5
= ( 1 (1 + 100 ∗ ) −1+ ) =1 + ( 100 (1 + 100 ∗ ) −1+ ) =

𝑐 - thogharkogh.cba.am · 𝑃= 𝑐 𝑓 ∑ (1 (1+ 𝑦 100∗𝑓) 𝑖−1+𝑡) 𝑁 𝑖=1 + (100 (1+ 𝑦 100∗𝑓) 𝑁−1+𝑡) 𝑡= 𝑆𝑁

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Page 1: 𝑐 - thogharkogh.cba.am · 𝑃= 𝑐 𝑓 ∑ (1 (1+ 𝑦 100∗𝑓) 𝑖−1+𝑡) 𝑁 𝑖=1 + (100 (1+ 𝑦 100∗𝑓) 𝑁−1+𝑡) 𝑡= 𝑆𝑁

𝐷𝑃 =𝑐

𝑓∑

(

1

(1 +𝑦

100 ∗ 𝑓)𝑖−1+𝑡

)

𝑁

𝑖=1

+

(

100

(1 +𝑦

100 ∗ 𝑓)𝑁−1+𝑡

)

𝑡 =𝐷𝑆𝑁

𝐷𝐶𝐶

Page 2: 𝑐 - thogharkogh.cba.am · 𝑃= 𝑐 𝑓 ∑ (1 (1+ 𝑦 100∗𝑓) 𝑖−1+𝑡) 𝑁 𝑖=1 + (100 (1+ 𝑦 100∗𝑓) 𝑁−1+𝑡) 𝑡= 𝑆𝑁
Page 3: 𝑐 - thogharkogh.cba.am · 𝑃= 𝑐 𝑓 ∑ (1 (1+ 𝑦 100∗𝑓) 𝑖−1+𝑡) 𝑁 𝑖=1 + (100 (1+ 𝑦 100∗𝑓) 𝑁−1+𝑡) 𝑡= 𝑆𝑁
Page 4: 𝑐 - thogharkogh.cba.am · 𝑃= 𝑐 𝑓 ∑ (1 (1+ 𝑦 100∗𝑓) 𝑖−1+𝑡) 𝑁 𝑖=1 + (100 (1+ 𝑦 100∗𝑓) 𝑁−1+𝑡) 𝑡= 𝑆𝑁
Page 5: 𝑐 - thogharkogh.cba.am · 𝑃= 𝑐 𝑓 ∑ (1 (1+ 𝑦 100∗𝑓) 𝑖−1+𝑡) 𝑁 𝑖=1 + (100 (1+ 𝑦 100∗𝑓) 𝑁−1+𝑡) 𝑡= 𝑆𝑁