice_14 is not ready: its value is still in the process of being computed by the codice_15 (integer divide) instruction. Finally, assume that registers codice_10 and codice_11 hold the same value, and thus all the loads and stores in the snippet access the same memory word.