Question

Panamint Systems Corporation is estimating activity costs associated with producing disk drives, tapes drives, and wire drives. The indirect labor can be traced to four separate activity pools. The budgeted activity cost and activity base data by product are provided below.Activity Cost Activity BaseProcurement $370,000 Number of purchase ordersScheduling 250,000 Number of production ordersMaterials handling 500,000 Number of movesProduct development 730,000 Number of engineering changesProduction 1,500,000 Machine hoursNumber of Number of Number of Number of Machine Number ofPurchase Orders Production Orders Moves Engineering Changes Hours UnitsDisk drives 4,000 300 1,400 10 2,000 2,000Tape drives 4,000 150 800 10 8,000 4,000Wire drives 12,000 800 4,000 25 10,000 2,50035. Determine the activity rate for procurement per purchase order.a $43.53b $18.50c $15.42d $37.0036. Determine the activity rate per production order for scheduling.a $200.00b $20.00c $29.41d $10.4237. Determine the activity rate for materials handling per move.a $58.82b $50.00c $20.83d $80.6538. Determine the activity rate for product development per change.a $73,000b $8,588c $30,417d $16,22239. Determine the activity rate for production per machine hour.a $62.50b $150.00c $75.00d $176.4740. Determine the activity-based cost for each disk drive unit.a $92.25b $130.69c $394.12d $279.5741. Determine the activity-based cost for each wire drive unit.a $204.13b $173.51c $744.06d $394.1242. Determine the activity-based cost for each tape drive unit.a $97.73b $232.69c $394.12d $103.84

Answer 1

Answer:

a.- activity rate for producrement:

370,000 / 18,000 = 18.20

b.- for scheduling:

250,000 / 1,250 = 200

c.- for materials handling:

500,000 / 6,200 = 80,65

d.- for development:

730,000 / 45 = 16222,22222222222

e.- for machine hours:

1,500,000 / 20,000 = 75

f.-

disk   $  279.57

tape  $  232.69

wire  $   744.06

Explanation:

The budgeted activity cost and activity base data by product are provided below.

Procurement $370,000 Number of purchase orders

Scheduling $250,000 Number of production orders

Materials handling $500,000 Number of moves

Product development $730,000 Number of engineering changes

Production 1,500,000 Machine hours

          purchase /production / moves/engineering /machine hours/ units

Disk drives   4,000   300   1,400       10    2,000       2,000

Tape drives  4,000    150     800       10     8,000       4,000

Wire drives 12,000   800  4,000      25    10,000       2,500

Total:           20,000   1,250 6,200     45    20,000    

activity rate:

cost pool over cost driver:

a.- activity rate for producrement:

370,000 / 18,000 = 18.20

b.- for scheduling:

250,000 / 1,250 = 200

c.- for materials handling:

500,000 / 6,200 = 80,65

d.- for development:

730,000 / 45 = 16222,22222222222

e.- for machine hours:

1,500,000 / 20,000 = 75

f.- activity cost for disk drive:

Disk drives  

4,000 x 18.2            =   72,800

   300 x 200            =   60,000

1,400  x 80.65         =   112, 910

     10  x 16,222.22   = 162,222.22

2,000  x 75               = 150,000    

Total Cost                    557,932.22‬

Units: 2,000

Cost per unit: 557,932.22 / 2,000 = 278.96611

For the other two units we repeat the process for Disk Driver

Wire tapes  

12,000 x 18.2            =  218,400

   800 x 200            =  160,000

4,000 x 80.65         =  322,600

     25 x 16,222.22   = 405,555.5

1,000  x 75               = 750,000    

Total Cost                 1,856,555.5

units: 2,500

1,856,555.5 / 2,500 = 742,62

Tape drives  

4,000 x 18.2            =   72,800

   150 x 200            =   30,000

  800 x 80.65         =   64,520

     10  x 16,222.22   = 162,222.22

10,000 x 75               = 600,000    

Total Cost                    929.542,22‬

units: 4,000

929,542.22 / 4,000 = 232,39