Question

(a) Best Brains College wishes to buy a new machine for printing its examination materials. Details of the two machines under consideration are as follows:

• C14 produces 7,500 copies of printed material in one hour.

• C20 produces 7,800 copies of the material in one hour.  

• The monthly energy requirement for C14 is 1,200 kilowatts and that for C20 is 1,400 kilowatts.

• Each machine is programmed to work a maximum of 8 hours daily while the university works 25 days in a month.

Required:

(i). Compute the productivity of energy per day for each machine.

8. marks)

      (ii) If the decision to buy is based solely on the productivity of the machines, determine the one that should be bought by management of the college. (1 mark)

(c ) The smooth flow of activities in an operations system is of primary concern to managers. However, these activities are often slowed down by bottlenecks.

     (i) With the aid of a diagram, explain how a bottleneck may reduce the flow of activities in an operations system.

(ii) Describe how each of the following may act as a bottleneck in an organization and suggest one way of removing each one of them:

• Electricity supply

• Shortage of raw materials

• Management

Answer 1

a)

1)

Solution:- Here,

Machine C14 = produces 7,500 copies in one hour.

Monthly energy consumption by C14 = 1200kw

Monthy working days = 25 days

Total number of hours in a month = 25 * 8 = 200

Energy consumption per hour = Energy consumed in a month/ Total number of hours in a month

                                                                          = 1200/200 = 6kw

Thus,

Productivity of C14 per day = Output per hour/energy consumption per hour

= 7500/6 = 1250

Then,

Machine C20 = produces 7800 copies in 1 hour

Monthy energy consumption by C20 = 1400kw

Monthy working days = 25 days

Total number of hours in a month = 200

Energy consumption per hour = Energy consumed in a month/ Total number of hours in a month

                                                        1400/200 = 7kw

Thus,

The productivity of C20 per day = 7800/7=1114

Hence, the productivity of energy per day for C12 is 1250 and C20 is 1114

2.

If the decision to buy the machine is based on productivity then management of the college should buy C14 because the productivity of C14 is more than C20.