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Start in period 4: C(1) = 200 C(2) = [200 + (300)(0.3)]/2 = 145 C(3) = [(2)(145) + (2)(200)(0.3)]/3 = 136.67 Stop. Hence y1= 335

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Start in period 4: C(1) = 200 C(2) = [200 + (300)(0.3)]/2 = 145 C(3) = [(2)(145) + (2)(200)(0.3)]/3 = 136.67 Stop. Hence y1= 335 + 200 + 140 = 675, y4= 440 + 300 + 200 = 940 Question 6 Over the last five questions, you applied the methods which were explained in the videos. Now, it is your turn to research! In this section, students are required to use Dynamic Programming based on the 'Wagner-Whitin' Algorithm to develop a schedule to show when

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