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臺北高等行政法院 高等庭(含改制前臺北高等行政法院)九十一年度訴字第五○○七號

關鍵資訊

  • 裁判案由
    營利事業所得稅
  • 案件類型
    行政
  • 審判法院
    臺北高等行政法院 高等庭(含改制前臺北高等行政法院)
  • 裁判日期
    93 年 02 月 25 日

                                        調 便 退 便 使 使 便 使 使 使 便 使 退 使 調 ( 15,991,000+154,669,108) X50,186,166/ 50,186,166+1,464,501+1,911,941,924+ 7,965,277=4,255,074 (15,991,000+154,669,108) X 1,911,941,924/1,911,941,924+ 1,464,501+50,186,166+7,965,277=162,105,523 便 調 (50,186,166-4,255,074=45,931,092) 退 100*25%=00 000-00=75 25/75=0.33... 100*25%=25 75*10%=7.0 000-00-0.5=67.5 (25+7.5)/67.5=0.48148.....0.4815 25/65=0.3846... 使 (A*3/4)-X>0 (B-Y)>0 (B-Y-X)>0 T2-T1=X/4 X/4 T1=(A*3/4)-X+(B-Y)3/4 K1=(A/4)+(B-Y)/4 (A/4)+(B-Y)/4 1 > (A*3/4)-X+(B-Y)3/4 3 T2=(A*3/4)+3/4(B-Y-X) K2=(A/4)+(B-Y-X)/4 (A/4)+(B-Y-X)/4 1 = (A*3/4)+(B-Y-X)3/4 3 X/4 (B-Y)<0 T2-T1=0 T1=(A*3/4)-X+(B-Y) K1=(A/4) (A/4) (A*3/4)-X+(B-Y) T2=(A*3/4)+(B-Y-X) K2=(A/4) (A/4) 1 >b-x-y (A*3/4)+(B-Y-X) 3 (B-Y)>0 (B-Y-X)<0 T2-T1=(B-Y)/4 T1=(A*3/4)-X+(B-Y)3/4 K1=(A/4)+(B-Y)/4 (A/4)+(B-Y)/4 1 > (A*3/4)-X+(B-Y)3/4 3 T2=(A*3/4)+(B-Y-X) K2=(A/4) (A/4) 1 >b-x-y (A*3/4)+(B-Y-X) 3 (B-Y)/4 調 使 使 使 (15,991,000+154,669,108)*(50,186,166/ 50,186,166+1,464,501+ 1,911,941,924+7,965,277)=4,255,074 50,186,166-4,255,074=45,931,092