Chapter 4B Demonstration Problem Solutions Page 1 Please Send Comments and Corrections to me at [email protected] Demo 4B-1 ANSWER (a) Direct Method Prod.Dept.1 Prod.Dept.2 Service Dept. 1 $15,000 (2000/3200 x $15,000) $9,375 (1200/3200 x $15,000) $5,625 Service Dept. 2 $40,000 (120/250 x $40,000 $19,200 (130/250 x $40,000) $20,800 Allocated Costs $28,575 $26,425 (b) Step Method (Starting With Highest Percentage) S1 S2 100/3300 = .030303 S2 S1 10/260 = .0384615 S2 does more work for S1 than S1 does for S2. So, we start with S2. S1 P1 P2 S1 $16,539 (15,000+1,539) (2000/3200 x $16,539) $10,337 (1200/3200 x $16,539) $6,202 S2 $40,000 (10/260 x 40,000) $1,539 (120/260 x $40,000) $18,462 (130/260 x $40,000) $20,000 Allocated Costs $1,539 $28,799 $26,202 (c) Reciprocal Method S1 = 15,000 + (10/260) x S2 S2 = 40,000 + (100/3300) x S1 S1 = 15,000 + (10/260) x (40,000 + (100/3300 x S1)) S1 = 15,000 + 1538 + .0011655 S1 .9988345 S1 = 16,538 S1 = 16,538 /.9988345 S1 = $16,557 S2 = 40,000 + (100/3300 x $16,557) S2 = 40,000 + 502 S2 = $40,502Chapter 4B Demonstration Problem Solutions Page 2 Please Send Comments and Corrections to me at [email protected] S1 S2 P1 P2 S1 (15,000 + 1,557) $16,557 (100/3300 x$16,557) X $ 502 (2000/3300 x $16,557) $10,035 (1200/3300 x $16,557) $6,020 S2 ($40,000 + 502) $40,502 (10/260 x $40,502) $1,557 X (120/260 x $40,502) $18,693 (130/260 x $40,502) $20,251 Allocated Costs $1,557 $502 $28,728 $26,271 Demo 4B-2 ANSWER (a) Direct Method Prod.Dept.1 Prod.Dept.2 Service Dept. 1 $145,600 (88/220 x $145,600) $58,240 (132/220 x $145,600) $87,360 Service Dept. 2 $226,200 (51,200/256000x $226,200) $45,240 (204800/256000 x $226,200) $180,960 Allocated Costs $103,480 $268,320 (b) Step Method (Starting With Highest Percentage) S1 S2 20/240 = .0833 S2 S1 20/276 = .0724 S1 does more work for S2 than S2 does for S1. So, we start with S1. S2 P1 P2 S1 $145,600 (20/240 x 145,600) $12,133 (88/240 x $145,600) $53,387 (132/240 x $145,600) $80,080 S2 (226,200+12,133) $238,333 (51200/256000 x $238,333) $47,667 (204800/256000x238,333) $190,666 Allocated Costs $12,133 $101,054 $270,746Chapter 4B Demonstration Problem Solutions Page 3 Please Send Comments and Corrections to me at [email protected] (c) Reciprocal Method S1 = 145,600 + (20/276) x S2 S2 = 226,200 + (20/240) x S1 S1 = 145,600 + (20/276) x (226,200 + (20/240 x S1)) S1 = 145,600 + 16,391 + .00603 S1 .99397 S1 = 161,991 S1 = 161,991 /.99397 S1 = $162,974 S2 = 226,200 + (20/240 x $162,974) S2 = 226,200 + 13,581 S2 = $239,781 S1 S2 P1 P2 S1 (145,600 + 17,375) $162,974 (20/240 x $162,974) X $13,581 (88/240 x $162,974) $59,757 (132/240x $162,974) $89,635 S2 ($226,200 + 13,581) $239,781 (20/276 x $239,781) $17,375 X (51.2/276 x 239781) $44,481 (204.8/276x239781) $177,925 Allocated Costs $17,375 $13,581 $104,238 $267,560 Demo 4B-3 ANSWER (a) Direct Method Assembly Finishing Maintenance $1,185 (80/155 x $1185) $612 (75/155 x $1185) $573 Legal $840 (800/1800 x $840) $373 (1000/1800 x $840) $467 Allocated Costs $985 $1,040 (b) Step Method M L 20/175 = .1143 L M 300/2100 = .1428 L does more work for M than M does for L. So, we start with L.Chapter 4B Demonstration Problem Solutions Page 4 Please Send Comments and Corrections to me at [email protected] Maintenance Assembly Finishing Legal $840 (300/2100 x 840) $120 (800/2100 x $840) $320 (1000/2100 x $840) $400 Maintenance $1305 (1185+120) (80/155 x $1305) $674 (75/155 x 1305) $631 Allocated Costs $12 $994 $1,031 (c) Reciprocal Method M = 1,185 + (300/2100) x L L = 840 + (20/175) x M M = 1185 + (300/2100) x (840 + (20/175 x M)) M = 1185 + 120 + .0163 M .9837 M = 1305 M = 1305 /.9837 M = $1326 L = 840 + (20/175 x 1326) L = 840 + 152 L = $992 Maintenance Legal Assembly Finish Maintenance $1326 (1185+141) (20/175 x $1326) X $152 (80/175 x $1326) $606 (75/175 x $1326) $568 Legal $992 ($840 + 152) (300/2100 x $992) $141 X (800/2100 x 992) $378 (1000/2100 x 992) $473 Allocated Costs $141 $152 $984
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