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Since the decisions to be made are xn = p...View Answer
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Since the decisions to be made are xn = p...
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Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let xn be the advertising dollars (in millions)spent on product n.Let sn be the amount of advertising budget remaining.Let 11ea84c6_c8f5_6e03_83dc_851a16cedca9_TB2462_11 be the increase in sales of product n when xn million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is 11ea84c6_c8f5_6e04_83dc_a3dd4f39bb2d_TB2462_11 .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3, 11ea84c6_c8f5_9515_83dc_c3088e9f746d_TB2462_00 For n = 2, 11ea84c6_c8f5_9516_83dc_3fbe43462541_TB2462_00 For n = 1, 11ea84c6_c8f5_bc27_83dc_e7490a764576_TB2462_00 Hence,since s2 = 6 - 11ea84c6_c8f5_bc28_83dc_615396b448c0_TB2462_11 and s3 = s2 - 11ea84c6_c8f5_e339_83dc_7b10ef04fb7b_TB2462_11 ,there are two optimal plans as given in the table below. 11ea84c6_c8f6_0a4a_83dc_95acde0f2665_TB2462_00
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Since the decisions to be made are the advertising expenditures on the three products,the stages for a dynamic programming formulation of this problem correspond to the three products.When making the decision for a particular product,the essential information is the amount of the advertising budget still remaining,so this becomes the current state in this formulation.Let xn be the advertising dollars (in millions)spent on product n.Let sn be the amount of advertising budget remaining.Let 11ea84c6_c8f5_6e03_83dc_851a16cedca9_TB2462_11 be the increase in sales of product n when xn million dollars are spent on product n,as given by the above table.Then,using the usual dynamic programming notation presented in Sec.10.2 of the textbook,the recursive relationship for this problem is 11ea84c6_c8f5_6e04_83dc_a3dd4f39bb2d_TB2462_11 .The solution procedure now starts at the end (stage 3)and moves backward stage by stage.For n = 3, 11ea84c6_c8f5_9515_83dc_c3088e9f746d_TB2462_00 For n = 2, 11ea84c6_c8f5_9516_83dc_3fbe43462541_TB2462_00 For n = 1, 11ea84c6_c8f5_bc27_83dc_e7490a764576_TB2462_00 Hence,since s2 = 6 - 11ea84c6_c8f5_bc28_83dc_615396b448c0_TB2462_11 and s3 = s2 - 11ea84c6_c8f5_e339_83dc_7b10ef04fb7b_TB2462_11 ,there are two optimal plans as given in the table below. 11ea84c6_c8f6_0a4a_83dc_95acde0f2665_TB2462_00
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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 11ea84c6_c901_ca63_83dc_1992569ec41c_TB2462_11 ,9 - 11ea84c6_c901_ca64_83dc_650189c8fb77_TB2462_00 ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \(\le\) x2\(\le\) min{R1,R2},so 11ea84c6_c901_f175_83dc_d5b564580514_TB2462_00 is easily solved,as shown in the following table. 11ea84c6_c901_f176_83dc_4b542b0e019d_TB2462_00 For stage 1 (the first variable),we now have f1(13,9, x1)= 11ea84c6_c902_1887_83dc_7954a5d9cfdf_TB2462_11 = 11ea84c6_c902_1888_83dc_e17f54016762_TB2462_11 .For 0 \(\le\) x1 \(\le\) 2, 11ea84c6_c902_3f99_83dc_d1b0fc253086_TB2462_00 \(\implies\) f1(13,9, x1)= 11ea84c6_c902_3f9a_83dc_61f7810aa53f_TB2462_11 .Since 11ea84c6_c902_66ab_83dc_bdac37f38edb_TB2462_11 for 0 ≤ x1 ≤ 2,we now have 11ea84c6_c902_66ac_83dc_6b683ccd2d61_TB2462_11 at x1 = 2.For x1 \(\le\) 2, 11ea84c6_c902_8dbd_83dc_0db9fabd8267_TB2462_11 \(\implies\) f1(13,9, x1)= 11ea84c6_c902_8dbe_83dc_cb4e668eae84_TB2462_11 Since 11ea84c6_c902_b4cf_83dc_2bb949c47aa7_TB2462_11 = 5 - 4x1 < 0 for x1 ≥ 0,it follows that 11ea84c6_c902_b4d0_83dc_9d415be54e18_TB2462_11 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, 11ea84c6_c902_b4d1_83dc_75e652fe9298_TB2462_11 (13,9)= 11ea84c6_c902_dae2_83dc_5d8afae9009c_TB2462_11 at 11ea84c6_c902_dae3_83dc_a57d75d48423_TB2462_11 = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 11ea84c6_c903_01f4_83dc_4376dbeca1d1_TB2462_11 = 5 and R2 = 9 - 11ea84c6_c903_01f5_83dc_dd46efd200f3_TB2462_11 = 5 in the two constraints,so 11ea84c6_c903_2906_83dc_f96d3c9256a1_TB2462_11 = min{R1,R2} = 5,the optimal solution for the overall problem is 11ea84c6_c903_2907_83dc_15f605169b8a_TB2462_11 = 2, 11ea84c6_c903_5018_83dc_75553210bff4_TB2462_11 = 5,with Z* = 15.
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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 11ea84c6_c901_ca63_83dc_1992569ec41c_TB2462_11 ,9 - 11ea84c6_c901_ca64_83dc_650189c8fb77_TB2462_00 ).For stage 2 (the second variable),we have f2 (R1,R2, x2)= x2,0 \(\le\) x2\(\le\) min{R1,R2},so 11ea84c6_c901_f175_83dc_d5b564580514_TB2462_00 is easily solved,as shown in the following table. 11ea84c6_c901_f176_83dc_4b542b0e019d_TB2462_00 For stage 1 (the first variable),we now have f1(13,9, x1)= 11ea84c6_c902_1887_83dc_7954a5d9cfdf_TB2462_11 = 11ea84c6_c902_1888_83dc_e17f54016762_TB2462_11 .For 0 \(\le\) x1 \(\le\) 2, 11ea84c6_c902_3f99_83dc_d1b0fc253086_TB2462_00 \(\implies\) f1(13,9, x1)= 11ea84c6_c902_3f9a_83dc_61f7810aa53f_TB2462_11 .Since 11ea84c6_c902_66ab_83dc_bdac37f38edb_TB2462_11 for 0 ≤ x1 ≤ 2,we now have 11ea84c6_c902_66ac_83dc_6b683ccd2d61_TB2462_11 at x1 = 2.For x1 \(\le\) 2, 11ea84c6_c902_8dbd_83dc_0db9fabd8267_TB2462_11 \(\implies\) f1(13,9, x1)= 11ea84c6_c902_8dbe_83dc_cb4e668eae84_TB2462_11 Since 11ea84c6_c902_b4cf_83dc_2bb949c47aa7_TB2462_11 = 5 - 4x1 < 0 for x1 ≥ 0,it follows that 11ea84c6_c902_b4d0_83dc_9d415be54e18_TB2462_11 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, 11ea84c6_c902_b4d1_83dc_75e652fe9298_TB2462_11 (13,9)= 11ea84c6_c902_dae2_83dc_5d8afae9009c_TB2462_11 at 11ea84c6_c902_dae3_83dc_a57d75d48423_TB2462_11 = 2.Therefore,since x1 = 2 leaves a slack of R1 = 13 - 2 11ea84c6_c903_01f4_83dc_4376dbeca1d1_TB2462_11 = 5 and R2 = 9 - 11ea84c6_c903_01f5_83dc_dd46efd200f3_TB2462_11 = 5 in the two constraints,so 11ea84c6_c903_2906_83dc_f96d3c9256a1_TB2462_11 = min{R1,R2} = 5,the optimal solution for the overall problem is 11ea84c6_c903_2907_83dc_15f605169b8a_TB2462_11 = 2, 11ea84c6_c903_5018_83dc_75553210bff4_TB2462_11 = 5,with Z* = 15.
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The three stages involve making the deci...View Answer
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The three stages involve making the deci...
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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 remaining slack in the constraint 11ea84c6_c8ff_5950_83dc_bfe2e2d77935_TB2462_11 ,so s1 = 4 and s2 = 4 - 11ea84c6_c8ff_5951_83dc_2fba474800a7_TB2462_11 .For n = 2, 11ea84c6_c8ff_8062_83dc_4f7168d1b773_TB2462_00 .For n = 1, 11ea84c6_c8ff_8063_83dc_ef8e5a0c5677_TB2462_00 .To maximize (2x1 + 4 - 11ea84c6_c8ff_a774_83dc_714b48c95a65_TB2462_11 )over 0 ≤ x1 ≤ 2,we take the first and second derivatives. 11ea84c6_c8ff_a775_83dc_912dc21b0e22_TB2462_11 (2x1 + 4 - 11ea84c6_c8ff_ce86_83dc_1f4c6419b6e3_TB2462_11 )= 2 - 2x1 = 0,so x1 = 1 is a critical point. 11ea84c6_c8ff_ce87_83dc_11a0fb3e035c_TB2462_11 (2x1 + 4 - 11ea84c6_c8ff_f598_83dc_f909734e56f8_TB2462_11 )= -2,so x1 = 1 is indeed the maximum over 0 ≤ x1 ≤ 2.Since x1 = 1 leaves a slack of 11ea84c6_c8ff_f599_83dc_678e77a3fd07_TB2462_11 in the constraint,the overall optimal solution is 11ea84c6_c900_1caa_83dc_6da46d7fe201_TB2462_11 = 1, 11ea84c6_c900_1cab_83dc_b3797bfe08f2_TB2462_11 ,with Z* = 5.
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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 remaining slack in the constraint 11ea84c6_c8ff_5950_83dc_bfe2e2d77935_TB2462_11 ,so s1 = 4 and s2 = 4 - 11ea84c6_c8ff_5951_83dc_2fba474800a7_TB2462_11 .For n = 2, 11ea84c6_c8ff_8062_83dc_4f7168d1b773_TB2462_00 .For n = 1, 11ea84c6_c8ff_8063_83dc_ef8e5a0c5677_TB2462_00 .To maximize (2x1 + 4 - 11ea84c6_c8ff_a774_83dc_714b48c95a65_TB2462_11 )over 0 ≤ x1 ≤ 2,we take the first and second derivatives. 11ea84c6_c8ff_a775_83dc_912dc21b0e22_TB2462_11 (2x1 + 4 - 11ea84c6_c8ff_ce86_83dc_1f4c6419b6e3_TB2462_11 )= 2 - 2x1 = 0,so x1 = 1 is a critical point. 11ea84c6_c8ff_ce87_83dc_11a0fb3e035c_TB2462_11 (2x1 + 4 - 11ea84c6_c8ff_f598_83dc_f909734e56f8_TB2462_11 )= -2,so x1 = 1 is indeed the maximum over 0 ≤ x1 ≤ 2.Since x1 = 1 leaves a slack of 11ea84c6_c8ff_f599_83dc_678e77a3fd07_TB2462_11 in the constraint,the overall optimal solution is 11ea84c6_c900_1caa_83dc_6da46d7fe201_TB2462_11 = 1, 11ea84c6_c900_1cab_83dc_b3797bfe08f2_TB2462_11 ,with Z* = 5.
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