Question: Cost estimation, cumulative average-time learning curve The Nautilus Company, which is under contract to the U.S. Navy, assembles troop deployment boats. As part of its

Cost estimation, cumulative average-time learning curve The Nautilus Company, which is under contract to the U.S. Navy, assembles troop deployment boats. As part of its research program, it completes the assembly of the first of a new model (PT109) of deployment boats. The Navy is impressed with the PT109. It requests that Nautilus submit a proposal on the cost of producing another six PT109s. Nautilus reports the following cost information for the first PT109 assembled and uses a 90% cumulative average-time learning model as a basis for forecasting direct manufacturing labor-hours for the next six PT109s. (A 90% learning curve means b = ???0.152004.)

Direct material Direct manufacturing labor time for first boat Direct manufacturing labor

Required1. Calculate predicted total costs of producing the six PT109s for the Navy. (Nautilus will keep the first deployment boat assembled, costed at $1,575,000, as a demonstration model for potential customers.)2. What is the dollar amount of the difference between (a) the predicted total costs for producing the six PT109s in requirement 1, and (b) the predicted total costs for producing the six PT109s, assuming that there is no learning curve for direct manufacturing labor? That is, for (b) assume a linear function for units produced and direct manufacturinglabor-hours.

Direct material Direct manufacturing labor time for first boat Direct manufacturing labor rate Variable manufacturing overhead cost Other manufacturing overhead Tooling costs" Leaming curve for manufacturing labor time per boat $200,000 15,000 labor-hours 40 per direct manufacturing 25 per direct manufacturing labor-hour 20% Jof direct manufacturing labor costs $280,000 90% cumulative average time Tooling can be reused at no extra cost because all of its cost has been assigned to the first deployment boat. In 0.9 -0.105361 Using the formula (p. 359), for a 90% learning curve, b= In 2 =-0.152004 0.693147

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