Case Study 2

Case Study 2 Springfield Express is a luxury passenger carrier in Texas. All seats are first class, and the following data are available: Number of seats per passenger train car 90 Average load factor (percentage of seats filled) 70% Average full passenger fare $ 160 Average variable cost per passenger $ 70 Fixed operating cost per month $3,150,000 Formula : Revenue = Units Sold * Unit priceContribution Margin = Revenue – All Variable Cost Contribution Margin Ratio = Contribution Margin/Selling Price Break Even Points in Units = (Total Fixed Costs + Target Profit )/Contribution Margin Break Even Points in Sales = (Total Fixed Costs + Target Profit )/Contribution Margin Ratio Margin of Safety = Revenue – Break Even Points in Sales Degree of Operating Leverage = Contribution Margin/Net Income Net Income = Revenue – Total Variable Cost – Total Fixed Cost Unit Product Cost using Absorption Cost = (Total Variable Cost + Total Fixed Cost)/# of units a. Contribu tion margin per passenger =$160 – $70 = $90Contribution margin ratio =$90/$160=56. 25% Break-even point in passengers = Fixed costs/Contribution Margin = $ 3,150,000/$90 Passengers =35,000 Break-even point in dollars = Fixed Costs/Contribution Margin Ratio = $ 3,150,000/56. 25% $ 5,600,000 b. Compute # of seats per train car (remember load factor? )= 90 * 70% = 63 Seats filled Compute # of train cars (rounded) = 35,000/63 = 556 train cars filled c. Contribution margin = $190 – $70 = $120 Break-even point in passengers = fixed costs/ contribution margin =$ 3,150,000/$120 Passengers = 26,250 BE = 90 seats *60% = 54Train cars (rounded) = 26,250/54 = 486 d. Contribution margin = $190 – $90 = $70 Break-even point in passengers = Fixed costs/Contribution Margin = $ 3,150,000/$70 Passengers =45,000 BE = 90 seats *70% = 63 Train cars (rounded) = 45,000/63 = 714 e. Contribution margin = $205 – $85 = $120 (P = Passengers) Sales205*P Variable Exp. 085*P Contribution M. 120*P Fixed Exp. 3,600,000 PretaxX Tax Exp. :X*30% Net Income Op. 750,000 750,000 = X – 0. 3X (X (1 – 0. 3) => 750,000/(1-0. 3) = X X=$ 1,071,428. 57 (Pre-Tax) $ 1,071,429= 120P – $ 3,600,000 => $ 1,071,429 + $ 3,600,000= 120P => 4,671,429/120= P P =38,928 f. Contribution margin = $120 – $70 = $50 # of discounted seats = 90*70%; 90*80% ( Difference is 10%; 90*10% = 9 Seats Contribution margin for discounted fares X #discounted seats = $50 * 9 Seats = $450 50 Train *$ 450 train cars per day * 30 days per month= $675,000 $ 675,000 (-) $ 180,000 additional fixed costs = $495,000 pretax income. g. 1. Compute Contribution margin Route 1 Route2 Overall Mix Sales160*P175*P335 *p Variable Exp. 070*P070*P140 *p Contribution M. 090*P105*P195 *P Route 1 Contribution Margin Ratio =$90/$160=56. 5% Route 2 Contribution Margin Ratio =$105/$175=60% Overall Contribution Margin Ratio =$195/$335=58. 20% Answer: Yes, it should, because the CMR is greater with the two rou tes. 2. BE = 90 * 60% = 54 Seats filled Contribution margin = $175 – $70 = $105 (P = Passengers) Sales175*P (54 Seats) Variable Exp. 070*P Contribution M. 105*P Fixed Exp. 3,150,000+250,000=3,400,000 Pretax120,000 120,000 = (105P*(54 Seats)) – 3,400,000 => 3,520,000 = 5,670P => 3,520,000/5,670 = P P=621 621/54 =12 train cars 3. Contribution margin = $175 – $70 = $105 BE = 90 seats *75% = 68Contribution margin = $175 – $70 = $105 (P = Passengers) Sales175*P (68 Seats) Variable Exp. 070*P Contribution M. 105*P Fixed Exp. 3,150,000+250,000=3,400,000 Pretax120,000 120,000 = (105P*(68 Seats)) – 3,400,000 => 3,520,000 = 7,140P => 3,520,000/7,140= P P=493 493/68 = 7 train cars 4. Springfield should consider Qualitative factors such as: (1) effect on employee morale, schedules and other internal elements; (2) relationships with and commitments to older and new suppliers; (3) effect on present and future customers; and (4) long-term future effect on profita bility and new businesses. Case Study 2 Chapter 2 Case Study Summary 1: 21-year old woman that has had type 1 diabetes for the past 8 years, was brought to the hospital in a coma. She was prescribed to take 92 units of insulin a day to maintain her sugar levels within normal limits and prevent excess sugar in her urine. Upon admission she was hypontensive, tachycardic and hyperventilating.Her labs show she is acidonic, arterial blood carbon dioxide levels were low, blood oxygen tension is normal, bicarbonate levels are really low indicating metabolic acidosis, low sodium levels, slightly high level of potassium, Chloride level is on the low end of normal, very high levels of blood urea and nitrogen, total carbon dioxide levels are really low, extremely high sugar levels and high creatinine levels. She tested positive for ketones. She recieved 8 units of regular insulin through an IV and 8 units per hour by IV infusion pump.Her blood sugar levels began to drop at about 100 mg/dL each hour. After seven hours her breathing an d pH went back to normal, following an injection of intravenous sodium bicarbonate to raise her pH and vigorous IV fluids and electrolyte replacement. 1. It seems her type 1 diabetes is uncontrolled. As her body could not use the sugar and there is not enough insulin, fat was used for fuel instead. During fat breakdown, byproducts called ketones are developed. Ketone bodies are acidic and dangerous when it build up in the body causing all her symptoms upon admission. . Yes, her pH levels became normal. Meaning her bicarbonate levels increased to compensate for the increased hydrogen proton levels in her blood. 3. When the body uses fat for energy instead of sugar, the body creates a byproduct called ketones. Since this is a byproduct, it flows through the renal system waiting to be expelled from the body. Normally, the existence of ketone bodies are detected through a urine sample. 4. The potassium results were high which can indicate some problem with her kidneys. 5.The low sodium results are based on the fact that sodiums job in the body is to keep proper acid-base equilibrium (homeostasis). Sodium has alkaline properties so if the levels are low the acid levels will be higher. 6. Diabetes can affect normal control of BP and can cause damage to the nerves supplying the blood vessels. When the blood pressure lowers the glomerular filtration rate decreases. 7. Anion gap measures of anions in the arterial blood. Anion gap equals chloride plus bicarbonate minus sodium Na-(Cl + HCO3-).The patient has a anion gap of 30. Normal levels are 7 to 16. 8. Osmolality measures the concentration of all chemical particles found in the fluid part of blood. Normal values range from 275 to 295. The patient has a osmolality of 351. 1 Summary 2: 14 year old boy that was never vaccinated against poliomyelitis got the disease late summer. He was hospitalized and needed a respirator during the severity of the illness. Once he began to recover, they took him off the respirator with no apparent effects. Days later a blood analysis revealed the following. H level is slightly acidic, carbon dioxide levels are high and indicate some respiratory acidosis, blood oxygen level is low, bicarbonate level is high, sodium levels are normal, potassium is normal, chloride level is slightly low, and total carbon dioxide levels are high. 1. It seems the patient has respiratory acidosis. Production of carbon dioxide occurs fast and the failure of proper ventilated increases the CO2 in the blood. 2. Buffers are normal compensatory mechanisms to respond to the acidosis. 3.Yes, the HCO3 (bicarbonate) test is elevated and bicarbonate is a buffer. 4. Acute respiratory acidosis is when a abrupt failure of ventilation occurs. Chronic respiratory acidosis may be secondary to many disorders. 5. Total CO2 measures the serum bicarbonate and available forms of carbon dioxide. Bicarbonate takes up about 95% of the total. They take the bicarbonate measurements by the sample of the venous bl ood and arterial blood gas analysis. 6. Chloride levels are slightly lower due to respiratory muscle weakness. Case Study 2 Chapter 2 Case Study Summary 1: 21-year old woman that has had type 1 diabetes for the past 8 years, was brought to the hospital in a coma. She was prescribed to take 92 units of insulin a day to maintain her sugar levels within normal limits and prevent excess sugar in her urine. Upon admission she was hypontensive, tachycardic and hyperventilating.Her labs show she is acidonic, arterial blood carbon dioxide levels were low, blood oxygen tension is normal, bicarbonate levels are really low indicating metabolic acidosis, low sodium levels, slightly high level of potassium, Chloride level is on the low end of normal, very high levels of blood urea and nitrogen, total carbon dioxide levels are really low, extremely high sugar levels and high creatinine levels. She tested positive for ketones. She recieved 8 units of regular insulin through an IV and 8 units per hour by IV infusion pump.Her blood sugar levels began to drop at about 100 mg/dL each hour. After seven hours her breathing an d pH went back to normal, following an injection of intravenous sodium bicarbonate to raise her pH and vigorous IV fluids and electrolyte replacement. 1. It seems her type 1 diabetes is uncontrolled. As her body could not use the sugar and there is not enough insulin, fat was used for fuel instead. During fat breakdown, byproducts called ketones are developed. Ketone bodies are acidic and dangerous when it build up in the body causing all her symptoms upon admission. . Yes, her pH levels became normal. Meaning her bicarbonate levels increased to compensate for the increased hydrogen proton levels in her blood. 3. When the body uses fat for energy instead of sugar, the body creates a byproduct called ketones. Since this is a byproduct, it flows through the renal system waiting to be expelled from the body. Normally, the existence of ketone bodies are detected through a urine sample. 4. The potassium results were high which can indicate some problem with her kidneys. 5.The low sodium results are based on the fact that sodiums job in the body is to keep proper acid-base equilibrium (homeostasis). Sodium has alkaline properties so if the levels are low the acid levels will be higher. 6. Diabetes can affect normal control of BP and can cause damage to the nerves supplying the blood vessels. When the blood pressure lowers the glomerular filtration rate decreases. 7. Anion gap measures of anions in the arterial blood. Anion gap equals chloride plus bicarbonate minus sodium Na-(Cl + HCO3-).The patient has a anion gap of 30. Normal levels are 7 to 16. 8. Osmolality measures the concentration of all chemical particles found in the fluid part of blood. Normal values range from 275 to 295. The patient has a osmolality of 351. 1 Summary 2: 14 year old boy that was never vaccinated against poliomyelitis got the disease late summer. He was hospitalized and needed a respirator during the severity of the illness. Once he began to recover, they took him off the respirator with no apparent effects. Days later a blood analysis revealed the following. H level is slightly acidic, carbon dioxide levels are high and indicate some respiratory acidosis, blood oxygen level is low, bicarbonate level is high, sodium levels are normal, potassium is normal, chloride level is slightly low, and total carbon dioxide levels are high. 1. It seems the patient has respiratory acidosis. Production of carbon dioxide occurs fast and the failure of proper ventilated increases the CO2 in the blood. 2. Buffers are normal compensatory mechanisms to respond to the acidosis. 3.Yes, the HCO3 (bicarbonate) test is elevated and bicarbonate is a buffer. 4. Acute respiratory acidosis is when a abrupt failure of ventilation occurs. Chronic respiratory acidosis may be secondary to many disorders. 5. Total CO2 measures the serum bicarbonate and available forms of carbon dioxide. Bicarbonate takes up about 95% of the total. They take the bicarbonate measurements by the sample of the venous bl ood and arterial blood gas analysis. 6. Chloride levels are slightly lower due to respiratory muscle weakness.