can you please explain this to me . Thank you!
Project by Matlab 0. The System | Description 1. N, M and T are any given positive integers. 2. A N-dimessional vector: m = T1, T2, ..., AN-1, an] 21,1 (2,1 (1,2 12,2 01,N-1 02,N-1 21,N 02,N 3. A N x N matrix: A = (ai,j)nxn = aN-1,1 an,1 AN-1,2 an,2 an-1,N-1 ON,N-1 ON-1,N ANN bi(1) 61(2) bi(M 1) bi(M) b2(1) b2(2) b2(M 1) b2(M) 4. ANXM matrix: B = (bi(j))nxm = bn-1(1) bn-1(2) bn-1(M 1) bp-1(M) bn(1) bn(2) bn(M 1) bn(M) 5. A T-dimensional vector: 0 [01,02, ..., OT-1,OT], where 0t (d, D) for any te {1,2, ..., T} and the given values d and D. 6. Purpose: Numerical results, 1 and 2-dimensional diagram. 1 Phase I 1. Task 1-1 (Forward Induction Method): (a) Initialization: For any i E {1,2, ..., N}, find ai(i) = 7;b(01). (b) Recursion: For any j e {1, 2, ..., N} and te {2, ..., T} , find N 04(j) = at-1(i)ai,jb;(Or). i=1 (c) Termination: find PO|A) = at(j). ar j=1 2 Numerical Results in Phase I 21,1 a 1,2 = 1. For the special values, for example, (a) N=3, M=4, and T=6; (b) A 3-dimensional vector: A = [71, 72, 713] [x,0.6 X, 0.4). 01,N 0.1 0.2 0.7 (c) A 3 x 3 matrix: A = = (ai,j)3x3 = 02,1 12,2 02, N 0.2 0.3 0.5 an,1 an,2 ANN 0.3 0.4 0.3 bi(1) b(2) b(3) b(4) 0.4 0.1 0.2 0.3 (d) A 3x4 matrix: B = (bi(j))3x4 b2(1) b2(2) b2(3) b2(4) 0.3 0.2 0.2 0.4 63(1) 63(2) 63(3) 63(4) 0.4 0.2 0.1 (e) A 6-dimensional vector: 0 = [01,02, ...,O5, Oc] = (2, 1, 3, 4, 9, 8]. 1-1 0.3 2. For the above given special values, if x=0,1, 0.3, 0.5, respectively, find the following Termination value, 3 P(0|1) (;). (2) j=1 3. For the above given special values, draw the curve of above termination value for x from 0 to 0.5. Project by Matlab 0. The System | Description 1. N, M and T are any given positive integers. 2. A N-dimessional vector: m = T1, T2, ..., AN-1, an] 21,1 (2,1 (1,2 12,2 01,N-1 02,N-1 21,N 02,N 3. A N x N matrix: A = (ai,j)nxn = aN-1,1 an,1 AN-1,2 an,2 an-1,N-1 ON,N-1 ON-1,N ANN bi(1) 61(2) bi(M 1) bi(M) b2(1) b2(2) b2(M 1) b2(M) 4. ANXM matrix: B = (bi(j))nxm = bn-1(1) bn-1(2) bn-1(M 1) bp-1(M) bn(1) bn(2) bn(M 1) bn(M) 5. A T-dimensional vector: 0 [01,02, ..., OT-1,OT], where 0t (d, D) for any te {1,2, ..., T} and the given values d and D. 6. Purpose: Numerical results, 1 and 2-dimensional diagram. 1 Phase I 1. Task 1-1 (Forward Induction Method): (a) Initialization: For any i E {1,2, ..., N}, find ai(i) = 7;b(01). (b) Recursion: For any j e {1, 2, ..., N} and te {2, ..., T} , find N 04(j) = at-1(i)ai,jb;(Or). i=1 (c) Termination: find PO|A) = at(j). ar j=1 2 Numerical Results in Phase I 21,1 a 1,2 = 1. For the special values, for example, (a) N=3, M=4, and T=6; (b) A 3-dimensional vector: A = [71, 72, 713] [x,0.6 X, 0.4). 01,N 0.1 0.2 0.7 (c) A 3 x 3 matrix: A = = (ai,j)3x3 = 02,1 12,2 02, N 0.2 0.3 0.5 an,1 an,2 ANN 0.3 0.4 0.3 bi(1) b(2) b(3) b(4) 0.4 0.1 0.2 0.3 (d) A 3x4 matrix: B = (bi(j))3x4 b2(1) b2(2) b2(3) b2(4) 0.3 0.2 0.2 0.4 63(1) 63(2) 63(3) 63(4) 0.4 0.2 0.1 (e) A 6-dimensional vector: 0 = [01,02, ...,O5, Oc] = (2, 1, 3, 4, 9, 8]. 1-1 0.3 2. For the above given special values, if x=0,1, 0.3, 0.5, respectively, find the following Termination value, 3 P(0|1) (;). (2) j=1 3. For the above given special values, draw the curve of above termination value for x from 0 to 0.5
