Toyco problem: TOYCO assembles three types of toys—trains, trucks, and cars—using three operations. Available assembly t

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899604
Joined: Mon Aug 02, 2021 8:13 am

Toyco problem: TOYCO assembles three types of toys—trains, trucks, and cars—using three operations. Available assembly t

Post by answerhappygod »

Toyco problem: TOYCO assembles three types of toys—trains,
trucks, and cars—using three operations. Available assembly times
for the three operations are 430, 460, and 420 minutes per day,
respectively, and the revenues per toy train, truck, and car are
$3, $2, and $5, respectively. The assembly times per train for the
three operations are 1, 3, and 1 minutes, respectively. The
corresponding times per truck and per car are (2, 0, 4) and (1, 2,
0) minutes (a zero time indicates that the operation is not
used).
Toyco Problem Toyco Assembles Three Types Of Toys Trains Trucks And Cars Using Three Operations Available Assembly T 1
Toyco Problem Toyco Assembles Three Types Of Toys Trains Trucks And Cars Using Three Operations Available Assembly T 1 (39.81 KiB) Viewed 24 times
Q4-2. In the original TOYCO model, toy trains are not part of
the optimal product mix. The company recognizes that market
competition will not allow raising the unit price of the toy.
Instead, the company wants to concentrate on improving the assembly
operation itself. This entails reducing the assembly time per unit
in each of the three operations to a specified percentage, p% of
the original time. Determine the value of p that will make toy
trains just profitable.
TOYCO primal Maximize z = 3x1 +222 + 523 subject to 21 + 232 + 23 < 430 (Operation 1) 3х1 + 273 < 460 (Operation 2) 21 + 4x2 < 420 (Operation 3) 21, 22, 23 > 0 Optimal solution: 21 = 0,22 = 100, 23 =230, z 100, 23 =230, z = $1350 =
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply