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In: Accounting
An engineering company which uses job costing to attribute costs to individual products and services provided...
An engineering company which uses job costing to attribute costs to individual products and services provided to its customers has began the preparation of its fixed production overhead cost budget for 2017. Cost has been identified as follows:
Machining Assembly Finishing Stores Maintenance
K’000 K’000 K’000 K’000 K’000
Apportioned cost 6,000 2,500 1,500 1,000 800
The stores and maintenance departments are production department and provide services as follows:
Machining Assembly Finishing Stores Maintenance
Stores 40% 30% 20% - 10%
Maintenance 55% 20% 20% 5% -
The number of machine and labour hours budgeted for 2018 is:
Machining Assembly Finishing
Machine 50,000 4,000 5,000
Labour hours 10,000 30,000 20,000
Required:
(a) Calculate appropriate overhead absorption rates for each production department for 2017.
(b) Prepare a quotation for job K34 to be commenced early in 2017 assuming it has:
Direct materials Costing K24,000
Direct labour Costing K15,000
And requires:
Machine Labour
Machining department 45 10
Asembly 5 15
Finishing department 4 12
Add that profit is 20% of selling price
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| An engineering company | ||||||
| Workings for answer a | ||||||
| Reciprocal Method | Stores | Maintenance | Machining | Assembly | Finishing | Total |
| Stores % | 0.00% | 10.00% | 40.00% | 30.00% | 20.00% | 100.00% |
| Maintenance % | 5.00% | 0.00% | 55.00% | 20.00% | 20.00% | 100.00% |
| Overhead costs | 1,000.00 | 800.00 | 6,000.00 | 2,500.00 | 1,500.00 | 11,800.00 |
| Let cost of Stores is $ X and Maintenance is $ Y |
| So, X= 1000+ 5% of Y |
| So, X= 1000+ 0.05Y |
| Y= 800+ 10% of X |
| Y= 800+ 0.10X |
| Substituting the value of X in equation 2: |
| Y = 800 + 0.10(1000 + 0.05Y) |
| Y = 800 + 100+.005Y |
| Y – 0.005Y = 800+100 |
| 0.995Y = 900 |
| Y = 900/0.995 |
| Y = $ 905 |
| Substituting the value of Y in equation 1: |
| So, X= 1000+ 0.05Y |
| So, X= 1000+ 0.05*905 |
| So, X= 1000+ 45 |
| X= $ 1,045 |
| Allocation Summary | Stores | Maintenance | Machining | Assembly | Finishing | Total |
| Stores % | 0.00% | 10.00% | 40.00% | 30.00% | 20.00% | 100.00% |
| Maintenance % | 5.00% | 0.00% | 55.00% | 20.00% | 20.00% | 100.00% |
| Department cost | 1,000.00 | 800.00 | 6,000.00 | 2,500.00 | 1,500.00 | 11,800.00 |
| Total costs | ||||||
| Stores | (1,045.00) | 105.00 | 418.00 | 313.50 | 209.00 | 0.50 |
| Maintenance | 45.00 | (905.00) | 497.75 | 181.00 | 181.00 | (0.25) |
| Cost after allocation | - | - | 6,915.00 | 2,995.00 | 1,890.00 | 11,800.00 |
| Answer a | Machining | Note |
| Machine Hours | 50,000.00 | A |
| Overhead cost | 6,915.00 | B |
| Overhead absorption rates | 0.14 | C=B/A |
| Answer a | Assembly | Finishing | Note |
| Labor Hours | 30,000.00 | 20,000.00 | D |
| Overhead cost | 2,995.00 | 1,890.00 | E |
| Overhead absorption rates | 0.10 | 0.09 | F=E/D |
| Bid for Job K | |||||
| Answer b | Machining | Assembly | Finishing | Total | Note |
| Machine/ Labor hour used | 45.00 | 15.00 | 12.00 | G | |
| Departmental predetermined OH Rate | 0.14 | 0.10 | 0.09 | H= See C, F | |
| Overhead applied to Job K ('000) | 6.22 | 1.50 | 1.13 | 8,855.00 | I=G*H |
| Direct materials | 24,000.00 | ||||
| Direct Labor | 15,000.00 | ||||
| Total cost of Job K | 47,855.00 | J | |||
| Profit | 11,963.75 | K=J/80%*20% | |||
| Sell price | 59,818.75 | L=J+K |
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