Hello, thanks for the question. The line BH consists of two extreme points, i.e., B and H. Point B is located at x1=4 and x2=0. Point H is located at x1=4 and x2=1. When we compare the z-value between B and H, H has a higher z-value. Therefore, we choose the point H as the solution of Subproblem 7. I hope this explanation is clear.
The solution for subproblem 1 (Z-value = 165/4, x1 =15/4, x2= 9/4) is obtained by the graphical approach 4:17. The solutions for the following iterations are also obtained by applying the graphical approach, because we only have 2 variables in this problem. I hope this explanation is clear. Thanks!
Optimal Z for IP
Hello! IP's Z optimal is always worse than its LP Relaxation. So, for maximization, the IP's Z optimal
In sub problem 7, the point of intersection becomes the line BH which only represents line x1=4 and x2=0. Why did you take x2=1?
Hello, thanks for the question. The line BH consists of two extreme points, i.e., B and H. Point B is located at x1=4 and x2=0. Point H is located at x1=4 and x2=1. When we compare the z-value between B and H, H has a higher z-value. Therefore, we choose the point H as the solution of Subproblem 7.
I hope this explanation is clear.
Terima kasih banyak pakk.
Sama2, Vicky!
@@coverCell How to contact you, Sir? Can you please send me an email or whatsapp, I have a question I am confuse about one point in the normal LP.
@@ie95_Moh Hello, you may find my email address and Linkedin here ua-cam.com/play/PL0ZLI5mvN8kCsgjrK-YrTC81KY00rQKCK.html
Can you please explain how you got the Z-value = 165/7, x1 =15/4, x2= 9/4 (and for the rest of the other iterations)
The solution for subproblem 1 (Z-value = 165/4, x1 =15/4, x2= 9/4) is obtained by the graphical approach 4:17.
The solutions for the following iterations are also obtained by applying the graphical approach, because we only have 2 variables in this problem.
I hope this explanation is clear. Thanks!