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Homework 5 Due: October 4, 2019
- (10 points) Recall the following problem from Homework 2: Consider a window company with only three employees which makes two different kinds of hand-crafted windows: a wood-framed and an aluminum-framed window. They earn $60 profit for each wood-framed window and $ profit for each aluminum-framed window. Employee 1 makes the wood frames, and can make 6 per day. Employee 2 makes the aluminum frames, and can make 4 per day. Employee 3 forms and cuts the glass, and can make 48 square feet of glass per day. Each wood-framed window uses 6 square feet of glass and each aluminum-framed window uses 8 square feet of glass. Let x 1 and x 2 be the number of wood-framed and aluminum-framed windows, respectively. The linear programming model can be formulated as:
maximize 60x 1 + 30x 2 subject to 6 x 1 + 8x 2 ≤ 48 x 1 ≤ 6 x 2 ≤ 4 x 1 , x 2 ≥ 0
The optimal solution to this problem is (x 1 , x 2 ) = (6, 1 .5), with an optimal objective function value of 405. (a) Suppose that one day, Employee 1 offers to work overtime to produce up to 9 wood frames instead of 6, but asks that he be paid an extra fixed amount of $90. Can his offer be evaluated without re-solving the model? Explain why or why not.
Solution: Let us first evaluate the allowable increase in the right hand side of the second constraint. Re-writing the problem in standard form, we get:
maximize 60x 1 + 30x 2 subject to 6 x 1 + 8x 2 + s 1 = 48 x 1 + s 2 = 6 + ∆ x 2 + s 3 = 4 x 1 , x 2 , s 1 , s 2 , s 3 ≥ 0
Plugging in the optimal solution of (x 1 , x 2 ) = (6, 1 .5) into the constraint set allows us to determine that the variables s 1 and s 2 are the nonbasic variables. Setting these to zero leave us with the following system of equations:
6 x 1 + 8x 2 = 48 x 1 = 6 + ∆ x 2 + s 3 = 4
Solving this gives us
x 1 = 6 + ∆ ≥ 0
x 2 =
s 3 =
Homework 5 Due: October 4, 2019
from which we conclude the allowable range for ∆ without changing the optimal basis is − 3. 33 ≤ ∆ ≤ 2 Since the employee’s offer increases the right hand side by 3, which is beyond the range for ∆, we cannot calculate the associated increase in profit without resolving the model. So we would need to resolve the model to determine whether increasing the right hand side to 9 at a cost of $90 is a good idea for the company.
(b) Suppose that on a different day, Employee 1 offers to work overtime to produce up to 7 wood frames instead of 6, but asks that he be paid an extra fixed amount of $30. Is this a good deal from the company’s point of view?
Solution: Plugging the expressions x 1 = 6 + ∆ and x 2 = 6 − 4 3∆ from part a) into the objective function gives us 405 + 37.5∆, which tells us that the shadow price of the x 1 ≤ 6 constraint is 37.5. This means increasing the right hand side of the constraint from 6 to 7 would increase our profit by $37.5. Since the employee asks for only an extra $30, this is a good idea for the company.
(c) What would be the change in total profit if the company agreed to Employee 1’s offer from part b)?
Solution: The total profit would increase by $7.50.
(d) Suppose that now Employee 2 offers to work overtime to produce up to 6 aluminum frames instead of 4, but asks that he be paid an extra fixed amount of $1. Is this a good deal from the company’s point of view?
Solution: No, increasing the RHS of the third constraint does not affect the optimal solution since the constrain is non-binding so we will not be willing to pay any amount of money for this additional resource.
Homework 5 Due: October 4, 2019
- (10 points) Recall the problem from Homework 1 where a medical equipment company produces a piece of diagnostic equipment at two plants, and must supply this product to three medical centers. The problem was to determine how much of the product to ship from each plant to each customer, such that all demand is satisfied. Let xij be the number of units shipped from Plant i to Customer j. This problem can then be formulated as:
minimize 600x 11 + 800x 12 + 700x 13 + 400x 21 + 900x 22 + 600x 23 subject to x 11 + x 12 + x 13 ≤ 400 (Production capacity at Plant 1) x 21 + x 22 + x 23 ≤ 500 (Production capacity at Plant 2) x 11 + x 21 = 300 (Demand from Customer 1) x 12 + x 22 = 200 (Demand from Customer 2) x 13 + x 23 = 400 (Demand from Customer 3) x 11 , x 12 , x 13 , x 21 , x 22 , x 23 ≥ 0
(Note: when using AMPL, entering the command ‘display varname, var;’ after solving the problem will display the names and values of all your variables at the optimal solution. Enter- ing the command ‘display conname, con;’ will display the names and shadow prices of your constraints.)
a) Formulate and solve this problem in AMPL. State the optimal solution and optimal ob- jective function value. b) Suppose Customer 1 decreases their order by 50 units. What impact will this have on the company’s total cost? (The optimal basis does not change). c) Suppose the company has the option of purchasing a new machine that can add 100 units of production, at only one plant, at a cost of $12,000 (and suppose you know that making this purchase will not change the optimal basis). Which plant, if any, should this machine be purchased for? Provide a justification for your decision that does not involve resolving the problem. (Consider this separately from the change in part b), that is, assume Customer 1’s demand is still 300). d) What would the cost of the machine in part c) have to be in order for your decision to change? Explain why briefly. e) Suppose now that tweaking the production process at Plant 1 increases its capacity from 400 to 410. Without resolving the problem, can you determine what the slack variable for Plant 1’s production constraint will be in the new optimal solution? Explain why briefly. (Consider this change on the original formulation, separately from the changes considered in parts b, c and d).
Solution: Solution 3.
a) This problem can be formulated in AMPL as:
Medical Equipment company
Homework 5 Due: October 4, 2019
var x11; var x12; var x13; var x21; var x22; var x23;
minimize cost: 600x11 + 800x12 + 700x13 + 400x21 + 900x22 + 600x23;
subject to
Plant1: x11 + x12 + x13 <= 400; Plant2: x21 + x22 + x23 <= 500; Demand1: x11 + x21 = 300; Demand2: x12 + x22 = 200; Demand3: x13 + x23 = 400; Nonneg11: x11 >=0; Nonneg21: x21 >=0; Nonneg12: x12 >=0; Nonneg22: x22 >=0; Nonneg13: x13 >=0; Nonneg23: x23 >=0;
The optimal solution is (x 11 , x 12 , x 13 , x 21 , x 22 , x 23 ) = (0, 200 , 200 , 300 , 0 , 200) with an optimal cost of 540,000.
b) Since the shadow price of Customer 1’s constraint is 500, a decrease in demand of 50 will decrease the optimal objective function by 25,000.
c) The shadow prices for the production constraints at Plant 1 and Plant 2 are 0 and -100, respectively. Thus increasing the production capacity at Plant 1 would not improve the objective function. From the shadow price, we see that the value of an additional unit of production at Plant 1 is 100, thus the value of 100 additional units (knowing that the optimal basis does not change) is 10,000. Since the price of the machine is greater than 10,000, it is not worth it for the company to purchase it.
d) If the price of the machine decreased to below $10,000, then it would be profitable to purchase it, since the improvement (reduction) in the objective function would outweight the total cost of the machine – a net gain for the company.
e) Since the left hand side of the constraint evaluates to exactly 400 for the given optimal solution, the value of the slack variable in that constraint must be 0. Also, since the shadow price of Plant 1’s production capacity constraint is 0, increasing the right hand side of that constraint by a small amount will not change the optimal solution, meaning the value of the slack variable in the new optimal solution will be 10.
