Following the 1-1 draw against Napoli, AC Milan will be looking to grab all three points against Parma this weekend, building on their recent positive performances. In attack, there could be a change, according to a report.

Kris Piatek was disappointing yet again last time out, failing to make the most of his chances. Therefore, as reported by today's edition of Tuttosport (via MilanNews.it), Stefano Pioli could decide to drop the Polish international from the line-up.
In fact, the manager is considering playing Rafael Leao from start instead, as he needs to get the goalscoring up and running. With this change, the Rossoneri's attack would be a lot more dynamic, as the Portuguese youngster often drifts out wide.
No matter who starts up front come Sunday, the Rossoneri's attackers need to start scoring goals. Against Napoli, Bonaventura rescued his side from disaster, scoring the crucial equaliser with an absolute thunderbolt on the edge of the box.
bauadps
3
for me Milan problem that we don't have good options available on the bench to cover any player absent 😤 as we saw in Napoli match when Bennacer was suspended Biglia and Kessie didn't step up for him as they lost many balls in middle and our defense suffer from counterattack because of it 😠 and Piatek now struggle front of the goal and lost his last year form 😞 maybe because he's one season wonder or maybe that he get overwhelmed with the media and become selfish as he shoot any ball he get without even looking on his position also he can't cover his ball and dive for any simple touch from defender ☹ so maybe start given Leao chances is the right thing to do from now on 😉
yowellster21
2
With the come backs and current performances. While Paqueta has shown flair, he still needs to prove he can hold down a spot, which i think he can especially against Parma. Leao is up front until he either shows he’s not able or piatek finds some fucking form.
Minatrown
5
Despite not winning any games yet I can see that Pioli is a more experienced manager than the useless Giampolo.