what is the optimal solution? Autolgnite produces electronic ignition systems fo

Q: what is the optimal solution? Autolgnite produces electronic ignition systems fo

Number of units of component 1 produced = 2000B +600D number of units of componen 2 produced = 1000(1-B) +1,400(1-D) therefore: 2000B +600D = 1000 (1-B) + 1400 )1-D) 3000 B + 2000 D = 2,400 Max 200B + 600D In addittion B

ID: 2427065 • Letter: W

Question

what is the optimal solution?

Autolgnite produces electronic ignition systems for automobiles at a plant in Cleveland, Ohio. Each ignition system is assembled from two components produced at AutoIgnite's plants in Buffalo, New York, and Dayton, Ohio. The Buffalo plant can produce 2000 units of component 1, 1000 units of component 2, or any combination of the two components each day. For instance, 60% of Buffalo's production time could be used to produce component 1 and 40% of Buffalo's production time could be used to produce component 2; in this case, the Buffalo plant would be able to produce 0.6(2000) = 1200 units of component 1 each day and 0.4(1000) = 400 units of component 2 each day. The Dayton plant can produce 600 units of component 1, 1400 units of component 2, or any combination of the two components each day. At the end of each day, the component production at Buffalo and Dayton is sent to Cleveland for assembly of the ignition systems on the following workday. a. Formulate a linear programming model that can be used to develop a daily production schedule for the Buffalo and Dayton plants that will maximize daily production of ignition systems at Cleveland Let B = proportion of Buffalo's time used to produce component 1 D = proportion of Dayton's time used to produce component 1

Explanation / Answer

Number of units of component 1 produced = 2000B +600D

number of units of componen 2 produced = 1000(1-B) +1,400(1-D)

therefore:

2000B +600D = 1000 (1-B) + 1400 )1-D)

3000 B + 2000 D = 2,400

Max 200B + 600D

In addittion B<=1 and D<=1

Max 2000 B + 600D

st.

3000B +2000D = 2400

B <=1

D<= 1

B,D >=0

Optimal solution

Buffalo - component 1 .8(2000) = 1,600

Buffaloe component 2 .2(1000) = 200

Dayton component 1 0 (600) = 0

Daytion component 2 1 (1400) = 1400

Total units = 1600 per day