Question

blem 1. 60 points Gage blocks must be lapped for flatness and parallelism. Cluster of identical CNC lapping machines are available. Both sides of a block are lapped at a time; the lapping wheels are large enough to cover all gage blocks regardless of their lengths. Calculate the number of CNC lapping machines if we want to maximize the machine time by having only one single operator per cluster.  Machine availability: 85% (15% time for maintenance)  Rejection rate: 4%  Lapping time per block: 20 minutes  Time to walk between machines: 30 sec  Loading/unloading time: 90 sec

Answer 1

Answer:

a. 22.30 machines required

b. 22.22mm

c. 11 machines

Explanation:

Please see attachments for guide

Answer 2

Complete Question

The complete question is shown on the first uploaded image

Answer:

The number of CNC machine required is  $N_{CNC} = 12 \ Machines$

Explanation:

From the question awe are told the

         The machine availability is  $M_a =$ 85%

           The  Rejection time is  $R_t =$4%

           The lapping time is $L =$20 minutes

         The  time to work between machines is  $T_b =$$30s$ $= \frac{30}{60} = 0.5minutes$

          The loading and unloading time $k= 90sec = \frac{90}{60} = 1.5 minutes$

For one the loading and unloading time would be

         $k_{one\ block} = L + T_b +k$      

 Substituting values

       $k_{one \ block} =20+ 0.5+1.5$

                       $= 22 minutes$

for the 8 different sizes of the block the loading and unloading time would be

        $k_{8 \ different \ sizes } = 8*22$

                                  $= 176 \ minutes$    

From the the target is given as

                                    $T = 120 \ sets /week$

The available time is given as

$A = (5 \ day / week \ 8 \hr /day , 1 hr \ for \ lunch /break) M_a$

  $=( 5*8) -(5 \ hours \ of \ lunch \ break \ 1 \ hour \ per \ day) * 0.85$

  $= 29.95 hrs$

$A = 29.95 *60$

   $= 1797 \ minutes$    

The quantity of sets of block to be lapped by one machine per week would be

                  $=\frac{A}{k_{8 \ different \ sizes } }$

                   $= \frac{1797}{176} *(100- R_t)$%

                   $= \frac{1796}{176} * 0.96$

                   $=9.796$ sets

Hence

    The number of CNC lapping machine would be

                $N_{CNC} =\frac{120}{9.796}$

                           $\approx 12 \ machines$