Comments (1)
Hi @gYunTian,
Here's a response from @amahmudwave and @pierrelouisp:
When expanding:
lagrange_hard_shift * sum_d ( effort * sum_n q_i(n,d) - workforce ) ** 2
We get:
lagrange_hard_shift * effort ** 2 * sum_d sum_m sum_n q_i(n,d) q_j(m, d) - 2 * lagrange_hard_shift * sum_d sum_n q_i(n,d) * effort * workforce (Your answer)
+ lagrange_hard_shift * (effort ** 2 * sum_d sum_n q_i(n,d))
This is because
(sum_n q_i(n,d))** 2 == sum_n q_i(n,d)**2 + sum_m sum_n q_i(n,d) q_j(m, d) == sum_n q_i(n,d) + sum_m sum_n q_i(n,d) q_j(m, d)
When we take the summation of binary variables and square them, we get the sum of the different variables multiplied with each other, plus the sum of the square of individual variables. ie. (a + b)**2 == (a**2 + b**2 + 2ab)
.
The sum of the square of individual variables is equal to the sum of the original binary variables, since 0 ** 2 = 0 and 1 ** 2 = 1.
Hope this helps!
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