دروس مادة التسيير المالي و المحاسبي للفصل الثالث سنة ثالثة ثانوي

‫ﻓﻬﺭﺱ ﺍﻹﺭﺴﺎل ﺍﻟﺜﺎﻟﺙ‬ ‫ﺍﻟﻭﺤﺩﺓ )‪ :(15‬ﺍﻟﻔﻭﺍﺌﺩ ﺍﻟﻤﺭﻜﺒﺔ‬ ‫ﺍﻟﻭﺤﺩﺓ )‪ :(16‬ﺍﻟﺩﻓﻌﺎﺕ ﺍﻟﺜﺎﺒﺘﺔ‬ ‫ﺍﻟﻭﺤﺩﺓ )‪ :(17‬ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﻭﺽ‬ ‫ﺍﻟﻭﺤﺩﺓ )‪ :(18‬ﺍﻟﺘﻨﺒﺅﺍﺕ ﻁﻭﻴﻠﺔ ﺍﻷﺠل‬ ‫ﺍﻟﻭﺤﺩﺓ )‪ :(19‬ﺍﻟﺘﻨﺒﺅﺍﺕ ﻗﺼﻴﺭﺓ ﺍﻷﺠل‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫ﺍﻟﻤﺠﺎل ﺍﻟﻤﻔﺎﻫﻴﻤﻲ ﺍﻟﺭﺍﺒﻊ‪ :‬ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﻤﺎﻟﻴﺔ ﻁﻭﻴﻠﺔ ﺍﻷﺠل‬ ‫ﺍﻟﻭﺤﺩﺓ )‪ :(16‬ﺍﻟﺩﻓﻌﺎﺕ ﺍﻟﺜﺎﺒﺘﺔ‬ ‫‪ANNUITES CONSTANTES‬‬ ‫ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ‪:‬‬ ‫ـ ﻴﻘﻴﻡ ﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﻓﻲ ﺘﺎﺭﻴﺦ ﻤﺎ‪.‬‬ ‫ﺍﻟﻤﺩﺓ ﺍﻟﻼﺯﻤﺔ ‪ 10 :‬ﺴﺎﻋﺎﺕ‬ ‫ﺍﻟﻤﺭﺍﺠﻊ‪ :‬ﺍﻟﻜﺘﺏ ﺍﻟﻤﺩﺭﺴﻴﺔ ﺍﻟﻤﻘﺭﺭﺓ ‪.‬‬ ‫ﻤﺅﺸﺭﺍﺕ ﺍﻟﺘﻘﻭﻴﻡ ﺍﻟﺫﺍﺘﻲ‪:‬‬ ‫ ﻴﺤﺩﺩ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ‪.‬‬ ‫ ﻴﺤﺩﺩ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ‪.‬‬ ‫ ﻴﻘﻴﻡ ﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﻓﻲ ﺘﺎﺭﻴﺦ ﻤﺎ‪.‬‬ ‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫‪ .1‬ﺘﻌﺭﻴﻑ ﺍﻟﺩﻓﻌﺔ ﺍﻟﺜﺎﺒﺘﺔ‬ ‫‪ .2‬ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ‬ ‫‪ .3‬ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ‬ ‫‪ .4‬ﺘﻘﻴﻴﻡ ﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ‬ ‫‪ .5‬ﺃﻨﺸﻁﺔ ﺍﻹﻋﻼﻡ ﺍﻵﻟﻲ ﻭ ﺤﻠﻭﻟﻬﺎ‬ ‫‪ .6‬ﺃﺴﺌﻠﺔ ﺍﻟﺘﻘﻭﻴﻡ ﺍﻟﺫﺍﺘﻲ‬ ‫‪ .7‬ﺃﺠﻭﺒﺔ ﺍﻟﺘﻘﻭﻴﻡ ﺍﻟﺫﺍﺘﻲ‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪ .1‬ﺘﻌﺭﻴﻑ ﺍﻟﺩﻓﻌﺔ ﺍﻟﺜﺎﺒﺘﺔ‪:‬‬‫ﺍﻟﺩﻓﻌﺎﺕ ﻫﻲ ﻤﺒﺎﻟﻎ ﺘﺩﻓﻊ ﻋﻠﻰ ﻓﺘﺭﺍﺕ ﺯﻤﻨﻴﺔ ﺜﺎﺒﺘﺔ )ﻤﺘﺴﺎﻭﻴﺔ( ﻭﺍﻟﻔﺎﺭﻕ ﺍﻟﺯﻤﻨﻲ ﺍﻟﺫﻱ ﻴﻔﺼل ﺒـﻴﻥ ﺘﺴـﺩﻴﺩ‬ ‫ﺩﻓﻌﺘﻴﻥ ﻴﺴﻤﻰ ﺒﺎﻟﻤﺩﺓ‪ ،‬ﻓﻘﺩ ﺘﻜﻭﻥ ﺍﻟﻤﺩﺓ ﺴﻨﻭﻴﺔ‪ ،‬ﺴﺩﺍﺴﻴﺔ ﺜﻼﺜﻴﺔ ﺃﻭ ﺸﻬﺭﻴﺔ‪.‬‬ ‫ﻭ ﺍﻟﺩﻓﻌﺎﺕ ﻋﻠﻰ ﻋﺩﺓ ﺃﻨﻭﺍﻉ ﻭ ﻫﺫﺍ ﺤﺴﺏ ﺍﻟﻬﺩﻑ ﺍﻟﺫﻱ ﺘﺩﻓﻊ ﻤﻥ ﺃﺠﻠﻪ ‪:‬‬ ‫ﺃ ـ ﻓﺈﺫﺍ ﻜﺎﻥ ﺍﻟﻬﺩﻑ ﻤﻨﻬﺎ ﻫﻭ ﺘﻜﻭﻴﻥ ﺭﺃﺱ ﻤﺎل ﻓﻲ ﻨﻬﺎﻴﺔ ﻤﺩﺓ ﻤﺤﺩﺩﺓ ﻓﻬﻲ ﺇﺫﹰﺍ‪:‬‬‫‪ -‬ﺩﻓﻌﺎﺕ ﻋﺎﺩﻴﺔ ﻟﻼﺴﺘﺜﻤﺎﺭ ﺃﻭ ﺍﻟﺭﺴﻤﻠﺔ )‪ (Annuités de Capitalisation‬ﻋﻨﺩﻤﺎ ﺘـﺩﻓﻊ‬‫ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﻨﻬﺎﻴﺔ ﻜل ﻤﺩﺓ ﺃﻱ ﺍﻟﺩﻓﻌﺔ ﺍﻷﻭﻟﻰ ﺘﺩﻓﻊ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﺍﻷﻭﻟﻰ ﻭ ﻫﻜﺫﺍ‪ . ...‬ﻭ ﻤﻥ‬ ‫ﺃﻤﺜﻠﺔ ﻫﺫﻩ ﺍﻟﺩﻓﻌﺎﺕ ﺃﻗﺴﺎﻁ ﺍﻟﺘﺄﻤﻴﻥ ﻋﻠﻰ ﺍﻟﺤﻴﺎﺓ‪.‬‬‫‪ -‬ﺩﻓﻌﺎﺕ ﺍﻟﺘﻭﻅﻴﻑ )‪ (Annuités de placement‬ﻋﻨﺩﻤﺎ ﺘﺩﻓﻊ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﺒﺩﺍﻴﺔ ﻜل ﻤﺩﺓ‬ ‫ﺃﻱ ﺍﻟﺩﻓﻌﺔ ﺍﻷﻭﻟﻰ ﺘﺩﻓﻊ ﻋﻨﺩ ﺘﻭﻗﻴﻊ ﺍﻟﻌﻘﺩ‪.‬‬‫ﺏ ـ ﺃﻤﺎ ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻬﺩﻑ ﻤﻨﻬﺎ ﺘﺴﺩﻴﺩ ﺩﻴﻥ )ﻗﺭﺽ ﻤﺜ ﹰﻼ(‪ ،‬ﻓﻬﻲ ﺘﺩﻋﻰ ﺒﺩﻓﻌﺎﺕ ﺍﻟﺴـﺩﺍﺩ ) ‪Annuités de‬‬‫‪ (Remboursement‬ﺃﻭ ﺒﺩﻓﻌﺎﺕ ﺍﻻﺴﺘﻬﻼﻙ )‪ ،(Annuités d'amortissement‬ﻭﻋﺎﺩﺓ ﻤـﺎ‬ ‫ﺘﺩﻓﻊ ﻓﻲ ﻨﻬﺎﻴﺔ ﻜل ﻤﺩﺓ‪.‬‬ ‫‪ .2‬ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ‬ ‫‪ .1.2‬ﺍﻟﺼﻴﻐﺔ ﺍﻟﻌﺎﻤﺔ ﻟﻠﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ )ﺍﻟﺠﻤﻠﺔ(‬ ‫ﻨﻌﺘﺒﺭ ﺃﻥ ﺍﻟﺩﻓﻌﺔ ﺍﻷﺨﻴﺭﺓ ﺘﺩﻓﻊ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻭﺤﺩﺓ ﺍﻟﺯﻤﻨﻴﺔ ﺍﻷﺨﻴﺭﺓ ﺃﻱ ﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ‪.‬‬ ‫ﻨﺭﻤﺯ ﻟـ‪:‬‬ ‫‪ : a‬ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ﺍﻟﺜﺎﺒﺘﺔ‪ : n‬ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ‬ ‫‪ :‬ﻤﻌﺩل ﻓﺎﺌﺩﺓ ‪) 1DA‬ﻤﻌﺩل ﺍﻟﻔﺎﺌﺩﺓ ﺍﻟﻤﺭﻜﺒﺔ(‪i‬‬ ‫‪ :‬ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﺍﻟﺯﻤﻥ )‪ ،(n‬ﻋﻨﺩ ﺘﺴﺩﻴﺩ ﺁﺨﺭ ﺩﻓﻌﺔ‪A.‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ‬‫‪012345‬‬ ‫‪p‬‬ ‫‪n-1 n‬‬‫‪aaaaa‬‬ ‫‪a‬‬ ‫اﻟﻤﺪة‬ ‫‪aa‬‬ ‫‪A‬‬ ‫ﺍﻟﺠﻤﻠﺔ ﺘﺴﺎﻭﻱ ﻤﺠﻤﻭﻉ ﺍﻟﺠﻤل ﻟﻜل ﺩﻓﻌﺔ ﻤﻥ ﺍﻟﺩﻓﻌﺎﺕ ﺃﻱ ‪A:‬‬ ‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ‬ ‫ﻤﺩﺓ ﺍﻟﺘﻭﻅﻴﻑ‬ ‫ﺍﻟﺩﻓﻌﺎﺕ‬ ‫)‪(n − 1‬‬ ‫‪1‬‬ ‫‪a (1 + i )n −1‬‬ ‫‪a (1 + i )n −2‬‬ ‫)‪(n − 2‬‬ ‫‪2‬‬ ‫‪a (1 + i )n −3‬‬ ‫)‪(n − 3‬‬ ‫‪3‬‬ ‫‪a (1 + i )n −4‬‬ ‫)‪(n − 4‬‬ ‫‪4‬‬ ‫‪a (1 + i )n −5‬‬ ‫)‪(n − 5‬‬ ‫‪5‬‬ ‫‪............................................................‬‬ ‫) ‪a (1 + i )n − p (n − p‬‬ ‫‪p‬‬ ‫‪..............................................................‬‬ ‫) ‪a(1+ i‬‬ ‫‪1‬‬ ‫‪n −1‬‬ ‫‪a0‬‬ ‫‪n‬‬ ‫ﻴﻤﻜﻥ ﺃﻥ ﻨﻜﺘﺏ ﻤﺠﻤﻭﻉ ﺍﻟﺠﻤل ﻋﻠﻰ ﺍﻟﻨﺤﻭ ﺍﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪A = a + a(1+ i )+ .....+ a(1+ i )n −3 + a(1+ i )n −2 + a(1+ i )n −1‬‬‫ﻨﻼﺤﻅ ﺃﻥ ﻤﺠﻤﻭﻉ ﺍﻟﺠﻤل ﺘﺸﻜل ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل )‪ (a‬ﻭ ﺃﺴﺎﺴﻬﺎ ) ‪ (1 + i‬ﻭ ﺤﺩﻫﺎ ﺍﻷﺨﻴﺭ‬ ‫‪. a(1 + i )n −1‬‬ ‫ﻭ ﺤﻴﺙ ﺃﻥ ﻤﺠﻤﻭﻉ ﺤﺩﻭﺩ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻫﻭ ﺤﺴﺏ ﺍﻟﻘﺎﻨﻭﻥ ﺍﻟﺘﺎﻟﻲ‪:‬‬ ‫‪S=a‬‬ ‫×‬ ‫‪qn -1‬‬ ‫‪,‬‬ ‫ﺃﻭ‬ ‫‪S‬‬ ‫=‬ ‫‪d.q-a‬‬ ‫‪,‬‬ ‫‪q-1‬‬ ‫‪q-1‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺤﻴﺙ )‪ (q‬ﻫﻭ ﺍﻷﺴﺎﺱ ﻭ )‪ (d‬ﻫﻭ ﺍﻟﺤﺩ ﺍﻷﺨﻴﺭ ﻭ )‪ (a‬ﺍﻟﺤﺩ ﺍﻷﻭل ﻭ )‪ (n‬ﻋﺩﺩ ﺍﻟﺤﺩﻭﺩ‪.‬‬ ‫ﻭ ﻤﻨﻪ‪:‬‬ ‫‪A‬‬ ‫=‬ ‫‪a (1‬‬ ‫‪+‬‬ ‫‪i )n −1 (1 + i‬‬ ‫)‬ ‫‪−‬‬ ‫‪a‬‬ ‫‪(1 + i ) − 1‬‬ ‫‪A = a × (1 + i )n − 1‬‬ ‫‪i‬‬‫ﻴﻌﺒﺭ ﻋﻥ ﺠﻤﻠﺔ ‪ n‬ﺩﻓﻌﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺫﺍﺕ ‪ 1 DA‬ﻟﻜل ﺩﻓﻌﺔ ﻤﺴﺩﺩﺓ ﻓﻲ ﻨﻬﺎﻴﺔ‬ ‫‪(1 + i )n‬‬ ‫‪−1‬‬ ‫ﺍﻟﻤﻘﺩﺍﺭ‬ ‫‪i‬‬‫ﺍﻟﻤﺩﺓ‪ .‬ﻭ ﻨﺴﺘﺨﺭﺝ ﻗﻴﻤﺘﻪ ﺇﻤﺎ ﺒﺎﻟﺤﺴﺎﺏ ﺍﻟﻌـﺎﺩﻱ ﻭ ﺍﻻﺴﺘﻌﺎﻨﺔ ﺒﺎﻵﻟﺔ ﺍﻟﺤﺎﺴﺒﻴﺔ ﺍﻟﻌﻠﻤﻴﺔ ﺃﻭ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺠﺩﺍﻭل‬ ‫ﺍﻟﻤﺎﻟﻴﺔ‪.‬‬ ‫‪ .2.2‬ﺍﺴﺘﻌﻤﺎل ﺍﻟﺼﻴﻐﺔ ﺍﻟﻌﺎﻤﺔ ﻟﻠﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ‬ ‫ﺃ ـ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ )ﺍﻟﺠﻤﻠﺔ(‬ ‫ﺃ‪ 1.‬ـ ﺠﻤﻠﺔ ﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ‪:‬‬ ‫ﺘﻁﺒﻴﻕ ‪:‬‬‫ﺃﺤﺴﺏ ﺠﻤﻠﺔ ‪ 10‬ﺩﻓﻌﺎﺕ ﻋﻨﺩ ﺘﺴﺩﻴﺩ ﺁﺨﺭ ﺩﻓﻌﺔ‪ ،‬ﻗﻴﻤﺔ ﻜل ﺩﻓﻌﺔ ‪ 10.000DA‬ﺒﻤﻌﺩل ‪ %6‬ﺴﻨﻭﻴﹰﺎ‪.‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪A = a × (1 + i )n − 1‬‬ ‫‪i‬‬ ‫‪A‬‬ ‫=‬ ‫× ‪10.000‬‬ ‫‪(1, 06)10 − 1‬‬ ‫‪0, 06‬‬ ‫‪A = 10.000 × 13,180795‬‬ ‫‪A = 131.807, 95DA‬‬ ‫ﺃ‪ 2.‬ـ ﺍﺴﺘﻨﺘﺎﺝ ﺍﻟﺼﻴﻐﺔ ﺍﻟﻌﺎﻤﺔ ﻟﺠﻤﻠﺔ ﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ ‪:‬‬‫ﻓﻲ ﻫﺫﻩ ﺍﻟﺤﺎﻟﺔ ﺘﺩﻓﻊ ﺍﻟﺩﻓﻌﺔ ﺍﻷﺨﻴﺭﺓ ﻓﻲ ﺒﺩﺍﻴﺔ ﺍﻟﻭﺤﺩﺓ ﺍﻟﺯﻤﻨﻴﺔ ﺍﻷﺨﻴﺭﺓ ﺃﻱ ﺃﻥ ﺍﻟﺩﻓﻌﺎﺕ ﺘﺩﻓﻊ ﻓﻲ ﺒﺩﺍﻴﺔ ﻜـل‬ ‫ﻤﺩﺓ ﺃﻭ ﻜل ﻓﺘﺭﺓ‪.‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫‪012345‬‬ ‫‪p‬‬ ‫‪n-1 n‬‬ ‫‪aa a a a a‬‬ ‫‪a‬‬ ‫اﻟﻤﺪة ‪a‬‬ ‫‪A′‬‬ ‫ﺍﻟﺠﻤل ﺍﻟﻤﺘﺤﺼل ﻋﻠﻴﻬﺎ ﻓﻲ ﻨﻬﺎﻴﺔ ﻜل ﻓﺘﺭﺓ ﺃﻱ ﻋﻨﺩ ﺍﻟﻤﺩﺓ ‪ n‬ﻫﻲ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ‬ ‫ﻤﺩﺓ ﺍﻟﺘﻭﻅﻴﻑ‬ ‫ﺍﻟﺩﻓﻌﺎﺕ‬ ‫) ‪(n‬‬ ‫‪1‬‬ ‫‪a (1 + i )n‬‬ ‫‪a (1 + i )n −1‬‬ ‫)‪(n − 1‬‬ ‫‪2‬‬ ‫‪a (1 + i )n −2‬‬ ‫)‪(n − 2‬‬ ‫‪3‬‬ ‫‪a (1 + i )n −3‬‬ ‫)‪(n − 3‬‬ ‫‪4‬‬ ‫‪............................................................‬‬ ‫‪a(1 + i )2‬‬ ‫‪2 n −1‬‬ ‫) ‪a (1 + i‬‬ ‫‪1n‬‬ ‫ﻟﺘﻜﻥ ‪ A ′‬ﺍﻟﺠﻤﻠﺔ ﺍﻟﻤﺤﺼل ﻋﻠﻴﻬﺎ ﺒﻌﺩ ﺠﻤﻊ ﺠﻤل ﺍﻟﺩﻓﻌﺎﺕ ﺒﻌﺩ ﺍﻟﻤﺩﺓ ‪ n‬ﻓﻨﺤﺼل ﻋﻠﻰ ‪:‬‬‫‪A′ = a(1 + i ) + a(1 + i )2 + .... + a(1 + i )n −3 + a(1 + i )n −2 + a(1 + i )n −1‬‬‫ﻨﻼﺤﻅ ﺃﻥ ﻤﺠﻤﻭﻉ ﺍﻟﺠﻤل ﺘﺸﻜل ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل ) ‪ a(1 + i‬ﻭﺃﺴﺎﺴـﻬﺎ ) ‪ (1 + i‬ﻭ ﺤـﺩﻫﺎ‬ ‫ﺍﻷﺨﻴﺭ ‪ ، a(1 + i )n −1‬ﻭ ﺤﺴﺏ ﻗﺎﻨﻭﻥ ﻤﺠﻤﻭﻉ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻬﻨﺩﺴﻴﺔ ﻓﺈﻥ‪:‬‬ ‫‪A′ = a(1 + i ) (1 + i )n − 1‬‬ ‫‪i‬‬ ‫ﺃﻭ ﻨﺴﺘﺨﺩﻡ ﺍﻟﺼﻴﻐﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺒﻌﺩ ﻨﺸﺭ ﺍﻟﻤﻘﺩﺍﺭ ) ‪: (1 + i‬‬ ‫⎡‬ ‫‪(1‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪)n‬‬ ‫‪+1‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫⎥⎤‪1‬‬ ‫⎢‬ ‫‪i‬‬ ‫⎦‬ ‫‪A‬‬ ‫‪′‬‬ ‫=‬ ‫‪a‬‬ ‫⎣‬ ‫‪−‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫ﺇﻻ ﺠﻤﻠﺔ ﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ‬ ‫ﻨﻼﺤﻅ ﻤﻥ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺴﺎﺒﻘﺔ ﺃﻥ ﺠﻤﻠﺔ ﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ ﻤﺎ ﻫﻲ‬ ‫ﺍﻟﻤﺩﺓ ﻤﻀﺭﻭﺒﺔ ﻓﻲ ﺍﻟﻤﻘﺩﺍﺭ ) ‪(1 + i‬‬ ‫ﺃﻱ ‪A ′ = A (1 + i ) :‬‬ ‫ﺘﻁﺒﻴﻕ ‪:‬‬‫ﻭﻅﻑ ﺸﺨﺹ ﻓﻲ ﺒﺩﺍﻴﺔ ﻜل ﺴﻨﺔ ﻤﺒﻠﻎ ‪ 10.000 DA‬ﻟﻤﺩﺓ ‪ 10‬ﺴﻨﻭﺍﺕ‪ ،‬ﺤﻴﺙ ﺍﻟﻔﻭﺍﺌﺩ ﺘﺤﺴﺏ ﻓـﻲ ﻨﻬﺎﻴـﺔ‬ ‫ﺍﻟﺴﻨﺔ‪ .‬ﻤﺎ ﻫﻭ ﻤﻘﺩﺍﺭ ﺍﻟﻤﺒﻠﻎ ﺍﻟﺫﻱ ﺘﺤﺼل ﻋﻠﻴﻪ ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻤﻌﺩل ‪ %7‬ﺴﻨﻭﻴﹰﺎ‪.‬‬ ‫ﺍﻟﺤل‪:‬‬ ‫ﺍﻟﺠﻤﻠﺔ ﺍﻟﺘﻲ ﺤﺼل ﻋﻠﻴﻬﺎ ﺍﻟﺸﺨﺹ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟـ ‪ 10‬ﺴﻨﻭﺍﺕ ﻫﻲ‪:‬‬ ‫‪A′ = a (1 + i ) (1+i )n −1‬‬ ‫‪i‬‬ ‫‪A′=10.000(1,07)(1,007,)0170 −1‬‬ ‫‪A ′ =10.000×1,07×13,816447‬‬ ‫‪A′=147.835,98DA‬‬ ‫ﺏ ـ ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ )‪(a‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪A = a × (1 + i )n − 1‬‬ ‫‪i‬‬ ‫‪a‬‬ ‫=‬ ‫‪A‬‬ ‫×‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪−1‬‬ ‫‪i )n‬‬ ‫ﺘﻁﺒﻴﻕ ‪:‬‬‫ﺠﻤﻠﺔ ‪ 14‬ﺩﻓﻌﺔ ﺜﺎﺒﺘﺔ ﺩﻓﻌﺕ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﻫﻲ ‪ 2.421.492 DA‬ﻭ ﺍﻟﻤﻌﺩل ﺍﻟﺴﻨﻭﻱ ﻫﻭ ‪ .%8‬ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ‬ ‫ﺍﻟﺩﻓﻌﺔ ؟‪.‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪2.421.492‬‬ ‫=‬ ‫‪a‬‬ ‫×‬ ‫‪(1,‬‬ ‫‪0 8 )14‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪0, 08‬‬ ‫‪a‬‬ ‫=‬ ‫× ‪2.421.492‬‬ ‫‪0, 08‬‬ ‫‪(1, 08)14 − 1‬‬ ‫‪a = 2.421.492 × 0, 041297‬‬ ‫‪a = 100.000DA‬‬ ‫ﺠـ ـ ﺤﺴﺎﺏ ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ ) ‪(n‬‬ ‫ﻣﻦ أﺟﻞ ﺣﺴﺎب ﻋﺪد اﻟﺪﻓﻌﺎت ﻧﺴﺘﺨﺪم ﺧﻮاص اﻟﻤﻌﺎدﻻت اﻟﻠﻮﻏﺎﺭﻴﺘﻤﻴﺔ ﻜﻤﺎ ﻴﻠﻲ‪:‬‬ ‫‪A = a × (1 + i )n − 1‬‬ ‫‪i‬‬ ‫‪(1 + i )n − 1 = A‬‬ ‫‪ia‬‬ ‫‪(1 + i )n − 1 = i .A‬‬ ‫‪a‬‬ ‫‪(1 + i )n = i .A + 1‬‬ ‫‪a‬‬ ‫ﻴﻜﻭﻥ ﻤﻌﻠﻭﻤﹰﺎ ﻭ ﻟﻴﻜﻥ )‪ (b‬ﻓﺈﻥ ‪:‬‬ ‫‪i .A‬‬ ‫‪+1‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ﺍﻟﻤﻘﺩﺍﺭ‬ ‫‪a‬‬ ‫‪(1 + i )n = b‬‬ ‫) ‪log(1 + i )n = log(b‬‬ ‫) ‪n . log(1 + i ) = log(b‬‬ ‫‪n‬‬ ‫=‬ ‫) ‪log(b‬‬ ‫) ‪log(1 + i‬‬ ‫ﺘﻁﺒﻴﻕ ‪:‬‬‫ﻜﻡ ﺩﻓﻌﺔ ﺫﺍﺕ ‪ 20.000 DA‬ﻟﻠﺩﻓﻌﺔ ﻴﺠﺏ ﺃﻥ ﺘﺩﻓﻊ ﻓﻲ ﻨﻬﺎﻴﺔ ﻜل ﺴﻨﺔ ﻟﻠﺤﺼﻭل ﻋﻠـﻰ ﺠﻤﻠـﺔ ﻗـﺩﺭﻫﺎ ‪DA‬‬ ‫‪ 300.516,10‬ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻤﻌﺩل ﺍﻟﺴﻨﻭﻱ ﺍﻟﻤﻁﺒﻕ ‪.%4‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺍﻟﺤل‪:‬‬ ‫‪300.516,10‬‬ ‫=‬ ‫× ‪20.000‬‬ ‫‪(1, 04)n −‬‬ ‫‪1‬‬ ‫‪0, 04‬‬ ‫‪(1, 04)n − 1‬‬ ‫=‬ ‫‪300.516,10‬‬ ‫‪0, 04‬‬ ‫‪20.000‬‬ ‫‪(1, 04)n‬‬ ‫‪−1‬‬ ‫=‬ ‫‪0, 04 × 300.516,10‬‬ ‫‪20.000‬‬ ‫‪(1 + i )n‬‬ ‫=‬ ‫‪0, 04 × 300.516,10‬‬ ‫‪+1‬‬ ‫‪20.000‬‬ ‫‪(1, 04)n = 0,6010322‬‬ ‫) ‪log(1,04 )n = log(0,6010322‬‬ ‫)‪n . log(1,04)= log(0,6010322‬‬ ‫) ‪log( 0,6010322‬‬ ‫)‪n = log(1,04‬‬ ‫‪n = 12‬‬ ‫ﺃﻱ ‪ 12‬ﺩﻓﻌﺔ‬ ‫ﺩ ـ ﺤﺴﺎﺏ ﺍﻟﻤﻌﺩل ) ‪(i‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪A = a × (1 + i )n − 1‬‬ ‫‪i‬‬ ‫‪(1 + i )n − 1 = A‬‬ ‫‪ia‬‬ ‫ﺒﻤﺎ ﺃﻥ ﺍﻟﻌﻨﺎﺼﺭ ‪ n ,a , A‬ﻤﻌﻠﻭﻤﺔ ﻓﺈﻥ ﺍﻟﻌﻨﺼﺭ ﺍﻟﺫﻱ ﺴﻨﺒﺤﺙ ﻋﻨﻪ ﻫﻭ ﺍﻟﻤﻌﺩل‬ ‫) ‪ (i‬ﻭ ﻤﻥ ﺤل ﻫﺫﻩ ﺍﻟﻤﻌﺎﺩﻟﺔ ﻴﻌﺘﻤﺩ ﻋﻠﻰ ﻁﺭﻴﻘﺘﻴﻥ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺍﻟﻁﺭﻴﻘﺔ )‪ : (1‬ﻁﺭﻴﻘﺔ ﺍﻟﺘﺠﺭﺒﺔ ﺃﻭ ﺍﻟﺘﻜﺭﺍﺭ)‪: (Itération‬‬ ‫ﺤﺘﻰ ﻨﺤﺼل ﻋﻠﻰ ﻗﻴﻤﺔ ﻤﺴـﺎﻭﻴﺔ ﻟﻘﻴﻤـﺔ‬ ‫) ‪(i‬‬ ‫ﺒﺈﻋﻁﺎﺀ ﻗﻴﻡ ﻟـ‬ ‫‪(1 + i )n − 1‬‬ ‫ﺃﻱ ﺤﺴﺎﺏ ﺍﻟﻤﻘﺩﺍﺭ‬ ‫‪i‬‬ ‫ﻤﺴﺘﺨﺩﻤﺔ ﻜﺜﻴـﺭﺍ ﻓـﻲ‬ ‫ﺍﻟﻁﺭﻴﻘﺔ‬ ‫) ‪ (i‬ﻫﻲ ﺍﻟﻤﻌﺩل‪ .‬ﻫﺫﻩ‬ ‫ﺍﻟﻤﻌﻁﺎﺓ ﻟـ‬ ‫ﺍﻟﻘﻴﻤﺔ‬ ‫ﺘﻜﻭﻥ ﺒﺫﻟﻙ‬ ‫ﻭ‬ ‫)‪(A‬‬ ‫ﺍﻟﻤﻘﺩﺍﺭ‬ ‫‪a‬‬ ‫ﺒﺭﺍﻤﺞ ﺍﻟﺤﺎﺴﻭﺏ ﻤﺜل ﺍﻟﻤﺠﺩﻭل ‪ Excel‬ﺍﻟﺫﻱ ﻴﻘﻭﻡ ﺒﺤﺴﺎﺏ ﺍﻟﻤﻌﺩل ﺒﺘﻜﺭﺍﺭ ﺍﻟﺘﺠﺭﺒﺔ‪.‬‬ ‫ﺍﻟﻁﺭﻴﻘﺔ ﺍﻟﺜﺎﻨﻴﺔ ‪ :‬ﻁﺭﻴﻘﺔ ﺍﻻﺴﺘﻜﻤﺎل ﺍﻟﺨﻁﻲ)‪(Interpolation linière‬‬ ‫ﺒﻤﻌﺭﻓﺔ ﻤﻌﺩﻟﻴﻥ ﻴﻜﻭﻥ ﺍﻟﻤﻌﺩل ) ‪ (i‬ﻤﺤﺼﻭﺭﹰﺍ ﺒﻴﻨﻬﻤﺎ‪ ،‬ﻴﻤﻜﻥ ﺤﺴﺎﺏ ﺍﻟﻤﻌﺩل ﺒﺎﺴﺘﺨﺩﺍﻡ ﺒﺎﺴﺘﻜﻤﺎل ﻟﻸﺠﺯﺍﺀ‬ ‫ﺍﻟﻤﺘﻨﺎﺴﺒﺔ ﻜﻤﺎ ﻴﻭﻀﺤﻬﺎ ﺍﻟﺸﻜل ﺍﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪i0 < i < i1‬‬ ‫ﺍﻟﻔﺭﻕ= ) ‪(i1 − i0‬‬ ‫اﻟﻔﺮق= ) ‪ii 0 (i − i 0‬‬ ‫‪i1‬‬ ‫اﻟﻔﺮق= ) ‪(Ai − Ai0‬‬ ‫‪(1 + i 0 )n‬‬ ‫‪−1‬‬ ‫‪Ai‬‬ ‫=‬ ‫‪(1 + i )n‬‬ ‫‪−1‬‬ ‫‪i0‬‬ ‫‪i‬‬ ‫‪Ai 0‬‬ ‫=‬ ‫‪Ai1‬‬ ‫=‬ ‫‪(1 + i1 )n‬‬ ‫‪−1‬‬ ‫‪i1‬‬ ‫‪( A − A )( A i1‬اﻟﻔﺮق=‬ ‫ﺒﺘﻁﺒﻴﻕ ﺍﻟﻘﺎﻋﺩﺓ ﺍﻟﺜﻼﺜﻴﺔ ‪:‬‬ ‫)‪i1‬‬ ‫) ‪(ii0 1 − i 0‬‬ ‫‪−‬‬ ‫‪A‬‬ ‫‪i‬‬ ‫→‬ ‫‪0‬‬ ‫) ‪(A i − A i0 ) → (i − i 0‬‬ ‫‪(i‬‬ ‫= )‪− i0‬‬ ‫‪(Ai‬‬ ‫) ‪− A i0 ) × (i 1 − i 0‬‬ ‫) ‪( A i1 − A i0‬‬‫‪i‬‬ ‫=‬ ‫‪i0 +‬‬ ‫‪(Ai‬‬ ‫) ‪− Ai0 )× (i1 − i 0‬‬ ‫) ‪(Ai1 − Ai0‬‬ ‫ﺘﻁﺒﻴﻕ‪:‬‬ ‫ﺠﻤﻠﺔ ‪ 8‬ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﺘﻘﺩﺭ ﺒـ ‪ 16.740 DA‬ﻭ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ‪ . 1.200DA‬ﻤﺎ ﻫﻭ ﺍﻟﻤﻌﺩل ﺍﻟﻤﻁﺒﻕ‬ ‫ﻤﻊ ﺍﻟﻌﻠﻡ ﺃﻨﻪ ﻤﺤﺼﻭﺭ ﺒﻴﻥ ﻤﻌﺩل ‪ %15‬ﻭ ‪.%16‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺍﻟﺤل‬ ‫ﻤﻥ ﺍﻟﻤﻌﻁﻴﺎﺕ ‪%15 < i < 16 :‬‬ ‫‪16.740 = 1.200 × (1+ i )8 −1‬‬ ‫‪i‬‬ ‫‪(1‬‬ ‫‪+‬‬ ‫‪i )8‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫=‬ ‫‪16.740‬‬ ‫‪i‬‬ ‫‪1.200‬‬ ‫‪(1+ i )8 −1 = 13, 95‬‬ ‫‪i‬‬ ‫‪Ai‬‬ ‫‪= (1+ i )8 −1 = 13, 95‬‬ ‫ﺇﺫﻥ ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪i‬‬ ‫ﻨﺤﺴﺏ ﺍﻟﺠﻤﻠﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﻗﻴﻤﺔ ﻜل ﺩﻓﻌﺔ ‪ 1 DA‬ﺒﺎﻟﻤﻌﺩﻟﻴﻥ ‪ %15‬ﻭ‪.%16‬‬ ‫‪A i15‬‬ ‫=‬ ‫‪(1,15)8 − 1‬‬ ‫‪= 13,726819‬‬ ‫‪0,15‬‬ ‫‪A i16‬‬ ‫=‬ ‫‪(1,16)8 − 1‬‬ ‫‪= 14, 240093‬‬ ‫‪0,16‬‬ ‫ﻨﻁﺒﻕ ﻋﻼﻗﺔ ﺍﻻﺴﺘﻜﻤﺎل ﺍﻟﺨﻁﻲ ﻟﺤﺴﺎﺏ ﺍﻟﻤﻌﺩل‪:‬‬ ‫‪i‬‬ ‫‪= i0‬‬ ‫‪+‬‬ ‫‪( Ai‬‬ ‫) ‪− Ai 0 )×( i1 − i 0‬‬ ‫) ‪(Ai1 − Ai0‬‬ ‫‪i‬‬ ‫=‬ ‫‪0‬‬ ‫‪,15 +‬‬ ‫‪(13‬‬ ‫) ‪,95 − 13,726819 )( 0,16 − 0,15‬‬ ‫) ‪(14 ,240093 − 13 ,726819‬‬ ‫‪i = 0 ,15 + 0 ,00438‬‬ ‫‪i = 0 ,15438‬‬ ‫‪i = 15, 438%‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪ .3‬ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ‬ ‫‪ .1.3‬ﺍﻟﺼﻴﻐﺔ ﺍﻟﻌﺎﻤﺔ ﻟﻠﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‬‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﺘﻌﻨﻲ ﺍﻟﻌﻭﺩﺓ ﺒﺎﻟﺩﻓﻌﺎﺕ ﺇﻟﻰ ﺘﺎﺭﻴﺦ ﺘﻭﻗﻴﻊ ﺍﻟﻌﻘﺩ )ﺍﻟﺘﺎﺭﻴﺦ ﺼﻔﺭ( ﺃﻱ‬ ‫ﻓﺘﺭﺓ ﻗﺒل ﺩﻓﻊ ﺍﻟﺩﻓﻌﺔ ﺍﻷﻭﻟﻰ‪.‬‬ ‫ﻭ ﻨﺭﻤﺯ ﺇﻟﻰ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻓﻲ ﺍﻟﺘﺎﺭﻴﺦ ﺼﻔﺭ ﺒﺎﻟﺭﻤﺯ ) ‪(V 0‬‬‫‪012345‬‬ ‫‪p‬‬ ‫‪n-1 n‬‬‫‪aaaaa‬‬ ‫‪a‬‬ ‫اﻟﻤﺪة ‪a a‬‬‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﺘﺴﺎﻭﻱ ﻤﺠﻤﻭﻉ ﺍﻟﻘﻴﻡ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻜل ﺩﻓﻌﺔ ﻤﻥ ﺍﻟﺩﻓﻌﺎﺕ ‪V0‬‬ ‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‬ ‫ﺍﻟﺩﻓﻌﺔ‬ ‫‪a (1 + i )−1‬‬ ‫‪1‬‬ ‫‪a (1 + i )−2‬‬ ‫‪2‬‬ ‫‪a (1 + i )−3‬‬ ‫‪3‬‬ ‫‪.................................................‬‬ ‫)‪a(1 + i )−(n −1‬‬ ‫‪n −1‬‬ ‫‪a (1 + i )−n‬‬ ‫‪n‬‬ ‫ﻭ ﻤﻨﻪ ﻨﻜﺘﺏ ‪:‬‬‫‪V0 = a (1 + i )−n + a (1 + i )−(n −1) + ... + a (1 + i )−3 + ... + a (1 + i )−2 + a (1 + i )−1‬‬‫ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﻘﻴﻡ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻠﺩﻓﻌﺎﺕ ﺘﺸﻜل ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل ‪ a(1 + i )−n‬ﻭ ﺤﺩﻫﺎ ﺍﻷﺨﻴﺭ‬ ‫‪ a(1 + i )−1‬ﻭ ﺃﺴﺎﺴﻬﺎ ) ‪ (1 + i‬ﻭ ﺤﺴﺏ ﻗﺎﻨﻭﻥ ﻤﺠﻤﻭﻉ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻓﺈﻥ ‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a(1 +‬‬ ‫‪i‬‬ ‫‪)−1 (1 + i ) − a(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪(1 + i ) − 1‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ‪(1 + i )−1 (1 + i ) = 1 :‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪×1−‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫ﻓﺈﻥ ‪:‬‬ ‫‪i‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪ ‬ﺍﺴﺘﻨﺘﺎﺝ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻓﻲ ﺤﺎﻟﺔ ﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ‬‫‪0 1 23 4 5‬‬ ‫‪p‬‬ ‫‪n-1 n‬‬‫‪a‬‬ ‫‪a‬‬ ‫‪a‬‬ ‫‪a‬‬ ‫‪aa‬‬ ‫‪a‬‬ ‫اﻟﻤﺪة ‪a‬‬ ‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ‪ V ′ 0‬ﺘﺴﺎﻭﻱ ﻤﺠﻤﻭﻉ ﺍﻟﻘﻴﻡ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺩﻓﻌﺎﺕ‪.‬‬‫)‪′V V0 =′ 0a + a (1 + i )−1 + a (1 + i )−2 + a (1 + i )−3 + ... + a (1 + i )− (n −1‬‬ ‫ﻭ ﺤﺴﺏ ﻗﺎﻨﻭﻥ‬ ‫ﻨﻼﺤﻅ ﺃﻥ ‪ V ′ 0‬ﺘﺸﻜل ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل ‪ a‬ﻭ ﺃﺴﺎﺴﻬﺎ ‪(1 + i )−1‬‬ ‫ﻤﺠﻤﻭﻉ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻬﻨﺩﺴﻴﺔ ﺍﻟﺘﺎﻟﻲ ‪:‬‬ ‫‪S=a‬‬ ‫×‬ ‫‪qn -1‬‬ ‫‪,‬‬ ‫‪q-1‬‬ ‫ﺤﻴﺙ )‪ (a‬ﻫﻭ ﺍﻟﺤﺩ ﺍﻷﻭل )‪ (q‬ﻫﻭ ﺍﻷﺴﺎﺱ ﻭ )‪ (n‬ﻋﺩﺩ ﺍﻟﺤﺩﻭﺩ ﻓﺈﻥ‪:‬‬ ‫‪′V‬‬ ‫‪0‬‬ ‫=‬ ‫‪a‬‬ ‫×‬ ‫‪((1 + i )−1 )n −‬‬ ‫‪1‬‬ ‫‪(1 + i )−1 − 1‬‬ ‫ﺒﻌﺩ ﺘﺒﺴﻴﻁ ﺍﻟﻌﻼﻗﺔ ﻨﺤﺼل ﻋﻠﻰ‪:‬‬ ‫‪′V 0 = a(1 + i ) 1 − (1 + i )−n‬‬ ‫‪i‬‬ ‫ﺃﻱ ﺃﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ ﺘﺴﺎﻭﻱ ‪:‬‬ ‫) ‪′V 0 = V 0 ( 1 + i‬‬ ‫‪ .2.3‬ﺍﺴﺘﻌﻤﺎل ﺍﻟﺼﻴﻐﺔ ﺍﻟﻌﺎﻤﺔ ﻟﻠﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‬ ‫ﺃ ـ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺘﻁﺒﻴﻕ)‪:(1‬‬‫ﻤﺎ ﻫﻲ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺠﻤﻭﻉ ‪ 8‬ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﺤﻴﺙ ﻜـل ﺩﻓﻌـﺔ ﺘﺴـﺎﻭﻱ ‪120.000 DA‬‬ ‫ﺒﻤﻌﺩل ‪ %7‬ﺴﻨﻭﻴﹰﺎ‪.‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫×‬ ‫‪1 − (1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬ ‫‪V0‬‬ ‫=‬ ‫× ‪120.000‬‬ ‫‪1−‬‬ ‫‪(1, 07 )−8‬‬ ‫‪0, 08‬‬ ‫‪V 0 = 120.000 × 5, 971298‬‬ ‫‪V 0 = 716.555, 8 DA‬‬ ‫ﺘﻁﺒﻴﻕ )‪: (2‬‬‫ﻤﺎ ﻫﻲ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟـ ‪ 6‬ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﻓﻲ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ ﻗﻴﻤﺔ ﻜل ﺩﻓﻌﺔ ‪ 50.000DA‬ﺒﻤﻌﺩل ‪ %4‬ﺴﻨﻭﻴﹰﺎ‪.‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫‪′V 0 = a ( 1 + i ) 1 − ( 1 + i ) − n‬‬ ‫‪i‬‬ ‫‪′V‬‬ ‫‪0‬‬ ‫=‬ ‫‪50.000(1, 04 ) 1 −‬‬ ‫‪(1, 04 )−6‬‬ ‫‪0,04‬‬ ‫‪V ′ 0 = 50.000 × 1, 04 × 5, 242136‬‬ ‫‪′V 0 = 2 7 2 .5 9 1 , 1 0 D A‬‬ ‫ﺏ ـ ﺤﺴﺎﺏ ﺍﻟﺩﻓﻌﺔ )‪(a‬‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫×‬ ‫‪1−‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬ ‫‪a‬‬ ‫‪=V0‬‬ ‫×‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪i‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪(1 +‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺘﻁﺒﻴﻕ‪:‬‬‫ﻤﺘﺘﺎﻟﻴﺔ ‪ 8‬ﺩﻓﻌﺎﺕ ﻤﻘﻴﻤﺔ ﺴﻨﺔ ﻗﺒل ﺍﻟﺩﻓﻌﺔ ﺍﻷﻭﻟﻰ ﺒﻤﻌﺩل ‪ %3,5‬ﻫﻲ ‪ . 1.374.791DA‬ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ‬ ‫ﺍﻟﺜﺎﺒﺘﺔ ؟‬ ‫ﺍﻟﺤل‪:‬‬ ‫‪a‬‬ ‫× ‪=V 0‬‬ ‫‪i‬‬ ‫‪i )−n‬‬ ‫‪1 − (1 +‬‬ ‫‪a‬‬ ‫=‬ ‫× ‪1.374.791‬‬ ‫‪0;035‬‬ ‫‪1−(1,035)−8‬‬ ‫‪a = 1.374.791 × 0, 1454766  200.000DA‬‬ ‫ﺠـ ـ ﺤﺴﺎﺏ ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ )‪(n‬‬ ‫ﻤﻥ ﺃﺠل ﺤﺴﺎﺏ ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ ﻨﺴﺘﺨﺩﻡ ﺨﻭﺍﺹ ﺍﻟﻤﻌﺎﺩﻻﺕ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻤﻴﺔ ﻜﻤﺎ ﻴﻠﻲ‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪× 1 − (1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬ ‫‪1 − (1 + i )−n = V 0‬‬ ‫‪ia‬‬ ‫‪1 − (1 + i )−n = i ×V 0‬‬ ‫‪a‬‬ ‫‪−(1 + i )−n = i ×V 0 − 1‬‬ ‫‪a‬‬ ‫‪(1 + i )−n = 1 − i ×V 0‬‬ ‫‪a‬‬ ‫ﺍﻟﻤﻘﺩﺍﺭ ‪ 1 − i ×V 0‬ﻴﻤﻜﻥ ﺤﺴﺎﺒﻪ ﻭ ﻟﻴﻜﻥ ) ‪(b‬‬ ‫‪a‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪(1 + i )−n = b‬‬ ‫) ‪log(1 + i )−n = log(b‬‬ ‫) ‪−n log(1 + i ) = log(b‬‬ ‫‪−n‬‬ ‫=‬ ‫) ‪log(b‬‬ ‫)‬ ‫‪log(1 + i‬‬ ‫‪n‬‬ ‫=‬ ‫‪−‬‬ ‫) ‪log(b‬‬ ‫)‬ ‫‪log(1 + i‬‬ ‫ﺘﻁﺒﻴﻕ‪:‬‬‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﻫﻲ ‪ 200.000 DA‬ﻭ ﻗﻴﻤﺔ ﻜل ﺩﻓﻌﺔ ‪. 20.028,80 DA‬‬ ‫ﺃﺤﺴﺏ ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ ﻋﻠﻤﹰﺎ ﺃﻥ ﺍﻟﻤﻌﺩل ﺍﻟﺴﻨﻭﻱ ﺍﻟﻤﻁﺒﻕ ﻫﻭ ‪.%4‬‬ ‫ﺍﻟﺤل‪:‬‬ ‫× ‪V0 = a‬‬ ‫‪1−(1+ i )− n‬‬ ‫‪i‬‬ ‫‪2‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪.0‬‬ ‫‪00‬‬ ‫‪2‬‬ ‫‪0‬‬ ‫‪.0‬‬ ‫‪2‬‬ ‫‪8‬‬ ‫‪,‬‬ ‫×‪8‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫(‬ ‫) ‪1,04‬‬ ‫‪−‬‬ ‫‪n‬‬ ‫‪0,04‬‬ ‫=‬ ‫‪1−‬‬ ‫‪(1,04‬‬ ‫‪)−‬‬ ‫‪n‬‬ ‫=‬ ‫‪0,04× 200.000‬‬ ‫‪2 0 .0 2 8 ,8‬‬ ‫‪−(1,04)− n‬‬ ‫=‬ ‫‪0,04× 200.000‬‬ ‫‪−1‬‬ ‫‪2 0 .0 2 8 ,8‬‬ ‫‪(1, 04)−n‬‬ ‫=‬ ‫‪1−‬‬ ‫‪0,04× 200.000‬‬ ‫‪2 0 .0 2 8 ,8‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪(1, 04)− n = 0,600575‬‬ ‫)‪log(1,04)− n = log(0,600575‬‬ ‫)‪−n log(1,04)= log(0,600575‬‬ ‫‪−‬‬ ‫‪n‬‬ ‫=‬ ‫‪log(0,600575‬‬ ‫)‬ ‫) ‪log(1,04‬‬ ‫) ‪log( 0,600575‬‬ ‫)‪n = − log(1,04‬‬ ‫‪n =13‬‬ ‫ﺩ ـ ﺤﺴﺎﺏ ﺍﻟﻤﻌﺩل ) ‪:( i‬‬‫ﻴﺘﻡ ﺍﻟﻠﺠﻭﺀ ﺇﻟﻰ ﻁﺭﻴﻘﺔ ﺍﻟﺘﺠﺭﻴﺏ ﻤﻥ ﺃﺠل ﺤﺼﺭ ﺍﻟﻤﻌﺩل ﺒﻴﻥ ﻤﻌﺩﻟﻴﻥ ﻤﺘﻘﺎﺭﺒﻴﻥ ﺃﻭ ﹰﻻ ﺘـﻡ ﺘﺤﺩﻴـﺩﻩ ﺜﺎﻨﻴـﹰﺎ‬ ‫ﺒﺎﺴﺘﺨﺩﺍﻡ ﻁﺭﻴﻘﺔ ﺍﻻﺴﺘﻜﻤﺎل ﺍﻟﺨﻁﻲ ﻟﻸﺠﺯﺍﺀ ﺍﻟﻤﺘﻨﺎﺴﺒﺔ‪.‬‬ ‫ﺘﻁﺒﻴﻕ ‪:‬‬‫ﻗﻴﻤﺔ ﻤﺤل ﺘﺠﺎﺭﻱ ﻨﻘﺩﹰﺍ ‪ 1.880.000 DA‬ﻭ ﻫﻲ ﺘﻤﺜل ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ‪ 10‬ﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﻗﻴﻤـﺔ‬ ‫ﻜل ﺩﻓﻌﺔ ‪. 240.000 DA‬‬ ‫ﻤﺎ ﻫﻭ ﺍﻟﻤﻌﺩل ﺍﻟﻤﻁﺒﻕ ؟‪.‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫× ‪V0 = a‬‬ ‫‪1−(1+i )−n‬‬ ‫‪i‬‬ ‫‪1.880.000= 240.000×1−(1+ i )−10‬‬ ‫‪i‬‬ ‫‪1−‬‬ ‫‪(1+ i‬‬ ‫‪)−10‬‬ ‫=‬ ‫‪1.880.000‬‬ ‫‪i‬‬ ‫‪240.000‬‬ ‫‪1−(1+ i )−10 = 7,833333‬‬ ‫‪i‬‬ ‫ﺒﻌﺩ ﺘﺠﺭﺒﺔ ﻋﺩﺓ ﻤﻌﺩﻻﺕ ﻨﺠﺩ ﺃﻥ ﺍﻟﻤﻌﺩل )‪ (i‬ﻤﺤﺼﻭﺭﹰﺍ ﺒﻴﻥ ﺍﻟﻤﻌﺩﻟﻴﻥ ‪ %4,5‬ﻭ ‪.%4,75‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺍﻟﻔﻭﺍﺭﻕ‬ ‫‪0,079385‬‬ ‫‪7,816347‬‬ ‫ﻤﻥ ﺍﺠل ﻤﻌﺩل ‪%4,75‬‬‫‪0,096371‬‬ ‫‪7,833333‬‬ ‫ﻤﻥ ﺃﺠل ﻤﻌﺩل ) ‪(i‬‬ ‫‪7,912718‬‬ ‫ﻤﻥ ﺃﺠل ﻤﻌﺩل‪%4,5‬‬ ‫ﺒﺘﻁﺒﻴﻕ ﺍﻟﻘﺎﻋﺩﺓ ﺍﻟﺜﻼﺜﻴﺔ ﻨﺠﺩ‪:‬‬ ‫‪(0,0475 - 0,045) → 0,096371‬‬ ‫‪(i - 0,045) → 0,079385‬‬ ‫‪i‬‬ ‫‪- 0,045‬‬ ‫=‬ ‫)‪0,079385 × (0,0475 - 0,045‬‬ ‫‪0, 096371‬‬ ‫‪i‬‬ ‫=‬ ‫‪0,045 +‬‬ ‫‪0,079385 × 0,0025‬‬ ‫‪0, 096371‬‬ ‫‪i = 0,045 + 0,00205‬‬ ‫‪i = 0,04705‬‬ ‫‪i = 4,705%‬‬ ‫ﻤﻼﺤﻅﺔ‪:‬‬‫ﺒﺎﻟﻨﺴﺒﺔ ﻟﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ ﻓﺈﻥ ﺤﺴﺎﺏ ﻤﺨﺘﻠﻑ ﺍﻟﻌﻨﺎﺼﺭ )‪ (i , n ,a,V 0′‬ﻴﺘﻡ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤـﺔ‬ ‫ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ )‪ (V 0′‬ﺒﻨﻔﺱ ﺍﻟﻜﻴﻔﻴﺔ ﺍﻟﺴﺎﺒﻘﺔ‪.‬‬ ‫‪ .4‬ﺘﻘﻴﻴﻡ ﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ‬‫‪j‬‬ ‫‪0123‬‬ ‫‪p n-1 n‬‬ ‫‪m‬‬ ‫اﻟﻤﺪة‬ ‫‪aa a a‬‬ ‫‪aa‬‬‫‪V−j V0‬‬ ‫‪V p An Am‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫ﻋﻨﺩ ﺩﺭﺍﺴﺘﻨﺎ ﻟﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﻗﻴ ‪‬ﻤﻨﺎ ﻫﺫﻩ ﺍﻷﺨﻴﺭﺓ ﻓﻲ ﺍﻟﺯﻤﻥ )‪ (0‬ﻭ ﺍﻟﺯﻤﻥ )‪ (n‬ﺒﺤﻴﺙ ﺤﺼﻠﻨﺎ ﻋﻠﻰ‬ ‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ‪ V0‬ﻭ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ‪ An‬ﻜﻤﺎ ﻴﻠﻲ‪:‬‬‫‪An‬‬ ‫=‬ ‫‪a‬‬ ‫‪(1 + i )n‬‬ ‫‪−1‬‬ ‫و‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪1−‬‬ ‫‪(1 +‬‬ ‫‪i )−n‬‬ ‫‪i‬‬ ‫‪i‬‬ ‫ﻭ ﻨﻔﺱ ﺍﻟﺸﻲﺀ ﻓﻴﻤﺎ ﻴﺘﻌﻠﻕ ﺒﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ‪.‬‬‫ﻭ ﺒﺘﻁﺒﻴﻕ ﻤﺒﺩﺃ ﺍﻟﺘﻜﺎﻓﺅ ﻴﻤﻜﻥ ﺃﻥ ﻨﻘﻴﻡ ﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﻓﻲ ﺃﻱ ﺘﺎﺭﻴﺦ ﻜﺎﻥ‪ .‬ﻭﺴﻨﻘﺘﺼﺭ ﻓﻲ ﺍﻟﺘﻘﻴﻴﻡ ﻋﻠﻰ ﺩﻓﻌﺎﺕ‬ ‫ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﻤﻊ ﺍﻹﺸﺎﺭﺓ ﺇﻟﻰ ﺃﻥ ﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ ﺘﻌﺎﻟﺞ ﺒﻨﻔﺱ ﺍﻟﻜﻴﻔﻴﺔ‪.‬‬ ‫‪ .1.4‬ﺍﻟﺘﻘﻴﻴﻡ ﻓﻲ ﺍﻟﺯﻤﻥ ) ‪ ( j‬ﺤﻴﺙ )‪( j < 0‬‬ ‫ﻨﻀﻊ ‪ V − j‬ﻗﻴﻤﺔ ﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﺍﻟﺯﻤﻥ ‪. j‬‬ ‫ﻴﻤﻜﻥ ﺘﺤﺩﻴﺩ ﻗﻴﻤﺔ ‪ V − j‬ﻜﻤﺎ ﻴﻠﻲ‪:‬‬ ‫ﺃ ـ ﺍﻨﻁﻼﻗﹰﺎ ﻤﻥ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‪:‬‬ ‫‪V − j = V 0 (1 + i )− j‬‬ ‫‪V−j‬‬ ‫‪= a 1 − (1 + i )−n‬‬ ‫‪(1 + i )− j‬‬ ‫‪i‬‬ ‫ﺏ ـ ﺍﻨﻁﻼﻗﹰﺎ ﻤﻥ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ‪:‬‬ ‫) ‪V − j = An (1 + i )− (n + j‬‬ ‫‪V −j‬‬ ‫‪= a (1 + i )n‬‬ ‫) ‪− 1 (1 + i )− (n + j‬‬ ‫‪i‬‬ ‫ﺘﻁﺒﻴﻕ ‪:‬‬‫ﻗﺎﻤﺕ ﺍﻟﻤﺅﺴﺴﺔ )ﺱ( ﺒﺸﺭﺍﺀ ﻤﺨﺯﻥ ﺘﺴﺩﺩ ﺜﻤﻨﻪ ﺒﻭﺍﺴﻁﺔ ‪ 6‬ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ‪ 15.000 DA‬ﻭ ﺍﻟﺩﻓﻌﺔ‬ ‫ﺍﻷﻭل ﺘﺴﺘﺤﻕ ﺴﻨﺔ ﺒﻌﺩ ﺍﺴﺘﻼﻡ ﺍﻟﻤﺨﺯﻥ‪.‬‬ ‫ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺎﺕ ‪ 3‬ﺴﻨﻭﺍﺕ ﻗﺒل ﺍﻟﺩﻓﻊ ﺍﻷﻭل ﻋﻠﻤﹰﺎ ﺃﻥ ﺍﻟﻤﻌﺩل ﻫﻭ ‪ %7‬؟‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪-2 -1‬‬ ‫ﺍﻟﺤل‪:‬‬ ‫‪0 1 2 3 4 56‬‬ ‫اﻟﻤﺪة‬ ‫اﻟﺪﻓﻊ اﻷول ‪V −2‬‬‫ﻭ ‪ 3‬ﺴﻨﻭﺍﺕ ﻗﺒـل ﺍﻟـﺩﻓﻊ‬ ‫ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﻤﺩﺓ ﺍﻟﻔﺎﺼﻠﺔ ﺒﻴﻥ ﺘﺎﺭﻴﺦ ﺍﺴﺘﻼﻡ ﺍﻟﻤﺨﺯﻥ )ﺍﻟﺯﻤﻥ ﺼﻔﺭ(‬ ‫ﺍﻷﻭل ﻫﻲ ﺴﻨﺘﻴﻥ‪.‬‬ ‫‪V j = V 0 (1 + i )− j‬‬ ‫‪V −2‬‬ ‫=‬ ‫‪15.000 1 −‬‬ ‫‪(1, 07 )−6‬‬ ‫‪(1, 07 )−2‬‬ ‫‪0, 07‬‬ ‫‪V −2 = 15.000 × 4, 766539 × 0,873438‬‬ ‫‪V −2 = 62.449,14DA‬‬ ‫ﻨﻔﺱ ﺍﻟﻨﺘﻴﺠﺔ ﻨﺤﺼل ﻋﻠﻴﻬﺎ ﺒﺘﻁﺒﻴﻕ ﺍﻟﻌﻼﻗﺔ ‪:‬‬ ‫‪V−j‬‬ ‫‪= a (1 + i )n‬‬ ‫) ‪− 1 (1 + i )−(n + j‬‬ ‫‪i‬‬ ‫‪V −2‬‬ ‫=‬ ‫)‪15.000 (1, 07 )6 − 1 (1, 07 )−(6+ 2‬‬ ‫‪0, 07‬‬ ‫‪V −2‬‬ ‫=‬ ‫‪15.00‬‬ ‫‪0‬‬ ‫‪(1,‬‬ ‫‪07)6‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪(1,‬‬ ‫‪07‬‬ ‫‪)−8‬‬ ‫‪0, 07‬‬ ‫‪V −2 = 15.000 × 7,15329 × 0,582009‬‬ ‫‪V −2 = 62.449,14DA‬‬ ‫‪ .2.4‬ﺍﻟﺘﻘﻴﻴﻡ ﻓﻲ ﺍﻟﺯﻤﻥ ) ‪ (p‬ﺤﻴﺙ ) ‪(0 < p < n‬‬ ‫ﻨﻀﻊ ‪ V p‬ﻗﻴﻤﺔ ﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﺍﻟﺯﻤﻥ ‪. p‬‬ ‫ﻴﻤﻜﻥ ﺘﺤﺩﻴﺩ ﻗﻴﻤﺔ ‪ V p‬ﻜﻤﺎ ﻴﻠﻲ‪:‬‬ ‫ﺃ ـ ﺍﻨﻁﻼﻗﹰﺎ ﻤﻥ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪V p = V 0 (1 + i )p‬‬ ‫‪Vp‬‬ ‫‪= a 1 − (1 + i )− n‬‬ ‫‪(1 + i )p‬‬ ‫‪i‬‬ ‫ﺏ ـ ﺍﻨﻁﻼﻗﹰﺎ ﻤﻥ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ‪:‬‬ ‫) ‪V p = A n (1 + i )− (n − p‬‬ ‫‪Vp‬‬ ‫‪= a (1 + i )n‬‬ ‫) ‪− 1 (1 + i )− (n − p‬‬ ‫‪i‬‬‫ﺠـ ـ ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﻗﻴﻤﺔ ‪ V p‬ﻋﻥ ﻁﺭﻴﻕ ﺠﻤﻊ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﻟﻠﺩﻓﻌﺎﺕ ﺍﻟﻤﺴﺩﺩﺓ ﻟﻐﺎﻴﺔ ﺍﻟـﺯﻤﻥ‬ ‫)‪ (p‬ﻭ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻠﺩﻓﻌﺎﺕ ﺍﻟﻤﺘﺒﻘﻴﺔ ﺒﻌﺩ ﺍﻟﺯﻤﻥ )‪ (p‬ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬‫‪Vp‬‬ ‫‪= a (1 + i )p‬‬ ‫) ‪− 1 + a 1 − (1 + i )− (n − p‬‬ ‫‪i‬‬ ‫‪i‬‬ ‫ﺘﻁﺒﻴﻕ‪:‬‬ ‫ﺘﺭﻏﺏ ﺇﺤﺩﻯ ﺍﻟﻤﺅﺴﺴﺎﺕ ﻓﻲ ﺸﺭﺍﺀ ﺘﺠﻬﻴﺯﺍﺕ ﻭﻗﺩ ﺍﻗﺘﺭﺡ ﻋﻠﻴﻬﺎ ﺍﻟﺒﺎﺌﻊ ﻁﺭﻴﻘﺘﻴﻥ ﻟﻠﺘﺴﺩﻴﺩ‪.‬‬‫ـ ﺍﻟﻁﺭﻴﻘﺔ )‪ :(1‬ﺘﺴﺩﻴﺩ ‪ 12‬ﺩﻓﻌﺎﺕ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ‪ 300.000 DA‬ﺍﻷﻭﻟﻰ ﺘﺩﻓﻊ ﺴﻨﺔ ﺒﻌﺩ ﺘﺎﺭﻴﺦ ﺍﻟﺸﺭﺍﺀ‪،‬‬ ‫ـ ﺍﻟﻁﺭﻴﻘﺔ )‪ : (2‬ﺘﺴﺩﻴﺩ ﻤﺒﻠﻐﹰﺎ ﻭﺤﻴﺩﹰﺍ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﺴﻨﺔ ‪ 4‬ﻤﻥ ﺘﺎﺭﻴﺦ ﺍﻟﺸﺭﺍﺀ‪.‬‬ ‫ﺇﺫﺍ ﺃﺨﺫﻨﺎ ﺒﻌﻴﻥ ﺍﻻﻋﺘﺒﺎﺭ ﺍﻟﻔﻭﺍﺌﺩ ﺍﻟﻤﺭﻜﺒﺔ ﺒﻤﻌﺩل ‪ %4‬ﺴﻨﻭﻴﹰﺎ ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻊ ﺍﻟﻭﺤﻴﺩ؟‬ ‫ﺍﻟﺤل ‪:‬‬ ‫ـ ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻊ ﺍﻟﻭﺤﻴﺩ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‪:‬‬ ‫‪V 4 = V 0 (1 + i )p‬‬ ‫‪V4‬‬ ‫=‬ ‫‪300.000 1 −‬‬ ‫‪(1, 04)−12‬‬ ‫‪(1, 04)4‬‬ ‫‪0, 04‬‬ ‫‪V 4 = 300.000 × 9, 385074 × 1,169859‬‬ ‫‪V 4 = 3.293.762, 4DA‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ـ ﻨﻔﺱ ﺍﻟﻨﺘﻴﺠﺔ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺼﻴﻐﺔ ﺍﻟﺠﻤﻠﺔ‪:‬‬ ‫) ‪V p = An (1 + i )−(n − p‬‬ ‫‪V4‬‬ ‫=‬ ‫‪300.000‬‬ ‫)‪(1, 04)12 − 1 (1, 04)−(12−4‬‬ ‫‪0, 04‬‬ ‫‪V4‬‬ ‫=‬ ‫‪300.000‬‬ ‫‪(1, 04)12 − 1 (1, 04)−8‬‬ ‫‪0, 04‬‬ ‫‪V 4 = 300.000 × 15,025805 × 0,7306902‬‬ ‫‪V 4 = 3.293.762,4‬‬‫ـ ﻴﻤﻜﻥ ﺃﻥ ﻨﺤﺼل ﻋﻠﻰ ﻨﻔﺱ ﺍﻟﻨﺘﻴﺠﺔ ﺇﺫﺍ ﺠﻤﻌﻨﺎ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﻟﻠﺩﻓﻌﺎﺕ ﺍﻟﻤﺴﺩﺩﺓ ﻟﻐﺎﻴﺔ ﺍﻟﺴﻨﺔ ‪ 4‬ﻭ ﺍﻟﻘﻴﻤﺔ‬ ‫ﺍﻟﺤﺎﻟﻴﺔ ﻟﻠﺩﻓﻌﺎﺕ ﺍﻟﻤﺘﺒﻘﻴﺔ ﺒﻌﺩ ﺍﻟﺴﻨﺔ‪ 4‬ﻜﻤﺎ ﻴﻠﻲ‪:‬‬‫‪V4‬‬ ‫=‬ ‫‪300.000 (1, 04)4 − 1‬‬ ‫‪+‬‬ ‫‪300.000 1 −‬‬ ‫)‪(1, 04)− (12− 4‬‬ ‫‪0, 04‬‬ ‫‪0, 04‬‬‫‪V4‬‬ ‫=‬ ‫‪300.000 (1, 04)4 − 1‬‬ ‫‪+‬‬ ‫‪300.000 1 −‬‬ ‫‪(1, 04)−8‬‬ ‫‪0, 04‬‬ ‫‪0, 04‬‬‫)‪V 4 = (300.000 × 4,246464)+(300.000 × 6,732744‬‬‫‪V 4 = 3.293.762,4‬‬ ‫‪ .3.4‬ﺍﻟﺘﻘﻴﻴﻡ ﻓﻲ ﺍﻟﺯﻤﻥ ) ‪ (m‬ﺤﻴﺙ )‪(m > n‬‬ ‫ﻨﻀﻊ ‪ Am‬ﻗﻴﻤﺔ ﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﺍﻟﺯﻤﻥ ‪. m‬‬ ‫ﻴﻤﻜﻥ ﺘﺤﺩﻴﺩ ﻗﻴﻤﺔ ‪ Am‬ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫ﺃ ـ ﺍﻨﻁﻼﻗﹰﺎ ﻤﻥ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ‪:‬‬ ‫‪A m = V 0 (1 + i )m‬‬ ‫‪Am‬‬ ‫‪= a 1 − (1 + i )−n‬‬ ‫‪(1 + i )m‬‬ ‫‪i‬‬ ‫ﺏ ـ ﺍﻨﻁﻼﻗﹰﺎ ﻤﻥ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫) ‪Am = An (1 + i )(m −n‬‬ ‫‪Am‬‬ ‫‪= a (1 + i )n‬‬ ‫) ‪− 1 (1 + i )(m −n‬‬ ‫‪i‬‬ ‫ﺘﻁﺒﻴﻕ‪:‬‬‫ﻟﺘﺴﺩﻴﺩ ﺜﻤﻥ ﺘﺠﻬﻴﺯﺍﺕ ﻜﺎﻥ ﻋﻠﻰ ﺍﻟﻤﺸﺘﺭﻱ ﺃﻥ ﻴﺩﻓﻊ ﻓﻲ ﻨﻬﺎﻴﺔ ﻜل ﺴﻨﺔ ﺩﻓﻌﺔ ﺘﻘﺩﺭ ﺒـ ‪ 300.000 DA‬ﻟﻤﺩﺓ‬ ‫‪ 4‬ﺴﻨﻭﺍﺕ‪ .‬ﻏﻴﺭ ﺃﻨﻪ ﺍﻗﺘﺭﺡ ﻋﻠﻰ ﺍﻟﺒﺎﺌﻊ ﺘﺴﺩﻴﺩ ﺜﻤﻥ ﺍﻟﺘﺠﻬﻴﺯﺍﺕ ﺒﺩﻓﻊ ﻭﺤﻴﺩ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﺴﻨﺔ ﺍﻟﺴﺎﺩﺴﺔ‪.‬‬ ‫ﺒﻤﻌﺩل ﺴﻨﻭﻱ ‪ %10‬ﻣﺎ هﻮ ﻣﻘﺪار اﻟﺪﻓﻊ اﻟﻮﺣﻴﺪ ؟‬ ‫ﺍﻟﺤل‪:‬‬‫‪0 1 2 34‬‬ ‫‪56‬‬ ‫اﻟﻤﺪة‬ ‫‪A4 A6‬‬ ‫ـ ﺍﻨﻁﻼﻗﹰﺎ ﻤﻥ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‪:‬‬ ‫‪A6 =V 0 (1 + i )m‬‬ ‫‪A6‬‬ ‫=‬ ‫‪300.000 1 −‬‬ ‫‪(1, 1)−4‬‬ ‫‪(1, 1)6‬‬ ‫‪0,1‬‬ ‫‪A6 = 300.000 × 3,169865 × 1,771561‬‬ ‫‪A6 = 1.684.683DA‬‬ ‫ـ ﺍﻨﻁﻼﻗﹰﺎ ﻤﻥ ﺼﻴﻐﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫) ‪Am = An (1 + i )(m −n‬‬ ‫‪A6‬‬ ‫=‬ ‫× ‪300.000‬‬ ‫)‪(1,1)4 − 1 (1,1)(6−2‬‬ ‫‪0,1‬‬ ‫‪A6‬‬ ‫=‬ ‫× ‪300.000‬‬ ‫‪(1,1)4 − 1 (1,1)2‬‬ ‫‪0,1‬‬ ‫‪A6 = 300.000× 4,641× 1,21‬‬ ‫‪A6 = 1.684.683‬‬ ‫‪ .5‬ﺃﻨﺸﻁﺔ ﺍﻹﻋﻼﻡ ﺍﻵﻟﻲ‬‫ﻴﺤﺘﻭﻱ ﺍﻟﻤﺠﺩﻭل ‪ Excel‬ﻋﻠﻰ ﺼﻴﻎ ﻟﺩﻭﺍل ﻤﺎﻟﻴﺔ ﻭ ﻤﻨﻬﺎ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤـﺔ ﺍﻟﺤﺎﻟﻴـﺔ )‪ (VA‬ﻭ ﺍﻟﻘﻴﻤـﺔ‬ ‫ﺍﻟﻤﻜﺘﺴﺒﺔ )‪ (VC‬ﻟﻤﺘﺘﺎﻟﻴﺔ ﺩﻓﻌﺎﺕ ﺍﻟﺴﺩﺍﺩ‪.‬‬ ‫ﺍﻟﻌﻤل ﺍﻟﻤﻁﻠﻭﺏ‪:‬‬‫ﺃﺤﺴﺏ ﻟﻠﺠﺩﻭل ﺍﻟﺘﺎﻟﻲ ﻜل ﻤﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻭ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺼﻴﻎ ﺍﻟﻤﺎﻟﻴﺔ ﻟﻠﻤﺠﺩﻭل ‪.Excel‬‬‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ‬ ‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ‬ ‫ﺍﻟﻤﻌﺩل )‪(i‬‬ ‫ﺍﻟﺩﻓﻌﺔ‬ ‫ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ‬ ‫)‪(VC‬‬ ‫)‪(VA‬‬ ‫)‪(a‬‬ ‫)‪(n‬‬ ‫‪8% 100.000‬‬ ‫‪10‬‬ ‫‪6% 200.000‬‬ ‫‪15‬‬ ‫‪10% 50.000‬‬ ‫‪20‬‬ ‫‪7% 30.000‬‬ ‫‪8‬‬ ‫‪4% 5.000‬‬ ‫‪6‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺍﻟﺤﻠﻭل ﺍﻟﻤﻘﺘﺭﺤﺔ ﻷﻨﺸﻁﺔ ﺍﻹﻋﻼﻡ ﺍﻵﻟﻲ‬ ‫ﺨﻁﻭﺍﺕ ﺍﻟﺤل ‪:‬‬ ‫‪ .1‬ﺇﻋﺎﺩﺓ ﺭﺴﻡ ﺍﻟﺠﺩﻭل ﺍﻟﺴﺎﺒﻕ ﻋﻠﻰ ﻭﺭﻗﺔ ﺍﻟﻤﺼﻨﻑ‪.‬‬ ‫‪ .2‬ﻜﺘﺎﺒﺔ ﺼﻴﻐﺔ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ)‪ (VA‬ﻓﻲ ﺍﻟﺨﻠﻴﺔ )‪ (D2‬ﻭ ﻨﺴﺨﻬﺎ ﻋﻠﻰ ﺒﻘﻴﺔ ﺍﻟﺨﻼﻴﺎ ﻜﻤﺎ ﻴﻠﻲ‪:‬‬‫ﺃ ـ ﻤﻥ ﺸﺭﻴﻁ ﺍﻟﻘﻭﺍﺌﻡ ﺍﺨﺘﺭ ﻗﺎﺌﻤﺔ ﺇﺩﺭﺍﺝ )‪ (Insertion‬ﺜﻡ ﺩﺍﻟﺔ )‪ ،(fonction‬ﺜﻡ ﺍﺨﺘﺭ ﻤﻥ ﺍﻟﻨﺎﻓـﺫﺓ ﺍﻟﺘـﻲ‬ ‫ﺘﻅﻬﺭ ﻨﻭﻉ ﺍﻟﺩﻭﺍل ﻤﺎﻟﻴﺔ‪ ،‬ﻭ ﻤﻥ ﺍﻟﺩﻭﺍل ﺍﻟﻤﺎﻟﻴﺔ ﺍﺨﺘﺭ ﺍﻟﺩﺍﻟﺔ )‪ (VA‬ﻜﻤﺎ ﻫﻭ ﻤﺒﻴﻥ ﻓﻲ ﺍﻟﺼﻭﺭﺓ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫ﺏ ـ ﻓﻲ ﺍﻟﻨﺎﻓﺫ ﺍﻟﺘﻲ ﺘﻅﻬﺭ ﻴﻁﻠﺏ ﻤﻨﻙ ﺇﺩﺨﺎل ﺍﻟﺒﻴﺎﻨﺎﺕ ﻓﻲ ﺍﻟﻤﺭﺒﻌﺎﺕ ﺍﻟﺘﻲ ﺘﻌﺒﺭ ﻋﻥ ﻤﺨﺘﻠـﻑ ﺍﻟﻌﻨﺎﺼـﺭ‬ ‫ﺍﻟﺘﻲ ﻴﺘﻡ ﺒﻬﺎ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻭ ﻫﻲ ﻜﺎﻟﺘﺎﻟﻲ‪:‬‬ ‫ﺍﻟﻤﻌﺩل‬ ‫‪Taux‬‬ ‫ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ )‪(n‬‬ ‫‪Npm‬‬ ‫ﻤﺒﻠﻎ ﺍﻟﺩﻓﻌﺔ )‪ (a‬ﻭ ﺘﺴﺒﻘﻬﺎ ﺩﻭﻤﹰﺎ ﺍﻹﺸﺎﺭﺓ ﺍﻟﺴﺎﻟﺒﺔ )‪(-‬‬ ‫‪Vpm‬‬ ‫ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﻭ ﻨﻀﻌﻬﺎ ﺘﺴﺎﻭﻱ ﺍﻟﺼﻔﺭ‬ ‫‪VC‬‬ ‫ﻨﻀﻊ ﺍﻟﻘﻴﻤﺔ )‪ (0‬ﻟﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﻭ ﺍﻟﻘﻴﻤﺔ )‪ (1‬ﻟﺩﻓﻌﺎﺕ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ‬ ‫‪Type‬‬ ‫ﺇﺩﺨﺎل ﺍﻟﺒﻴﺎﻨﺎﺕ ﻴﻜﻭﻥ ﻜﻤﺎ ﻫﻭ ﻤﺒﻴﻥ ﻓﻲ ﺍﻟﺼﻭﺭﺓ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺠـ ـ ﺒﻌﺩ ﻜﺘﺎﺒﺔ ﺼﻴﻐﺔ )‪ (VA‬ﻓﻲ ﺍﻟﺨﻠﻴﺔ )‪ (D2‬ﻴﺘﻡ ﻨﺴﺨﻬﺎ ﺇﻟﻰ ﺒﻘﻴﻤﺔ ﺍﻟﺨﻼﻴﺎ )‪ D3‬ﺇﻟﻰ ‪.(D6‬‬‫‪ .2‬ﻜﺘﺎﺒﺔ ﺼﻴﻐﺔ ﺤﺴﺎﺏ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻤﻜﺘﺴﺒﺔ ﻓﻲ ﺍﻟﺨﻠﻴﺔ )‪ (E2‬ﻭ ﻨﺴﺨﻬﺎ ﻋﻠﻰ ﺒﻘﻴﺔ ﺍﻟﺨﻼﻴﺎ ﻤﻊ ﺇﺘﺒﺎﻉ ﻨﻔﺱ‬‫ﺍﻟﺨﻁﻭﺍﺕ ﺍﻟﺴﺎﺒﻘﺔ ﻤﻊ ﺍﻟﺘﻨﺒﻴﻪ ﺇﻟﻰ ﺍﺨﺘﻴﺎﺭ ﺍﻟﺩﺍﻟﺔ ﺍﻟﻤﺎﻟﻴﺔ )‪ (VC‬ﻭ ﺠﻌل ﻗﻴﻤﺔ )‪ (Va‬ﻤﻌﺩﻭﻤﺔ ﻓﻲ ﻨﺎﻓﺫﺓ ﺇﺩﺨﺎل‬ ‫ﺍﻟﺒﻴﺎﻨﺎﺕ‪.‬‬ ‫‪ .3‬ﺍﻟﺤل ﺍﻟﻨﻬﺎﺌﻲ ﺍﻟﻤﻘﺘﺭﺡ ﻴﻅﻬﺭ ﻜﻤﺎ ﻴﻠﻲ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪ .6‬ﺃﺴﺌﻠﺔ ﺍﻟﺘﻘﻭﻴﻡ ﺍﻟﺫﺍﺘﻲ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ)‪:(1‬‬‫ﺭﺃﺱ ﻤﺎل ﺒـ ‪ 1.000.000 DA‬ﻴﺴﺩﺩ ﺒـﻭﺍﺴﻁﺔ ‪ 10‬ﺩﻓﻌﺔ ﺜﺎﺒﺘﺔ ‪،‬ﺘﺩﻓﻊ ﺍﻷﻭﻟﻰ ﺴﻨﺔ ﺒﻌﺩ ﺘﺎﺭﻴﺦ ﺍﻟﺘﻌﺎﻗﺩ‪ .‬ﺇﺫﺍ‬ ‫ﻜﺎﻥ ﻤﻌﺩل ﺍﻟﻔﺎﺌﺩﺓ ﺍﻟﻤﺭﻜﺒﺔ ‪ %6‬ﺴﻨﻭﻴﹰﺎ‪.‬‬ ‫ﺍﻟﻌﻤل ﺍﻟﻤﻁﻠﻭﺏ ‪:‬‬ ‫ﺃ ـ ﺃﺤﺴﺏ ﻤﺒﻠﻎ ﺍﻟﺩﻓﻌﺔ ﺍﻟﺜﺎﺒﺘﺔ‪.‬‬‫ﺏ ـ ﺒﻌﺩ ﺘﺴﺩﻴﺩ ﺍﻟﺩﻓﻌﺔ ‪ 6‬ﻭﺍﻓﻕ ﺍﻟﻤﺩﻴﻥ ﻋﻠﻰ ﻤﻀﺎﻋﻔﺔ ﻤﺒﻠﻎ ﺍﻟﺩﻓﻌﺔ ﻤﺎ ﺩﺍﻤﺕ ﺍﻟﻔﻭﺍﺌﺩ ﻗﺩ ﺍﻨﺨﻔﻀﺕ ﺇﻟـﻰ‬ ‫‪ % 4,5‬ﺴﻨﻭﻴﹰﺎ‪ .‬ﺃﺤﺴﺏ ﺍﻟﻤﺩﺓ ﺍﻟﻼﺯﻤﺔ ﻟﻠﺘﺨﻠﺹ ﻤﻥ ﺒﻘﻴﺔ ﺍﻟﺩﻴﻥ‪ ).‬ﺒﺎﻟﺘﻘﺭﻴﺏ ﺒﺎﻟﺯﻴﺎﺩﺓ(‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ)‪:(2‬‬‫ﻴﺭﻏﺏ ﺸﺨﺹ ﻓﻲ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﺘﺠﻬﻴﺯﺍﺕ ﺒـ )‪ ، (X DA‬ﻭﻟﻪ ﺃﻥ ﻴﺨﺘﺎﺭ ﻤـﻥ ﺒـﻴﻥ ﺍﻟﺤﻠـﻭل ﺍﻟﺘﺎﻟﻴـﺔ‬ ‫ﺍﻟﻤﻘﺘﺭﺤﺔ ﻟﻠﺘﺴﺩﻴﺩ ‪:‬‬ ‫‪ - 1‬ﺘﺴﺩﻴﺩ ‪ 3‬ﺩﻓﻌﺎﺕ ﻗﻴﻤﺔ ﻜل ﺩﻓﻌﺔ ‪ 3.603 DA‬ﺘﺩﻓﻊ ﺨﻼل ‪ 3 ،2 ،1‬ﺴﻨﻭﺍﺕ‪.‬‬ ‫ـ ﺒﻔﺎﺌﺩﺓ ﻤﺭﻜﺒﺔ ‪ %4‬ﺃﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﺘﺠﻬﻴﺯﺍﺕ ‪) .‬ﺘﻘﺭﺏ ﺇﻟﻰ ﺍﻷﻟﻑ ﺍﻷﻗﺭﺏ(‪.‬‬‫‪ - 2‬ﺘﺴﺩﻴﺩ ﻋﻨﺩ ﺘﺎﺭﻴﺦ ﺍﻟﺸﺭﺍﺀ ﻤﺒﻠﻎ ‪ 5.800 DA‬ﻭ ﻫﺫﺍ ﺍﻟﻤﺒﻠﻎ ﻴﻤﺜل ‪ %58‬ﻤﻥ ﻗﻴﻤﺔ ﺍﻟﺘﺠﻬﻴﺯﺍﺕ ﻭ ﺍﻟﺒـﺎﻗﻲ‬ ‫ﻴﺴﺩﺩ ﺒﻌﺩ ‪ 3‬ﺴﻨﻭﺍﺕ ﺒﻤﺒﻠﻎ ‪.5002,2672 DA‬‬ ‫ـ ﻤﺎ ﻫﻭ ﻤﻌﺩل ﺍﻟﻔﺎﺌﺩﺓ ﺍﻟﻤﺭﻜﺒﺔ ﺍﻟﻤﻁﺒﻕ‪.‬‬‫‪ - 3‬ﺍﻟﺘﺴﺩﻴﺩ ﺒﻭﺍﺴﻁﺔ ‪ 6‬ﺴﺩﺍﺴﻴﺎﺕ ﻤﺒﻠﻎ ‪ 1.815,50 DA‬ﻟﻠﺴﺩﺍﺴﻲ‪ ،‬ﺍﻟﺴﺩﺍﺴﻴﺔ ﺍﻷﻭﻟﻰ ﺘﺩﻓﻊ ﺒﻌﺩ ‪ 6‬ﺃﺸﻬﺭ ﻤﻥ‬ ‫ﺘﺎﺭﻴﺦ ﺍﻟﺸﺭﺍﺀ‪.‬‬ ‫ـ ﺒﻤﻌﺩل ﻓﺎﺌﺩﺓ ‪ %8‬ﺴﻨﻭﻴﹰﺎ ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﺍﻟﺴﺩﺍﺴﻴﺎﺕ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ‪.‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ)‪:(3‬‬‫ﻗﺎﻡ ﺸﺨﺹ ﺒﻔﺘﺢ ﺤﺴﺎﺏ ﺩﻓﺘﺭ ﺘﻭﻓﻴﺭ ﻟﻭﻟﺩﻴﻪ ﻟﺩﻯ ﺍﻟﺼﻨﺩﻭﻕ ﺍﻟﻭﻁﻨﻲ ﻟﻠﺘﻭﻓﻴﺭ ﻭ ﺍﻻﺤﺘﻴﺎﻁ ﻋﻨﺩﻤﺎ ﻜﺎﻥ ﻋﻤﺭ‬‫ﺍﻟﻤﺒﻠﻎ ﺍﻟ ‪‬ﻤﻭﺩﻉ‬ ‫‪1‬‬ ‫‪ 16‬ﺴﻨﺔ‪ ،‬ﺤﻴﺙ ﻭﻀﻊ ﻓﻲ ﺤﺴﺎﺏ ﺍﻷﻭل‬ ‫ﺍﻟﻭﻟﺩ ﺍﻷﻭل ‪ 6‬ﺴﻨﻭﺍﺕ ﻭ ﻋﻤﺭ ﺍﻟﻭﻟﺩ ﺍﻟﺜﺎﻨﻲ‬ ‫‪3‬‬ ‫ﻓﻲ ﺤﺴﺎﺏ ﺍﻟﺜﺎﻨﻲ ﻭﻗﻴﻤﺔ ﺍﻟﻤﺒﻠﻐﻴﻥ ﻤﻌﹰﺎ ‪. 320.000 DA‬‬‫ﺍﺴﺘﻤﺭ ﻫﺫﺍ ﺍﻟﺸﺨﺹ ﻓﻲ ﺇﻴﺩﺍﻉ ﻨﻔﺱ ﺍﻟﻤﺒﻠﻎ ﺍﻟﻤﻭﺩﻉ ﻓﻲ ﺤﺴﺎﺏ ﻜل ﻭﻟﺩ ﻤﻥ ﻭﻟﺩﻴﻪ ﻓﻲ ﺒﺩﺍﻴﺔ ﻜل ﺴﻨﺔ ﺤﺘـﻰ‬ ‫ﺒﻠﻎ ﻜل ﻭﺍﺤﺩ ﻤﻨﻬﻤﺎ ﺴﻥ ‪ 25‬ﺴﻨﺔ‪.‬‬ ‫ﺍﻟﻌﻤل ﺍﻟﻤﻁﻠﻭﺏ‪:‬‬‫ﺇﺫﺍ ﻜﺎﻥ ﻤﻌﺩل ﺍﻟﻔﺎﺌﺩﺓ ﺍﻟﻤﺭﻜﺒﺔ ‪ %7‬ﺴﻨﻭﻴﺎ ﻓﻤﺎ ﻫﻭ ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﺫﻱ ﺘﺤﺼل ﻋﻠﻴﻪ ﻜل ﻭﻟﺩ ﻋﻨﺩ ﺒﻠﻭﻏﻪ ﺴﻥ‬ ‫‪ 25‬ﺴﻨﺔ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ)‪:(4‬‬‫ﺍﻗﺘﺭﻀﺕ ﻤﺅﺴﺴﺔ ‪ 1.000.000 DA‬ﺒﺘﺎﺭﻴﺦ ‪ .2000/01/01‬ﻭ ﺍﻗﺘﺭﺡ ﻋﻠﻴﻬﺎ ﺍﻟﺒﻨﻙ ﺜﻼﺜﺔ ﻁﺭﻕ ﻟﻠﺘﺴﺩﻴﺩ ‪:‬‬ ‫ـ ﺍﻟﻁﺭﻴﻘﺔ )‪:(1‬‬ ‫ﺘﺴﺩﻴﺩ ﺩﻓﻌﺔ ﻭﺤﻴﺩﹰﺓ ﺒﻌﺩ ‪ 10‬ﺴﻨﻭﺍﺕ‪.‬‬ ‫ـ ﺍﻟﻁﺭﻴﻘﺔ )‪:(2‬‬ ‫ﺍﻟﺘﺴﺩﻴﺩ ﺒﻭﺍﺴﻁﺔ ‪ 5‬ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﺘﺴﺘﺤﻕ ﺍﻷﻭﻟﻰ ﺒﺘﺎﺭﻴﺦ ‪.2001/01/01‬‬ ‫ـ ﺍﻟﻁﺭﻴﻘﺔ )‪:(3‬‬‫ﺍﻟﺘﺴﺩﻴﺩ ﺒﻭﺍﺴﻁﺔ ‪ 5‬ﺩﻓﻌﺎﺕ ﻤﺘﺴﺎﻭﻴﺔ ﺘﺴﺘﺤﻕ ﻜـل ﺩﻓﻌـﺔ ﻤﻨﻬـﺎ ﺒﻌـﺩ ﺴـﻨﺘﻴﻥ ﺃﻱ‪ :‬ﺍﻷﻭﻟـﻰ ﻓـﻲ‬ ‫‪ 2002/01/01‬ﻭ ﺍﻟﺜﺎﻨﻴﺔ ﻓﻲ ‪... 2004/01/01‬ﺍﻟﺦ‪.‬‬ ‫ﺍﻟﻌﻤل ﺍﻟﻤﻁﻠﻭﺏ ‪:‬‬ ‫ﻤﻌﺩل ﻓﺎﺌﺩﺓ ﻤﺭﻜﺒﺔ ‪ %7‬ﺴﻨﻭﻴﹰﺎ ﺃﺤﺴﺏ ﻗﻴﻤﺔ ﻜل ﺩﻓﻌﺔ ﻭﻓﻕ ﺍﻟﻁﺭﻕ ﺍﻟﺜﻼﺙ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ)‪:(5‬‬‫ﺒﺘﺎﺭﻴﺦ ‪ 2006/01/01‬ﺍﺸﺘﺭﺕ ﺇﺤﺩﻯ ﺍﻟﻤﺅﺴﺴﺎﺕ ﻋﻘﺎﺭﹰﺍ ﹸﺘﺴﺩﺩ ﺜﻤﻨﻪ ﺒﻭﺍﺴﻁﺔ ‪ 8‬ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﺘﺴﺩﺩ ﺍﻷﻭﻟﻰ‬ ‫ﺒﺘﺎﺭﻴﺦ ‪ ،2008/01/01‬ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ‪ 341.396,33DA‬ﺒﻤﻌﺩل ﻓﺎﺌﺩﺓ ﻤﺭﻜﺒﺔ ‪ %6‬ﺴﻨﻭﻴﹰﺎ‪.‬‬ ‫ﺍﻟﻌﻤل ﺍﻟﻤﻁﻠﻭﺏ ‪:‬‬ ‫‪ .1‬ﺃﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﻌﻘﺎﺭ ﻓﻲ ﺘﺎﺭﻴﺦ ﺍﻟﺸﺭﺍﺀ‪.‬‬‫‪ .2‬ﺇﺫﺍ ﺃﺭﺍﺩﺕ ﺘﺴﺩﻴﺩ ﺍﻟﻤﺒﻠﻎ ﺩﻓﻌﺔ ﻭﺍﺤﺩﺓ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ﻓﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﻫﺫﻩ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﻨﻬﺎﻴﺔ‬ ‫ﺍﻟﻤﺩﺓ ؟‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫ﺇﺫﺍ ﺃﺭﺍﺩﺍﺕ ﺃﻥ ﺘﺴﺩﺩ ﺍﻟﻤﺒﻠﻎ ﺩﻓﻌﺔ ﻭﺍﺤﺩﺓ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﺴﻨﺔ ‪ 15‬ﻓﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﻫﺫﺍ ﺍﻟﺩﻓﻊ‬ ‫‪.3‬‬ ‫ﺍﻟﻭﺤﻴﺩ ؟‬ ‫‪.4‬‬ ‫ﻤﺎ ﻫﻲ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺎﺕ ‪ 3‬ﺴﻨﻭﺍﺕ ﻗﺒل ﺍﻟﺩﻓﻊ ﺍﻷﻭل ؟‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪ .7‬ﺃﺠﻭﺒﺔ ﺍﻟﺘﻘﻭﻴﻡ ﺍﻟﺫﺍﺘﻲ‬ ‫ﺤل ﺍﻟﺘﻤﺭﻴﻥ)‪:(1‬‬ ‫ﺃ ـ ﺤﺴﺎﺏ ﻤﺒﻠﻎ ﺍﻟﺩﻓﻌﺔ ﺍﻟﺜﺎﺒﺘﺔ ‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪1 − (1 +‬‬ ‫‪i )−n‬‬ ‫‪i‬‬ ‫‪a‬‬ ‫‪=V 0‬‬ ‫‪1−‬‬ ‫‪i‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪(1 +‬‬ ‫‪a‬‬ ‫=‬ ‫× ‪1.000.000‬‬ ‫‪1−‬‬ ‫‪0, 06‬‬ ‫‪(1 + 0, 06)−10‬‬ ‫‪a = 135.868DA‬‬ ‫ﺏ ـ ﺤﺴﺎﺏ ﺍﻟﻤﺩﺓ ﺍﻟﻼﺯﻤﺔ ﻟﻠﺘﺨﻠﺹ ﻤﻥ ﺒﻘﻴﺔ ﺍﻟﺩﻴﻥ ﺒﻌﺩ ﺘﺴﺩﻴﺩ ﺍﻟﺩﻓﻌﺔ ‪: 6‬‬ ‫ـ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻠﺩﻓﻌﺎﺕ ‪ 4‬ﺍﻟﺒﺎﻗﻴﺔ ‪:‬‬ ‫‪V4‬‬ ‫=‬ ‫‪a‬‬ ‫‪1 − (1 +‬‬ ‫‪i )−n‬‬ ‫‪i‬‬ ‫‪V4‬‬ ‫=‬ ‫‪135.868‬‬ ‫×‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪(1 +‬‬ ‫‪0, 06‬‬ ‫‪)−4‬‬ ‫‪0,‬‬ ‫‪06‬‬ ‫‪V4‬‬ ‫=‬ ‫‪135.868‬‬ ‫×‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪(1, 06)−4‬‬ ‫‪0, 06‬‬ ‫‪V 4 = 135.868 × 3,465105‬‬ ‫‪V4 = 470.796DA‬‬ ‫ـ ﺤﺴﺎﺏ ﺍﻟﻤﺩﺓ ‪:‬‬‫‪V4‬‬ ‫=‬ ‫‪2a‬‬ ‫‪1−‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬‫‪470.796‬‬ ‫=‬ ‫× ‪2 × 135.868‬‬ ‫‪1−‬‬ ‫‪(1 + 0, 045)−n‬‬ ‫‪0, 045‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪470.796‬‬ ‫=‬ ‫‪271.736‬‬ ‫×‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪(1, 045)−n‬‬ ‫‪0, 045‬‬ ‫‪1 − (1, 045)−n‬‬ ‫=‬ ‫‪470.796‬‬ ‫‪0, 045‬‬ ‫‪271.736‬‬ ‫‪(1, 045)−n‬‬ ‫‪= 1−‬‬ ‫‪0, 045 × 470.796‬‬ ‫‪271.736‬‬ ‫‪(1, 045)−n = 0,922036‬‬ ‫ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﻠﻭﻏﺎﺭﻴﺘﻡ ﺍﻟﻌﺸﺭﻱ ﻨﺠﺩ‪:‬‬ ‫‪(1, 045)−n = 0,922036‬‬ ‫)‪log(0,922036‬‬ ‫)‪n = − log(1,045‬‬ ‫‪n=2‬‬ ‫ﺤل ﺍﻟﺘﻤﺭﻴﻥ)‪:(2‬‬‫‪ .1‬ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺘﺠﻬﻴﺯﺍﺕ )‪ (X DA‬ﻤﺎ ﻫﻲ ﺇﻻ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﺍﻟﺯﻤﻥ )‪:(0‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪1−‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬ ‫‪V0‬‬ ‫=‬ ‫× ‪3.603‬‬ ‫‪1−‬‬ ‫‪(1 + 0, 04)−3‬‬ ‫‪0, 04‬‬ ‫‪V0‬‬ ‫=‬ ‫× ‪3.603‬‬ ‫‪1−‬‬ ‫‪(1, 04)−3‬‬ ‫‪0, 04‬‬ ‫‪V0 = 3.603 × 2,775091‬‬ ‫‪V 0 = 10.000DA‬‬ ‫‪ 2‬ـ ﺤﺴﺎﺏ ﻤﻌﺩل ﺍﻟﻔﺎﺌﺩﺓ ﺍﻟﻤﺭﻜﺒﺔ ﺍﻟﻤﻁﺒﻕ‪:‬‬ ‫ﺍﻟﻤﺒﻠﻎ ﺍﻟﺒﺎﻗﻲ ﻤﻥ ﺍﻟﺘﺠﻬﻴﺯﺍﺕ ‪10.000 − 5.800 = 4.200 :‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪A = a (1 + i )n‬‬ ‫‪5.002, 2672 = 4.200(1 + i )3‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)3‬‬ ‫=‬ ‫‪5.002, 2672‬‬ ‫‪4.200‬‬ ‫‪(1 + i )3 = 1,191016‬‬ ‫‪1+i= 3 1,191016‬‬ ‫‪1‬‬ ‫‪i = (1,191016)3 − 1‬‬ ‫‪i = 0,06‬‬ ‫‪i = 6%‬‬ ‫ـ ﺠﻤﻠﺔ ‪ 6‬ﺴﺩﺍﺴﻴﺎﺕ ‪:‬‬ ‫▪ ﺍﻟﻤﻌﺩل ﺍﻟﺴﺩﺍﺴﻲ ﺍﻟﻤﻜﺎﻓﺊ ﻟﻠﻤﻌﺩل ﺍﻟﺴﻨﻭﻱ ‪%8‬‬ ‫‪(1 + i a ) = (1 + i s )2‬‬ ‫‪i s = (1 + i a ) − 1‬‬ ‫‪i s = 1, 08 − 1 = 0,03923‬‬ ‫‪i s = 3,923%‬‬ ‫▪ ﺠﻤﻠﺔ ‪ 6‬ﺴﺩﺍﺴﻴﺎﺕ ‪:‬‬ ‫‪A = a (1 + i s )n − 1‬‬ ‫‪is‬‬ ‫‪A‬‬ ‫=‬ ‫× ‪1.815, 5‬‬ ‫‪(1 +‬‬ ‫‪0, 03923)6‬‬ ‫‪−1‬‬ ‫‪0, 03923‬‬ ‫‪A = 1.815, 5 × 6,620149‬‬ ‫‪A =12.018,88 DA‬‬ ‫ﺤل ﺍﻟﺘﻤﺭﻴﻥ)‪: (3‬‬ ‫‪ .1‬ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﻤﺒﻠﻎ ﺍﻟ ‪‬ﻤﻭﺩﻉ ﻓﻲ ﺤﺴﺎﺏ ﻜل ﻭﻟﺩ‪:‬‬ ‫ـ ﺍﻟﻤﺒﻠﻎ ﺍﻟﻤﻭﺩﻉ ﻓﻲ ﺤﺴﺎﺏ ﺍﻟﻭﻟﺩ ﺍﻷﻭل ) ‪ (a1‬ﻭ ﺍﻟﻭﻟﺩ ﺍﻟﺜﺎﻨﻲ ) ‪: (a2‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪a1‬‬ ‫=‬ ‫‪a2‬‬ ‫⇔‬ ‫‪3a1 = a 2‬‬ ‫‪3‬‬ ‫‪a 1 + 3 a 1 = 3 2 0 .0 0 0‬‬ ‫‪4 a 1 = 3 2 0 .0 0 0‬‬ ‫= ‪a1‬‬ ‫‪3 2 0 .0 0 0‬‬ ‫‪4‬‬ ‫‪a 1 = 8 0 .0 0 0 D A‬‬ ‫‪a 2 = 2 4 0 .0 0 0 DA‬‬ ‫ـ ﺤﺴﺎﺏ ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﺫﻱ ﺤﺼل ﻋﻠﻴﻪ ﻜل ﻭﻟﺩ‪:‬‬ ‫ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﺩﻓﻌﺎﺕ ﺘﺩﻓﻊ ﻓﻲ ﺒﺩﺍﻴﺔ ﺍﻟﻤﺩﺓ‪.‬‬‫ﻟﺘﻜﻥ ‪ A1‬ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﺫﻱ ﺤﺼل ﻋﻠﻴﻪ ﺍﻟﻭﻟﺩ ﺍﻷﻭل ﻭ ‪ n1‬ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ ﻭﻫﻲ ﺘﺴـﺎﻭﻱ )‪19 = 6-25‬‬ ‫ﺩﻓﻌﺔ(‬ ‫‪A1‬‬ ‫=‬ ‫‪a1 (1 +‬‬ ‫‪i‬‬ ‫‪) (1 +‬‬ ‫‪i )n1‬‬ ‫‪−1‬‬ ‫‪i‬‬ ‫‪A1‬‬ ‫=‬ ‫‪80.000 × (1 +‬‬ ‫× )‪0, 07‬‬ ‫‪(1 +‬‬ ‫‪0, 07)19‬‬ ‫‪−1‬‬ ‫‪0, 07‬‬ ‫‪A1 = 80.000 × 1, 07 × 37,378964‬‬ ‫‪A 1 = 3.199.639,31DA‬‬‫ﻟﺘﻜﻥ ‪ A2‬ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﺫﻱ ﺤﺼل ﻋﻠﻴﻪ ﺍﻟﻭﻟﺩ ﺍﻷﻭل ﻭ ‪ n2‬ﻋﺩﺩ ﺍﻟﺩﻓﻌﺎﺕ ﻭﻫﻲ ﺘﺴﺎﻭﻱ )‪9 = 16-25‬‬ ‫ﺩﻓﻌﺎﺕ(‪.‬‬ ‫‪A2‬‬ ‫=‬ ‫‪a 2 (1 +‬‬ ‫‪i ) (1 +‬‬ ‫‪i )n 2‬‬ ‫‪−1‬‬ ‫‪i‬‬ ‫‪A1‬‬ ‫=‬ ‫‪240.000 × (1 +‬‬ ‫× ) ‪0, 07‬‬ ‫‪(1 +‬‬ ‫‪0, 07)9‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪0,07‬‬ ‫‪A 2 = 240.000 × 1, 07 × 11,977988‬‬ ‫‪A 2 = 3.075.947,31DA‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺣﻞ اﻟﺘﻤﺮﻳﻦ)‪:(4‬‬ ‫ـ ﺍﻟﻁﺭﻴﻘﺔ )‪ :(1‬ﺤﺴﺎﺏ ﺍﻟﺠﻤﻠﺔ‬ ‫‪A = a (1 + i )n‬‬ ‫‪A = 1.000.000(1 + 0, 07)10‬‬ ‫‪A = 1.000.000 × 1,967151‬‬ ‫‪A = 1 .9 6 7 .1 5 1DA‬‬ ‫ـ ﺍﻟﻁﺭﻴﻘﺔ )‪ :(2‬ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ﺍﻟﺜﺎﺒﺘﺔ‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪1 − (1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬ ‫‪a‬‬ ‫‪=V 0‬‬ ‫‪1−‬‬ ‫‪i‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪(1 +‬‬ ‫‪a‬‬ ‫=‬ ‫× ‪1.000.000‬‬ ‫‪1−‬‬ ‫‪0, 07‬‬ ‫‪(1, 07)−5‬‬ ‫‪a = 1.000.000 × 0,24389‬‬ ‫‪a=243.890DA‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ـ ﺍﻟﻁﺭﻴﻘﺔ )‪ :(3‬ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ﺍﻟﺜﺎﺒﺘﺔ‬ ‫‪0 1 2 3 4 5 6 7 8 9 10‬‬‫‪V0‬‬ ‫=‬ ‫‪a (1 +‬‬ ‫‪i )−2‬‬ ‫‪a‬‬ ‫‪(1‬‬ ‫‪+‬‬ ‫‪a‬‬ ‫)‬ ‫‪−‬‬ ‫‪4‬‬ ‫‪+‬‬ ‫‪a‬‬ ‫‪(1‬‬ ‫‪+‬‬ ‫‪i a) − 6‬‬ ‫‪+‬‬ ‫‪a a(1 +‬‬ ‫‪i )−8‬‬ ‫‪+‬‬ ‫‪a (1 +‬‬ ‫‪i )−10‬‬ ‫‪+a‬‬ ‫‪i‬‬ ‫‪a‬‬‫⎦⎤ ‪V 0 = a ⎣⎡ (1 + i ) − 2 + (1 + i ) − 4 + (1 + i ) − 6 + (1 + i ) − 8 + (1 + i ) − 10‬‬ ‫ﻨﻼﺤﻅ ﺃﻥ ﺍﻟﻤﻘﺩﺍﺭ ﺒﻴﻥ ﺍﻟﻤﺠﺎﻟﻴﻥ ﻋﺒﺎﺭﺓ ﻋﻥ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﺤﺩﻫﺎ ﺍﻷﻭل ‪ (1 + i )−2‬ﻭ ﺃﺴﺎﺴﻬﺎ ‪(1 + i )−2‬‬ ‫ﻭ ﻋﺩﺩ ﺤﺩﻭﺩﻫﺎ ‪ 5‬ﻭ ﻤﻨﻪ ﺤﺴﺏ ﺼﻴﻐﺔ ﻤﺠﻤﻭﻉ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﻬﻨﺩﺴﻴﺔ ﻓﺈﻥ ‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫⎡‬ ‫(‬ ‫‪1‬‬ ‫‪+‬‬ ‫× ‪i )−2‬‬ ‫⎤ ‪((1 + i )−2 )−5 − 1‬‬ ‫⎢‬ ‫⎥‬ ‫⎣‬ ‫‪(1 + i )−2 − 1‬‬ ‫⎦‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫⎡‬ ‫(‬ ‫‪1‬‬ ‫‪+‬‬ ‫× ‪i )−2‬‬ ‫⎤ ‪(1 + i )−10 − 1‬‬ ‫⎢‬ ‫⎥‬ ‫⎣‬ ‫‪(1‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪)−2‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫⎦‬ ‫ﻭ ﻤﻨﻪ ﺒﺎﻟﺘﻌﻭﻴﺽ ﻨﺠﺩ‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫×‬ ‫⎡‬ ‫‪(1‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪)−2‬‬ ‫×‬ ‫⎤ ‪(1 + i )−10 − 1‬‬ ‫⎢‬ ‫⎥‬ ‫⎣‬ ‫‪(1‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪)−2‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫⎦‬ ‫‪1.000.000‬‬ ‫=‬ ‫‪a‬‬ ‫×‬ ‫⎡‬ ‫‪(1,‬‬ ‫‪07)−2‬‬ ‫×‬ ‫‪(1, 07)−10‬‬ ‫‪−1‬‬ ‫⎤‬ ‫⎢‬ ‫‪(1, 07)−2‬‬ ‫‪−1‬‬ ‫⎥‬ ‫⎣‬ ‫⎦‬ ‫‪1.000.000 = a × 3,393034‬‬ ‫=‪a‬‬ ‫‪1.000.000‬‬ ‫‪3,393034‬‬ ‫‪a = 294.721,48DA‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺤل ﺍﻟﺘﻤﺭﻴﻥ)‪:(5‬‬ ‫‪ .1‬ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﻌﻘﺎﺭ ﻓﻲ ﺘﺎﺭﻴﺦ ﺍﻟﺸﺭﺍﺀ‬ ‫‪0 1 2 3 4 5 6 7 8 9 10‬‬ ‫‪a aa a aa a a‬‬ ‫‪)−8‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪1− (1+ i‬‬ ‫‪−1‬‬ ‫‪(1‬‬ ‫‪+‬‬ ‫‪i‬‬ ‫‪)−1‬‬ ‫‪i‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫‪1‬‬ ‫‪.3‬‬ ‫‪9‬‬ ‫‪6‬‬ ‫‪,3‬‬ ‫‪3‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫(‬ ‫‪1‬‬ ‫‪,06 )−‬‬ ‫‪8‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫(‬ ‫‪1‬‬ ‫‪,‬‬ ‫‪0‬‬ ‫‪6‬‬ ‫)‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪0,06‬‬ ‫‪V‬‬ ‫‪0‬‬ ‫=‬ ‫×‬ ‫‪V 0 = 341.396,33×6,209793×0,943396‬‬ ‫‪V0 = 2.000.000 D A‬‬ ‫‪ .2‬ﺤﺴﺎﺏ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ )ﺍﻟﺠﻤﻠﺔ(‬ ‫‪(1+ i )8 −1‬‬‫‪A =a‬‬ ‫‪i‬‬‫‪A‬‬ ‫=‬ ‫‪341‬‬ ‫‪.39‬‬ ‫×‪6,33‬‬ ‫‪(1,06 )8‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪0,06‬‬‫‪A = 341.396,33×9,897468‬‬‫‪A = 3.378.959,22DA‬‬ ‫ﻴﻤﻜﻥ ﺍﻟﺤﺼﻭل ﻋﻠﻰ ﻨﻔﺱ ﺍﻟﻨﺘﻴﺠﺔ ﺘﻘﺭﻴﺒﹰﺎ ﻜﻤﺎ ﻴﻠﻲ ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪A = V0 (1 + i )9‬‬ ‫‪A =2.000.000×1,68948‬‬ ‫‪A = 3.378.960DA‬‬ ‫‪ .3‬ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺎﺕ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﺴﻨﺔ ‪.15‬‬‫‪0 1 2 3 4 5 6 7 8 9 10‬‬ ‫‪15‬‬‫‪a‬‬ ‫‪a‬‬ ‫‪a‬‬ ‫=‪a‬‬ ‫‪a‬‬ ‫‪(1a‬‬ ‫‪+‬‬ ‫‪ia)15a−9‬‬ ‫‪A15‬‬ ‫‪A15‬‬ ‫‪A‬‬ ‫‪A15 = 3.378.959, 22(1, 06)6‬‬ ‫‪A15 = 3.378.959, 22×1,418519‬‬ ‫‪A15 = 4.793.117,85DA‬‬ ‫‪ .4‬ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺎﺕ ‪ 3‬ﺴﻨﻭﺍﺕ ﻗﺒل ﺍﻟﺩﻓﻊ ﺍﻷﻭل ‪:‬‬‫‪-1 0 1 2 3 4 5 6 7 8 9 10‬‬ ‫‪V‬‬ ‫‪a‬‬ ‫‪=aVoa(1‬‬ ‫‪+a‬‬ ‫‪i‬‬ ‫‪a)−1 a‬‬ ‫‪aa‬‬ ‫‪−3‬‬ ‫‪V −3 = 2.000.000 × (1, 06)−1‬‬ ‫‪V −3 = 2.000.000 × 0,943396‬‬ ‫‪V −3 = 1.886.792DA‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫ﺍﻟﻤﺠﺎل ﺍﻟﻤﻔﺎﻫﻴﻤﻲ ﺍﻟﺭﺍﺒﻊ‪ :‬ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﻤﺎﻟﻴﺔ ﻁﻭﻴﻠﺔ ﺍﻷﺠل‬ ‫ﺍﻟﻭﺤﺩﺓ )‪ :(17‬ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﻭﺽ‬ ‫ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ‪:‬‬ ‫ـ ﻴﻨﺠﺯ ﺠﺩﻭل ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ‪.‬‬ ‫ﺍﻟﻤﺩﺓ ﺍﻟﻼﺯﻤﺔ ‪ 08 :‬ﺴﺎﻋﺎﺕ‬ ‫ﺍﻟﻤﺭﺍﺠﻊ‪ :‬ﺍﻟﻜﺘﺏ ﺍﻟﻤﺩﺭﺴﻴﺔ ﺍﻟﻤﻘﺭﺭﺓ ‪.‬‬ ‫ﻤﺅﺸﺭﺍﺕ ﺍﻟﺘﻘﻭﻴﻡ ﺍﻟﺫﺍﺘﻲ‪:‬‬ ‫ ﹸﺘﺤﺩﺩ ﺍﺴﺘﻬﻼﻙ ﻭ ﻓﺎﺌﺩﺓ ﻜل ﺩﻓﻌﺔ ﺜﺎﺒﺘﺔ‪.‬‬ ‫ ﹸﺘﺤﺩﺩ ﺍﻟﻌﻼﻗﺎﺕ ﺒﻴﻥ ﻋﻨﺎﺼﺭ ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ‪.‬‬ ‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫‪ .1‬ﺘﻌﺭﻴﻑ ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ‬ ‫‪ .2‬ﺠﺩﻭل ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ‬ ‫‪ .3‬ﺍﻟﻌﻼﻗﺎﺕ ﺒﻴﻥ ﻋﻨﺎﺼﺭ ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ‬ ‫‪ .4‬ﺃﻨﺸﻁﺔ ﺍﻹﻋﻼﻡ ﺍﻵﻟﻲ ﻭ ﺤﻠﻭﻟﻬﺎ‬ ‫‪ .5‬ﺃﺴﺌﻠﺔ ﺍﻟﺘﻘﻭﻴﻡ ﺍﻟﺫﺍﺘﻲ‬ ‫‪ .6‬ﺃﺠﻭﺒﺔ ﺍﻟﺘﻘﻭﻴﻡ ﺍﻟﺫﺍﺘﻲ‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪ .1‬ﺘﻌﺭﻴﻑ ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ‬‫ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ ﻫﻭ ﺍﻟﻘﺭﺽ ﺍﻟﺫﻱ ﻴﺘﻡ ﺒﻴﻥ ﺸﺨﺼﻴﻥ )ﻁﺒﻴﻌﻴﻴﻥ ﺃﻭ ﺍﻋﺘﺒﺎﺭﻴﻴﻥ( ﻭﻴﺴﻤﻰ ﻫـﺫﺍ ﺍﻟﻨـﻭﻉ ﻤـﻥ‬‫ﺍﻟﻘﺭﻭﺽ ﺒﺎﻟﻘﺭﻭﺽ ﻏﻴﺭ ﺍﻟﻤﺠﺯﺌﺔ)‪ (Emprunts Indivis‬ﺤﻴﺙ ﻴﺤﺘﻭﻱ ﻋﻘﺩ ﺍﻟﻘﺭﺽ ﻋﻠﻰ ‪‬ﻤﻘﺭﺽ ﻭﺍﺤﺩ‪.‬‬ ‫ﻓﻲ ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ ﻴﻠﺘﺯﻡ ﺍﻟﻤﻘﺘﺭﺽ ﻋﺎﺩﺓ ﺒﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫ﺃ ـ ﺩﻓﻊ ﻓﻭﺍﺌﺩ ﺒﺼﻔﺔ ﺩﻭﺭﻴﺔ ﻋﻠﻰ ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﻤﻘﺘﺭﺽ ﻭ ﻏﻴﺭ ﺍﻟﻤﺴﺩﺩ‪.‬‬‫ﺏ ـ ﺘﺴﺩﻴﺩ ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﻤﻘﺘﺭﺽ‪ :‬ﻴﺩﻋﻰ ﻫﺫﺍ ﺍﻟﺘﺴﺩﻴﺩ ﺒـ \" ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ \" ﻭ ﻗﺩ ﻴﺘﻡ ﺘﺴﺩﻴﺩ ﺍﻟﻘـﺭﺽ‬ ‫ﺩﻓﻌﺔ ﻭﺍﺤﺩﺓ ﺃﻭ ﻋﻠﻰ ﻋﺩﺓ ﺩﻓﻌﺎﺕ ﻓﻲ ﺃﻏﻠﺏ ﺍﻷﺤﻴﺎﻥ‪.‬‬ ‫ﻓﺎﻟﻤﺩﻴﻥ ﻴﻘﻭﻡ ﺩﻭﺭﻴﹰﺎ ﺒﺘﺴﺩﻴﺩ ﺩﻓﻌﺔ ﺘﺤﺘﻭﻱ ﻋﻠﻰ‪:‬‬ ‫ﺍﻟﺩﻓﻌﺔ = ﻓﺎﺌﺩﺓ ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﻤﺘﺒﻘﻲ ‪ +‬ﺍﻻﺴﺘﻬﻼﻙ‬ ‫ﻭ ﺴﺘﻘﺘﺼﺭ ﺩﺭﺍﺴﺘﻨﺎ ﻋﻠﻰ ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﻭﺽ ﺒﻭﺍﺴﻁﺔ ﺍﻟﺩﻓﻌﺎﺕ ﺍﻟﺜﺎﺒﺘﺔ‪.‬‬ ‫‪ .2‬ﺠﺩﻭل ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ‬ ‫‪ .1.2‬ﻋﻨﺎﺼﺭ ﺍﻟﺠﺩﻭل‬ ‫ﺇﺫﺍ ﺤﻠﹼﻠﻨﺎ ﻋﻨﺎﺼﺭ ﻜل ﺩﻓﻌﺔ ﻤﻥ ﺍﻟﺩﻓﻌﺎﺕ ﻴﻤﻜﻨﻨﺎ ﻭﻀﻊ ﺠﺩﻭل ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ ﺍﻟﻌﺎﺩﻱ‪.‬‬ ‫ﻓﺈﺫﺍ ﺭﻤﺯﻨﺎ ﺒـ ‪ : V0 :‬ﻟﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﻤﻘﺘﺭﺽ )ﺃﺼل ﺍﻟﻘﺭﺽ( ﻓﻲ ﺍﻟﺯﻤﻥ ﺼﻔﺭ‪.‬‬‫‪ : an , ..., a3 , a2 , a1‬ﺍﻟﺩﻓﻌﺎﺕ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺘﺩﻓﻊ ﺍﻷﻭﻟﻰ ﺴﻨﺔ ﺒﻌﺩ ﺇﻤﻀﺎﺀ ﺍﻟﻌﻘﺩ ﻭﺍﻟﺜﺎﻨﻴﺔ ﺴﻨﺔ ﻤـﻥ ﺒﻌـﺩ ﻭ‬ ‫ﻫﻜﺫﺍ ﺃﻱ ﺃﻨﻨﺎ ﺃﻤﺎﻡ ﺩﻓﻌﺎﺕ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ‪.‬‬ ‫‪ : An , ..., A3 , A2 , A1‬ﺍﻻﺴﺘﻬﻼﻜﺎﺕ ﺍﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺘﻲ ﺘﺤﺘﻭﻴﻬﺎ ﺍﻟﺩﻓﻌﺔ ﺍﻷﻭل‪ ،‬ﺍﻟﺜﺎﻨﻴﺔ ‪ .....‬ﺇﻟـﻰ ﻏﺎﻴـﺔ‬ ‫ﺍﻟﺩﻑﻋﺔ ﺍﻷﺨﻴﺭﺓ)‪.(n‬‬ ‫‪ :V n , ...,V 3 ,V 2 ,V1‬ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﻤﺘﺒﻘﻲ ﺒﻌﺩ ﺘﺴﺩﻴﺩ ﺍﻟﺩﻓﻌﺔ ﺍﻷﻭﻟﻰ‪ ،‬ﺍﻟﺜﺎﻨﻴﺔ‪ ،‬ﺍﻟﺜﺎﻟﺜﺔ‪.... ،‬‬ ‫ﺍﻟﺩﻓﻌﺔ)‪.(n‬‬ ‫‪ : i‬ﺍﻟﻤﻌﺩل ﺍﻻﺴﻤﻲ ﻟﻠﻘﺭﺽ‪.‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪ :‬ﻤﺩﺓ ﺍﻟﻘﺭﺽ )ﻤﺩﺓ ﺍﻟﺘﺴﺩﻴﺩ(‪n.‬‬ ‫‪.2.2‬ﺒﻨﺎﺀ ﺍﻟﺠﺩﻭل‬ ‫ﺒﺎﺴﺘﺨﺩﺍﻡ ﺍﻟﺭﻤﻭﺯ ﺍﻟﺴﺎﺒﻘﺔ ﹸﻨﻌﺩ ﺠﺩﻭل ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ ﻜﻤﺎ ﻴﻠﻲ‪:‬‬ ‫اﻟﻘﺮض‬ ‫اﻟﻤﺘﺒﻘﻲ‬‫اﻟﻘﺮض اﻟﻤﺘﺒﻘﻲ ﻓﻲ‬ ‫اﻟﺪﻓﻌﺔ‬ ‫اﻻﺳﺘﻬﻼك‬ ‫اﻟﻔﺎﺋﺪة‬ ‫ﻓﻲ ﺏﺪایﺔ‬ ‫اﻟﻮﺣﺪات‬‫ﻥﻬﺎیﺔ اﻟﻮﺣﺪة اﻟﺰﻣﻨﻴﺔ‬ ‫اﻟﻮﺣﺪة‬ ‫اﻟﺰﻣﻨﻴﺔ‬ ‫‪V1 =V 0 − A1‬‬ ‫اﻟﺰﻣﻨﻴﺔ‬ ‫‪V 2 =V1 − A2‬‬ ‫‪V 3 =V 2 − A3‬‬ ‫‪a1 =V 0i + A1‬‬ ‫‪A1 V 0i V 0‬‬ ‫‪1‬‬ ‫‪a2 =V1i + A2‬‬ ‫‪A2 V1i V1‬‬ ‫‪2‬‬ ‫‪a3 =V 2i + A3‬‬ ‫‪A3 V 2i V 2‬‬ ‫‪3‬‬‫‪V p −1 =V p −2 − Ap −1 ap −1 =V p −2i + Ap −1‬‬ ‫‪Ap −1‬‬ ‫‪V p −2i‬‬ ‫‪V p −2‬‬ ‫‪p-1‬‬ ‫‪P‬‬‫‪V p =V p −1 − Ap‬‬ ‫‪ap =V p −1i + Ap‬‬ ‫‪Ap‬‬ ‫‪V p −1i‬‬ ‫‪V p −1‬‬ ‫‪P+1‬‬‫‪V p +1 =V p − Ap +1‬‬ ‫‪ap +1 =V p i + Ap +1‬‬ ‫‪Ap +1‬‬ ‫‪Vpi‬‬ ‫‪Vp‬‬ ‫‪n-1‬‬ ‫‪An −1‬‬ ‫‪Vn−2i‬‬ ‫‪Vn−2‬‬ ‫‪n‬‬‫‪V n −1 =V n −2 − An −1 an −1 =V n −2i + An −1‬‬‫‪V n =V n −1 − An = 0‬‬ ‫‪an =V n −1i + An‬‬ ‫‪An‬‬ ‫‪V n −1i‬‬ ‫‪V n −1‬‬ ‫ﻤﻼﺤﻅﺎﺕ ‪:‬‬‫‪ .1‬ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﻤﺘﺒﻘﻲ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ‪ Vn‬ﺃﻱ ﺒﻌﺩ ﺘﺴﺩﻴﺩ ﺍﻟﺩﻓﻌﺔ ﺍﻷﺨﻴﺭﺓ ﻴﺴﺎﻭﻱ ﺍﻟﺼﻔﺭ ﺃﻱ ‪:‬‬ ‫‪V n −1 = An‬‬ ‫‪ .2‬ﺍﻟﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﺃﻱ ‪an = ... = a3 = a2 = a1‬‬ ‫‪ .3‬ﺃﺼل ﺍﻟﻘﺭﺽ ﻴﺴﺎﻭﻱ ﻤﺠﻤﻭﻉ ﺍﻻﺴﺘﻬﻼﻜﺎﺕ ﺃﻱ ‪:‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪V 0 = A1 + A 2 + A 3 + ... + An‬‬ ‫‪n‬‬ ‫= ‪∑V 0‬‬ ‫‪Ap‬‬ ‫‪p =1‬‬ ‫‪ .4‬ﻤﺠﻤﻭﻉ ﺍﻟﺩﻓﻌﺎﺕ ﺘﺴﺎﻭﻱ ﻤﺠﻤﻭﻉ ﺍﻟﻔﻭﺍﺌﺩ ﺯﺍﺌﺩ ﻤﺠﻤﻭﻉ ﺍﻻﺴﺘﻬﻼﻜﺎﺕ ‪:‬‬ ‫‪nn n‬‬ ‫‪∑ap = ∑ Ap + ∑ I p‬‬ ‫‪p =1 p =1 p =1‬‬ ‫‪ .3‬ﺍﻟﻌﻼﻗﺎﺕ ﺒﻴﻥ ﻋﻨﺎﺼﺭ ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ‬ ‫‪ .1.3‬ﺃﺼل ﺍﻟﻘﺭﺽ ﻭ ﺍﻻﺴﺘﻬﻼﻜﺎﺕ‬ ‫ﺃ ـ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﺴﺘﻬﻼﻜﻴﻥ ﻤﺘﺘﺎﻟﻴﻴﻥ‬ ‫) ‪a n − a n − 1 = (V n − 1 i + A n ) − (V n − 2 i + A n − 1‬‬ ‫‪an − an −1 = V n −1i + An −V n −2i − An −1‬‬ ‫ﻤﻥ ﺠﺩﻭل ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ ﻟﺩﻴﻨﺎ‪V n −1 =V n − 2 − An −1 :‬‬ ‫ﻭ ﻤﻨﻪ‪:‬‬‫‪a n − a n −1 = (V n − 2 − An −1 )i + An −V n − 2 i − An −1‬‬‫‪an − an −1 = V n − 2i − An −1i + An −V n − 2i − An −1‬‬‫‪an − an −1 = − An −1i + An − An −1‬‬ ‫ﻭ ﻤﺎ ﺒﻤﺎ ﺃﻥ ﺍﻟﺩﻓﻌﺎﺕ ﻤﺘﺴﺎﻭﻴﺔ ﻓﺈﻥ ‪an − an −1 = 0 :‬‬ ‫‪− A n −1i + A n − A n −1 = 0‬‬ ‫‪A n = A n −1i + A n −1‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫) ‪A n = A n −1 (1 + i‬‬ ‫) ‪A n = A n −1 (1 + i‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺃﻱ ﺍﺴﺘﻬﻼﻙ ﺃﻱ ﺴﻨﺔ ﻴﺴﺎﻭﻱ ﺠﻤﻠﺔ ﺍﺴﺘﻬﻼﻙ ﺍﻟﺴﻨﺔ ﺍﻟﺘﻲ ﻗﺒﻠﻬﺎ ﻟﺴﻨﺔ ﻭﺍﺤﺩﺓ‬‫ﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ)‪ :(1‬ﻴﺴﺩﺩ ﻗﺭﺽ ﺒﻭﺍﺴﻁﺔ ‪ 5‬ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﺒﻤﻌﺩل ‪ %6‬ﺴﻨﻭﻴﹰﺎ‪ ،‬ﻤﺠﻤﻭﻉ ﺍﻻﻫﺘﻼﻜﻴﻥ ﺍﻟﺜﺎﻟﺙ ﻭ‬ ‫ﺍﻟﺭﺍﺒﻊ ‪50.000 DA‬‬ ‫ﺍﻟﻤﻁﻠﻭﺏ ‪ :‬ﺃﺤﺴﺏ ‪A5 , A4 , A3‬‬ ‫ﺍﻟﺤل ‪:‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪A3 + A4 = 50.000‬‬ ‫) ‪A4 = A3 (1 + i‬‬ ‫ـ ﺤﺴﺎﺏ ‪: A3‬‬ ‫‪A3 + A3 (1 + i ) = 50.000‬‬ ‫‪A3 (1 + 1 + 0, 06) = 50.000‬‬ ‫‪2, 06A3 = 50.000‬‬ ‫‪A3‬‬ ‫=‬ ‫‪50.000‬‬ ‫=‬ ‫‪2 4 .2 7 1 ,8 5 DA‬‬ ‫‪2,06‬‬ ‫‪A3 = 24.271,85DA‬‬ ‫ـ ﺤﺴﺎﺏ ‪: A4‬‬ ‫‪A4 = 24.271,85 × 1,06‬‬ ‫‪A4 = 25.728,15DA‬‬ ‫ـ ﺤﺴﺎﺏ ‪: A5‬‬ ‫) ‪A5 = A4 (1 + i‬‬ ‫‪A5 = 25.728,15 × 1, 06‬‬ ‫‪A5 = 27.271,84DA‬‬ ‫ﺏ ـ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻻﺴﺘﻬﻼﻜﺎﺕ ﻭ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻷﻭل‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫) ‪A2 = A1 (1 + i‬‬ ‫‪A3 = A2 (1 + i ) = A1 (1 + i )2‬‬ ‫‪A4 = A3 (1 + i ) = A1 (1 + i )3‬‬ ‫‪..............................................‬‬ ‫‪An = An −1 (1 + i ) = A1 (1 + i )n −1‬‬ ‫‪An = A1 (1 + i )n −1‬‬ ‫ﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ)‪ :(2‬ﺃﺤﺴﺏ ﻟﻠﻤﺜﺎل ﺍﻟﺴﺎﺒﻕ ‪. A1‬‬ ‫‪A5 = A1 (1 + i )4‬‬ ‫‪A1‬‬ ‫=‬ ‫‪A5‬‬ ‫‪)4‬‬ ‫‪(1 + i‬‬ ‫‪A1 = A5 (1 + i )−4‬‬ ‫‪A1 = 27.271, 84(1, 06)−4‬‬ ‫‪A1 = 27.271, 84 × 0,792093‬‬ ‫‪A1 = 21.601,85DA‬‬ ‫ﺟـ ـ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﺴﺘﻬﻼﻜﻴﻥ ﻤﺘﻌﺎﻗﺒﻴﻥ‬ ‫ﻭ ﺒﺼﻔﺔ ﻋﺎﻤﺔ ﻭ ﻤﻬﻤﺎ ﻜﺎﻥ ﺍﻻﺴﺘﻬﻼﻙ ﻓﺈﻥ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﺴﺘﻬﻼﻜﻴﻥ ﻤﺘﻌﺎﻗﺒﻴﻥ ﻫﻲ‪:‬‬ ‫)‪An = A1 (1 + i )n −1 ........(1‬‬ ‫‪Ap = A1 (1 + i )p −1‬‬ ‫)‪A1 = Ap (1 + i )−(p −1) .....(2‬‬ ‫ﺒﺘﻌﻭﻴﺽ )‪ (2‬ﻓﻲ )‪ (1‬ﻨﺠﺩ‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫‪An = Ap (1 + i )−( p −1) (1 + i )n −1‬‬ ‫‪An = Ap (1 + i )− p +1 (1 + i )n −1‬‬ ‫‪An = Ap (1 + i )− p +1+n −1‬‬ ‫‪An = Ap (1 + i )n −p‬‬ ‫ﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ)‪:(3‬‬‫‪‬ﻴﺴﺩﺩ ﻗﺭﺽ ﺒﻭﺍﺴﻁﺔ ‪ 12‬ﺩﻓﻌﺔ ﺜﺎﺒﺘﺔ ﺒﻤﻌﺩل ‪ %5‬ﺴﻨﻭﻴﹰﺎ‪ ،‬ﺇﺫﺍ ﻜﺎﻥ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻟﺜﺎﻟﺙ ‪20.000 DA‬‬ ‫ﺍﻟﻤﻁﻠﻭﺏ‪ :‬ﺃﺤﺴﺏ ‪ A12 , A8 , A5‬ﺒﺩﻻﻟﺔ ‪A3‬‬ ‫ﺍﻟﺤل‪:‬‬ ‫ـ ﺤﺴﺎﺏ ‪: A12‬‬ ‫‪An = Ap (1 + i )n −p‬‬ ‫‪A12 = A3 (1 + i )12−3‬‬ ‫‪A12 = A3 (1 + i )9‬‬ ‫‪A12 = 20.000(1, 05)9 = 31.026,56DA‬‬ ‫ـ ﺤﺴﺎﺏ ‪: A8‬‬ ‫‪A8 = A3 (1 + i )8−3‬‬ ‫‪A8 = A3 (1 + i )5‬‬ ‫‪A12 = 20.000(1, 05)5 = 25.525,63DA‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ـ ﺤﺴﺎﺏ ‪: A5‬‬ ‫‪A5 = A3 (1 + i )5−3‬‬ ‫‪A5 = A3 (1 + i )2‬‬ ‫‪A5 = 20.000 × (1, 06)2 = 22.472DA‬‬ ‫ﺏ ـ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺃﺼل ﺍﻟﻘﺭﺽ ﻭ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻷﻭل‬ ‫ﺃﺼل ﺍﻟﻘﺭﺽ ﻴﺴﺎﻭﻱ ﻤﺠﻤﻭﻉ ﺍﻻﺴﺘﻬﻼﻜﺎﺕ ﺃﻱ‪:‬‬‫‪V0 = A1 + A2 + A3 + ... + An‬‬‫‪V0 = A1 + A1(1+ i ) + A2(1+ i )2 + A3(1+ i )3 + ... + A1(1+ i )n−1‬‬ ‫ﻨﻼﺤﻅ ﺃﻥ ﺃﺼل ﺍﻟﻘﺭﺽ ﻴﺴﺎﻭﻱ ﻤﺘﺘﺎﻟﻴﺔ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻷﻭل ﻭ ﻫﻲ ﻤﺘﺘﺎﻟﻴﺔ ﻫﻨﺩﺴﻴﺔ ﻤﺠﻤﻭﻋﻬﺎ ﻫﻭ‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪A1‬‬ ‫‪(1 +‬‬ ‫‪i )n‬‬ ‫‪−1‬‬ ‫‪i‬‬ ‫ﺃﻱ ﺃﺼل ﺍﻟﻘﺭﺽ ﻫﻭ ﺠﻤﻠﺔ ﻤﺘﺘﺎﻟﻴﺔ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻷﻭل ﺨﻼل ﺍﻟﻤﺩﺓ )‪.(n‬‬ ‫ﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ)‪ :(4‬ﺃﺤﺴﺏ ﺃﺼل ﺍﻟﻘﺭﺽ ﻟﻠﻤﺜﺎل ﺍﻟﺘﻭﻀﻴﺤﻲ)‪(2‬‬‫‪V0‬‬ ‫=‬ ‫‪21.601,85‬‬ ‫‪(1, 06)5 − 1‬‬ ‫‪0, 06‬‬‫‪V 0 = 21.601,85 × 5,637093=121.771,63DA‬‬ ‫‪ .2.3‬ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﺩﻓﻌﺎﺕ ﻭ ﺍﻻﺴﺘﻬﻼﻜﺎﺕ‬ ‫ﺃ ـ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﺩﻓﻌﺔ ﻭ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻷﺨﻴﺭ‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫ﺭﺃﺱ ﺍﻟﻤﺎل ﺍﻟﻤﺘﺒﻘﻲ ﻓﻲ ﻨﻬﺎﻴﺔ ﺍﻟﻤﺩﺓ ‪ Vn‬ﺃﻱ ﺒﻌﺩ ﺘﺴﺩﻴﺩ ﺍﻟﺩﻓﻌﺔ ﺍﻷﺨﻴﺭﺓ ﻴﺴﺎﻭﻱ ﺍﻟﺼﻔﺭ ﺃﻱ ‪:‬‬ ‫‪V n −1 = An‬‬ ‫ﻟﺩﻴﻨﺎ ﻤﻥ ﺠﺩﻭل ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ‪:‬‬ ‫‪a =V n-1i + An‬‬ ‫ﻭ ﻤﻨﻪ‪:‬‬ ‫‪a = Ani + An‬‬ ‫) ‪a = An (1 + i‬‬ ‫) ‪a = An (1 + i‬‬ ‫ﺃﻱ ﻤﺒﻠﻎ ﺍﻟﺩﻓﻌﺔ ﻴﺴﺎﻭﻱ ﺠﻤﻠﺔ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻷﺨﻴﺭ ﻟﻤﺩﺓ ﺴﻨﺔ‪.‬‬ ‫ﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ)‪:(5‬‬ ‫ﺃﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ﻟﻠﻤﺜﺎل ﺍﻟﺘﻭﻀﻴﺤﻲ)‪(1‬‬ ‫) ‪a = A5(1 + i‬‬ ‫‪a = 27.271,84 × 1,06 = 28.908,15DA‬‬ ‫ﺏ ـ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﺩﻓﻌﺔ ﻭ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻷﻭل‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫)‪a = An (1 + i ).............(1‬‬ ‫)‪An = A1(1 + i )n−1......(2‬‬ ‫ﺒﺘﻌﻭﻴﺽ )‪ (2‬ﻓﻲ )‪ (1‬ﻨﺠﺩ‪a = A1(1 + i )n−1(1 + i ) :‬‬ ‫‪a = A1 (1 + i )n‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺃﻱ ﻤﺒﻠﻎ ﺍﻟﺩﻓﻌﺔ ﻴﺴﺎﻭﻱ ﺠﻤﻠﺔ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻷﻭل ﻟـ )‪ (n‬ﺴﻨﺔ‪.‬‬ ‫ﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ)‪: (6‬‬ ‫ﺃﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ﻟﻠﻤﺜﺎل ﺍﻟﺘﻭﻀﻴﺤﻲ)‪(2‬‬ ‫‪a = A1 (1 + i )5‬‬ ‫‪a = 21.601, 85 × (1, 06)5‬‬ ‫‪a = 21.601, 85 × 1,338225=28.908,15DA‬‬ ‫ﺠـ ـ ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﺩﻓﻌﺔ ﻭ ﺍﺴﺘﻬﻼﻙ ﻤﺎ‬ ‫ﻟﺩﻴﻨﺎ‪:‬‬ ‫)‪a = An (1 + i )..................(1‬‬ ‫) ‪A n = A p (1 + i )n − p ..........(2‬‬ ‫ﺒﺘﻌﻭﻴﺽ )‪ (2‬ﻓﻲ )‪ (1‬ﻨﺠﺩ‪:‬‬ ‫) ‪a = Ap (1 + i )n − p (1 + i‬‬ ‫‪a = A p (1 + i )n − p +1‬‬ ‫ﺃﻱ ﻤﺒﻠﻎ ﺍﻟﺩﻓﻌﺔ ﻴﺴﺎﻭﻱ ﺠﻤﻠﺔ ﺍﺴﺘﻬﻼﻙ ﺍﻟﺴﻨﺔ ) ‪ ( p‬ﺨﻼل ﺍﻟﻤﺩﺓ )‪(n − p + 1‬‬ ‫ﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ)‪:(7‬‬‫‪‬ﻴﺴﺩﺩ ﻗﺭﺽ ﺒﻭﺍﺴﻁﺔ ‪ 10‬ﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﺒﻤﻌﺩل ‪ %7‬ﺴﻨﻭﻴﹰﺎ‪ ،‬ﺇﺫﺍ ﻜﺎﻥ ﺍﻻﺴﺘﻬﻼﻙ ﺍﻟﺴﺎﺒﻊ ‪30.000 DA‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬ ‫ﺍﻟﻤﻁﻠﻭﺏ‪:‬‬ ‫ﺃﺤﺴﺏ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ‬ ‫ﺍﻟﺤل‪:‬‬ ‫‪a = Ap (1 + i )n−p +1‬‬ ‫‪a = A7 (1 + i )10−7+1‬‬ ‫‪a = A7 (1 + i )4‬‬ ‫‪a = 30.000 × (1,07)4 = 39.323,88DA‬‬ ‫‪ .3.3‬ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺃﺼل ﺍﻟﻘﺭﺽ ﻭ ﺍﻟﺩﻓﻌﺎﺕ ﺍﻟﺜﺎﺒﺘﺔ‬ ‫‪V0‬‬ ‫=‬ ‫‪A1‬‬ ‫‪(1 +‬‬ ‫‪i )n‬‬ ‫‪−‬‬ ‫)‪1 ...........(1‬‬ ‫ﻨﻌﻠﻡ ﺃﻥ‪:‬‬ ‫‪i‬‬ ‫ﻭ ﻨﻌﻠﻡ ﻜﺫﻟﻙ ﺃﻥ‪:‬‬ ‫‪a = A1(1 + i )n‬‬ ‫)‪A1 = a(1 + i )−n ..................(2‬‬ ‫ﻨﻌﻭﺽ )‪ (2‬ﻓﻲ )‪ (1‬ﻓﻨﺠﺩ‪:‬‬‫‪V0‬‬ ‫=‬ ‫‪a(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪(1 +‬‬ ‫‪i )n‬‬ ‫‪−‬‬ ‫)‪1 .......................(3‬‬ ‫‪i‬‬‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)−n (1 +‬‬ ‫‪i‬‬ ‫‪)n‬‬ ‫‪−‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬

‫ﺍﻹﺭﺴﺎل ‪3‬‬ ‫‪ 3‬ﺜﺎﻨﻭﻱ ﺘﺴﻴﻴﺭ ﻭﺍﻗﺘﺼﺎﺩ ﺍﻟﺘﺴﻴﻴﺭ ﺍﻟﻤﺤﺎﺴﺒﻲ ﻭ ﺍﻟﻤﺎﻟﻲ‬‫ﺃﻱ ﺃﺼل ﺍﻟﻘﺭﺽ ﻋﺒﺎﺭﺓ ﻋﻥ ﺍﻟﻘﻴﻤﺔ ﺍﻟﺤﺎﻟﻴﺔ ﻟﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺩﻓﻌﺎﺕ ﺍﻟﺜﺎﺒﺘﺔ‪ ،‬ﻭ ﻤﻨﻪ ﻗﻴﻤﺔ ﺍﻟﺩﻓﻌﺔ ﺘﺴﺎﻭﻱ‪:‬‬ ‫‪a‬‬ ‫‪=V 0‬‬ ‫‪1−‬‬ ‫‪i‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪(1 +‬‬ ‫ﻤﻥ ﺍﻟﻌﻼﻗﺔ )‪ (3‬ﻴﻤﻜﻥ ﺃﻥ ﻨﺴﺘﺨﺭﺝ ﻜﺫﻟﻙ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺒﻴﻥ ﺃﺼل ﺍﻟﻘﺭﺽ ﻭ ﺍﻟﺩﻓﻌﺎﺕ‪.‬‬ ‫‪V 0 (1 +‬‬ ‫‪i )n‬‬ ‫=‬ ‫‪a‬‬ ‫‪(1 +‬‬ ‫‪i )n‬‬ ‫‪−‬‬ ‫‪1‬‬ ‫‪i‬‬ ‫ﺃﻱ ﺠﻤﻠﺔ ﺃﺼل ﺍﻟﻘﺭﺽ ﺘﺴﺎﻭﻱ ﺠﻤﻠﺔ ﻤﺘﺘﺎﻟﻴﺔ ﺍﻟﺩﻓﻌﺎﺕ‪.‬‬ ‫ﻤﺜﺎل ﺘﻭﻀﻴﺤﻲ)‪:(8‬‬ ‫ﺃﺤﺴﺏ ﺃﺼل ﺍﻟﻘﺭﺽ ﻟﻠﻤﺜﺎل ﺍﻟﺘﻭﻀﻴﺤﻲ)‪. (7‬‬ ‫ﺍﻟﺤل‪:‬‬ ‫‪V0‬‬ ‫=‬ ‫‪a‬‬ ‫‪1‬‬ ‫‪−‬‬ ‫‪(1 +‬‬ ‫‪i‬‬ ‫‪)−n‬‬ ‫‪i‬‬ ‫‪V0‬‬ ‫=‬ ‫‪39.323, 88 1‬‬ ‫‪−‬‬ ‫‪(1, 07 )−10‬‬ ‫‪0,07‬‬ ‫‪V 0 = 39.323, 88 × 7,023581‬‬ ‫‪V0 = 276.194,45DA‬‬ ‫‪ .4.3‬ﺍﻟﻌﻼﻗﺔ ﺒﻴﻥ ﺍﻟﻔﻭﺍﺌﺩ ﻭ ﺍﻻﺴﺘﻬﻼﻜﺎﺕ‬‫ﺍﻟﻔﻭﺍﺌﺩ ﺘﺸﻜل ﻤﺘﺘﺎﻟﻴﺔ ﺤﺴﺎﺒﻴﺔ ﻤﺘﻨﺎﻗﺼﺔ ﺃﻱ ﻓﺎﺌﺩﺓ ﺍﻟﺴﻨﺔ ﺍﻷﻭﻟﻰ ﺃﻜﺒﺭ ﻤﻥ ﻓﺎﺌﺩﺓ ﺍﻟﺴﻨﺔ ﺍﻟﺜﺎﻨﻴﺔ ﻷﻥ ﺭﺃﺱ ﺍﻟﻤﺎل‬ ‫ﺍﻟﻤﺘﺒﻘﻲ ﻴﺘﻨﺎﻗﺹ ﻤﻥ ﻓﺘﺭﺓ ﻷﺨﺭﻯ‪.‬‬ ‫ﺇﺫﺍ ﻭﻀﻌﻨﺎ ﺍﻟﻔﺭﻕ ﺒﻴﻥ ﻓﺎﺌﺩﺓ ﺴﻨﺘﻴﻥ ﻤﺘﺘﺎﻟﻴﺘﻴﻥ ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ ﺤﺴﺏ ﺠﺩﻭل ﺍﺴﺘﻬﻼﻙ ﺍﻟﻘﺭﺽ‪:‬‬ ‫) ‪I n−1 − I n = (an−1 − An−1 ) − (an − An‬‬ ‫‪I n−1 − I n = an−1 − An−1 − an + An−1‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ﺍﻟﺩﻓﻌﺎﺕ ﺜﺎﺒﺘﺔ ﻓﺈﻥ‪:‬‬ ‫‪I n−1 − I n = − An−1 + An‬‬ ‫‪I n−1 − I n = An − An−1‬‬ ‫‪I n −1 − I n = An −1 (1 + i ) − An −1‬‬ ‫‪I n −1 − I n = An −1 + An −1i − An −1‬‬ ‫‪I n −1 − I n = An −1i‬‬