Started on	Monday, 23 October 2023, 11:15 AM
State	Finished
Completed on	Monday, 23 October 2023, 11:45 AM
Time taken	30 mins
Marks	3.40/7.00
Grade	4.86 out of 10.00 (48.57%)

Question 1

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Consider the algebraic LP model for the APPLES problem (with 4 decision variables) in standard form. The number of functional constraints involving 3 decision variables with non-zero coefficients is Answer 1 Question 1. The constraint for the apple juice total demand is Answer 2 Question 1 X_as + Answer 3 Question 1 X_aj + Answer 4 Question 1 X_aas + Answer 5 Question 1 X_aaj <= Answer 6 Question 1, where X_as is apple syrup produced, X_aj is apple juice produced, X_aas is advertising spent on apple syrup, X_aaj is advertising spent on apple juice.

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Feedback: written in the standard form, 4 functional constraints involve 2 decision variables with non-zero coefficient, 1 functional constraint involves 4 decision variables with non-zero coefficient, none involves only 3 decision variables with non-zero coefficient.

Question 2

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Consider the optimal solution to the LP model for the APPLES problem. The product with the largest total demand including advertising is Answer 1 Question 2. The number of non-binding functional constraints is Answer 2 Question 2.

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Feedback: the total demand for apple syrup is 7914.29 while the total demand for apple juice is 11871.43; the functional constraint on 20% of production is non-binding.

Question 3

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Consider the optimal solution to the IP model for the APPLES problem. The total production cost is Answer 1 Question 3. The number of binding functional constraints is Answer 2 Question 3. The optimal solution is of Answer 3 Question 3 quality compared to the LP optimal solution.

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Feedback: there are 2 binding constraints on the IP optimal solution, the total budget and the total demand for apple juice; the IP optimal solution has a lower total profit compared to the LP optimal solution.

Question 4

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Consider the following modification to the original LP model for the APPLES problem. The following two constraints are added to the model: X_aas + X_aaj <= 10000 and X_aas + X_aaj >= 20000, where X_aas is advertising spent on apple syrup, X_aaj is advertising spent on apple juice. This results in Answer 1 Question 4. These two constraints can be reformulated as X_aas + X_aaj Answer 2 Question 4 U <= 10000 and X_aas + X_aaj Answer 3 Question 4 V >= 20000, where U and V are auxiliary non-negative decision variables.

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Feedback: entering the two new constraints as given makes the model infeasible; the reformulation with slack and surplus decision variables allows the model to be solved; although using – for both U and V also solves the model, it is conceptually incorrect (redundant) to use – for V in the >= constraint, we don’t need to subtract anything if 20000 is exceeded because the constraint is satisfied already.

Question 5

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Consider the optimization model for the LIGHTS PRODUCTION problem. Solving the model as IP gives an optimal solution with objective function value Z = Answer 1 Question 5 and the total number of all non-binding constraints equal to Answer 2 Question 5. Before applying optimization, the company had a reference solution with total profit of 7500250 for exactly the same problem. Now you can tell them that the optimal profit you found is the Answer 3 Question 5 and the profit they had before is a Answer 4 Question 5.

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Feedback: the IP optimal solution has the same quality as the LP solution hence it is the upper bound; the reference solution of poorer quality is a lower bound.

Question 6

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Consider the following notation for the optimization model for the LIGHTS PRODUCTION problem:

Si is the selling price for each product

Mi is the material cost per unit for each product

Ri is the production cost in regular time per unit for each product

Oi is the production cost in overtime per unit for each product

Xir is units produced of product i in regular time

Xio is units produced of product i in overtime

Xia is the advertising expenditure in product I

i = 1 (table lamp), 2 (floor lamp), 3 (ceiling fixture), 4 (chandelier)

The algebraic expression in compact notation using sigma (sum) for the objective function is as follows:

Select one:

Maximize Z = SUM[i=1 to 4] (Si – Mi – Ri – Oi) (Xir + Xio) – SUM[i=1 to 4] Xia

Maximize Z = SUM[i=1 to 4] (Si – Mi – Ri) Xir + SUM[i=1 to 4] (Si – Mi – Oi) Xio – SUM[i=1 to 4] Xia

Maximize Z = SUM[i=1 to 4] SUM[i=1 to 4] ((Si – Mi – Ri) Xir + (Si – Mi – Oi) Xio) – SUM[i=1 to 4] Xia

Maximize Z = (Si – Mi – Ri) Xir + (Si – Mi – Oi) Xio – Xia for i = 1 to 4

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Feedback: Maximize Z = SUM[i=1 to 4] (Si – Mi – Ri) Xir + SUM[i=1 to 4] (Si – Mi – Oi) Xio – SUM[i=1 to 4] Xia

The correct answer is: Maximize Z = SUM[i=1 to 4] (Si – Mi – Ri) Xir + SUM[i=1 to 4] (Si – Mi – Oi) Xio – SUM[i=1 to 4] Xia

Question 7

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Use LP-Solve for this question. The following modifications are made to the LIGHTS PRODUCTION problem. The total advertising budget reduces from 18000 to 15500. The total production of table lamps must be equal to the total production of floor lamps. The lp-relaxation optimal solution for this modified problem has an objective function value of Answer 1 Question 7 with the total production of table lamps = total production of floor lamps = Answer 2 Question 7. The IP optimal solution to this modified problem has an objective function value of Answer 3 Question 7, that is, Answer 4 Question 7 compared to the Answer 5 Question 7 bound.

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Feedback: the new constraint is Xtr + Xto – Xfr – Xro = 0; solving the new model as LP gives a non-integer solution with Z=7295125; solving the new model as IP in Excel takes much longer than in lp-solve; the IP solution has Z=7295113 with the total production of table lamps = total production of floor lamps = 21999; for a maximization model the quality of the lp-relaxation solution is an upper bound.

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Workshop 3 In-class Test

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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