Consider the Following Nonlinear Programming Problem

Question

Examlex · Accepted Answer

The Answer is The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state s_n be the vector (R₁,R₂),where R_i is the slack in the i-th constraint,so s₁ = (13,9)and s₂ = (13 - 2 $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ ,9 - $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ ).For stage 2 (the second variable),we have f₂ (R₁,R₂,x₂)= x₂,0

\le

x₂

\le

min{R₁,R₂},so $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ is easily solved,as shown in the following table. $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ For stage 1 (the first variable),we now have f₁(13,9,x₁)= $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ = $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ .For 0

\le

x₁

\le

2, $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$

\implies

f₁(13,9,x₁)= $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ .Since $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ for 0 ≤ x₁ ≤ 2,we now have $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ at x₁ = 2.For x₁

\le

2, $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$

\implies

f₁(13,9,x₁)= $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ Since $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ = 5 - 4x₁ < 0 for x₁ ≥ 0,it follows that $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ at x₁ = 2.Since f₁(13,9,x₁)is maximized at x₁ = 2 over both the range 0 ≤ x₁ ≤ 2 and the range x₁ ≥ 2,the maximum must be at x₁ = 2 over the entire range of x₁ ≥ 0.Consequently, $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ (13,9)= $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ at $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ = 2.Therefore,since x₁ = 2 leaves a slack of R₁ = 13 - 2 $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ = 5 and R₂ = 9 - $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ = 5 in the two constraints,so $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ = min{R₁,R₂} = 5,the optimal solution for the overall problem is $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ = 2, $The two stages involve making the decisions on the values of the two variables in order.At stage n,let the state sn be the vector (R1,R2),where Ri is the slack in the i-th constraint,so s1 = (13,9)and s2 = (13 - 2 ,9 - ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \le x2 \le min{R1,R2},so is easily solved,as shown in the following table. For stage 1 (the first variable),we now have f1(13,9, x1)= = .For 0 \le x1 \le 2, \implies f1(13,9, x1)= .Since for 0 ≤ x1 ≤ 2,we now have at x1 = 2.For x1 \le 2, \implies f1(13,9, x1)= Since = 5 - 4x1 < 0 for x1 ≥ 0,it follows that at x1 = 2.Since f1(13,9,x1)is maximized at x1 = 2 over both the range 0 ≤ x1 ≤ 2 and the range x1 ≥ 2,the maximum must be at x1 = 2 over the entire range of x1 ≥ 0.Consequently, (13,9)= at = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 = 5 and R2 = 9 - = 5 in the two constraints,so = min{R1,R2} = 5,the optimal solution for the overall problem is = 2, = 5,with Z* = 15.$ = 5,with Z* = 15.

Consider the Following Nonlinear Programming Problem

Definitions: