Question: Management of High Tech Services (HTS) would like to develop a model that will help allocate its technicians time between service calls to regular contract

Management of High Tech Services (HTS) would like to develop a model that will help allocate its technicians time between service calls to regular contract customers and new customers. A maximum of 80 hours of technician time is available over the two-week planning period. To satisfy cash flow requirements, at least $800 in revenue (per technician) must be generated during the two-week period. Technician time for regular customers generates $25 per hour. However, technician time for new customers only generates an average of $8 per hour because in many cases a new customer contact does not provide billable services. To ensure that new customer contacts are being maintained, the technician time spent on new customer contacts must be at least 60% of the time spent on regular customer contacts. Given these revenue and policy requirements, HTS would like to determine how to allocate technician time between regular customers and new customers so that the total number of customers contacted during the two-week period will be maximized. Technicians require an average of 50 minutes for each regular customer contact and 1 hour for each new customer contact.

(a) Develop a linear programming model that will enable HTS to allocate technician time between regular customers and new customers.

If required, round your answers to one decimal place. For those boxes in which you must enter subtractive or negative numbers use a minus sign. (Example: -300)
Let R = time allocated to regular customer service
N = time allocated to new customer service

- Select your Answer -MaxMinItem 1 R + N
s.t.
R + N - Select your Answer -=Item 6 Technician Hours Available
R + N - Select your Answer -=Item 10 Revenue
R + N - Select your Answer -=Item 14 Time Spent on New Customers
R, N - Select your Answer -=Item 16

(b) Find the optimal solution.

The optimal solution is R = , N = , value = .

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