Exercise 3.4.6

Answers

Follow example 3, we need minimize NMF = A XF2 + I+(C) + I+(R) with X = CR, so we can compute

  • $(X_1,C_1) = argmin [A XF2 + 1 2ρX CR0 + U0F2] = $withX ≥ 0,C ≥ 0

Let C = [c11c12 c21c22 ] ,X = [x11x12 x21x22 ] , we have

A XF2 + 1 2ρX CR0 + U0F2 = [ 2 x11 1 x12 1 x212 x22 ] F2 + 1 2ρ [x11 2c11 + 1 x12 c12 x21 2c21 x22 c22 + 1 ]F2 = (2 x11)2 + (1 x 12)2 + (1 x 21)2 + (2 x 22)2 + 1 2ρ ((x11 2c11 + 1)2 + (x 12 c12)2 + (x 21 2c21)2 + (x 22 c22 + 1)2)

Now take derivatives w.r.t. to X,C, we have

NMF x11 = 2(2 x11) + ρ(x11 2c11 + 1) = (ρ + 2)x11 2ρc11 + ρ 4 = 0 NMF x12 = 2(1 x12) + ρ(x12 c12) = (ρ + 2)x12 ρc12 2 = 0 NMF x21 = 2(1 + x21) + ρ(x21 2c21) = (ρ + 2)x21 2ρc21 + 2 = 0 NMF x22 = 2(2 x22) + ρ(x22 c22 + 1) = (ρ + 2)x22 ρc22 + ρ 4 = 0 NMF c11 = 2ρ(x11 2c11 + 1) = 0 NMF c12 = ρ(x12 c12) = 0 NMF c21 = ρ(x21 2c21) = 0 NMF c22 = ρ(x22 c22 + 1) = 0

Note, we also need satisfy X 0 and C 0, Solve the equations we have X1 = [21 0 2 ] and C1 = [1.51 0 3 ]

  • Now solve for R1 = argminX1 C1R + U0F2 with R 0, that is

X1 C1R + U0F2 = [2 1.5R11 R211 1.5R12 R22 3R21 2 3R22 ] F2 = (2 1.5R11 R21)2 + (1 1.5R 12 R22)2 + (3R 21)2 + (2 3R 22)2

Take derivatives w.r.t. R, we have

R11 = 2(1.5)(2 1.5R11 R21) = 0 R12 = 2(1.5)(1 1.5R12 R22) = 0 R21 = 2(2 1.5R11 R21) + 18R21 = 0 R22 = 2(1 1.5R12 R22) 6(2 3R22) = 0

Solve for the equations we have R = [4 32 9 02 3 ] 0

  • Lastly, we have U1 = U0+X1C1R1 = [51 0 5 ]

Note X1 = C1R1 = [21 0 2 ] which is what we have solved in problem 5.

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2020-03-20 00:00
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