# Case Problem "toys for Boys"

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Case Problem "Toys for Boys"

Managerial Report for the case "Toys for Boys"

Case Description: "Toys for Boys" is a company that produces toys and would like to optimize its production plan for four of its products - T1, T2, T3, T4. Information about profit for each type of the products are provided: \$4.8 for T1,\$12.00 for T2, \$6.00 for T3 and \$7.20 for T4. The required production time for each type of products in three production departments (in hours per unit) and available production time for the departments (in hours) are provided below:

Operation T1 T2 T3 T4 Available hours

Machining 0.70 0.75 0.55 0.34 400

Assembly 0.55 0.82 0.80 0.55 480

Packaging 0.24 0.32 0.45 0.27 220

Raw Materials 1.9 2.5 1.8 2.0 1500

Additional data about needs for the raw materials for each product (in units per product) and all available units of raw materials are placed after that. The company has one advanced order of 100 units of T1.

Objective functions and constrains: Since the objective is to maximize the production plan for the four products and production operations, the objective functions is to maximizeits total profit considering the profit coming from each type of toy. In this case, variables are the following:

X1 - stands for units produced of T1 toy

X2 - stands for units produced of T2 toy

X3 - stands for units produced of T3 toy

X4 - stands for units produced of T4 toy

Thus, the objective function would be:

Max 4.8X1 +12X2 + 6X3 +7.2X4

Constrains are the following:

1. X1>=100 (Demand for T1)

2. 0.7X1+0.75X2+0.55X3+0.35X4<=400 (Machining Department Capacity)

3. 0.55X1+0.82X2+0.80X3+o.55X4<=480 (Assembly Department Capacity)

4. 0.24X1+0.32X2+0.45X3+0.27X4<=220 (Packaging Department Capacity)

5. 1.9X1 + 2.5X2 + 1.8 X3+ 2 X4<=1500 (Raw material Availability)

6. X1, X2, X3, X4>=0 (non-negativity constrains)

The whole linear programming model should be:

Max 4.8X1 +12X2 + 6X3 +7.2X4

s.t.

X1>=100

0.7X1+0.75X2+0.55X3+0.35X4<=400

0.55X1+0.82X2+0.80X3+o.55X4<=480

0.24X1+0.32X2+0.45X3+0.27X4<=220

1.9X1 + 2.5X2 + 1.8 X3+ 2 X4<=1500

X1, X2, X3, X4>=0

Solution, general recommendations

B. Solution: Specialized software for linear programming models, the Management scientist software, was used to solve this case. For full solution of the case, please see Appendix 1.

According to the solution in Appendix 1, the objective function value (or amount of the maximized profit for the current model) is \$ 6186 (after rounding). This profit would be achieved if the company produce 100 units of T1, 330 units if T2 (after rounding) and 242 units of T4 (after rounding). Production of the T3 is not advisable for this production plan and could be considered as an alternative if the profit for this product is increased by \$2.732 or its minimum is \$8.73 .

C.Constrains: According to the solution, constrain 1 (Demand), constrain 2 (Capacity of Machining Operations) and Constrain 5 (Availability of the Raw materials) are binding. In other words, limitations for this constrains are fulfilled in their maximum/minimum and they don't result in slack/surplus for the production plan.

Constrain 3 and constrain 4 has slack of respectively 21 and 24 (after rounding) hours, which are not used for the current production plan. These hours remain available and the management can use them to other production plan or for another managerial decision (to sell to other company or use for other assignments, for example).

D. Dual Prices: There are only three dual processes which correspond to Constrain 1, Constrain 2, and Constrain 3.

The dual price for Constrain 1 shows that if the company increase produced products T1 with one unit, the value of objective function will decrease with \$5.52 and this kind change is not recommended (the units of T1 should be keep on its minimum).

The dual price for Constrain 2 shows that increasing of the available hours of the Machining department will result in increasing the profit with \$9.23 and it is recommended as possible change of the production plan till reaching 451 hours before new solution model will be required (according to the values in the Right-hand side ranges column)

The dual price for Constrain 5 shows that if the company increases its raw material availability with one unit, this would result in increasing of \$2.03 of the total profit and it is advisable till reaching 1602 units of raw material before the new model would be required.

E. Change one right-hand side of one constrain: According

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